Problems in Building Physics 3031476670, 9783031476679

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Marko Pinterić

Problems in Building Physics

Problems in Building Physics

Marko Pinterić

Problems in Building Physics

123

Marko Pinterić Maribor, Slovenia

ISBN 978-3-031-47667-9 ISBN 978-3-031-47668-6 https://doi.org/10.1007/978-3-031-47668-6

(eBook)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Contents Introduction

1

I

3

Problems

P1 Basics of thermodynamics

5

P2 Heat transfer

9

P3 Heat transfer in building components

11

P4 Moisture in building components

29

P5 Basics of waves

41

P6 Sound propagation

43

P7 Building acoustics

45

P8 Illumination

51

II Solutions

55

P1 Basics of thermodynamics

57

P2 Heat transfer

69

P3 Heat transfer in building components

77

P4 Moisture in building components

107

P5 Basics of waves

169

P6 Sound propagation

171

P7 Building acoustics

179

P8 Illumination

193

PA Appendix

205

Bibliography

2 11 v

Preface When I was entrusted with giving lectures on building physics for students of civil engineering and architecture, I was faced with the problem of finding a suitable introductory literature for future civil engineers and architects. Optimally, the literature should cover both the physical principles underlying the problems of heat, moisture, sound and light transfer, and their application in the form of international standards. As I was not satisfied with the available options, I embarked on the project of writing my own book. These efforts led to the publication of the first edition of the book Building Physics: From physical principles to international standards in 2017. In just a few years, significant new developments in international standards have occurred, and the need arose to cover several topics more comprehensively, so the second edition of the book was published in 2021. Both editions were very well received by readers. Interestingly, many readers expressed a desire for the solutions to the end of chapter problems. This fact encouraged me to start a new project, the Problem Book, which not only contains the solutions to the problems already published but also adds new ones with the aim of not only providing the solutions but also relating them to real situations and applications. I am confident that the result of these efforts, which is now before you, will help to deepen the knowledge and awareness of building physics. Nevertheless, there is always room for improvement, and I look forward to constructive suggestions and criticism of this book. Contact information, errata, updates and supplementary content can be accessed at http://www.pinteric.com/books/. Marko Pinteri´c Maribor, 1st August 2023

vii

Introduction This problem book is intended as a companion to the 2021 second edition of Building Physics: From physical principles to international standards. It contains not only the solutions to the problems already included in the 2021 book but also many new problems. To maintain the proper order of the topics covered, the problems had to be renumbered. To avoid confusion, a special mark is introduced n.m, where n.m is the number of the same problem in the 2021 book. Furthermore, to better connect the problems to theory, the number of every expression from the 2021 book used in the solutions is consistently linked back. Because there is no comprehensive problem book on building physics in English, at least to my knowledge, vast majority of problems are novel. Great efforts have been made to make the problems as realistic as possible. However, a few problems are considerable simplifications of real situations in building physics for two reasons. First, the real situations are often too extensive and complex to be solved in a reasonable amount of time; for example, the complete calculation of the heat losses of a building is a huge project in itself. Second, the logic of the solutions to the real problems and the simplified version is usually the same; for example, the calculation procedure for determining characteristic temperatures does not change if only the most consequential layers of a wall are taken into account instead of all layers. Note that the problems and solutions are separated, dividing the book into two parts. This was done on purpose so that the user would not be tempted to check the solution before making a serious attempt to solve the problem. It is very important that the user finds the solutions themself, if possible. If the user only reads the solution, the learning effect is rather limited. The problems in the first part of the book are given with the answers in brackets. At the beginning of the chapter there is sometimes also additional material, which has been omitted from the 2021 book but is the subject of some additional problems. To distinguish between different types of problems, three additional marks have been developed. The mark identifies informative, usually simpler problems that are intended to promote understanding of building physics concepts, whereas the marks and identify difficult and very difficult problems, respectively.

1

Problems in Building Physics

2

The solutions in the second part of the book are often accompanied by detailed illustrations; a good understanding of the problem, reinforced by a suitable illustration, sometimes represents half the effort required to solve it. In addition, the solutions are written in great detail so that even those who are not skilled in solving problems can understand them. The solutions are also accompanied by the following marks, placed in the margins: This mark indicates the identification and explanation of an error that regularly appears in the solutions of beginners.

This mark indicates a particularly important step in the solution.

! This mark indicates particularly difficult parts of the solution.

i

This mark indicates important information that is related to the problem or solution, deepens the understanding of the concepts of building physics or relates to real situations.

In the problem-solving procedure, intermediate results are written down if they are informative or simplify the procedure. Note that the calculation continues with the exact values, even though the intermediate values are rounded. Finally, the solutions to the psychrometric problems are given using both the Mollier diagram and the ashrae-style chart.

I. Problems

P1 Basics of thermodynamics 1.1 When the copper bar length is measured using a steel tape at −10 ○ C, the measured value is 2000.0 mm. What is the measurement value at 30 ○ C? Take the linear expansion coefficients of copper and steel to be 1.7×10−5 K−1 and 1.2×10−5 K−1 , respectively. (2000.4 mm)

P1.1

1.2 The volume of alcohol in a thermometer glass bulb at 0 ○ C is 200 mm3 . The bulb is attached to a narrow glass tube of diameter 0.50 mm. Calculate the ‘length’ of degree Celsius on the glass tube. Assume that the glass does not expand. Take the cubic expansion coefficient of alcohol to be 1.1 × 10−3 K−1 . (1.1 mm)

P1.2

1.3 We want to put 2.0 kg of oxygen gas at 20 ○ C into a gas cylinder of volume 100 L. What minimum pressure should the gas cylinder be able to endure? Take the molar mass of oxygen to be 0.032 kg/mol. (1.5 × 106 Pa)

P1.3

1.4 Calculate the density of dry air at pressure 1.00 bar and (a) at 10.0 ○ C and (b) at 20.0 C. Take the molar mass of dry air to be 29.0 g/mol. (1.23 kg/m3 , 1.19 kg/m3 )

P1.4

○

1.5 A gas mixture of mass 10.0 kg and molar mass 0.026 kg/mol contains oxygen of mass 2.0 kg and molar mass 0.032 kg/mol. If the pressure of the gas mixture amounts to 2.5 bar, what is the partial pressure of the oxygen? (0.41 bar)

P1.5

1.6 Air at atmospheric pressure 1.013 × 105 Pa and at 10 ○ C is in a room of volume 10 m . Calculate the heat that should be provided to heat up the air to 30 ○ C if (a) the room is well sealed, (b) the room is leaky. Take the molar mass of air to be 29 g/mol and the specific heat capacity at constant volume to be 720 J/(kg K). (180 kJ, 250 kJ)

P1.6

3

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Pinterić, Problems in Building Physics, https://doi.org/10.1007/978-3-031-47668-6_1

5

Problems in Building Physics

6

1.7 A stainless steel vessel of mass 600 g contains 2.0 L of liquid water, both at 20 C. A 1.0 kW heater is inserted into the vessel and turned on. Calculate the time needed to bring the vessel and water to the temperature of boiling water. Take the specific heat capacities of stainless steel and water to be 600 J/(kg K) and 4200 J/(kg K), respectively. (12 min)

P1.7

○

1.8 A glass sphere of radius 2.5 cm at 150 ○ C is tossed into water of mass 2.0 kg at 18 ○ C. Calculate the equilibrium temperature. Take the density of glass to be 2.5 × 103 kg/m3 , as well as the specific heat capacities of water and glass to be 4200 J/(kg K) and 800 J/(kg K), respectively. (20 ○ C)

P1.8

1.9 An aluminium cube of mass 1.0 kg at 600 ○ C is tossed into water of mass 1.0 kg at 20 ○ C. Calculate the mass of vaporised water. Take the specific heat of vaporisation of water to be 2.26 × 106 J/kg, as well as the specific heat capacities of water and aluminium to be 4200 J/(kg K) and 900 J/(kg K), respectively. (50 g)

P1.9

1.10 Ice of mass 500 g at −10 ○ C and steel of mass 1.0 kg at 500 ○ C are placed into a well-insulated steel vessel of mass 450 g at 25 ○ C. Calculate the equilibrium temperature. Take the specific heat of fusion of water to be 336 kJ/kg, as well as the specific heat capacities of water, ice and steel to be 4200 J/(kg K), 2100 J/(kg K) and 470 J/(kg K), respectively. (22 ○ C)

P1.10

1.11 A well-insulated vessel contains 2.0 kg of liquid water at 0 ○ C. Water is frozen by pumping out the water vapour until all liquid either vaporises or freezes. Calculate the mass of the obtained ice. Take the specific heat of fusion to be 336 kJ/kg and the specific heat of vaporisation at 0 ○ C to be 2.5 MJ/kg. (1.8 kg)

P1.11

1.12 A room of a rectangular floor plan of dimensions 10.0 m × 20.0 m and height 3.0 m contains dry air of density 1.165 kg/m3 at 30 ○ C. In the room, we place a bucket with 12 L of liquid water at 5 ○ C. After a while, the temperatures of air and water equalise at 20 ○ C. Calculate the mass of the evaporated water. Take the specific heat capacity of dry air to be 1000 J/(kg K), the specific heat capacity of liquid water to be 4200 J/(kg K) and the specific heat of vaporisation of water at 20 ○ C to be 2450 kJ/kg. (2.5 kg)

P1.12

1.13 An average adult human being consumes most of their energy on heat losses, through radiation, convection, perspiration, respiration and conduction (listed in order of importance). Calculate the power due to the perspiration and respiration under the following assumptions:

P1.13

P1 Basics of thermodynamics

• For perspiration and respiration, the average person consumes 0.8 L of liquid water at 20 ○ C per day, with the water being converted into vapour at 37 ○ C. The specific heat capacity of water is 4200 J/(kg K) and the specific heat of vaporisation of water at 37 ○ C is 2410 kJ/kg. • The average person inhales and exhales 11 m3 of air in a room, with the air being heated from 20 ○ C to 37 ○ C. The density of inhaled air is 1.2 kg/m3 and the specific heat capacity is 1000 J/(kg K). We will calculate other heat losses in problem P2.6 in Chapter P2. (26 W)

P1.14 An instantaneous electric water heater, which is a water heater that heats water instantly as it flows through the appliance, heats a maximum of 1.8 L of water per minute from 12 ○ C to 38 ○ C. What is the maximum electrical power of the appliance if its efficiency is 94 %? The specific heat capacity of water is 4200 J/(kg K). (3.5 kW) A central heating system with radiators is supplied with water at 55 ○ C. The heated water can be provided either by a furnace of 100 % efficiency or by an air-towater heat pump whose actual coefficient of performance is half the ideal coefficient of performance. Calculate the external temperature below which heating with a furnace is cheaper, assuming that the price of electricity per kWh is 2.5 times larger than the price of gas per kWh. (−10.6 ○ C)

P1.15

P1.16 An air-to-water heat pump supplies hot water to a central heating system. The supply water temperature must be 55 ○ C for a system with radiators and 35 ○ C for an underfloor heating system. For the external temperature −2 ○ C, calculate the energy savings expressed as a percentage if an underfloor heating system is used instead of a system with radiators. For the heat pump, assume that the actual coefficient of performance is half the ideal coefficient of performance. (31 %)

7

P2 Heat transfer 2.1 A freezer of height 150 cm, width 60 cm and depth 60 cm has walls of 50 mm thick expanded polystyrene of thermal conductivity 0.040 W/(m K). The freezer maintains the internal temperature −18 ○ C, and the external temperature is 30 ○ C. Calculate the electrical power of the freezer if the efficiency of the heat pump is 300 %. (55 W)

P2.1

2.2 A cuboid vessel, whose edges measure 30 cm, 35 cm and 45 cm, has walls of 20 mm thick expanded polystyrene of thermal conductivity 0.040 W/(m K). The vessel contains 2.0 kg of ice at 0 ○ C, and the external temperature is 30 ○ C. Calculate the melting time of the ice. We want to prolong melting time to 7.0 h. Calculate the thickness of the additional layer of extruded polystyrene of thermal conductivity 0.035 W/(m K). Take the specific heat of fusion of water to be 336 kJ/kg. (3.9 h, 1.4 cm)

P2.2

2.3 A wall has two layers. The outer layer of thermal conductivity 0.038 W/(m K) is 8.0 cm thick, and inner layer of thermal conductivity 0.16 W/(m K) is 15.0 cm thick. What is the interface temperature (temperature at the boundary of the two layers) if the temperature on the external surface of the wall is −3.0 ○ C and the temperature on the internal surface of the wall is 17.0 ○ C? Calculate the distance between the external surface of the wall and the wall interior at temperature 0.0 ○ C. (10.8 ○ C, 1.7 cm)

P2.3

2.4 What temperature would gain a flat black surface on the Earth’s surface, neglecting heat exchange with Earthly objects, if (a) the black surface is perpendicular to the sunbeams, (b) the angle between the black surface and the sunbeams is 60○ ? For the Sun at zenith, the solar density of heat flow rate at the surface of the Earth is approximately 1000 W/m2 . What temperature would gain the flat grey surface, whose emittance is independent of wavelength? (91 ○ C, 78 ○ C; the same)

P2.4

9

Problems in Building Physics

10

2.5 What temperature would gain a flat black surface on the Earth’s surface, if the black surface is perpendicular to the sunbeams? The environment temperature is 25 ○ C, for the Sun at zenith the density of heat flow rate at the surface of the Earth is approximately 1000 W/m2 , and the average wind speed is 2.0 m/s. What temperature would gain the flat grey surface of emittance 0.10 independent of wavelength for the same conditions? (54 ○ C, 29 ○ C)

P2.5

P2.6 2.6 In problem P1.13 in Chapter P1, we calculated the heat flow rate due to the perspiration and respiration of an adult human being. Now calculate the net heat flow rate of radiation, and the heat flow rates of convection and conduction of a clothed inactive person in a conditioned room. For radiation and convection, assume that a person is a vertical convex surface of emittance 0.97 and surface area 1.8 m2 , so that surface coefficients typical for building physics can be used. Take the surface temperature of a clothed person to be 28 ○ C and the temperature of the room to be 20 ○ C. For conduction, assume that the surface area of the shoe sole is 5 dm2 and that the thermal resistance of the shoe sole is 0.25 m2 K/W. Take the temperature of the foot sole skin to be 33 ○ C and the temperature of the floor to be 20 ○ C. (83 W, 36 W, 2.6 W)

P3 Heat transfer in building components Airspaces In Section 3.1.3 of the book Building Physics, we examined the basics of heat losses through airspaces, in particular unventilated ones. However, the standard iso 6946 [1] prescribes practical engineering calculations for three types of airspaces: 1. unventilated air layer; 2. well-ventilated air layer; and 3. slightly ventilated air layer. Here, we take a closer look at the procedures for all three types.

Unventilated air layer An unventilated air layer is a layer that is not expressly designed for the flow of air through it. An air layer with openings to the external environment of area Ave is also considered an unventilated air layer if • Ave ≤ 500 mm2 per 1 m of length (in the horizontal direction) for vertical air layers, or • Ave ≤ 500 mm2 per 1 m2 of surface area for horizontal air layers. In this case, the radiative surface coefficient is given by expression (3.11): hra =

4 σ T3 . 1 + ε12 − 1 ε1

(P3.1)

11

Problems in Building Physics

12

Table P3.1: Convective surface coefficients for unventilated air layers [1]. hca / mW2 K

Upward

∆T ≤ 5 K ∆T > 5 K

Direction of Heat Flow Horizontal Downward

1.95 1.25 0.12 d −0.44 1.14 (∆T)1/3 0.73 (∆T)1/3 0.09 (∆T)0.187 d −0.44 — or 0.025/d, if larger than the preceding values —

On the other hand, the convective surface coefficient depends on the direction of the heat flow, the temperature difference within the air layer ∆T, and the thickness of the air layer d, as shown in Table P3.1.

Well-ventilated air layer An air layer with openings to the external environment of area Ave is a well-ventilated air layer if • Ave ≥ 1500 mm2 per 1 m of length (in the horizontal direction) for vertical air layers, or • Ave ≥ 1500 mm2 per 1 m2 of surface area for horizontal air layers. In this case, the thermal resistance of the air layer and all other layers between the air layer and the external environment is replaced by either the external surface resistance corresponding to the still air or the internal surface resistance from Table 3.1. In this book, we use the latter option.

Slightly ventilated air layer An air layer with openings to the external environment of area Ave is a slightly ventilated air layer if • 1500 mm2 > Ave > 500 mm2 per 1 m of length (in the horizontal direction) for vertical air layers, or • 1500 mm2 > Ave > 500 mm2 per 1 m2 of surface area for horizontal air layers. The total thermal resistance Rtot is calculated as an interpolation between the total thermal resistance of the unventilated air layer Rtot,nve and the well-ventilated air layer Rtot,ve : Rtot =

1500 mm2 − Ave Ave − 500 mm2 R + Rtot,ve . tot,nve 1000 mm2 1000 mm2

(P3.2)

P3 Heat transfer in building components

13

Table P3.2: Thermal resistance of roof spaces [1]. 2

Properties of Roof Tiled roof with no felt, boards or similar Sheeted roof, or tiled roof with felt or boards or similar under the tiles As preceding, but with aluminium cladding or other low emittance surface at underside of roof Roof lined with boards and felt

Ru / mWK 0.06 0.2 0.3 0.3

Roof spaces According to iso 6946 [1], for a roof structure consisting of a flat, insulated ceiling and a pitched roof, the roof space can be regarded as a thermally homogeneous layer with a thermal resistance Ru , as given in Table P3.2.

Heat losses through the ground In Section 3.3.2 of the book Building Physics, we examined the basics of heat losses through the ground using the standard iso 13370 [3]. In what follows, we will take a closer look at the procedures for all four situations described in the standard: 1. slab on the ground; 2. heated (conditioned) basement; 3. suspended floor; and 4. unheated (unconditioned) basement. Some of the symbols and equations from the standard have been adapted for pedagogical reasons. The common parameters for all situations are the area of the floor A, the exposed perimeter and the characteristic dimension of the floor. The exposed perimeter P (m) is the total length of the external wall separating a building or part of a building from the external environment or from an unheated (unconditioned) space, such as porches, attached garages, or storage rooms. In particular, for a single flat in a building with multiple flats, such as a row of terraced houses as parts of a building, the exposed parameter P excludes the length of the walls separating the flat under consideration from other flat(s). The characteristic dimension of the floor B (m) is simply calculated as B=

A . 0.5 P

(P3.3)

Problems in Building Physics

14

Table P3.3: Thermal conductivity of the ground [3]. λg / mWK

Clay, Silt 1.5

Sand, Gravel 2.0

Homogeneous Rock 3.5

Note that heat losses also depend on the thermal conductivity of the ground λg , whose value is given in Table P3.3. If the ground type is unknown, λg = 2.0 W/(m K) should be used. Related quantities are the total equivalent thicknesses, that is, the thicknesses of the ground whose thermal resistance is equal to the total thermal resistance of the shortest thermal paths through the ground. In the case of the floor (Fig. P3.1, left), the total thermal resistance and the total equivalent thickness of the floor df (m) are Rtot = Rsi + Rf +

w + Rse , λg

df = w + λg (Rsi + Rf + Rse ),

(P3.4)

where w is the thickness of the wall, Rf is the thermal resistance of the floor and Rsi is the internal surface resistance for downward (cooling season) or upward (heating season) heat flow. According to the standard, the thermal resistance of dense concrete slabs and thin floor coverings can be omitted. To ensure the consistency of calculations, we do not omit them in this book. In the case of the basement wall (Fig. P3.1, right), the total thermal resistance and the total equivalent thickness of the basement wall dbw (m) are Rtot = Rsi + Rbw + Rse , dbw = λg (Rsi + Rbw + Rse ),

(P3.5)

where Rbw is the thermal resistance of the basement wall and Rsi is the internal surface resistance for horizontal heat flow.

w

Rbw

z

Rf

Rf w Figure P3.1: Slab-on-ground (left) and heated basement (right). The arrows show the shortest thermal paths through the ground in a heating season.

P3 Heat transfer in building components

15

Slab-on-ground Slab-on-ground floors (Fig. P3.1, left) include all floors consisting of a slab in contact with the ground over its entire surface. For df < B (uninsulated and moderately insulated floors), the thermal transmittance of the floor is 2 λg πB Uf = ln ( + 1) . (P3.6) π B + df df For df ≥ B (well-insulated floors), the thermal transmittance of the floor is Uf =

λg . 0.457 B + df

(P3.7)

As stated in the book Building Physics (3.38), the ground heat transfer coefficient (without thermal bridges) is Hg = A Uf . (P3.8)

Heated (conditioned) basement The following procedure calculates only the heat losses below a horizontal plane at the level of the external ground surface; that is, heat losses through the basement wall in contact with external air must be calculated separately. The additional parameter is the depth of the basement floor below the ground level z (Fig. P3.1, right). For (df + 0.5 z) < B (uninsulated and moderately insulated floors), the thermal transmittance of the basement floor is Ubf =

2 λg πB ln ( + 1) . π B + df + 0.5 z df + 0.5 z

(P3.9)

For (df + 0.5 z) ≥ B (well-insulated floors), the thermal transmittance of the basement floor is λg Ubf = . (P3.10) 0.457 B + df + 0.5 z The thermal transmittance of the basement wall is Ubw =

2 λg 0.5 df z + 1) . (1 + ) ln ( πz df + z dbw

(P3.11)

If dbw < df , df should be replaced by dbw in the preceding expression for Ubw . As stated in the book Building Physics (3.39), the ground heat transfer coefficient (without thermal bridges) for the heated (conditioned) basement combines the heat losses through

Problems in Building Physics

16 both the floor and the walls as Hg = Abf Ubf + Abw Ubw ,

(P3.12)

where Abf is the area of the floor and Abw = P z is the area of the walls in contact with the ground.

Suspended floor A suspended floor is any type of floor held off the ground (Fig. P3.2, left). The procedure specified in iso 13370 [3] assumes that the underfloor space is naturally ventilated with external air. The suspended floor is actually a special case of the slab-on-floor, with an additional suspended floor. Because the space between the two floors is not heated (conditioned), both of the floors and the intermediate space are considered a single heat-transferring element. In a heating season, all the heat flows through the suspended floor of thermal transmittance Usf = 1/Rtot,sf into the underfloor space. The heat flow then splits: part flows through the ground, characterised by thermal transmittance Uf = 1/Rtot,f ; part through the walls of the underfloor space of thermal transmittance Uw = 1/Rtot,w ; and part is transported through ventilation, characterised by the heat transfer coefficient Hve (Section 3.3.4). If we put the total resistance of the heat flows through the ground, the walls of the underfloor space and ventilation in parallel (2.11) and the result in series with the total

h

Rsf

h

Rsf Rw

Rw

w

Rbw

z

Rf

Rf w Figure P3.2: Suspended floor (left) and unheated basement (right). The arrows show the thermal path in a heating season. The blue arrows correspond to the ventilation heat losses.

P3 Heat transfer in building components

17

resistance of the heat flow through the suspended floor (2.14), we get: Rtot = Rtot,sf +

A A

R tot,f

Ô⇒

+

1 1 = + U Usf Uf +

Aw R tot,w

+ Hve

1 hP Uw A

+ A1 Hve

.

(P3.13)

Here, the area ratio can be written differently, as Aw h P 2h = = , A A B where h is the height of the (upper) surface of the suspended floor above the ground level (Fig. P3.2, left). The thermal transmittance Uf is calculated as for the slab-on-floor, assuming uninsulated and moderately insulated floors (P3.6). Because the thermal resistance of dense concrete slabs can be neglected, only thermal insulation is to be included in (P3.4). The thermal transmittances Uw and Usf are calculated in the same way as for the direct heat losses in iso 6946 [1], that is, according to equation (3.3). The ventilation heat transfer coefficient is calculated as Hve = 725 ε v fw P = 1450

ε v fw A, B

where ε is the area of the ventilation openings per exposed perimeter length of the underfloor space, v is the average wind speed 10 m above the ground level, and fw is the wind shielding factor. The values for the shielding factor are 0.02, 0.05 and 0.10 for sheltered, average and exposed locations, respectively.

Unheated (unconditioned) basement The unheated (unconditioned) basement (Fig. P3.2, right) is actually a special case of the heated (conditioned) basement. Because the space below the basement ceiling is not heated (conditioned), the basement, along with its floor, walls and ceiling, is considered a single heat-transferring element. In a heating season, all the heat flows through the basement ceiling of thermal transmittance Usf = 1/Rtot,sf into the basement. The heat flow then splits: part flows through the basement floor, characterised by thermal transmittance Ubf = 1/Rtot,bf ; part through the basement walls below the ground level of thermal transmittance Ubw = 1/Rtot,bw ; part through the basement walls above the ground level of thermal transmittance Uw = 1/Rtot,w ; and part is transported through ventilation, characterised by the heat transfer coefficient Hve (Section 3.3.4). If we put the total resistance of the heat flows through the basement floor, the basement walls below and above the ground level, and ventilation in parallel (2.11) and the result in

Problems in Building Physics

18

series with the total resistance of the heat flow through the basement ceiling (2.14), we get: Rtot = Rtot,sf +

A

R tot,bf

Ô⇒

1 1 = + U Usf Ubf +

+

A bw R tot,bw

zP Ubw A

A +

Aw R tot,w

+ Hve

1 . + hAP Uw + A1 Hve

(P3.14)

Here, both area ratios can be written differently as Abw z P Aw h P = , = , A A A A where h is the height of the basement walls above the ground level (Fig. P3.2, right). The thermal transmittances Ubf and Ubw are calculated as for the heated basement, (P3.9), (P3.10) and (P3.11). The thermal transmittances Uw and Usf are calculated in the same way as those for direct heat losses in iso 6946 [1], that is, according to equation (3.3). The heat transfer coefficient Hve is calculated as in iso 13789 [5], that is, according to equation (3.44).

Slab-on-ground with edge insulation

dn

D

D

D

dn

In Section 3.3.2 of the book Building Physics, we also pointed out that additional insulation along the exposed perimeter of the floor for slab-on-ground (as well as suspended floor) significantly reduces the heat flow rate through the ground. For these cases, the standard iso 13370 [3] prescribes the calculation of the (negative) linear thermal transmittance Ψg,e that takes into account the effect of this additional insulation (Fig. P3.3).

dn

Figure P3.3: Slab-on-ground with edge insulation: horizontal edge insulation (left), vertical edge insulation (middle) and low-density beam foundation (right).

P3 Heat transfer in building components

19

The additional equivalent thickness resulting from the edge insulation is d ′ = λg ∆R = λg ( Ô⇒ d ′ = dn (

dn dn − ) λn λg

λg − 1) , λn

(P3.15)

where λn is the thermal conductivity and dn is the thickness of the edge insulation (or the foundation). The linear thermal transmittance Ψg,e also depends on the width or depth of the edge insulation D, the total equivalent thickness of the floor df and the thermal conductivity of the ground λg . For the horizontal edge insulation (Fig. P3.3, left), the linear thermal transmittance is Ψg,e = −

λg D D + 1)] . [ln ( + 1) − ln ( π df df + d ′

(P3.16)

For the vertical edge insulation (Fig. P3.3, middle), the linear thermal transmittance is Ψg,e = −

λg 2D 2D + 1) − ln ( + 1)] . [ln ( π df df + d ′

(P3.17)

For the low-density beam foundation (Fig. P3.3, middle), the expression for vertical edge insulation is used, with foundation thickness and depth substituted for insulation thickness and depth. The expression for the ground heat transfer coefficient (3.38) is then modified to Hg = A Uf + P Ψg,e .

(P3.18)

Problems in Building Physics

20

3.1 A vertical glass pane of thickness 6 mm separates the internal environment at 20 C and the external environment at −1 ○ C. Calculate the temperatures at the surfaces of the glass pane. Take the thermal conductivity of glass to be 0.80 W/(m K). (3.7 ○ C, 4.6 ○ C)

P3.1

○

3.2 A vertical double glazing with glass panes of thickness 4.0 mm, thermal conductivity 0.80 W/(m K) and airspace of thickness 16.0 mm between the glass panes, separates the internal space at 20 ○ C from the external environment. Take the thermal resistance of the airspace to be 0.19 m2 K/W, and calculate the external temperature at which the temperature of the internal surface of the double glazing is 15 ○ C. Calculate the other characteristic temperatures. (5.8 ○ C; 7.3 ○ C, 7.5 ○ C, 14.8 ○ C)

P3.2

3.3 Vertical walls of the thermal envelope consist of a concrete layer of thickness 20.0 cm and thermal conductivity 1.0 W/(m K) and a fac¸ade stone layer of thickness 2.0 cm and thermal conductivity 2.0 W/(m K). Calculate the thermal transmittance of the uninsulated and insulated walls, where the latter is obtained by adding an extra layer of expanded polystyrene (eps) of thickness 15.0 cm and thermal conductivity 0.035 W/(m K). For the external temperature −5 ○ C and internal temperature 20 ○ C, calculate the characteristic temperatures for all three situations (∣stone∣concrete∣, ∣stone∣concrete∣eps∣, ∣stone∣eps∣concrete∣). Plot the temperature as a function of layer thickness and as a function of thermal resistance. (2.63 W/(m2 K), 0.214 W/(m2 K); −2.4 ○ C, −1.7 ○ C, 11.5 ○ C; −4.8 ○ C, −4.7 ○ C, −3.7 ○ C, 19.3 ○ C; −4.8 ○ C, −4.7 ○ C, 18.2 ○ C, 19.3 ○ C)

P3.3

3.4 The total area of a room’s thermal envelope is 50 m2 : The vertical windows of thermal transmittance 1.1 W/(m2 K) make up 20 m2 of the envelope, and three wall types from problem P3.3 make up the rest of the area. Assume that the external and internal temperatures are constants with values −5 ○ C and 20 ○ C, respectively, and that the heat value of natural gas (amount of heat released during the combustion per cubic meter of gas) is about 33 MJ/m3 . Calculate the daily gas volume consumption for heating the room for uninsulated and insulated walls. What are the gas savings expressed as a percentage for the insulated wall? Is this percentage dependent on the internal and external temperatures? (6.6 m3 , 1.9 m3 ; 72 %; no)

P3.4

3.6 Calculate the external surface resistance for a fac¸ade of emittance 0.50. Take the average of external and surface temperatures to be 10 ○ C and the average wind speed to be 10 m/s. (0.021 m2 K/W)

P3.5

3.5 A vertical unventilated airspace of thickness 5.0 mm is enclosed by surfaces of emittance 0.90. Calculate the convective and radiative surface coefficients if the thermal resistance of the airspace is 0.11 m2 K/W. Take the average temperature to be 275 K. (3.9 W/(m2 K), 5.2 W/(m2 K))

P3.6

P3 Heat transfer in building components

P3.7 Determine the type of air layer of the ceiling structure above the flat according to the criteria of the size of the ventilation openings. The ceiling structure of area 70 m2 has five grilles of size 10 cm × 20 cm on the left and right sides. The cross-sectional area of the openings is 80 % of the cross-sectional area of the grilles. (well-ventilated)

P3.8

3.7 The vertical walls of a thermal envelope consist of

• fac¸ade bricks of thickness 100 mm and thermal conductivity 0.76 W/(m K); • an unventilated airspace of thickness 50 mm, on one side enclosed with aluminium foil; • fibre cement boards of thickness 16 mm and thermal conductivity 0.35 W/(m K); • mineral wool of thickness 150 mm and thermal conductivity 0.04 W/(m K); and • plasterboards of thickness 15 mm and thermal conductivity 0.21 W/(m K). Take the convective surface coefficient of the airspace to be 1.25 W/(m2 K) and emittances of the surfaces to be 0.90 and 0.05. Assume that the average temperature within the airspace is −2 ○ C. Calculate the thermal transmittance of the wall, the characteristic temperatures and the temperature factor of the internal surface if the internal and external temperatures are 21 ○ C and −5 ○ C, respectively. (0.206 W/(m2 K); −4.8 ○ C, −4.1 ○ C, −0.4 ○ C, −0.2 ○ C, 19.9 ○ C, 20.3 ○ C, 0.97)

P3.9 The vertical wall consists of • fac¸ade bricks of thickness 12 cm and thermal conductivity 0.76 W/(m K); • an unventilated airspace of thickness 4 cm; • eps of thickness 5 cm and thermal conductivity 0.04 W/(m K); • bricks of thickness 19 cm and thermal conductivity 0.19 W/(m K); and • plaster of thickness 1 cm and thermal conductivity 0.5 W/(m K). The emittance of the airspace surfaces is 0.9. Calculate the external temperature and all characteristic temperatures if the internal temperature is 20 ○ C and the temperature at the internal surface of the wall is 19 ○ C. Assume that the temperature difference in the airspace is smaller than 5 ○ C and the average temperature in the airspace is 1 ○ C. (−1.5 ○ C, −1.2 ○ C, 0.0 ○ C, 1.5 ○ C, 11.2 ○ C, 18.8 ○ C)

P3.10 The roof structure consists of • a roof covering of thickness 2.5 cm and thermal conductivity 0.76 W/(m K); • an airspace of thickness 5.0 cm; • timber panels of thickness 2.5 cm and thermal conductivity 0.14 W/(m K); • thermal insulation of thickness 12.0 cm and thermal conductivity 0.04 W/(m K); and • timber panels of thickness 2.5 cm and thermal conductivity 0.14 W/(m K).

21

Problems in Building Physics

22

The area of the ventilation openings is 300 mm2 per 1 m2 of roof surface and the emittance of the airspace surfaces is 0.9. Calculate the thermal transmittance and the characteristic temperatures if the external temperature is 0.0 ○ C and the internal temperature is 20.0 ○ C. Assume that the temperature difference in the airspace is smaller than 5 ○ C and the average temperature is 1 ○ C. (0.270 W/(m2 K); 0.2 ○ C, 0.4 ○ C, 1.3 ○ C, 2.3 ○ C, 18.5 ○ C, 19.5 ○ C)

P3.11 Calculate the thermal transmittance of the vertical wall of the thermal envelope, which consists of • fac¸ade panels of thickness 1.0 cm and thermal conductivity 0.92 W/(m K); • an airspace of thickness 5.0 cm; • thermal insulation of thickness 15.0 cm and thermal conductivity 0.041 W/(m K); • silicate bricks of thickness 20.0 cm and thermal conductivity 1.1 W/(m K); and • thermal mortar of thickness 1.0 cm and thermal conductivity 0.2 W/(m K). The airspace has 800 mm2 of openings per 1 m of the airspace length in the horizontal direction and the emittance of the airspace surfaces is 0.9. Assume that the temperature difference in the airspace is smaller than 5 ○ C and the average temperature in the airspace is 2 ○ C. (0.236 W/(m2 K))

P3.12 The roof structure consists of • a pitched sheeted roof; • plywood panels of thickness 1.2 cm and thermal conductivity 0.12 W/(m K); • expanded polystyrene of thickness 20.0 cm and thermal conductivity 0.04 W/(m K); • reinforced concrete of thickness 15.0 cm and thermal conductivity 2.3 W/(m K); and • plaster of thickness 1.0 cm and thermal conductivity 0.5 W/(m K). Calculate the thermal transmittance and the characteristic temperatures if the external temperature is −2.0 ○ C and the internal temperature is 20.0 ○ C. (0.181 W/(m2 K); −1.8 ○ C, −1.0 ○ C, −0.6 ○ C, 19.3 ○ C, 19.5 ○ C, 19.6 ○ C)

3.8 The vertical timber frame construction substructure, whose floor plan is shown in the figure, has two rows of vertical timber beams of dimensions 60 mm ×60 mm and of thermal conductivity 0.14 W/(m K) separated for r 1 = 565 mm along the wall and for r 2 = 40 mm perpendicular to the wall. The substructure is closed by plasterboards of thickness 25 mm and thermal conductivity 0.21 W/(m K) on one side, and with fibre cement boards of thickness 16 mm and thermal conductivity 0.35 W/(m K) plastered with

r2

P3.13

r1

P3 Heat transfer in building components

23

mortar of thickness 8 mm and thermal conductivity 0.50 W/(m K) on the other side. The intermediate space is filled with mineral wool of thermal conductivity 0.040 W/(m K). Calculate the upper limit and lower limit of the total thermal resistance and the thermal transmittance of the wall using the simplified calculation. (3.98 m2 K/W, 3.77 m2 K/W, 0.258 W/(m2 K))

P3.14 A vertical outer wall of length ltot = 230 cm, whose floor plan is shown in the figure, consists of • a fac¸ade layer of thickness 2.0 cm and thermal conductivity 1.5 W/(m K); • graphite polystyrene of thickness 10.0 cm and thermal conductivity 0.033 W/(m K); • osb boards of thickness 1.2 cm and thermal conductivity 0.13 W/(m K); • a timber load-bearing structure made of planks of rectangular cross section of dimensions r 1 = 5.0 cm × r 2 = 14.0 cm and thermal conductivity 0.14 W/(m K), filled up with mineral wool of thickness 14.0 cm and thermal conductivity 0.038 W/(m K); • osb boards of thickness 1.2 cm and thermal conductivity 0.13 W/(m K); and • plasterboards of thickness 1.25 cm and thermal conductivity 0.21 W/(m K).

r2

Calculate the upper and lower limits of the total thermal resistance and the thermal transmittance using the simplified calculation. (6.79 m2 K/W, 6.39 m2 K/W, 0.152 W/(m2 K))

r1

ltot

P3.15 A horizontal roof structure, whose vertical section is shown in the figure, consists

r2

of timber beams of rectangular cross section of dimensions r 1 = 15 cm × r 2 = 20 cm and thermal conductivity 0.14 W/(m K), spaced l ′ = 55 cm apart. On the top side, there are planks of thickness 2 cm and thermal conductivity 0.14 W/(m K), a well-ventilated air layer of thickness 2 cm, and a metal roofing of thickness 1 mm and thermal conductivity 50 W/(m K). On the bottom side, there are plasterboards of thickness 1.25 cm and thermal conductivity 0.28 W/(m K). The intermediate space is filled with mineral wool of thermal conductivity 0.04 W/(m K). Calculate the upper and lower limits of the total thermal resistance and the thermal transmittance of the structure using the simplified calculation in a heating season. (3.79 m2 K/W, 3.64 m2 K/W, 0.269 W/(m2 K))

l′

r1

Problems in Building Physics

24

3.9 A window with dimensions 150 cm × 100 cm consists of the 10 cm wide frame of thermal transmittance 1.4 W/(m2 K) and the glazing of thermal transmittance 1.1 W/(m2 K). The linear thermal transmittance on the junction of the frame and glazing is 0.070 W/(m K). Calculate the thermal transmittance of the entire window. (1.39 W/(m2 K))

P3.16

3.10 A room of height h = 2.5 m has two external mutually perpendicular walls of lengths a = 4.0 m and b = 7.0 m, as shown in the figure. The wider wall has two windows of height 1.5 m and width 2.0 m. Take the thermal transmittance of the walls to be 0.25 W/(m2 K), the thermal transmittance of the windows to be 1.1 W/(m2 K), the linear thermal transmittance at the junction of the two walls to be −0.030 W/(m K) and the linear thermal transmittance at the junction of the windows and the wall to be 0.050 W/(m K). Calculate the required power for heating the room, assuming the internal and external temperatures to be 20 ○ C and −10 ○ C, respectively. (378 W)

P3.17

b

h

a

P3.18 A lecture room has one external wall of width 15.0 m and height 4.0 m, with five windows of width 2.0 m and height 3.0 m. The thermal transmittances of the walls and the windows are 0.25 W/(m2 K) and 1.1 W/(m2 K), respectively, and the linear thermal transmittance at the junction of the windows and the wall is 0.070 W/(m K). Calculate the power of a heat pump required to heat the room for the internal temperature 20.0 ○ C and the external temperature −5.0 ○ C, if the coefficient of performance of the heat pump under these conditions is 200 %. (550 W) 3.11 Two-storey house in the shape of a cuboid of width a = 10.0 m, depth b = 8.0 m and height h = 6.0 m with a flat roof is erected on the ground, as shown in the figure. The house has windows of total area 30 m2 . Take the thermal transmittance of the roof to be 0.25 W/(m2 K), the thermal transmittance of the vertical walls to be 0.21 W/(m2 K), the thermal transmittance of the windows to be 1.1 W/(m2 K), the linear thermal transmittance at the junction of the vertical walls to be −0.050 W/(m K) and the linear thermal transmittance at the junction of the roof and the vertical walls to be 0.20 W/(m K). The remaining thermal bridges can be neglected. Calculate the direct heat transfer coefficient. Calculate the heat flow rate due to the direct heat losses if the internal and external temperatures are 21 ○ C and 5 ○ C, respectively. (98.1 W/K; 1570 W)

P3.19

P3 Heat transfer in building components

h

25

a

b

considered thermal bridges negligible thermal bridges

P3.20 A two-storey house of height h = 6.0 m, whose floor plan is in the shape of a rect-

angle of dimensions a = 10.0 m × b = 9.0 m and with a rectangular cutout of dimensions a′ = 5.0 m × b ′ = 2.0 m, as shown in the figure, is erected on the ground. The house has a flat roof and 22 windows of dimensions 1.0 m × 1.5 m. The thermal transmittances of the roof, vertical walls and windows are 0.25 W/(m2 K), 0.21 W/(m2 K) and 1.10 W/(m2 K), respectively. The linear thermal transmittances at the external junction of the vertical walls, at the internal junction of the vertical walls, at the junction of the roof and the vertical walls and at the junction of the windows and the vertical walls are −0.05 W/(m K), 0.05 W/(m K), 0.20 W/(m K) and 0.04 W/(m K), respectively (external dimensions). The contributions of the other thermal bridges can be neglected. Calculate the direct heat transfer coefficient. Calculate the heat flow rate due to the direct heat losses for the internal temperature 21 ○ C and external temperature −4 ○ C. (108 W/K; 2700 W) a

h

b

b′

a′

P3.21 The floor plan of three terraced houses of common breath b = 12.0 m and widths

l 1 = 7.0 m and l 2 = 8.0 m, and with slab-on-ground floors on the gravel ground is shown in the figure.

b

l1

l2

l1

Problems in Building Physics

26 The floor consists of

• a reinforced concrete slab of thickness 20.0 cm and thermal conductivity 1.0 W/(m K); • extruded polystyrene of thickness 5.0 cm and thermal conductivity 0.033 W/(m K); • a floating floor of thickness 6.5 cm and thermal conductivity 1.2 W/(m K); and • a parquet floor of thickness 1.0 cm and thermal conductivity 0.16 W/(m K). The thickness of the external wall is 45 cm. Calculate the thermal transmittance of the floor for the end terrace house and the mid terrace house, in a heating season. (0.274 W/(m2 K), 0.211 W/(m2 K))

P3.22 A house of a rectangular floor plan of dimensions 10.0 m × 7.5 m has a heated (conditioned) basement, whose floor is 1.5 m below the level of the sandy ground. The floor of the basement consists of • a reinforced concrete slab of thickness 20.0 cm and thermal conductivity 1.0 W/(m K); • extruded polystyrene of thickness 5.0 cm and thermal conductivity 0.033 W/(m K); and • a wood plank floor of thickness 1.0 cm and thermal conductivity 0.19 W/(m K). The walls of the basement in contact with the ground consist of • a waterproofing layer; • expanded polystyrene of thickness 8.0 cm and thermal conductivity 0.040 W/(m K); • concrete bricks of thickness 19.0 cm and thermal conductivity 0.49 W/(m K); and • plaster of thickness 1.0 cm and thermal conductivity 0.51 W/(m K). Calculate the thermal transmittances of the basement floor and the basement walls and the ground heat transfer coefficient in a heating season. (0.288 W/(m2 K), 0.297 W/(m2 K), 37.2 W/K)

P3.23 A house, whose floor plan in the shape of a rectangle of dimensions a = 12.0 m × b = 10.0 m and with a rectangular cutout of dimensions a ′ = 5.0 m × b ′ = 4.0 m is shown in the figure, has slab-on-ground floors on the clay ground. The floor consists of

• a reinforced concrete slab of thickness 20.0 cm and thermal conductivity 1.0 W/(m K); • a screed of thickness 5.0 cm and thermal conductivity 1.2 W/(m K); and • a parquet floor of thickness 1.0 cm and thermal conductivity 0.16 W/(m K).

P3 Heat transfer in building components

27

The edge of the slab is coated from below (horizontally) with extruded polystyrene of thickness 3.0 cm, thermal conductivity 0.033 W/(m K) and width 1.0 m. The thickness of the external wall is 40 cm. Calculate the thermal transmittance of the floor, the linear thermal transmittance of the floor edge, the ground heat transfer coefficient and the heat flow rate due to the heat losses through the ground. The internal temperature is 20 ○ C and the internal temperature is −2 ○ C. (0.501 W/(m2 K), −0.134 W/(m K), 44.2 W/K, 970 W) a

b′

b

a′

P3.24 A residential building has a heated (conditioned) basement, whose floor is 1.5 m below the level of the clay ground. The building, whose floor plan is shown in the figure, has two flats of breath b = 8.0 m and length l = 12 m. The floor of the basement consists of • a reinforced concrete slab of thickness 20.0 cm and thermal conductivity 1.0 W/(m K); • extruded polystyrene of thickness 3.0 cm and thermal conductivity 0.035 W/(m K); • a floating floor of thickness 6.5 cm and thermal conductivity 1.2 W/(m K); and • a ceramic tiling of thickness 1.0 cm and thermal conductivity 1.3 W/(m K). The walls of the basement in contact with the ground consist of • a waterproofing layer; • expanded polystyrene of thickness 10.0 cm and thermal conductivity 0.040 W/(m K); • concrete bricks of thickness 19.0 cm and thermal conductivity 0.49 W/(m K); and • plaster of thickness 1.0 cm and thermal conductivity 0.51 W/(m K). Calculate the heat flow rate for heat losses through the ground for one of the flats when the external and internal temperatures are −6 ○ C and 21 ○ C, respectively. (1.00 kW)

b

l

l

Problems in Building Physics

28

P3.25

In a passive house live four people.

• The volume of the house is 300 m3 and the air change rate at 50 Pa is 0.60 1/h. Calculate the heat losses due to the air leakage by assuming that the air change rate at 50 Pa is 20 times larger than the (natural) air change rate. • The house uses the recuperator of 75 % efficiency for a forced ventilation of air flow rate 120 m3/h. Calculate the heat losses due to the forced ventilation. The internal temperature is 20 ○ C, the external temperature is 4 ○ C, the density of air is 1.2 kg/m3 and the specific heat capacity of air is 1000 J/(kg K). (48 W, 160 W)

P3.26 The essential building elements of a two-storey house are • a pitched roof of area 60.0 m2 and thermal transmittance 0.26 W/(m2 K); • windows of area 46.5 m2 and thermal transmittance 1.2 W/(m2 K); • walls of area 130.6 m2 and thermal transmittance 0.21 W/(m2 K); • doors of area 2.1 m2 and thermal transmittance 1.0 W/(m2 K); • a slab-on-ground of area 60.0 m2 and thermal transmittance 0.28 W/(m2 K); and • a wall-balcony junction of length 6.0 m and linear thermal transmittance 0.40 W/(m K). The estimated air change rate is 0.20 1/h and the volume of the building is 336 m3 . Calculate the direct, ground, ventilation and transmission heat transfer coefficients. Calculate the required heating power if the external and internal design temperatures are −20 ○ C and 20 ○ C, respectively. The density of air is 1.2 kg/m3 and the specific heat capacity of air is 1000 J/(kg K). (103.3 W/K, 16.8 W/K, 22.4 W/K, 120.1 W/K; 5.7 kW)

P3.27 The transmission and ventilation heat transfer coefficients for a building are 250 W/K and 40 W/K, respectively. An air-to-water heat pump of electrical power 4.0 kW supplies a central heating system with radiators with water at 55 ○ C. At low temperatures, the heat pump cannot compensate for all heat losses, so an auxiliary heating source provides extra heat. Calculate the external temperature at which the auxiliary heating source turns on for the internal temperature 21 ○ C. Assume that the actual coefficient of performance of the heat pump is half the ideal coefficient of performance. (−12.5 ○ C)

P4 Moisture in building components Psychrometric charts In the book Building Physics, we mentioned the existence of two types of psychrometric charts, but showed their use only for the Mollier diagram. In this book, the solutions to the psychrometric problems are given using both the Mollier diagram and the ashrae-style chart. Therefore we start with some additional information and a comparison. There is no standard for drawing psychrometric charts, but they always display lines representing a constant (dry bulb) temperature, a constant relative humidity, a constant mass ratio and a constant specific enthalpy (Table P4.1). The Mollier diagram often also displays lines representing a constant density ρ, while the ashrae-style chart instead displays lines representing a constant specific volume vd (m3/kg), defined as vd =

V , md

(P4.1)

where md is the mass of dry air and V is the volume of humid air. The relationship between the quantities ρ and vd is not intuitive; to find it, we first establish that the mass of humid air ma is the sum of the mass of dry air and the mass of water vapour m, as in ma = md + m = md + x md ,

Table P4.1: Quantities displayed in different psychrometric charts.

Name (dry bulb) temperature relative humidity mass ratio specific enthalpy density specific volume wet bulb temperature

Symbol θ φ x h ρ vd

Unit ○ C % g/kg kJ/kg kg/m3 m3/kg ○ C

Mollier ✓ ✓ ✓ ✓ ✓

ashrae ✓ ✓ ✓ ✓ ✓ ✓

29

Problems in Building Physics

30

where in the second step we have expressed the mass of water vapour in terms of the mass ratio (4.11). The mass of dry air is therefore md =

ma . 1+x

(P4.2)

If we divide the expression by the volume of humid air, we get 1 ρ = . vd 1 + x This relationship is one of the reasons why the calculation procedure for two types of psychrometric charts sometimes differs considerably; in these cases, the solutions to problems using the two types of psychrometric charts are completely separated in this book. In both psychrometric charts, one of the axes represents the mass ratio x. Contrary to the first impression, the other axis is not the (dry bulb) temperature θ, but the skewed specific enthalpy h. A closer look reveals that the lines representing a constant specific enthalpy are parallel and equidistant, but the lines representing a constant (dry bulb) temperature are not. Finally, the ashrae-style psychrometric charts usually display the wet bulb temperature. For consistency between the two types of psychrometric charts and to avoid crowded lines, we will not display and use them in this book. Nevertheless, you can find some useful information on the wet bulb temperature in the solution to problem P4.5 .

Moisture transfer through ventilation In Section 3.3.4 of the book Building Physics, we calculated the amount of heat transferred through ventilation. Now, we will calculate the amount of transferred moisture. Due to the ventilation, the internal air of volume Va and mass concentration of water vapour vi is replaced by the external air of equal volume and mass concentration of water vapour ve (4.8). The mass of water vapour was changed for ∆m = mi − me = Va (vi − ve ). If we divide the equation by the time in which the air is replaced, we get the mass flow rate of water vapour due to the ventilation (4.2): q m,ve = q V (vi − ve ).

(P4.3)

Here, q V is the air flow rate (3.42). Note that this expression is the moisture equivalent of the expression (3.41). In the preceding calculation, we have assumed that the temperatures of the external and internal air are equal. If this is not the case, the external air is also heated or cooled

P4 Moisture in building components

31

isobarically from the external to the internal temperature during ventilation, changing its volume to match the volume of the air it replaces Va . However, this process also changes the mass concentration of water vapour ve , which must be recalculated for the internal temperature before it is used in the expression (P4.3). An example of this recalculation can be found in the solution to problem P4.24 . The mass concentration of water vapour is usually calculated from the temperature and relative humidity using expressions (4.9) and (4.7).

Other substance transfers through ventilation The process of mass transfer through ventilation is universal for all air particles, so the expression (P4.3) can also be applied to other air components. However, for most air components and air pollutants, the concentration is usually expressed in terms of molar fraction with the dimensionless unit parts per million y (ppm), defined as y=

N′ , N

where N ′ and N are the number of particles of an air component and the total number of air particles, respectively. The mass concentration of the air component v (4.8) and the density of the air ρ are related to these numbers by the definition of the amount of substance, (1.7) and (1.9), as NA v V, M′ NA N= ρ V, M

N′ =

where M ′ and M = 0.029 kg/mol are the molar masses of the air component and the entire air, respectively. The relationship between the molar fraction and the mass concentration is thus M v, M′ ρ RT y = ′ v, M p y=

(P4.4) (P4.5)

where T and p are the temperature and the pressure of the air, respectively. The second expression is obtained using the ideal gas law (1.12). Certain substances, such as CO2 and SF6 , can also be used to measure air leakage rate by generating a known mass flow of the substance and then measuring its mass concentration in the building. An important air component is radon 222 (222 Rn), a radioactive element that occurs naturally in soils and rocks and is the largest natural contributor to the radiation dose to which the population is exposed. However, radon concentration is usually measured

Problems in Building Physics

32

and expressed in terms of radioactive activity with the unit becquerel A (Bq), which is the quotient of the number of radioactive decays by time. Without going into the theoretical background, the relationship between the radioactive activity and the radon mass m is A=

ln(2) NA m, M T1/2

(P4.6)

where M = 0.222 kg/mol and T1/2 = 3.825 d are the molar mass and the half-life of radon 222, respectively.

Graphical determination of the water vapour barrier The Glaser method [4] provides a simple and effective steady state assessment of interstitial condensation. One of the options not covered in the book Building Physics is the graphical determination of the water vapour barrier properties required to prevent condensation. Let us first observe the situation with interstitial condensation presented in the book Building Physics (Fig. 4.24), reproduced in Fig. P4.1, top. Because the water vapour enters the building component from the inside, we must apply the water vapour barrier of equivalent air thickness sd,4 on the internal side, as shown in Fig. P4.1, bottom. The physical thickness of this additional layer is very small and therefore has negligible thermal resistance. This implies that the characteristic temperatures and water vapour pressures at saturation do not change; within the water vapour barrier, θ and psat are constant and the psat profile is horizontal. The value of pi does not change either, so the pi point is shifted horizontally to the right, flattening the slope of the second (inner) section of the p profile. If we move this point far enough, the p profile becomes a single straight line. Because the slope representing the density of water vapour flow д is now equal on both sides of the wall, the water vapour flow entering the interface is equal to the flow leaving the interface, дin = дout , and no more water accumulates. It follows that we must extend the first (outer) section of the p profile and find the intersection with a horizontal line starting at the former pi point. Then we can easily read the equivalent air layer thickness of the water vapour barrier sd,4 .

P4 Moisture in building components

33

p дin

дout

p sat

pi

p

pc

pe sd,1

sd,2

sd,3

дin

p sat

sd

p дout

pi

p

pe sd,1

sd,2

sd,3

sd,4

sd

Figure P4.1: The Glaser diagram for interstitial condensation (top) and the fixed situation with an applied water vapour barrier (bottom). By adding a water vapour barrier layer, we have straightened the p profile so that the densities of water vapour flow rate for the water vapour flow entering and leaving the interface equalise as дin = дout .

Problems in Building Physics

34

4.1 Humid air of volume 1.0 m3 , temperature 20 ○ C and pressure 1.0 bar contains 9.0 g of water vapour. Air is compressed isothermally. At which pressure does the air become saturated with water vapour? Take the molar mass of water to be 18 g/mol. (1.9 bar)

P4.1

4.13 In a room of a rectangular floor plan of dimensions 4.0 m × 5.0 m and height 2.5 m, there is dry air of density 1.20 kg/m3 at 21 ○ C. In the room, we place a pot with 3.0 L of liquid water at 41 ○ C. What is the mass of evaporated water at the moment when temperatures of air and liquid water in the room drop to 16 ○ C? What is the relative humidity then? The specific heat capacity of dry air is 1.0 kJ/(kg K), the specific heat capacity of liquid water is 4.2 kJ/(kg K) and the specific heat of vaporisation of water at 16 ○ C is 2460 kJ/kg. The molar mass of water is 18 g/mol. (0.25 kg, 37 %)

P4.2

Calculate the water vapour pressure and the pressure of dry air in a room for humid air of pressure 100 kPa, temperature 21 ○ C and relative humidity 60 %. What are the masses of water vapour and dry air in 300 L of this humid air? When we open a freezer of volume 300 L, the air from the room completely replaces the air in the freezer. When the freezer is closed again, the air inside cools down to −18 ○ C. How much water vapour deposits on the surfaces of the freezer? What is the pressure of the cooled humid air inside the freezer? The molar masses of water vapour and dry air are 18 g/mol and 29 g/mol, respectively. (1.5 kPa, 98.5 kPa; 3.3 g, 350.8 g; 3.0 g; 85.6 kPa)

P4.3

Saturated air of temperature 2 ○ C and pressure 100 kPa rises to an altitude 3 km, with its temperature falling to −17 ○ C and pressure to 70 kPa. How many snowflakes form in 1.0 m3 of the cooled air if the mass of one snowflake is 3 mg? The molar mass of water is 18 g/mol. (1000)

P4.4

Find the highest temperature of dry air that can be cooled to 14 ○ C by evaporation of liquid water alone. Assume that the temperature of liquid water is also 14 ○ C and take the specific heat of vaporisation at 14 ○ C to be 2.47 × 106 J/kg, the specific heat capacity of dry air to be 1000 J/(kg K), the specific heat capacity of water vapour to be 1900 J/(kg K) and the atmospheric pressure to be 1.013 × 105 Pa. Using a psychrometric diagram, devise a method to find the highest temperature of humid air that can be cooled to 14 ○ C in the same way. (38.6 ○ C)

P4.5

4.2 Using the psychrometric chart, determine the dew point, that is the temperature of a glass pane at which condensation occurs, for humid air at temperature 30 ○ C and relative humidity 40 %. (15 ○ C)

P4.6

P4.7 Using the psychrometric chart, determine the wet bulb temperature, that is the lowest temperature to which humid air can be cooled by evaporation of liquid water alone, for humid air at temperature 30 ○ C and relative humidity 40 %. (20 ○ C)

P4 Moisture in building components

P4.8 The air in a room at temperature 20 ○ C and relative humidity 50 % is heated

to 38 ○ C with an electric heater. Determine the relative humidity and specific enthalpy of the heated air using the psychrometric chart. (18 %, 57 kJ/kg)

4.3 An air dryer, which is working on the principle of removing heat, is used to eliminate the excess moisture in a room of dimensions 8.0 m × 6.0 m × 3.0 m. The initial relative humidity is 70 %, the initial and final air temperatures are 30 ○ C and 11 ○ C, respectively. Using the psychrometric chart, determine the mass of eliminated water. (1.7 kg)

P4.9

4.4 In an air conditioner, air of mass 100 kg at temperature 30 ○ C and relative humidity 60 % is first cooled and then heated, so the final temperature is 20 ○ C and relative humidity is 65 %. Using the psychrometric chart, calculate the removed heat, the added heat and the mass of removed water. (3.35 MJ, 0.69 MJ; 640 g)

P4.10

P4.11 The air in a room of a rectangular floor plan of dimensions 6.2 m × 4.0 m and

height 2.85 m is at temperature 30 ○ C and relative humidity 65 %. In a dehumidifier, 3.40 MJ of heat is removed from the air and then 0.64 MJ of heat is added to the air. Plot the process on a psychrometric chart. What are the final temperature, relative humidity and mass of removed water? (19 ○ C, 60 %, 730 g)

The wood drying chamber is ventilated by blowing in 100 m3/h of external air at temperature 5.0 ○ C and relative humidity 75 %. The air is first heated to 30.0 ○ C and then circulated in the chamber until it is adiabatically humidified to saturation and discharged. Plot the process on a psychrometric chart. What is the mass flow rate of water evaporated from the wood and the heating power? (0.22 g/s, 890 W)

P4.12

An air handling unit draws in fresh air at temperature 4.0 ○ C and relative humidity 80 %, first heats it to a certain temperature, then humidifies it to saturation and finally heats it to temperature 20.0 ○ C. Using the psychrometric chart, determine the air temperature after the first heating if the final relative humidity is to be 65 %. Calculate the heat consumed in processing 79 m3 of fresh air. (27 ○ C, 3.0 MJ)

P4.13

P4.14 An air conditioner of electrical power 1.1 kW and coefficient of performance 3.8 cools the air in a room. The air conditioner draws in 12 m3/min of air at 27 ○ C. Using the psychrometric chart, determine the outlet temperature of the air when the inlet relative humidity of the air is (a) 30 % and (b) 50 %? (9.2 ○ C, 13.3 ○ C)

35

Problems in Building Physics

36

4.5 Air of mass 50 kg, which includes water vapour of mass 700 g, at temperature 20 C is mixed adiabatically with air of mass 33 kg at temperature 25 ○ C and relative humidity 30 %. Using the psychrometric chart, determine the final temperature and relative humidity. (22 ○ C, 66 %)

P4.15

○

Using the psychrometric chart, determine the air flow rate of external air at temperature 5 ○ C and relative humidity 75 % that is adiabatically mixed with internal air at temperature 25 ○ C and relative humidity 80 % and whose air flow rate is 11.0 m3/h, to obtain an air mixture of mass ratio 12 g/kg. (5.0 m3/h)

P4.16

P4.17 A saturated air of volume 32 m3 at temperature 1 ○ C is heated to 13 ○ C and adiabatically mixed with an air of volume 92 m3 at relative humidity 90 % and temperature 21 ○ C. Plot the process on a psychrometric chart. What are the relative humidity and temperature of the air mixture? (84 %, 19 ○ C)

Air at temperature 35 ○ C and relative humidity 30 % is introduced into a drying chamber through the inlet. The air is adiabatically humidified to relative humidity 80 % before leaving the drying chamber through the outlet. This outlet air and fresh air at temperature 4 ○ C and relative humidity 80 % are then mixed, heated to temperature 35 ○ C and returned to the drying chamber. Using the psychrometric chart, determine the humid mass ratio of fresh air to outlet air to obtain the inlet air at relative humidity 30 %. (40 % fresh air, 60 % outlet air)

P4.18

4.6 Consider damp timber of density 450 kg/m3 and mass ratio of water to dry matter 20 %. Calculate the mass concentration of water of damp timber and the density of completely dry timber. (75 kg/m3 , 375 kg/m3 )

P4.19

4.7 The thermal transmittance of a vertical building component is 2.0 W/(m2 K). Take the internal relative humidity to be 65 % and the internal and external temperatures to be 20 ○ C and −5 ○ C, respectively. Calculate the relative humidity on the internal surface of the building component. (98 %)

P4.20

4.8 A vertical building component separates the internal environment at temperature 20 ○ C and relative humidity 70 % from the external environment at temperature 0 ○ C. Calculate the minimum total thermal resistance of the building element that prevents mould growth. Mould growth occurs for relative humidities above 80 %. (1.22 m2 K/W)

P4.21

P4 Moisture in building components

4.9 The temperature factor of the internal surface for the thermal bridge is 0.71. Assuming that the internal temperature is 21 ○ C and internal relative humidity is 55 %, calculate the external temperature at which mould growth occurs. Mould growth occurs for relative humidities above 80 %. (0.4 ○ C)

P4.22

P4.23

The vertical wall consists of

• fac¸ade panels of thickness 1.2 cm and thermal conductivity 0.6 W/(m K); • eps of thermal conductivity 0.04 W/(m K); • bricks of thickness 18 cm and thermal conductivity 0.8 W/(m K); and • plaster of thickness 1 cm and thermal conductivity 0.5 W/(m K). The average monthly internal temperature is 22 ○ C and the relative humidity is 70 %; the average monthly external temperature is −5 ○ C and the relative humidity is 90 %. Calculate the minimum thickness of eps that prevents mould growth. Mould growth occurs for relative humidities above 80 %. (4.7 cm)

An office is used by ten persons. Each person emits about 50 g of water vapour per hour. Calculate the minimum outlet air flow rate of ventilation to maintain the internal relative humidity 65 %. The internal temperature is 20 ○ C, the external temperature is 4 ○ C and the external relative humidity is 80 %. Take into account that the external air is heated to the internal temperature in an isobaric process before it enters the room. (78 m3/h)

P4.24

A 50 m long and 25 m wide indoor swimming pool contains water at temperature 27 ○ C. A ventilation system of outlet air flow rate 7.0 m3/s exchanges the internal air at temperature 29 ○ C with external air at temperature 5 ○ C and relative humidity 90 %. The evaporation mass flow rate from the pool can be described as

P4.25

q m,ev = h m,ev A(vi − vs ), where A is the area of the water surface, vi is the mass concentration of water vapour of the internal air, vs is the mass concentration of water vapour at saturation at water temperature and h m,ev = 7.8 × 10−3 m/s is the evaporative surface coefficient of mass transfer. Calculate the relative humidity of the internal air. Take into account that the external air is heated to the internal temperature in an isobaric process before it enters the swimming pool hall. (60 %) To maintain temperature 23 ○ C and relative humidity 60 %, a room must be ventilated with air of temperature 2 ○ C and relative humidity 80 % at outlet air flow rate 30 L/s. Using the psychrometric chart, calculate the net mass flow rate of water vapour, the total (enthalpic) heat losses due to the ventilation and the heating power required for ventilation. (2.5 × 10−4 kg/s, 1380 W, 750 W)

P4.26

37

Problems in Building Physics

38

P4.27 Calculate the mass flow rate of carbon monoxide generated by a gas stove if the molar fraction of carbon monoxide in the building is 20 ppm. The air change rate of the building is 0.8 1/h, the volume of the building is 150 m3 and the molar mass of carbon monoxide is 0.028 kg/mol. Neglect the external mass concentration of carbon monoxide and take the density of air to be 1.2 kg/m3 . (2.8 g/h)

In a residential building whose volume is 300 m3 and whose air leakage rate at 50 Pa is 1500 m3/h, we measure the indoor radon 222 concentration 450 Bq/m3 . Calculate the mass flow rate of radon from the ground if the outdoor radon concentration is 30 Bq/m3 , assuming that the air leakage rate at 50 Pa is 20 times larger than the (natural) air leakage rate. What is the air flow rate of forced ventilation required to keep the indoor radon concentration below 150 Bq/m3 , without taking into account the contribution of the air leakage? (5.5 × 10−15 kg/h, 260 m3/h)

P4.28

4.10 For all three situations (∣stone∣concrete∣, ∣stone∣concrete∣eps∣, ∣stone∣eps∣concrete∣) from problem P3.3 in Chapter P3 calculate the total equivalent air layer thickness, the characteristic water vapour pressures at saturation and the characteristic water vapour pressures, all under the assumption that there is no condensation. The internal and external relative humidities are both 65 %, the external and internal temperatures are −5 ○ C and 20 ○ C, respectively. The water vapour resistance factors are 120 for concrete, 200 for fac¸ade stone and 60 for expanded polystyrene. Plot the water vapour pressure and water vapour pressure at saturation as a function of the layer thicknesses and as a function of the equivalent air layer thicknesses. In which situation does condensation appear and where? (28, 37, 37; 401 Pa, 501 Pa, 530 Pa, 1352 Pa, 2337 Pa; 401 Pa, 409 Pa, 410 Pa, 450 Pa, 2238 Pa, 2337 Pa; 401 Pa, 409 Pa, 410 Pa, 2093 Pa, 2238 Pa, 2337 Pa; 261 Pa, 441 Pa, 1519 Pa; 261 Pa, 397 Pa, 1213 Pa, 1519 Pa; 261 Pa, 397 Pa, 703 Pa, 1519 Pa)

P4.29

P4.30

4.11 Vertical walls of the thermal envelope consists of (from outside to

inside) • fibre cement boards of thickness 2.0 cm, thermal conductivity 1.5 W/(m K) and water vapour resistance factor 50; • concrete of thickness 20.0 cm, thermal conductivity 1.0 W/(m K) and water vapour resistance factor 120; and • expanded polystyrene of thickness 8.0 cm, thermal conductivity 0.035 W/(m K) and water vapour resistance factor 60. The average monthly internal and external temperatures are 20 ○ C and 0 ○ C, respectively, the average monthly internal and external relative humidities are 45 % and 80 %, respectively. Using the Glaser method, determine whether condensation appears. If so, calculate the mass of water vapour condensed in one month with 30 days and estimate the thickness of polyethylene sheet of water vapour resistance factor 100 000 that would prevent the condensation. (yes; 34 g/m2 , 0.37 mm)

P4 Moisture in building components

P4.31

4.12 The vertical walls of the thermal envelope consists of

• bricks of thickness 10 cm, thermal conductivity 0.16 W/(m K) and water vapour resistance factor 16; • mineral wool of thickness 10 cm, thermal conductivity 0.035 W/(m K) and water vapour resistance factor 1; and • bricks of thickness 10 cm, thermal conductivity 0.16 W/(m K) and water vapour resistance factor 16. The average monthly internal and external temperatures are 20 ○ C and 5 ○ C, respectively, and the average monthly internal and external relative humidities are 50 % and 80 %, respectively. If condensed water exists in the most sensitive interstice, determine the mass of the evaporated water in one month with 30 days using the Glaser method. (62 g/m2 )

P4.32

The vertical walls of the thermal envelope consist of (from outside to inside)

• granite slabs of thickness 1.5 cm, thermal conductivity 2.8 W/(m K) and water vapour resistance factor 10 000; • expanded polystyrene of thickness 8.0 cm, thermal conductivity 0.035 W/(m K) and water vapour resistance factor 150; and • reinforced concrete of thickness 20.0 cm, thermal conductivity 2.3 W/(m K) and water vapour resistance factor 130. The average monthly internal temperature is 20 ○ C and the relative humidity is 45 %; the average monthly external temperature is 0 ○ C and the relative humidity is 70 %. Using the Glaser method determine whether condensation appears. If so, calculate the mass of water vapour condensed in one month with 30 days and estimate the thickness of polyethylene sheet of water vapour resistance factor 100 000 that would prevent the condensation. (yes; 5.1 g/m2 , 2.8 mm)

P4.33 The vertical wall of the thermal envelope consists of (from outside to inside) • fac¸ade bricks of thickness 120 mm, thermal conductivity 0.76 W/(m K) and water vapour resistance factor 16; • an unventilated air layer of thickness 20 mm; • eps of thickness 50 mm, thermal conductivity 0.04 W/(m K) and water vapour resistance factor 60; • aerated concrete of thickness 200 mm, thermal conductivity 0.13 W/(m K) and water vapour resistance factor 8; and • plasterboards of thickness 12.5 mm, thermal conductivity 0.21 W/(m K) and water vapour resistance factor 10.

39

40

Problems in Building Physics

The emittances of the air layer surfaces are 0.9. Using the Glaser method determine whether condensation appears, and, if so, calculate the mass of water vapour condensed in one month with 30 days. The average monthly internal temperature is 20 ○ C and the relative humidity is 60 %; the average monthly external temperature is −1 ○ C and the relative humidity is 80 %. Take the equivalent air layer thickness of the unventilated airspace to be 0 m. Assume that the temperature difference in the air layer is smaller than 5 ○ C and the average temperature in the air layer is 1 ○ C. (yes, 39.5 g/m2 )

P5 Basics of waves 5.1 Calculate the frequency of electromagnetic waves of wavelengths 0.50 µm and 10 µm. (6.0 × 1014 Hz, 3.0 × 1013 Hz)

P5.1

5.2 Calculate the shortest distance between two parallel walls for which standing sound waves of frequency 110 Hz appear. (1.56 m)

P5.2

5.3 The ear canal is a tube of length 25 mm, open on one side and closed on the other side with the tympanic membrane. Calculate frequencies of all harmonics of the standing wave in the ear canal that can be heard. (3.4 kHz, 10.3 kHz, 17.2 kHz)

P5.3

41

P6 Sound propagation 6.7 The sound pressure level gain due to the physiological effects, such as the shape of the head and the outer ear, is 20 dB [7]. For the quietest sound still detectable, calculate the sound power entering the human body through the tympanic membrane, if the effective area of the tympanic membrane is 43 mm2 . (4.3 × 10−15 W)

P6.1

6.1 The sound power of an isotropic point source is 2.0×10−6 W. Calculate the sound power level as well as the sound intensity and sound pressure level at distance 10 m from the source. (63 dB, 1.6 × 10−9 W/m2 , 32 dB)

P6.2

6.2 The sound pressure level is 70 dB at distance 3.0 m from the isotropic linear sound source. Determine the distance from the source at which the sound pressure level is 58 dB. (48 m)

P6.3

P6.4 Two specifications of a loudspeaker sold in a shop are an electrical power 10 W and a sound pressure level 105 dB at distance 0.5 m from the loudspeaker. Calculate the sound power level of the loudspeaker, assuming that it is an isotropic point sound source. What is the efficiency of the conversion of electrical energy into sound energy? (110 dB, 1.0 %)

6.3 Three individual isotropic point sound sources of sound power levels 55 dB, 50 dB and 45 dB are positioned closely to each other. What is the overall sound power level? What is the smallest distance from these sources at which they are no longer audible? (56.5 dB, 189 m)

P6.5

6.4 The sound measurement location is at distances 15 m from an electric motor of sound power level 90 dB and 25 m from the road of linear sound power level 75 dB. What is the measured sound pressure level? (57.4 dB)

P6.6

43

Problems in Building Physics

44

P6.7 A building is positioned 20 m from a straight road of linear sound power 25 µW/m. A device is to be placed 10 m from the building. What is the largest sound power of the device so that the total sound pressure level at the building is not larger than 55 dB? (150 µW)

P6.8 Two loudspeakers of equal sound power levels are placed 5.0 m apart. At distance 2.5 m from both loudspeakers, the measured sound pressure level is 40 dB. What is the sound pressure level at distances 4.0 m from the first loudspeaker and 1.0 m from the second loudspeaker? (45.2 dB)

6.5 When two sound sources operate, the measured sound pressure level is 80 dB. When the first sound source is turned off, the measured sound pressure level is 75 dB. What is the sound pressure level due to the first sound source? (78.3 dB)

P6.9

6.6 The sound pressure level between 8:00 and 11:00 is 60 dB and between 11:00 and 18:00 is 50 dB. What is the average sound pressure level? (55.7 dB)

P6.10

P6.11 The sound pressure levels are 61 dB, 58 dB and 52 dB for daytime, evening and night periods, respectively. What is the average sound pressure level for the entire day? (59.0 dB)

P6.12 What is the increase in sound pressure level when five loudspeakers emit sound instead of just one? (7.0 dB)

P6.13 The annual average daily traffic on a road during the daytime period is 285 light vehicles, 27 medium heavy vehicles, and 13 heavy vehicles per hour. Calculate the linear sound power level of the road during the daytime period if the linear sound power levels are 55.4 dB, 59.8 dB and 62.7 dB for one light, one medium heavy and one heavy vehicle per hour, respectively. (81.7 dB)

P7 Building acoustics Image source method The wavelengths of sound waves are relatively large at 0.02 m–20 m and are usually larger than the dimensions of the surface imperfections. This implies that the reflections of the sound are specular, that is, the angle of the reflected wave is equal to the angle of the incident wave with respect to the surface normal. The path of the reflected sound wave between the source and the receiver is uniquely defined because there is only one point on the surface that satisfies the condition that the angles of incidence and reflection are equal. This point can be easily determined using the image source method: The image source is obtained by mirroring the real source over the reflecting surface (Fig. P7.1). In the calculation, it can be assumed that the reflected sound is actually generated by the image source and not by the real source.

image source

source θ

θ

wall

θ θ

receiver

Figure P7.1: The paths of direct and reflected sound waves for specular reflection. Because the angles of incidence and reflection θ are equal, the path of the reflected sound wave can be easily determined by the image source method. The wavefronts of the direct and reflected waves are shown by solid and dashed lines, respectively.

45

Problems in Building Physics

46

7.1 A receiver and transmitter are located 1.5 m above the floor in a hall of floor plan dimensions 30 m × 40 m and height 6 m, as shown in the figure. Calculate the time delays for sound waves reflected from the right wall and sound waves reflected from the ceiling. (53 ms, 3.6 ms)

P7.1

40 m

4m

30 m

15 m

right wall

transmitter 5m left wall

5m

receiver

7.2 The sound pressure level of sound that is reflected once should be smaller for 7.0 dB. Calculate the minimum absorbance of the wall. (0.80)

P7.2

7.3 Sound is reflected twice from the walls of absorbance 0.20. What is the sound pressure level reduction? Neglect the reduction due to the sound path. (1.9 dB)

P7.3

7.4 Calculate the sound pressure level difference between the direct and both reflected sounds in problem P7.1. Do the reflected sounds together have a larger sound pressure level than the direct sound? For how much? Take into account the geometrical divergence, and take the absorbance of the walls and ceiling to be 0.10. (4.28 dB, 0.78 dB; yes, 0.82 dB)

P7.4

P7.5 A speaker and a listener are in a lecture hall of floor plan dimensions 15 m × 20 m and height 5 m, as shown in the figure. Calculate the time delay due to the sound reflection from the right wall. Calculate the sound pressure level for the direct sound and for the sound reflected from the right wall, taking into account the geometrical divergence. Take the height of the speaker’s mouth and the listener’s ear to be 1.5 m above the floor, the speaker’s sound power level to be 60 dB, and the absorbance of the walls to be 0.10. (28 ms; 26.2 dB, 21.1 dB)

2m speaker

5m 3m

listener

20 m

15 m

7.5 m

right wall

P7 Building acoustics

47

7.9 The horizontal distance between the right traffic lane and the building is 5.3 m and the distance between the right traffic lane and the transparent noise barrier is 2.9 m, as shown in the figure. Calculate the sound pressure level on the fac¸ade of the building 4.0 m above the ground, if absorbance of the barrier is 0.20. Assume for the traffic lane that the sound source is 5.0 cm above the ground and take the linear sound power level to be 70 dB. Take into account the direct sound and the sound reflected from the barrier. (55.4 dB)

P7.6

2.9 m

0.05 m

4.0 m

5.3 m

7.5 A lecture room of a rectangular floor plan of dimensions 10 m × 20 m and height 4 m has concrete walls and ceiling, parquet on the floors, six windows each of area 10 m2 and 70 seats. Calculate the reverberation time. Take the absorbance of concrete to be 0.02, absorbance of timber to be 0.06, absorbance of glass to be 0.03 and equivalent absorption area of each empty seat to be 0.5 m2 . (2.3 s)

P7.7

Sixty people are in a hall of floor plan dimensions 10.0 m × 15.0 m and height 4.0 m. The speaker and the listener are positioned in the hall as shown in the figure. The absorbance of the walls and ceiling is 0.040, the absorbance of the floor is 0.060 and the equivalent absorption area of one standing person is 1.0 m2 . Calculate the sound pressure level for the direct and diffuse sound at the listener’s location, if the sound power level of the speech is 70 dB. What is the total sound pressure level at this location? Take the heights of the speaker’s mouth and the listener’s ear to be equal. (39.5 dB, 56.8 dB, 56.9 dB)

P7.8

1m speaker

listener 5m 10 m

5m

2m

15 m

Problems in Building Physics

48

P7.9 The reverberation time in an empty room of a rectangular floor plan of dimensions 20.0 m × 10.0 m and height 4.0 m is 2.0 s. What is the reverberation time if there are also 30 people in the room and the equivalent absorption area of one person is 0.70 m2 ? What is the reverberation time if the concrete ceiling of absorbance 0.05 is additionally covered with an absorber of absorbance 0.50? (1.51 s, 0.74 s)

P7.10 The reverberation time in an empty university auditorium of a rectangular floor plan of dimensions 10.8 m × 7.1 m and height 3.8 m is 2.0 s. What is the reverberation time if the lecture is half-full (33 students and one professor) and the equivalent absorption area of one sitting person is 0.44 m2 ? Calculate what area of the concrete wall of absorbance 0.06 must be covered with an absorber of absorbance 0.50 to reduce the reverberation time to 0.74 s. (1.23 s; 58 m2 )

7.6 A person in the open space is listening to the point sound source located 40 m away. At some moment, the barrier of sound reduction index 5.0 dB is positioned between the person and the sound source, perpendicularly to the sound path. At what distance from the sound source should the person move to hear the sound as loud as in the beginning? (22.5 m)

P7.11

P7.12 The sound pressure level at the building at distance 32 m from the road axis is 57 dB. Between the road and the building there is a noise barrier of noise reduction index 10 dB. What is the linear sound power level of the road? (90 dB)

P7.13 The horizontal distance between the right traffic lane and the left building is 5.3 m, the distance between the right traffic lane and the right building is 6.1 m, and the distance between the right traffic lane and the noise barrier is 2.9 m, as shown in the figure. Calculate the sound pressure level at the fac¸ade of the right building 3.0 m above the ground, if the absorbance of the fac¸ade of the left building is 0.20 and the sound reduction index of the noise barrier is 15 dB. Assume for the traffic lane that the sound source is 5 cm above the ground and take its linear sound power level to be 75 dB. Take into account the sound transmission through the noise barrier and the sound reflection from the fac¸ade of the left building and neglect the sound diffraction. (54.1 dB)

3.0 m

3.2 m 2.0 m

2.9 m 0.05 m

5.3 m

P7 Building acoustics

49

7.7 The sound source of sound power level 80 dB is located in the source room. What are the diffuse sound pressure levels in the source room and the receiving room? The area of the wall between the rooms is 8 m2 , and the apparent sound reduction index is 50 dB. Take the equivalent absorption area of the source room to be 12 m2 and the equivalent absorption area of the receiving room to be 20 m2 . (75 dB, 21 dB)

P7.14

7.8 Two rooms of floor plan dimensions l 1 × w 1 = 5.0 m × 3.0 m and l 2 × w 2 = 3.0 m × 2.5 m and of height 2.5 m are positioned as shown in the figure. The reverberation times measured in both rooms are T60,1 = 0.40 s and T60,2 = 0.25 s. After we turn on the sound source in the first room, we find that the diffuse sound pressure levels are L p,1 = 82 dB and L p,2 = 39 dB. What is the apparent sound reduction index of the wall between the rooms? (40.9 dB)

l2

w1

w2

P7.15

l1

7.10 A living room of floor plan dimensions l 2 × w 2 = 5.0 m × 3.0 m is located directly adjacent to the engine room of floor plan dimensions l 1 × w 1 = 2.0 m × 2.0 m, as shown in the figure. Both rooms are of height 2.6 m. The apparent sound reduction index between the rooms is 57 dB, whereas the reverberation time of the living room is 0.50 s. What is the largest acceptable sound pressure level in the engine room, if the largest permissible sound pressure level in the living room is 35 dB? There is a device of sound power 100 dB in the engine room. In order not to exceed the largest acceptable sound pressure level, sound absorbers are installed on all engine room surfaces. What is their smallest absorbance? Neglect the increase of the apparent sound reduction index due to the installation of absorbers. (95.9 dB; 0.36)

w1

P7.16

w2

l1

l2

P8 Illumination Calculate the ray displacement in a glass pane, that is, the distance between the axis of the incident ray and the axis of the ray emerging on the other side of the glass pane. Take the angle of incidence to be 60○ and the thickness of the glass pane to be 4.0 mm. The speed of light in glass is 1.5 times smaller than the speed of light in air. (2.0 mm)

P8.1

8.2 Calculate the solid angle of the Sun observed from the Earth, if the P8.2 Sun is about 1.50 × 108 km away from the Earth, and the Sun’s radius is 6.96 × 105 km. Assuming that the luminance of the Sun on the Earth’s surface is 1.44 × 109 cd/m2 at zenith on a sunny day, calculate the maximum illuminance of a small horizontal surface. (6.76 × 10−5 sr, 9.74 × 104 lx)

h

P8.3 8.1 Four identical vertical street lamps of height h = 8.0 m are positioned in a way that their projection on the ground forms a square with sides of a = 5.0 m, as shown in the figure. Calculate the luminous intensity of each isotropic light so that illuminance in the centre of the square (point 0) is 50 lx. (1050 cd)

0 a

a

51

Problems in Building Physics

52

P8.4 Four identical vertical street lamps of height h = 4 m are arranged in a straight

line at distance l = 10 m from each other at the edge of a promenade of width d = 6 m, as shown in the figure. Lamps of luminous flux 10 000 lm each direct all the light into the lower half of the space. What is the illuminance on the other edge of the promenade, opposite the centre of all lamps (point 0)? (21.6 lx)

l

h

0 d

8.4 An isotropic light of luminous flux 1500 lm is located 2.2 m above the floor and 2.0 m from the vertical wall with the mirror, as shown in the figure. Calculate the illumination of the floor 1.0 m from the wall. Take into account only the direct light and the light reflected from the mirror. (23.7 lx)

P8.5

2.2 m

2.0 m

1.0 m

P8.6 An isotropic light of luminous flux 750 lm is located in the centre of the corridor

1.0 m

2.6 m

0.2 m

of width 1.5 m and height 2.6 m at distance 20 cm from the ceiling, as shown in the figure. The left side of the corridor is covered by a mirror. Calculate the illuminance of the right side of the corridor 1.0 m above the floor. Take into account only the direct light and the light reflected from the mirror. (18.4 lx)

1.5 m

P8 Illumination

53

8.3 Two cuboid-shaped buildings of height 22.5 m are positioned on the horizontal surface at geographical latitude 47○ (Northern Hemisphere), as shown in the figure. Buildings are rotated with respect to the east-west direction for θ = 30○ . Using the sun path diagram, determine the sunlight exposure on 21st March for a window at two-thirds of the south fac¸ade (point 0) at height 1.5 m. For the calculation, take into account only the angles of elevation of the Sun larger than 15○ and use the equidistant points 1 to 4. (3 h 20 min)

P8.7

10 m

N

30

m

W

E

15 m

10 m S 0 θ 1 2 3

m

4

10 m

30

P8.8 Three cuboid-shaped buildings of floor plan dimensions 20.0 m × 20.0 m and

height 20.0 m are positioned on the horizontal surface at geographical latitude 47○ (Northern Hemisphere), as shown in the figure. The buildings are rotated with respect to eastwest direction for θ = 10○ . Using the sun path diagram, determine the periods of direct sunlight and the sunlight exposure on 21st March for a window in the centre of the south fac¸ade of the northernmost building (point 0) at height 1.5 m. For the calculation, take into account only the angles of elevation of the Sun larger than 15○ . (08:55–09:50 13:20–16:40; 4 h 15 min) N

20 m

W

E S

0

15 m

θ

20 m 20 m

15 m

20 m

Problems in Building Physics

54

Calculate the largest illuminance of a horizontal surface at geographical latitude 47 by the Sun on a winter solstice day. The luminous flux of the Sun is 3.58 × 1028 lm, and the distance between the Sun and the Earth is approximately 1.50 × 108 km. Use the sun path diagram and assume that the atmosphere transmits approximately 77 % of light. (3.3 × 104 lx)

P8.9

○

II. Solutions

P1 Basics of thermodynamics P1.1 θ = −10 ○ C 0

1995

2000 θ = 30 ○ C

0

1995

2000

Both the copper bar and the steel tape expand due to the temperature change (1.3), as shown in the preceding figure. The exact final length of the copper bar lc for the initial length l 0,c = 2000 mm and the temperature change ∆θ = 30 ○ C − (−10 ○ C) = 40 ○ C is lc = l 0,c + ∆lc = l 0,c + α l,c l 0,c ∆θ = 2001.36 mm, where α l,c = 1.7 × 10−5 K−1 is the linear expansion coefficient of copper. By measuring the copper bar, we relate the length of the bar to the length of one metre represented by the steel tape. However, the length of l 0 = 1 m = 1000 mm representation on the steel tape increased to l = l 0 + ∆l = l 0 + α l,s l 0 ∆θ = 1000.48 mm, where α l,s = 1.2 × 10−5 K−1 is the linear expansion coefficient of steel. The measurement value is actually the number of reference lengths that fit into the measured length, or in other words, the ratio of the rod length to the length of the one

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Pinterić, Problems in Building Physics, https://doi.org/10.1007/978-3-031-47668-6_2

57

Problems in Building Physics

58 metre representation on the steel tape: n′ =

lc = 2.0004, l0

Ô⇒ l ′ = 2.0004 m.

When alcohol of initial volume V0 = 2 × 10−7 m3 and cubic expansion coefficient α V = 1.1 × 10−3 K−1 is heated for ∆θ = 1 ○ C, it expands for (1.4)

x

P1.2

∆V

∆V = α V V0 ∆θ. Assuming that the expansion of glass is negligible, the alcohol can expand only along the tube of diameter d = 5 × 10−4 m. The volume of expansion is the product of the cross-sectional area of the tube A and the column height x

V0

2

∆V = A x = π ( d2 ) x. If we equate the expressions for expansion, we get: x=

i

4α V V0 ∆θ = 1.1 × 10−3 m. π d2

The ‘length’ of the degree Celsius is proportional to the volume of the liquid and inversely proportional to the square of the diameter of the tube. To achieve a better resolution of the temperature reading, the length should be as large as possible. Because the cubic expansion coefficient is very small, all thermometers based on liquid expansion must have large vessels and thin tubes.

P1.3 The ideal gas law (1.12) connects four physical quantities, two of which, the temperature T = 293 K (θ = 20 ○ C) and the volume V = 0.1 m3 are given by the problem. To calculate the pressure, we still need the amount of substance. However, in this problem, as in most practical situations, the mass of the gas m = 2 kg is known instead. Because we usually know or can calculate the molar mass, we use the relationship between the mass and the amount of substance (1.9), as in pV = nRT =

m R T. M

P1 Basics of thermodynamics

59

For oxygen of molar mass M = 0.032 kg/mol, the pressure is p=

mRT = 1.5 × 106 Pa. MV

When using the ideal gas law or an equation with the symbol ‘T’, it is very important to convert the temperature to the Kelvin temperature scale.

P1.4 The density of gas is derived from the ideal gas law (1.12): m RT M m pM Ô⇒ ρ = = . V RT

pV = nRT =

The pressure and the molar mass of the air are p = 1 × 105 Pa and M = 0.029 kg/mol, respectively. The densities at two temperatures, Ta = 283 K (θ a = 10 ○ C) and Tb = 293 K (θ b = 20 ○ C), are pM kg = 1.23 3 , R Ta m pM kg ρb = = 1.19 3 . R Tb m ρa =

We see that warmer air has a lower density. Due to the buoyancy, free objects of lower density move upwards. Warmer air thus rises and triggers convective heat transfer in the room.

P1.5 As we pointed out in Section 1.4.1, the ideal gas law (1.12) can be written for the entire gas mixture of mass m = 10 kg and molar mass M = 0.026 kg/mol; and for each of its components, in our case for oxygen of mass m 1 = 2 kg and molar mass M 1 = 0.032 kg/mol: m R T, M m1 p1 V = n1 R T = R T. M1 pV = nRT =

i

Problems in Building Physics

60

If we divide the preceding equations, we establish the relationship between the pressures of the gas mixture p = 2.5 bar and oxygen p 1 : p m M1 = p1 m1 M m1 M Ô⇒ p 1 = p = 0.41 bar. m M1

P1.6 The pressure p = 1.013 × 105 Pa, the temperature T = 283 K (θ = 10 ○ C) and the volume V = 10 m3 are sufficient to fully characterise a gas. The mass of the air is then determined using the ideal gas law (1.12) and the molar mass M = 0.029 kg/mol: pV = nRT = Ô⇒ m =

m RT M

pMV = 12.5 kg. RT

The heat that must be added to the air depends on the type of heating process. If the room is well sealed, the volume is constant and the specific heat capacity is c V = 720 J/(kg K). However, if the room has leaks, the pressure inside the room is equalised to the external pressure by these leaks. The pressure is thus constant, and the specific heat capacity is (1.27) R c p = cV + . M The required heats (1.20) in the case of the well sealed room Qws and the leaky room Ql are Qws = m c V (θ 2 − θ 1 ) = 1.8 × 105 J, R Ql = m (c V + ) (θ 2 − θ 1 ) = 2.5 × 105 J, M where θ 1 = 10 ○ C and θ 2 = 30 ○ C are the initial and final air temperatures, respectively. Note that the required heat is larger in the case of the leaky room. This is because we must supply the air not only with the energy to increase its temperature, but also with the energy to increase its volume, as we argued in Section 1.6.

i

In practical situations, the rooms are always leaky and the pressure is constant. When heated, the air expands and some of the internal air leaves the room through the leaks; when cooled, the air shrinks and some of the external air enters the room through the leaks. We are almost always confronted with isobaric situations, although in the calculations we usually neglect the effects of the air leaving and entering the room.

P1 Basics of thermodynamics

61

P1.7 To heat a vessel of the mass ms = 0.6 kg and the specific heat capacity cs = 600 J/(kg K) and water of the mass mw = 2 kg and the specific heat capacity cw = 4200 J/(kg K) from θ 1 = 20 ○ C to boiling, that is, θ 2 = 100 ○ C, we need to provide heat (1.20) Q = ms cs (θ 2 − θ 1 ) + mw cw (θ 2 − θ 1 ) = 7.0 × 105 J. A heater of power P = 1 × 103 W provides the heat to the vessel and water by converting the equal amount of electrical energy E into internal energy and transferring it to water as heat Q. From the expression for the power P=

E , t

we get the time t in which the converted energy is equal to the required heat: t=

E Q = = 700 s. P P

In this problem we have implicitly assumed that the efficiency of the heater is 100 %, which is usually very close to reality.

i

P1.8 In calorimetry problems, that is, problems dealing with the transfer of heat to or from substances, we are usually interested in the final equilibrium state when no more heat is transferred, that is, when all substances have the same temperature (Section 1.2). The solution usually requires three crucial steps: 1. Determine which substances absorb the heat and which release it. 2. Write down the heat absorbed by or released from each substance as a positive quantity by subtracting the higher temperature from the lower temperature. 3. Equate the total amount of absorbed heat with the total amount of released heat to satisfy the law of conservation of energy.

In this problem, heat is released from a glass sphere of radius r = 0.025 m, density ρ = 2500 kg/m3 and specific heat capacity cs = 800 J/(kg K). We must first calculate the volume V and the mass ms of the sphere as ms = ρ V = ρ 43 π r 3 = 0.164 kg

Q

!

Problems in Building Physics

62 to write down the released heat Qr (1.20) as Qr = ms cs (θ s − θ ′ ),

where θ s = 150 ○ C is the initial temperature of the sphere and θ ′ is the final temperature of all substances. On the other hand, heat is absorbed by water of mass mw = 2 kg and specific heat capacity cw = 4200 J/(kg K). The absorbed heat Qa is Qa = mw cw (θ ′ − θ w ), where θ w = 18 ○ C is the initial temperature of the water. If we equate the released heat and the absorbed heat, Qr = Qa , we get: θ′ =

ms cs θ s + mw cw θ w = 20 ○ C. ms cs + mw cw

P1.9 Heat is released from an aluminium cube of mass mc = 1 kg, and specific heat capacity cc = 900 J/(kg K). The released heat Qr (1.20) is Qr = mc cc (θ c − θ ′ ), where θ c = 600 ○ C is the initial temperature of the cube and θ ′ is the final temperature of the cube and the water.

Q

On the other hand, heat is absorbed by water of mass mw = 1 kg, specific heat capacity cw = 4200 J/(kg K) and specific heat of vaporisation qv = 2.26 × 106 J/kg. The expression for the absorbed heat depends on the final state of the water (liquid, gaseous or both). Because the phase transition of the water starts at θ 0 = 100 ○ C, we can determine the final state by comparing the heat released by the aluminium cube for the final temperature θ 0 Q 1 = mc cc (θ c − θ 0 ) = 450 kJ with the heat required to heat all the water from the initial temperature θ w = 20 ○ C to θ 0 Q 2 = mw cw (θ 0 − θ w ) = 336 kJ and with the heat required to vaporise all the water (1.30) Q 3 = mw qv = 2260 kJ. We see that the released heat is sufficient to heat all the water to θ 0 (Q 1 > Q 2 ), but not to vaporise all the water as well (Q 1 < Q 2 + Q 3 ). Only part of the water vaporises and, as

P1 Basics of thermodynamics

63

we argued in Section 1.7, the final temperature must be θ ′ = 100 ○ C because only at this temperature are liquid and gaseous water in equilibrium. The absorbed heat Qa is thus used to heat all the water from θ w to θ ′ and then to vaporise only part of the water of mass me : Qa = mw cw (θ ′ − θ w ) + me qv . If we equate the released heat and the absorbed heat, Qr = Qa , we get: me =

1 [mc cc (θ c − θ ′ ) − mw cw (θ ′ − θ w )] = 0.05 kg. qv

P1.10 Heat is released from steel of mass ms = 1 kg and specific heat capacity cs = 470 J/(kg K). The released heat Qs (1.20) is Qs = ms cs (θ s − θ ′ ),

w s v

○

where θ s = 500 C is the initial temperature of the steel and θ ′ is the final temperature of all substances. On the other hand, heat is absorbed by water of mass mw = 0.5 kg, specific heat of fusion qwf = 3.36 × 105 J/kg, specific heat capacity in the solid phase cws = 2100 J/(kg K) and specific heat capacity in the liquid phase cwl = 4200 J/(kg K). The expression for the absorbed heat depends on the final state of the water (solid, liquid or both), which in turn also depends on a vessel of mass mv = 0.45 kg and specific heat capacity cv = 470 J/(kg K) at initial temperature θ v = 25 ○ C. Because the phase transition of the water starts at θ 0 = 0 ○ C, we can determine the final state by comparing the heat released by the steel and vessel for the final temperature θ 0 Q 1 = ms cs (θ s − θ 0 ) + mv cv (θ v − θ 0 ) = 240 kJ with the heat required to heat all the water from the initial temperature θ w = −10 ○ C to θ 0 Q 2 = mw cws (θ 0 − θ w ) = 10.5 kJ and with the heat required to melt all the water (1.29) Q 3 = mw qwf = 168 kJ. We see that the released heat is sufficient to heat to θ 0 and melt all the water (Q 1 > Q 2 +Q 3 ). The heat absorbed by the water Qw is thus used to heat solid water to θ 0 , then melt it to liquid water and finally heat liquid water from θ 0 to θ ′ : Qw = mw cws (θ 0 − θ w ) + mw qwf + mw cwl (θ ′ − θ 0 ).

Problems in Building Physics

64

In the case of more than two substances, the substance with the highest temperature releases the heat and the substance with the lowest temperature absorbs the heat, but we cannot be sure about the role of other substances. In our case, the vessel either releases heat or absorbs heat, as shown in the preceding figure, depending on whether its initial temperature θ v = 25 ○ C is higher or lower than the final temperature θ ′ . Fortunately, if we make an incorrect assumption, the calculated heat is negative and the direction of heat flow reverses, automatically correcting our wrong assumption. Let us assume that the final temperature θ ′ is higher than the initial temperature θ v and that the vessel absorbs heat, whose value is Qv = mv cv (θ ′ − θ v ). If we equate the released heat and the absorbed heat, Qs = Qw + Qv , we get: θ′ =

mw cwl θ 0 + ms cs θ s + mv cv θ v − mw qwf − mw cws (θ 0 − θ w ) = 22 ○ C. mw cwl + ms cs + mv cv

P1.11 The actual process is complex, but can be rationalised as follows: We divide water of mass m = 2 kg into two parts. One part of the water of mass m′ releases the heat and freezes, whereas the other part of the water of mass m −m′ absorbs the heat and vaporises. The absorbed heat (1.30) and the released heat (1.29) are

m−m′

Q

m′

Qa = (m − m′ )qv , Qr = m′ qf , where qv = 2.5 × 106 J/kg and qf = 3.36 × 105 J/kg are specific heats of vaporisation and fusion, respectively. The released heat and the absorbed heat must be equal, Qr = Qa , so we get: m′ =

qv m = 1.8 kg. qv + qf

P1.12 The heat is released from dry air of specific heat capacity cda = 1000 J/(kg K) and the mass mda , which is calculated from the dimensions of the room a = 10 m, b = 20 m, h = 3 m and the density of dry air ρ = 1.165 kg/m3 as mda = ρ V = ρ a b h = 700 kg,

P1 Basics of thermodynamics

65

where V is the volume of the room. The released heat (1.20) is Qr = mda cda (θ da − θ ′ ), where θ da = 30 ○ C is the initial temperature of dry air and θ ′ = 20 ○ C is the final temperature of all substances. On the other hand, heat is absorbed by water of heat capacity cw = 4200 J/(kg K) and specific heat of vaporisation qv = 2.45 × 106 J/kg. The heat used to heat all the water of mass mw = 12 kg from θ w = 5 ○ C to θ ′ and to evaporate part of the water of mass m (1.30) is Qa = mw cw (θ ′ − θ w ) + m qv . If we equate the released heat and the absorbed heat, Qr = Qa , we get: m=

1 [mda cda (θ da − θ ′ ) − mw cw (θ ′ − θ w )] = 2.5 kg. qv

Note that evaporation, heating and cooling processes occur simultaneously, the former over the entire temperature range. Fortunately, only the initial and final states matter, so we can rationalise all three processes as if they occurred in separate, sequential steps.

P1.13 To calculate the water-related power of perspiration and respiration, we first calculate the heat required to transform mw = gas gas 0.8 kg of water from the liquid state at θ 1 = 20 ○ C to the gaseous 20 ○ C 37 ○ C (a) state at θ 2 = 37 ○ C. This process is complex and involves simultaneous heating and evaporation over the entire temperature (b) liquid liquid range. Fortunately, only the initial and final states matter. There20 ○ C 37 ○ C fore, we can rationalise the processes as two separate, sequential steps. There are two possible paths: (a) water is first evaporated at θ 1 and then the water vapour is heated to θ 2 , or (b) liquid water is first heated to θ 2 and then evaporated at θ 2 . Because the problem gives the specific heat capacity of liquid water cw = 4200 J/(kg K) and the specific heat of vaporisation of water at 37 ○ C, qv = 2.41 × 106 J/kg, we take path (b) and write the provided heat, (1.20) and (1.30), as Qw = mw cw (θ 2 − θ 1 ) + m qv = 1.99 × 106 J. The released water-related power of perspiration and respiration is obtained by dividing the provided energy Ew , that is, the heat Qw , by the time t = 1 d = 86 400 s in which heat is released: Ew Qw Pw = = = 23.0 W. t t To calculate the air-related power of respiration, we have to calculate the mass of air ma from its volume V = 11 m3 and density ρ = 1.2 kg/m3 and then use its specific heat capacity

i

Problems in Building Physics

66 ca = 1000 J/(kg K):

Qa = ma ca (θ 2 − θ 1 ) = ρ V ca (θ 2 − θ 1 ) = 0.23 × 106 J. The released air-related power of respiration, obtained by dividing the provided energy Ea , that is, the heat Qa , by the time t in which heat is released, amounts to Pa =

Ea Qa = = 2.6 W. t t

The total power of perspiration and respiration is therefore P = Pw + Pa = 26 W.

P1.14 The heat required to heat m = 1.8 kg of water from θ 1 = 12 ○ C to θ 2 = 38 ○ C (1.20) is Q = m c (θ 2 − θ 1 ), where c = 4200 J/(kg K) is the specific heat capacity of water. The required electrical energy is then calculated from the efficiency η = 0.94: Q E Q m c (θ 2 − θ 1 ) Ô⇒ E = = . η η η=

The required electrical power, obtained by dividing the result by the time it takes to heat the water, t = 60 s, is P=

E m c (θ 2 − θ 1 ) = = 3.5 × 103 W. t ηt

P1.15 To compensate for the fact that electricity is k = 2.5 times more expensive than gas, the coefficient of performance of the heat pump must be k times larger than the efficiency of the furnace η = 1: COP = k η = 2.5 = 250 %.

P1 Basics of thermodynamics

67

The actual coefficient of performance is f = 0.5 times the ideal coefficient of performance (1.44), so the coefficient of performance is COP = f

Tw , Tw − Te

where Tw = 328 K (θ w = 55 ○ C) and Te are the water and external temperature, respectively. Note that the coefficient of performance decreases as the external temperature decreases. The threshold temperature is Te = Tw −

f Tw = 262.4 K. COP

The actual coefficient of performance of heat pumps is a very volatile value and depends on many additional factors, such as the modulation level (ratio of actual to maximum heat pump output) and the temperature difference between the incoming and outgoing water. Part of the transferred heat can also be diverted to defrost the external unit and prepare the domestic hot water. Apart from this, the theoretical expression derived for the Carnot cycle describes the coefficient of performance quite well up to a certain constant factor. The ratio of electricity to gas prices per kWh of about 2.5 is quite common and reasonable ratio considering that the efficiency of thermal power plants is about 40 %, that is, about 2.5 kWh of gas energy is needed to produce 1.0 kWh of electrical energy.

P1.16 The actual coefficient of performance is f = 0.5 times the ideal coefficient of performance (1.44), so the coefficient of performance is COP = f

Tw . Tw − Te

For two water temperatures, Tw1 = 328 K (θ w1 = 55 ○ C) and Tw2 = 308 K (θ w2 = 35 ○ C) and the same external temperature Te = 271 K (θ e = −2 ○ C), we get: Tw1 = 2.88 = 288 %, Tw1 − Te Tw2 COP2 = f = 4.16 = 416 %. Tw2 − Te

COP1 = f

The electrical power P required to transfer the heat flow rate Φ is obtained from the definition of COP (1.34): Q Φ COP = = . A P

i

i

Problems in Building Physics

68

The energy savings expressed as a percentage are therefore P1 − P2 = P1

i

Φ COP 1

−

Φ COP 2

Φ COP 1

=

COP2 − COP1 = 0.31 = 31 %. COP2

Underfloor heating is as effective as radiators even at much lower water temperatures because (a) the heat transfer area is larger and (b) the convective surface coefficient for the upward heat flow is twice as large as that for the horizontal heat flow (Table 2.3). The latter is due to the fact that the warmer air near the floor has a smaller density and is pushed upwards in the room by buoyancy, making heat transfer much more efficient.

P2 Heat transfer P2.1 The heat flow rate entering a freezer through expanded polystyrene depends on its thickness d = 0.05 m, thermal conductivity λ = 0.04 W/(m K) and area. The area is calculated from the dimensions of the freezer h = 1.5 m, a = 0.6 m and b = 0.6 m as A = 2 a b + 2 a h + 2 b h = 4.32 m2 .

Φin

θe

θi

The heat flow rate is then (2.5) Φin = λ

A(θ e − θ i ) = 166 W, d

Φout

where θ e = 30 ○ C and θ i = −18 ○ C are the external and internal temperatures, respectively. If we want to keep the internal temperature constant, we must remove all the heat that has entered the freezer with a heat pump. Therefore, the heat flow rate pumped out Φout must be equal to the heat flow rate entering the freezer, as in Φin = Φout . Considering that the efficiency of the heat pump, that is, the coefficient of performance, is COP = 3, the actual electrical power P that must be supplied to the heat pump is smaller (1.33): Qout Φout = A P Φin Ô⇒ P = = 55 W. COP COP =

In this problem we have neglected the effects of radiation and convection, which will be examined in later sections. Furthermore, the calculated heat flow entering the freezer is underestimated, because we have not taken into account the heat flow through the freezer door seals.

i 69

Problems in Building Physics

70

P2.2 The heat flow rate into the vessel depends on its surface area, which is calculated from the edge lengths a = 0.3 m, b = 0.35 m and c = 0.45 m as A = 2 a b + 2 a c + 2 b c = 0.795 m2 . At the beginning, the heat flow rate entering the vessel through the wall that consists only of an eps layer of thickness dEPS = 0.02 m and thermal conductivity λEPS = 0.04 W/(m K) (2.5) is A(θ e − θ i ) Φ 1 = λEPS = 47.7 W, dEPS where θ e = 30 ○ C and θ i = 0 ○ C are the external and internal temperatures, respectively. The heat required to melt the ice of mass m = 2 kg and specific heat of fusion qf = 3.36 × 105 J/kg (1.29) is provided by the heat flow entering the vessel (2.2): Q = m qf = Φ t. The initial melting time is therefore t1 =

m qf = 1.41 × 104 s = 3.9 h. Φ1

To increase the melting time to t 2 = 7 h = 2.52 × 104 s, we have to reduce the heat flow rate entering the vessel to m qf Φ2 = = 26.7 W. t2 The heat flow rate is reduced by adding an xps layer of thermal conductivity λXPS = 0.035 W/(m K). To calculate the heat flow rate through two layers, we take another form of Fourier’s law (2.9) and use the rule for serial layers (2.14): Φ2 =

A(θ e − θ i ) A(θ e − θ i ) A(θ e − θ i ) = = d EPS R REPS + RXPS + dλXPS λ EPS XPS

Ô⇒ dXPS = λXPS (

i

A(θ e − θ i ) dEPS − ) = 0.014 m. Φ2 λEPS

Note that the temperature in the vessel is constant throughout the melting process. Also, in this problem we have neglected the effects of radiation and convection, which will be examined in later sections.

P2 Heat transfer

71

P2.3 θ1 θ0

θ2

θ3 q

x

d1

d2

As we argued in Section 2.2.2, the temperatures in the wall do not change with time, so the heat flow rate Φ through the wall must be constant and shared by all the layers. However, its value still depends on the area of the wall A, which is inconvenient because the temperatures do not depend on the area. We therefore move on to the density of heat flow rate q (2.3), which is also constant and shared by all the layers. We begin the solution to problem by determining its value. Because we know the temperatures on the external and internal surfaces of the wall, θ 1 = −3 ○ C and θ 3 = 17 ○ C, respectively, we can calculate the density of heat flow rate using a special form of Fourier’s law (2.10) for the entire wall and the rule for serial layers (2.14) as θ3 − θ1 θ3 − θ1 θ3 − θ1 W q= = = = 6.57 2 , R R 1 + R 2 dλ 1 + dλ 2 m 1

2

where d 1 = 0.08 m and d 2 = 0.15 m are the thicknesses; and λ 1 = 0.038 W/(m K) and λ 2 = 0.16 W/(m K) are the thermal conductivities of the two layers. To obtain the interface temperature θ 2 we write another form of Fourier’s law (2.6) for one of the layers: θ3 − θ2 d2 d2 Ô⇒ θ 2 = θ 3 − q = 10.8 ○ C. λ2 q = λ2

We can also write a Fourier’s law for only part of a layer. Because the temperature θ 0 = 0 ○ C is obviously in the outer layer, we can directly write Fourier’s law for the distance x between the external surface and the position of this temperature: θ0 − θ1 x θ0 − θ1 Ô⇒ x = λ 1 = 0.017 m. q q = λ1

!

Problems in Building Physics

72

P2.4

qsol

α

θi

The heat flow rate intercepted by the surface for the solar density of heat flow rate qsol = 1000 W/m2 (2.17) is Φin = A qsol cos θ i , where A is the surface area and θ i is the angle of incidence. Note that the angle of incidence is measured from the line perpendicular to the surface (Fig. 2.4). So its value for case (a) is θ i,a = 0○ and for case (b) is θ i,b = 90○ − α = 30○ , where α = 60○ is the angle between the surface and the sunbeams. The surface also emits radiation. The emitted heat flow rate for a black surface (2.34) is Φem = A q(T) = A σ T 4 , where T is the temperature of the surface. If the temperature T is steady, the absorbed heat flow rate must be equal to the emitted heat flow rate (Fig. 2.7): Φin = Φem , A qsol cos θ i = A σ T 4 . The gained temperature is therefore √ T=

4

qsol cos θ i , σ

so we get Ta = 364 K (θ a = 91 ○ C) and Tb = 351 K (θ b = 78 ○ C) for cases (a) and (b), respectively. On the other hand, a grey surface absorbs less radiation than a black one, as quantified by the absorptance α (2.36), and emits less radiation, as quantified by the emittance ε (2.39): α A qsol cos θ i = ε A σ T 4 . Taking into account Kirchhoff ’s law (2.44), α = ε, the absorptance and emittance cancel each other out, and the temperatures of the grey and black surfaces are equal.

i

This result is obviously incorrect, because in reality, the grey surface should gain a lower temperature. The error is the result of neglecting other effects, in particular the convective cooling of the surface. We account for this effect in problem P2.5.

P2 Heat transfer

73 θi

θi

The influence of the angle of incidence on temperature explains many important natural effects, such as the difference in temperature between the summer and winter seasons. Although the distance between the Sun and the Earth is practically the same throughout the year, the angle of incidence is much smaller in the summer season than in the winter season, resulting in temperature fluctuations. The same argument explains the difference in temperature between morning and noon.

P2.5 In problem P2.4, we determined that the intercepted radiative heat flow rate Φin and the emitted radiative heat flow rate Φem for a black surface are Φin = A qsol , Φem = A σ T 4 , where A is the surface area, T is the surface temperature and qsol = 1000 W/m2 is the solar density of heat flow rate. However, the surface also releases heat by convection (2.30), whose heat flow rate is Φcv = A hc (T − Te ), where Te = 298 K (θ e = 25 ○ C) is the environment temperature and the convective surface coefficient (Table 2.3) is hc = 4 + 4v, where v = 2 m/s is the average wind speed. If the temperature T is steady, the absorbed heat flow rate must be equal to the released heat flow rate (Fig. 2.7): Φin = Φem + Φcv , A qsol = A σ T 4 + A(4 + 4v)(T − Te ), σ T 4 + (4 + 4v)T − (4 + 4v)Te − qsol = 0. It is extremely difficult to solve this complex equation analytically, so we will solve it by iteration. First, we write the equation in the form Tn = Te +

4 qsol − σ Tn−1 , 4 + 4v

where the n-th iteration of the temperature value Tn is obtained from the (n − 1)-th iteration Tn−1 .

i

Problems in Building Physics

74

Then we choose a reasonably good estimate (guess value) of the temperature, such as T1 = Te , and then calculate T2 , T3 , T4 . . . until the difference between two successive iterations of the temperature value is smaller than the desired precision. In this way, we get T = 327 K (θ = 54 ○ C). A grey surface absorbs less radiation than a black one, as quantified by the absorptance α (2.36), but also emits less radiation, as quantified by the emittance ε = 0.1 (2.39): α A qsol = ε A σ T 4 + A(4 + 4v)(T − Te ), ε σ T 4 + (4 + 4v)T − (4 + 4v)Te − ε qsol = 0. In the second step, we have taken into account Kirchhoff ’s law (2.44), α = ε. We write the equation in the form Tn = Te + ε

4 qsol − σ Tn−1 , 4 + 4v

choose a reasonably good estimate (guess value) of the initial temperature and iterate until the difference between two successive iterations of the temperature value is smaller than the desired precision. In this way, we get T = 302 K (θ = 29 ○ C). The grey surface gains a lower temperature than the black surface, which is consistent with our expectations.

P2.6 T0 , A 0 , ε 0

Tp , Ap , εp

i

To obtain the correct estimate of the temperature, we should also take into account the radiative heat exchange with the environment, the contribution of the sky and the wavelength dependence of the emittance, which would complicate the calculation considerably.

We will show that the situation with respect to the radiative heat flow rate is equivalent to that in Section 2.4.3. We can assume that the temperatures of all objects in the room, including the air, except the person, are equal. Because the temperatures of all objects in the environment are equal, the net radiation exchange for each pair of objects is zero, and the only nonzero net radiation exchange occurs between the surface of the person and the environment. Thus, we can consider the environment as a single surface of area A 0 , average temperature T0 = 293 K (θ 0 = 20 ○ C) and average emittance ε 0 and

P2 Heat transfer

75

use the expression for the radiative exchange between two grey surfaces (2.46). On the other hand, the person is a convex surface of area Ap = 1.8 m2 , temperature Tp = 301 K (θ p = 28 ○ C) and emittance εp = 0.97. All the radiation that leaves the surface of the person is intercepted by the surface of the environment, so the view factor is Fp0 = 1. Taking into account that A 0 ≫ Ap and linearising Tp4 −T04 , we obtain the same expressions as before, (2.47) and (2.48): W , m2 K Φr = Ap hr (θ p − θ 0 ) = 83 W. hr = 4 σ εp T 3 = 5.76

Here, T = 297 K is the average of T0 and Tp . For the convective heat flow rate, we simply use the Newton’s law of cooling (2.30). Because the person is considered a vertical surface, we take the internal convective surface coefficient for the horizontal heat flow, hc = 2.5 W/(m2 K) (Table 2.3). The convective heat flow rate is therefore Φcv = Ap hc (θ p − θ 0 ) = 36 W. For the conductive heat flow rate we simply use Fourier’s law (2.9) and obtain: Φcd =

As (θ s − θ 0 ) = 2.6 W. R

Here, As = 0.05 m2 is the surface area of the shoe sole, θ s = 33 ○ C is the temperature of the foot sole skin, and R = 0.25 m2 K/W is the thermal resistance of the shoe sole. Combining the radiative, convective and conductive heat losses with the perspiration and respiration heat losses calculated in Problem P1.13, we obtain total heat losses of about 140 W. All the obtained values agree very well with the generally accepted values for human heat losses. These losses also represent a heat gain for the building in which a person resides. This is significant for buildings of small heat losses and should be included in the overall energy balance of the building. Obviously, a person is neither a completely vertical nor a convex surface. Radiation is also transmitted between different parts of the human body, legs, arms, torso, head. . . , so the actual view factors are Fpp > 0 and Fp0 < 1. However, this simplified calculation still gives a reasonably good result.

i

i

P3 Heat transfer in building components P3.1 θe

θ se

θ si

θi

q

As the preceding figure shows, in real situations, we must take into account the contribution of two air layers on either side of the glass pane, quantified by corresponding surface resistances. For a vertical building element, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). The total thermal resistance (3.2) is therefore Rtot = Rse +

d m2 K + Rsi = 0.178 , λ W

where λ = 0.8 W/(m K) and d = 6 × 10−3 m are the thermal conductivity and thickness of the glass pane, respectively. As in the solution to problem P2.3 , we find that all layers share the same density of heat flow rate value and we begin the solution to problem by determining its value. Because we know the external and internal temperatures, θ e = −1 ○ C and θ i = 20 ○ C, respectively, we calculate the density of heat flow rate using a special form of Fourier’s law (2.10) that relates these two temperatures as q=

θi − θe W = 118 2 . Rtot m

77

Problems in Building Physics

78

To obtain the remaining characteristic temperatures, we write two forms of Fourier’s law, (2.10) and (2.6), for two of the layers: θ se − θ e Ô⇒ θ se = θ e + Rse q = 3.7 ○ C, Rse θ si − θ se d q=λ Ô⇒ θ si = θ se + q = 4.6 ○ C. d λ q=

i

Note that the thermal resistance of the glass pane, d/λ, represents only 4 % of the total thermal resistance. The temperature drop across the glass pane is therefore small compared to the total temperature drop. As discussed in Section 3.1.4, the function of a glass pane is primarily to separate air layers.

P3.2 θe

θ se

θ ′1

θ ′2

θ si

θi

q

As the preceding figure shows, we must take into account the contribution of two air layers on either side of the glazing and one inside. For the vertical building element, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1), whereas the thermal resistance of the airspace is given as Ra = 0.19 m2 K/W . The total thermal resistance (3.12) is therefore Rtot = Rse +

d d m2 K + Ra + + Rsi = 0.37 , λ λ W

where λ = 0.8 W/(m K) and d = 4 × 10−3 m are the thermal conductivity and thickness of the individual glass pane, respectively. As in the solution to problem P3.1 , we begin the calculation by determining the density of heat flow rate q through the building component. Because we know the temperatures, θ si = 15 ○ C and θ i = 20 ○ C, we calculate the density of heat flow rate by applying a special form of Fourier’s law (2.10) to the layer between these temperatures, that is, the internal air layer, as θ i − θ si W q= = 38.5 2 . Rsi m To obtain the external temperature θ e and the remaining characteristic temperatures, we write various forms of Fourier’s law, (2.10) and (2.6), to the entire building component

P3 Heat transfer in building components

79

and the individual layers: θi − θe Ô⇒ θ e = θ i − Rtot q = 5.8 ○ C, Rtot θ se − θ e q= Ô⇒ θ se = θ e + Rse q = 7.3 ○ C, Rse θ ′ − θ se d q=λ 1 Ô⇒ θ ′1 = θ se + q = 7.5 ○ C, d λ θ ′2 − θ ′1 ′ ′ q= Ô⇒ θ 2 = θ 1 + Ra q = 14.8 ○ C. Ra q=

Note that the cumulative thermal resistance of the two glass panes, 2d/λ, represents only 3 % of the total thermal resistance. The temperature drop across the glass panes is therefore small compared to the total temperature drop. As discussed in Section 3.1.4, the function of glass panes is primarily to separate air layers.

P3.3 In the problem we examine the wall with three different combinations of a stone layer of thickness ds = 0.02 m and thermal conductivity λs = 2 W/(m K); a concrete layer of thickness dc = 0.2 m and thermal conductivity λc = 1 W/(m K); and an eps layer of thickness de = 0.15 m and thermal conductivity λe = 0.035 W/(m K). Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). The external and internal temperatures are θ e = −5 ○ C and θ i = 20 ○ C, respectively.

Situation ∣stone∣concrete∣ θ e θ se θ ′1

θ si θ i q

The thermal transmittance of the first combination (3.3) is U=

1 Rse +

ds λs

+

dc λc

+ Rsi

= 2.63

W . m2 K

To determine the characteristic temperatures, we use the computational method described in Section 3.1.2. We begin by calculating the density of heat flow rate (3.5) as q = U(θ i − θ e ) = 65.8

W . m2

i

Problems in Building Physics

80 Then we apply (3.6) to each individual layer:

θ se − θ e Ô⇒ θ se = θ e + Rse q = −2.37 ○ C, Rse θ ′ − θ se ds q = λs 1 Ô⇒ θ ′1 = θ se + q = −1.71 ○ C, ds λs θ si − θ ′1 d c q = λc Ô⇒ θ si = θ ′1 + q = 11.45 ○ C. dc λc q=

Note that the temperature at the internal surface, θ si = 11.45 ○ C, is very low because the wall does not have good thermal insulation.

Situation ∣stone∣concrete∣eps∣ θ e θ se θ ′1

θ ′2

θ si θ i

q

The thermal transmittance of the second combination (3.3) is U=

1 Rse +

ds λs

+

dc λc

+

de λe

+ Rsi

= 0.214

W . m2 K

We begin by calculating the density of heat flow rate (3.5) as q = U(θ i − θ e ) = 5.36

W . m2

Then we apply (2.6) and (2.10) to each individual layer: θ se − θ e Ô⇒ θ se = θ e + Rse q = −4.79 ○ C, Rse θ ′ − θ se ds q = λs 1 Ô⇒ θ ′1 = θ se + q = −4.73 ○ C, ds λs ′ ′ dc θ − θ1 Ô⇒ θ ′2 = θ ′1 + q = −3.66 ○ C, q = λc 2 dc λc θ si − θ ′2 de q = λe Ô⇒ θ si = θ ′2 + q = 19.30 ○ C. de λe q=

Note that the temperature at the internal surface, θ si = 19.30 ○ C, has increased considerably compared to the previous situation, which is due to good thermal insulation. However, the temperature inside the wall, θ ′2 = −3.66 ○ C, is very low because the thermal insulation is on the inner side of the wall.

P3 Heat transfer in building components

81

Situation ∣stone∣eps∣concrete∣ θ e θ se θ ′1

θ ′2

θ si θ i q

The thermal transmittance of the third combination (3.3) is U=

1 Rse +

ds λs

+

de λe

+

dc λc

+ Rsi

= 0.214

W . m2 K

We begin by calculating the density of heat flow rate (3.5) as q = U(θ i − θ e ) = 5.36

W . m2

Neither the thermal transmittance nor the density of heat flow rate differ between the second and the third combination, as the layer order does not change the overall heat transmit performance of the wall. Then we apply Fourier’s law, (2.6) and (2.10), to each individual layer: θ se − θ e Ô⇒ θ se = θ e + Rse q = −4.79 ○ C, Rse θ ′ − θ se ds q = λs 1 Ô⇒ θ ′1 = θ se + q = −4.73 ○ C, ds λs θ ′2 − θ ′1 d e q = λe Ô⇒ θ ′2 = θ ′1 + q = 18.23 ○ C, de λe θ si − θ ′2 dc q = λc Ô⇒ θ si = θ ′2 + q = 19.30 ○ C. dc λc q=

Note that the temperature inside the wall, θ ′2 = 18.23 ○ C, has increased considerably compared to the previous situation, which is due to the fact that the thermal insulation is now on the outer side of the wall.

P3.4 To obtain the daily heat demand Q, we multiply the heat flow rate of the heat losses Φ by t = 1 d = 86 400 s (2.2) as in Q = Φ t. On the other hand, to calculate the gas volume consumption V from the heat value of the natural gas HV = 3.3 × 107 J/m3 , we have to construct the appropriate expression as HV =

Q . V

Problems in Building Physics

82 The daily gas volume consumption is therefore V=

Φt . HV

The heat flow rate of the heat losses is the sum of the heat flow rates through the walls and through the windows. The total area of the thermal envelope is Atot = 50 m2 , the area of the windows is Awd = 20 m2 , so the area of the walls is Awl = Atot − Awd = 30 m2 . The thermal transmittances for the uninsulated and the insulated wall, Uwlu = 2.63 W/(m2 K) and Uwli = 0.214 W/(m2 K), respectively, were determined in the solution to problem P3.3 , whereas the thermal transmittance of the windows Uwd = 1.1 W/(m2 K) is given in this problem. The heat losses also depend on the external and internal temperatures, θ e = −5 ○ C and θ i = 20 ○ C, respectively. For the uninsulated situation, the heat flow rate (3.4) and the daily gas volume consumption are Φu = Awl Uwlu (θ i − θ e ) + Awd Uwd (θ i − θ e ) = 2520 W, Φu t Vu = = 6.6 m3 , HV while for the uninsulated situation, the heat flow rate and the daily gas volume consumption are Φi = Awl Uwli (θ i − θ e ) + Awd Uwd (θ i − θ e ) = 710 W, Φi t Vi = = 1.9 m3 . HV Gas savings expressed as a percentage are defined as the ratio of the reduction in consumption, Vu − Vi , to the initial consumption Vu : Vu − Vi = 72 %. Vu Combining the preceding equations we get: Vu − Vi Φu − Φi Awl Uwlu − Awl Uwli = = . Vu Φu Awl Uwlu + Awd Uwd The temperature difference θ i − θ e cancels out in the second step. Thus, the gas savings are independent of the temperatures.

P3 Heat transfer in building components

83

P3.5 The mechanisms that transfer heat to and from the wall surface are convection and radiation, so the calculation of the surface resistance requires the calculation of convective and radiative surface coefficients. The external convective surface coefficient (Table 2.3) is hce = 4 + 4v = 44

W , m2 K

where v = 10 m/s is the average wind speed. On the other hand, the radiative surface coefficient, calculated from the surface emittance ε = 0.5 and the average of the external and surface temperatures T = 283 K (θ = 10 ○ C) (2.47), is hre = 4 σ ε T 3 = 2.57

W . m2 K

The external surface resistance (3.1) is thus Rse =

1 m2 K = 0.021 . hce + hre W

P3.6 The radiative surface coefficient of the airspace calculated from the surface emittances ε 1 = ε 2 = 0.9 and the average temperature T = 275 K (3.11) is hra =

4σ T 3 W = 3.9 2 . 1 1 m K + ε2 − 1 ε1

The mechanisms that transfer heat in the airspace are convection and radiation, so the thermal resistance of the airspace Ra = 0.11 m2 K/W is combination of the convective and radiative surface coefficients (3.10): Ra =

1 . hca + hra

The convective surface coefficient is thus hca =

1 W − hra = 5.2 2 . Ra m K

Problems in Building Physics

84

P3.7 Because the openings account for f = 80 % = 0.8 of each of the N = 10 grilles of size a = 0.1 m × b = 0.2 m, the total area of the openings is Ave = f N a b = 0.16 m2 . Because we are dealing with a horizontal air layer, the area of openings must be related to the area of the air layer A = 70 m2 . The area of the openings per 1 m2 of air layer surface is Ave 0.0023 m2 2300 mm2 = = A m2 m2 and the air layer is well-ventilated.

P3.8 θ ′1

θ e θ se

θ ′2 θ ′3

θ ′4 θ si θ i q

To calculate all characteristic temperatures, the structure of the building component and two characteristic temperatures must be provided. The layer thicknesses are d 1 = 0.1 m, d 2 = 0.05 m, d 3 = 0.016 m, d 4 = 0.15 m and d 5 = 0.015 m; and the layer thermal conductivities are λ 1 = 0.76 W/(m K), λ 3 = 0.35 W/(m K), λ 4 = 0.04 W/(m K) and λ 5 = 0.21 W/(m K). For the unventilated air layer, the convective surface coefficient hca = 1.25 W/(m2 K) and the emittances ε 1 = 0.9, ε 2 = 0.05 are given. The external and internal temperatures are θ e = −5 ○ C and θ i = 21 ○ C, respectively. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). The average temperature in the air layer is assumed to be T = 271 K (θ = −2 K). We begin by determining the thermal resistance of the airspace Ra . Because the airspace is unventilated and the mechanisms that transfer heat in the airspace are convection and radiation, we have to calculate the radiative surface coefficient hra (3.11) and combine it with the convective surface coefficient hca (3.10): hra = Ra =

1 ε1

W 4 σ T3 = 0.224 2 , m K + ε12 − 1

1 m2 K = 0.678 . hca + hra W

P3 Heat transfer in building components

85

Next, using the computational method described in Section 3.1.2, we calculate the thermal transmittance (3.13) and the density of heat flow rate (3.5) as U=

1 Rse +

d1 λ1

+ Ra +

d3 λ3

+

d4 λ4

+

d5 λ5

+ Rsi

q = U(θ i − θ e ) = 5.36

= 0.206

W , m2 K

W . m2

Finally, we apply Fourier’s law (3.6) to each individual layer: θ se = θ e + Rse q = −4.79 ○ C, d1 q = −4.08 ○ C, λ1 θ ′2 = θ ′1 + Ra q = −0.44 ○ C, θ ′1 = θ se +

d3 q = −0.20 ○ C, λ3 d4 θ ′4 = θ ′3 + q = 19.92 ○ C, λ4 d5 θ si = θ ′4 + q = 20.30 ○ C. λ5 θ ′3 = θ ′2 +

The temperature factor of the internal surface (3.8) is f Rsi =

θ si − θ e = 0.97. θi − θe

In real situations, the average temperature T is not known in advance and must be determined by iteration. In the first step, the characteristic temperatures are calculated using an estimate (guess value) for T. In the next step, the characteristic temperatures are calculated using the value for T calculated in the previous step. The process is repeated until the difference between the input and output values is sufficiently small. One iteration is usually sufficient.

P3.9 θ e θ se

θ ′1

θ ′2

θ ′3

θ ′4 θ si θ i

q

To calculate all characteristic temperatures, the structure of the building component and two characteristic temperatures must be provided. The layer thicknesses are d 1 = 0.12 m, d 2 = 0.04 m, d 3 = 0.05 m, d 4 = 0.19 m and d 5 = 0.01 m; and the layer thermal conductivities are λ 1 = 0.76 W/(m K), λ 3 = 0.04 W/(m K), λ 4 = 0.19 W/(m K) and λ 5 = 0.5 W/(m K). For the unventilated air layer, the emittances ε 1 = ε 2 = 0.9 are given.

!

Problems in Building Physics

86

The temperatures at the internal surface and the internal temperature are θ si = 19 ○ C and θ i = 20 ○ C, respectively. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). The average temperature and the temperature difference in the air layer are assumed to be T = 274 K (θ = 1 ○ C) and ∆T < 5 K (∆θ < 5 ○ C), respectively. We begin by determining the thermal resistance of the airspace Ra . The mechanisms that transfer heat in the airspace are convection and radiation, so we must determine the airspace convective and radiative surface coefficients. The convective surface coefficient for ∆T < 5 K and the horizontal heat flow is read from Table P3.1, hca = 1.25 W/(m2 K). The radiative surface coefficient is calculated from (P3.1) and then combined with the convective surface coefficient to obtain the thermal resistance of airspace (3.10): hra = Ra =

1 ε1

4σ T 3 W = 3.82 2 , m K + ε12 − 1

1 m2 K = 0.197 . hca + hra W

The next step is to determine the density of heat flow rate. Because we know only the temperatures θ si and θ i , the density of heat flow rate must be calculated differently than in Section 3.1.2. In particular, we apply a special form of Fourier’s law (3.6) to the layer between the temperatures we know, that is, to the internal air layer: q=

θ i − θ si W = 7.69 2 . Rsi m

There are a number of ways to calculate the remaining characteristic temperatures. Here, we will show two ways that start with the calculation of the external temperature θ e . The first way is to calculate the thermal transmittance (3.13) and then use the expression connecting it to the density of heat flow rate (3.5): U=

1 Rse +

d1 λ1

+ Ra +

d3 λ3

d4 λ4

d5 λ5

= 0.358

W , m2 K

+ Rsi q q = U(θ i − θ e ) Ô⇒ θ e = θ i − = −1.50 ○ C. U +

+

The second way is to calculate the total thermal resistance (3.12) and then apply Fourier’s law (3.6) to the entire wall: d1 d3 d4 d5 m2 K + Ra + + + + Rsi = 2.80 , λ1 λ3 λ4 λ5 W θi − θe q= Ô⇒ θ e = θ i − q Rtot = −1.50 ○ C. Rtot

Rtot = Rse +

P3 Heat transfer in building components

87

Finally, we apply Fourier’s law (3.6) to each individual layer: θ se = θ e + Rse q = −1.19 ○ C, d1 q = 0.02 ○ C, λ1 θ ′2 = θ ′1 + Ra q = 1.54 ○ C, θ ′1 = θ se +

d3 q = 11.15 ○ C, λ3 d4 θ ′4 = θ ′3 + q = 18.85 ○ C. λ4 θ ′3 = θ ′2 +

In real situations, the average temperature T and the temperature difference ∆T are not known in advance and must be determined by iteration. In the first step, the characteristic temperatures are calculated using estimates (guess values) for T and ∆T. In the next step, the characteristic temperatures are calculated using the values for T and ∆T calculated in the previous step. The process is repeated until the difference between the input and output values is sufficiently small. One iteration is usually sufficient.

P3.10 To calculate all characteristic temperatures, the structure of the building component and two characteristic temperatures must be provided. The layer thicknesses are d 1 = 0.025 m, d 2 = 0.05 m, d 3 = 0.025 m, d 4 = 0.12 m and d 5 = 0.025 m; and the layer thermal conductivities are λ 1 = 0.76 W/(m K), λ 3 = 0.14 W/(m K), λ 4 = 0.04 W/(m K) and λ 5 = 0.14 W/(m K). For the air layer, the emittances ε 1 = ε 2 = 0.9 are given. The external and internal temperatures are θ e = 0 ○ C and θ i = 20 ○ C, respectively. Because the building component is horizontal and the heat flows upwards, the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.1 m2 K/W (Table 3.1). There are less than 500 mm2 of ventilation openings per 1 m2 of roof surface, so the air layer is unventilated. The average temperature and the temperature difference in the air layer are assumed to be T = 274 K (θ = 1 ○ C) and ∆T < 5 K (∆θ < 5 ○ C), respectively.

θe θ se θ ′1 θ ′2 θ ′3 q

θ ′4 θ si θi

We begin by determining the thermal resistance of the airspace Ra . The mechanisms that transfer heat in the airspace are convection and radiation, so we must determine the airspace convective and radiative surface coefficients. The convective surface coefficient for ∆T < 5 K and upward heat flow is read from Table P3.1, hca = 1.95 W/(m2 K). The radiative surface coefficient is calculated from (P3.1) and then combined with the

!

Problems in Building Physics

88

convective surface coefficient to obtain the thermal resistance of airspace (3.10): hra =

4 σ T3 W = 3.82 2 , 1 1 m K + ε2 − 1 ε1

Ra =

1 m2 K = 0.173 . hca + hra W

Next, using the computational method described in Section 3.1.2, we calculate the thermal transmittance (3.13) and the density of heat flow rate (3.5): U=

1 Rse +

d1 λ1

+ Ra +

d3 λ3

+

d4 λ4

+

d5 λ5

+ Rsi

q = U(θ i − θ e ) = 5.40

= 0.270

W , m2 K

W . m2

Finally, we apply Fourier’s law (3.6) to each individual layer: θ se = θ e + Rse q = 0.22 ○ C, d1 q = 0.39 ○ C, λ1 θ ′2 = θ ′1 + Ra q = 1.33 ○ C, θ ′1 = θ se +

d3 q = 2.29 ○ C, λ3 d4 θ ′4 = θ ′3 + q = 18.50 ○ C, λ4 d5 θ si = θ ′4 + q = 19.46 ○ C. λ5 θ ′3 = θ ′2 +

!

In real situations, the average temperature T and the temperature difference ∆T are not known in advance and must be determined by iteration. In the first step, the characteristic temperatures are calculated using estimates (guess values) for T and ∆T. In the next step, the characteristic temperatures are calculated using the values for T and ∆T calculated in the previous step. The process is repeated until the difference between the input and output values is sufficiently small. One iteration is usually sufficient.

P3.11 To calculate thermal transmittance, the structure of the building component must be provided. The layer thicknesses are d 1 = 0.01 m, d 2 = 0.05 m, d 3 = 0.15 m, d 4 = 0.2 m and d 5 = 0.01 m; and the layer thermal conductivities are λ 1 = 0.92 W/(m K), λ 3 = 0.041 W/(m K), λ 4 = 1.1 W/(m K) and λ 5 = 0.2 W/(m K). For the airspace the emittances ε 1 = ε 2 = 0.9 are given.

P3 Heat transfer in building components

There are Ave = 800 mm2 of ventilation openings per 1 m of the airspace length in the horizontal direction, so the air layer is slightly ventilated and both the unventilated and well-ventilated situations must be calculated. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). The average temperature and the temperature difference in the air layer are assumed to be T = 275 K (θ = 2 ○ C) and ∆T < 5 K (∆θ < 5 ○ C). We begin by determining the thermal resistance of the unventilated airspace Ra . The mechanisms that transfer heat in the airspace are convection and radiation, so we must determine the airspace convective and radiative surface coefficients. The convective surface coefficient for ∆T < 5 K and the horizontal heat flow is read from Table P3.1, hca = 1.25 W/(m2 K). The radiative surface coefficient is calculated from (P3.1) and combined with the convective surface coefficient to obtain the thermal resistance of the unventilated airspace (3.10): hra =

4 σ T3 W = 3.86 2 , 1 1 m K + ε2 − 1 ε1

Ra =

1 m2 K = 0.196 . hca + hra W

The total thermal resistances of the unventilated air layer Rtot,nve and the well-ventilated air layer Rtot,ve (3.12) are d1 d3 d4 d5 m2 K + Ra + + + + Rsi = 4.27 , λ1 λ3 λ4 λ5 W d3 d4 d5 m2 K = Rsi + + + + Rsi = 4.15 . λ3 λ4 λ5 W

Rtot,nve = Rse + Rtot,ve

Note that if the air layer is well-ventilated, the thermal resistance of the air layer and all other layers between the air layer and the external environment is replaced by the internal surface resistance. The total thermal resistance is calculated as an interpolation between these two values (P3.2): Rtot =

1500 mm2 − Ave Ave − 500 mm2 m2 K R + R = 4.23 . tot,nve tot,ve 1000 mm2 1000 mm2 W

The thermal transmittance (3.3) is therefore U=

1 W = 0.236 2 . Rtot m K

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P3.12 To calculate all characteristic temperatures, the structure of the building component and two characteristic temperatures must be provided. The layer thicknesses are d 2 = 0.012 m, d 3 = 0.2 m, d 4 = 0.15 m and d 5 = 0.01 m; and the layer thermal conductivities are λ 2 = 0.12 W/(m K), λ 3 = 0.04 W/(m K), λ 4 = 2.3 W/(m K) and λ 5 = 0.5 W/(m K).

θe θ se

The external and internal temperatures are θ e = −2 ○ C and θ i = 20 ○ C, respectively. Because the building component is horizontal and the heat flows upwards, the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.1 m2 K/W (Table 3.1). The thermal resistance of roof space for a pitched sheeted roof is Ru = 0.2 m2 K/W (Table P3.2). Using the computational method described in Section 3.1.2, we calculate the thermal transmittance (3.3) and the density of heat flow rate (3.5): U=

1 Rse + Ru +

d2 λ2

+

d3 λ3

+

d4 λ4

+

d5 λ5

θ ′1 θ ′2

q

W = 0.181 2 , m K + Rsi

q = U(θ i − θ e ) = 3.98

W . m2

θ ′3

Then we apply Fourier’s law (3.6) to each individual layer: θ se = θ e + Rse q = −1.84 ○ C, θ ′1 = θ se + Ru q = −1.04 ○ C, d2 q = −0.65 ○ C, λ2 d3 θ ′3 = θ ′2 + q = 19.26 ○ C, λ3 d4 θ ′4 = θ ′3 + q = 19.52 ○ C, λ4 d5 θ si = θ ′4 + q = 19.60 ○ C. λ5 θ ′2 = θ ′1 +

la

lb

d6 d5 d4 d3 d2 d1

P3.13

θ ′4 θ si θi

In the simplified calculation described in Section 3.2.1, we only need to consider one repeating element, shown in the preceding figure, with the section a containing the timber beams and the section b representing the rest.

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91

The substructure essentially consists of six layers: the first layer of thickness d 1 = 0.008 m and thermal conductivity λ 1 = 0.50 W/(m K); the second layer of thickness d 2 = 0.016 m and thermal conductivity λ 2 = 0.35 W/(m K); the third layer of thickness d 3 = 0.06 m, which is divided into a section of thermal conductivity λa3 = 0.14 W/(m K) and a section of thermal conductivity λb3 = 0.04 W/(m K); the fourth layer of thickness d 4 = r 2 = 0.04 m and thermal conductivity λ 4 = 0.04 W/(m K); the fifth layer of thickness d 5 = 0.06 m, which is divided into a section of thermal conductivity λa5 = 0.14 W/(m K) and a section of thermal conductivity λb5 = 0.04 W/(m K); and the sixth layer of thickness d 6 = 0.025 m and thermal conductivity λ 6 = 0.21 W/(m K). Because the building component is vertical and the heat flows horizontally, the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). To calculate fractional areas (3.33), we take the horizontal lengths of the sections a and b, la = 0.06 m and lb = r 1 = 0.565 m, respectively, and the height of the substructure h: fa =

Aa h la la = = = 0.096, Aa + Ab h la + h lb la + lb fb = 1 − fa = 0.904.

For the upper limit of the total thermal resistance we first calculate the thermal resistance of the sections: d1 d2 d3 d4 d5 d6 m2 K + + + + + + Rsi = 2.21 , λ 1 λ 2 λa3 λ 4 λa5 λ 6 W d1 d2 d3 d4 d5 d6 m2 K = Rse + + + + + + + Rsi = 4.35 . λ 1 λ 2 λb3 λ 4 λb5 λ 6 W

Rtot;a = Rse + Rtot;b

The expression for the upper limit of the total thermal resistance and the resulting value are: Aa + Ab Aa Ab = + Rtot;upper Rtot;a Rtot;b Ô⇒ Rtot;upper = 3.98

m2 K . W

For the lower limit of the total thermal resistance we first calculate equivalent thermal conductivities: W , mK W , = λ 2 = 0.35 mK

λeq;1 = λ 1 = 0.5 λeq;2

λeq;3 = fa λa3 + fb λb3 = 0.05 λeq;4 = λ 4 = 0.4

W , mK

W , mK

λeq;5 = fa λa5 + fb λb5 = 0.05

W , mK

Problems in Building Physics

92 λeq;6 = λ 6 = 0.21

W . mK

The lower limit of the total thermal resistance is Rtot;lower = Rse +

d1 d2 d3 d4 d5 d6 m2 K + + + + + + Rsi = 3.77 . λeq;1 λeq;2 λeq;3 λeq;4 λeq;5 λeq;6 W

The thermal transmittance is then U=

1 2 W = = 0.258 2 . Rtot Rtot;upper + Rtot;lower m K

P3.14 In the simplified calculation described in Section 3.2.1, the section a contains the timber planks and the section b represents the rest. The wall has six layers: the first layer of thickness d 1 = 0.02 m and thermal conductivity λ 1 = 1.5 W/(m K); the second layer of thickness d 2 = 0.1 m and thermal conductivity λ 2 = 0.033 W/(m K); the third layer of thickness d 3 = 0.012 m and thermal conductivity λ 3 = 0.13 W/(m K); the fourth layer of thickness d 4 = r 2 = 0.14 m, which is divided into a section of thermal conductivity λa4 = 0.14 W/(m K) and a section of thermal conductivity λb4 = 0.038 W/(m K); the fifth layer of thickness d 5 = 0.012 m and thermal conductivity λ 5 = 0.13 W/(m K); and the sixth layer of thickness d 6 = 0.0125 m and thermal conductivity λ 6 = 0.21 W/(m K). Because the building component is vertical and the heat flows horizontally, the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). To calculate fractional areas (3.33), we first determine the total horizontal lengths of the sections a and b as la = 4 r 1 and lb = ltot − 4 r 1 , respectively, and take the height of the substructure h: fa =

Aa h la 4r 1 = 0.087, = = Aa + Ab h la + h lb ltot fb = 1 − fa = 0.913.

For the upper limit of the total thermal resistance we first calculate the thermal resistance of the sections: d1 d2 d3 d4 d5 d6 m2 K + + + + + + Rsi = 4.46 , λ 1 λ 2 λ 3 λa4 λ 5 λ 6 W d1 d2 d3 d4 d5 d6 m2 K = Rse + + + + + + + Rsi = 7.14 . λ 1 λ 2 λ 3 λb4 λ 5 λ 6 W

Rtot;a = Rse + Rtot;b

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The expression for the upper limit of the total thermal resistance and the resulting value are: Aa + Ab Aa Ab = + Rtot;upper Rtot;a Rtot;b Ô⇒ Rtot;upper = 6.79

m2 K . W

For the lower limit of the total thermal resistance we first calculate equivalent thermal conductivities: W , mK W = λ 2 = 0.033 , mK W = λ 3 = 0.13 , mK

λeq;1 = λ 1 = 1.5 λeq;2 λeq;3

λeq;4 = fa λa4 + fb λb4 = 0.048

W , mK

W , mK W = λ 6 = 0.21 . mK

λeq;5 = λ 5 = 0.13 λeq;6

The lower limit of the total thermal resistance is Rtot;lower = Rse +

d1 d2 d3 d4 d5 d6 m2 K + + + + + + Rsi = 6.39 . λeq;1 λeq;2 λeq;3 λeq;4 λeq;5 λeq;6 W

The thermal transmittance is then U=

1 2 W = = 0.152 2 . Rtot Rtot;upper + Rtot;lower m K

P3.15 In the simplified calculation described in Section 3.2.1, we only need to consider one repeating element with the section a containing the timber beams and the section b representing the rest. Because the air layer is well-ventilated, the thermal resistance of the air layer and all other layers between the air layer and the external environment is replaced by the internal surface resistance. The remaining roof structure consists of three layers: the first layer of thickness d 1 = 0.02 m and thermal conductivity λ 1 = 0.14 W/(m K); the second layer of thickness d 2 = 0.2 m, which is divided into a section of thermal conductivity λa2 = 0.04 W/(m K) and a section of thermal conductivity λb2 = 0.14 W/(m K); and the third layer of thickness d 3 = 0.0125 m and thermal conductivity λ 3 = 0.28 W/(m K).

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Because the building component is horizontal and the heat flows upwards, the internal surface resistance is Rsi = 0.10 m2 K/W (Table 3.1). To calculate fractional areas (3.33), we take the transverse lengths of the sections a and b, la = 0.55 m and lb = 0.15 m, respectively, and the longitudinal length of the roof structure l: fa =

Aa l la la = = = 0.786, Aa + Ab l la + l lb la + lb fb = 1 − fa = 0.214.

For the upper limit of the total thermal resistance we first calculate the thermal resistance of the sections: d1 d2 d3 m2 K + + + Rsi = 5.39 , λ 1 λa2 λ 3 W d1 d2 d3 m2 K = Rsi + + + + Rsi = 1.82 . λ 1 λb2 λ 3 W

Rtot;a = Rsi + Rtot;b

The expression for the upper limit of the total thermal resistance and the resulting value are: Aa + Ab Aa Ab = + Rtot;upper Rtot;a Rtot;b Ô⇒ Rtot;upper = 3.79

m2 K . W

For the lower limit of the total thermal resistance we first calculate equivalent thermal conductivities: λeq;1 = λ 1 = 0.14

W , mK

λeq;2 = fa λa2 + fb λb2 = 0.061 λeq;3 = λ 3 = 0.28

W , mK

W . mK

The lower limit of the total thermal resistance is Rtot;lower = Rsi +

d2 d3 m2 K d1 + + + Rsi = 3.64 . λeq;1 λeq;2 λeq;3 W

The thermal transmittance is then U=

1 2 W = = 0.269 2 . Rtot Rtot;upper + Rtot;lower m K

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P3.16 a

d

b

d

To calculate the thermal transmittance of the entire window, the thermal transmittances and the dimensions of the constituent parts must be provided. The thermal transmittances of the glazing and frame, Ug = 1.1 W/(m2 K) and Uf = 1.4 W/(m2 K), respectively, and the linear thermal transmittance of the junction of the glazing and the frame Ψgf = 0.07 W/(m K) are already given. Their dimensions, however, must be calculated from the dimensions of the window, a = 1.5 m, b = 1 m, and the width of the frame d = 0.1 m. The area of the glazing Ag , the area of the frame Af and the length of the junction of the glazing and the frame lgf are Ag = (a − 2d)(b − 2d) = 1.04 m2 , Af = a b − Ag = 0.46 m2 , lgf = 2(a − 2d) + 2(b − 2d) = 4.2 m. The heat flow rate for the entire window is the sum of the heat flow rates through the intermediate building components (3.4) and the heat flow rates through the linear thermal bridges (3.34), as in Φ = Ag Ug (θ i − θ e ) + Af Uf (θ i − θ e ) + lgf Ψgf (θ i − θ e ). The definition of the thermal transmittance of the entire window Uw and the resulting value are: Φ ≡ Aw Uw (θ i − θ e ) Ô⇒ Uw =

Ag Ug + Af Uf + lgf Ψgf W = 1.39 2 . Ag + Af m K

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P3.17 To calculate the heat flow rate of heat losses through the external walls, the thermal transmittances and the dimensions of the constituent parts must be provided. The thermal transmittances of the windows and walls, Uwd = 1.1 W/(m2 K) and Uwl = 0.25 W/(m2 K), respectively, and the linear thermal transmittances of the junction of the window and wall and of the junction of the two walls, Ψwd-wl = 0.05 W/(m K) and Ψwl-wl = −0.03 W/(m K), respectively, are already given. Their dimensions, however, must be calculated. The area of the N = 2 windows Awd , each of dimensions d 1 = 1.5 m and d 2 = 2 m and the length of the junction of the windows and the wall lwd-wl are Awd = N d 1 d 2 = 6 m2 , lwd-wl = N(2d 1 + 2d 2 ) = 14 m. The area of the walls Awl and the length of the junction of the two walls lwl-wl are Awl = a h + b h − Awd = 21.5 m2 , lwl-wl = h = 2.5 m. The heat flow rate through the walls is the sum of the heat flow rates through the intermediate building components (3.4) and the heat flow rates through the linear thermal bridges (3.34) as in Φ = Awd Uwd (θ i − θ e ) + Awl Uwl (θ i − θ e ) + lwd-wl Ψwd-wl (θ i − θ e ) + lwl-wl Ψwl-wl (θ i − θ e ) = 378 W, where θ i = 20 ○ C and θ e = −10 ○ C are the internal and external temperature, respectively. To achieve an energy balance, the power for heating must be equal to the heat flow rate of the heat losses through the external walls.

P3.18

b

d2

a

d1

First we have to calculate the heat flow rate through the external wall. For this we need the thermal transmittances and the dimensions of the constituent parts. The thermal transmittances of the windows and walls, Uwd = 1.1 W/(m2 K) and Uwl = 0.25 W/(m2 K),

P3 Heat transfer in building components

respectively, and the linear thermal transmittance of the junction of the windows and the wall, Ψwd-wl = 0.07 W/(m K) are already given. Their dimensions, however, must be calculated from the dimensions of the N = 5 windows, d 1 = 2 m and d 2 = 3 m, and the dimensions of the external wall, a = 15 m and b = 4 m. The area of the windows Awd , the area of the wall Awl and the length of the junction of the windows and the wall lwd-wl are Awd = N d 1 d 2 = 30 m2 , Awl = a b − Awd = 30 m2 , lwd-wl = N(2d 1 + 2d 2 ) = 50 m. The heat flow rate is the sum of the heat flow rates through the intermediate building components (3.4) and the heat flow rates through the linear thermal bridges (3.34) as in Φ = Awd Uwd (θ i − θ e ) + Awl Uwl (θ i − θ e ) + lwd-wl Ψwd-wl (θ i − θ e ) = 1100 W, where θ i = 20 ○ C and θ e = −5 ○ C are the internal and external temperature, respectively. If we want to keep the internal temperature constant, the energy we pump in and the energy that leaves the room through the external wall must be equal. So the calculated heat flow rate of the heat losses is equal to the heat flow rate of the pump. On the other hand, the actual electric power P that must be supplied to the heat pump also depends on the efficiency of the heat pump, that is, on the coefficient of performance COP = 2 (1.34): Q Φ = A P Φ Ô⇒ P = = 550 W. COP COP =

P3.19 To calculate the heat flow rate due to the direct heat losses, the thermal transmittances and the dimensions of the constituent parts must be provided. The thermal transmittances of the windows, roof and walls, Uwd = 1.1 W/(m2 K), Urf = 0.25 W/(m2 K) and Uwl = 0.21 W/(m2 K), respectively; and the linear thermal transmittances of the junction of the roof and the wall and of the junction of the walls, Ψrf-wl = 0.2 W/(m K) and Ψwl-wl = −0.05 W/(m K), respectively, are already given. The area of the windows Awd = 30 m2 is given too, but the other dimensions must be calculated. The areas of the roof Arf and walls Awl are Arf = a b = 80 m2 , Awl = 2 a h + 2 b h − Awd = 186 m2 .

97

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The length of the junction of the roof and the walls lrf-wl and the length of the junction of the walls lwl-wl are lrf-wl = 2 a + 2 b = 36 m, lwl-wl = 4 h = 24 m. The direct heat transfer coefficient (3.37) and the corresponding heat flow rate are then Hd = Awd Uwd + Arf Urf + Awl Uwl + lrf-wl Ψrf-wl + lwl-wl Ψwl-wl = 98.1

W , K

Φd = Hd (θ i − θ e ) = 1570 W, where θ i = 21 ○ C and θ e = 5 ○ C are the internal and external temperatures, respectively.

P3.20 To calculate the heat flow rate due to the direct heat losses, the thermal transmittances and the dimensions of the constituent parts must be provided. The thermal transmittances of the windows, roof and walls, Uwd = 1.10 W/(m2 K), Urf = 0.25 W/(m2 K) and Uwl = 0.21 W/(m2 K), respectively; and the linear thermal transmittances of the junction of the window and the wall, of the junction of the roof and the wall, of the external junction of the walls and of the internal junction of the walls, Ψwd-wl = 0.04 W/(m K), Ψrf-wl = 0.2 W/(m K), Ψwl-ex = −0.05 W/(m K) and Ψwl-in = 0.05 W/(m K), respectively, are already given. Their dimensions, however, must be calculated. We calculate the area of windows Awd and the length of the junction of the window and the wall lwd-wl from the dimensions of the N = 22 windows, d 1 = 1 m and d 2 = 1.5 m, as Awd = N d 1 d 2 = 33 m2 , lwd-wl = N(2d 1 + 2d 2 ) = 110 m. The area of the roof Arf and the area of the walls Awl , the length of the junction of the roof and the wall lrf-wl , the length of the external junction of the walls lwl-ex and the length of the internal junction of the walls lwl-in are Arf = a b − a ′ b ′ = 80 m2 , Awl = (2 a + 2 b)h − Awd = 195 m2 , lrf-wl = 2 a + 2 b = 38 m, lwl-ex = 5 h = 30 m, lwl-in = h = 6 m.

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The direct heat transfer coefficient (3.37) and the corresponding heat flow rate are then Hd = Awd Uwd + Arf Urf + Awl Uwl + lwd-wl Ψwd-wl + lrf-wl Ψrf-wl + lwl-ex Ψwl-ex + lwl-in Ψwl-in = 108 Φd = Hd (θ i − θ e ) = 2700 W, where θ i = 21 ○ C and θ e = −4 ○ C are the internal and external temperatures, respectively. All linear thermal transmittances for the house were given in terms of external dimensions. There is therefore no doubt about which edges of the house correspond to the external and internal junctions of the walls (Table 3.21).

P3.21 The calculation of heat losses through the ground for slab-on-ground floors requires the structure of the floor: the layer thicknesses d 1 = 0.2 m, d 2 = 0.05 m, d 3 = 0.065 m and d 4 = 0.01 m; the layer thermal conductivities λ 1 = 1 W/(m K), λ 2 = 0.033 W/(m K), λ 3 = 1.2 W/(m K) and λ 4 = 0.16 W/(m K); and the thickness of the external wall w = 0.45 m. In a heating season the heat flows downwards through the floor, and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.17 m2 K/W (Table 3.1). The total equivalent thickness of the floor (P3.4) is then df = w + λg (Rsi +

d1 d2 d3 + + + Rse ) = 4.53 m, λ1 λ2 λ3

where λg = 2 W/(m K) is the thermal conductivity of the gravel ground (Table P3.3). The geometry of the floor is used to calculate the characteristic dimensions of the end terrace house B 1 and the mid terrace house B 2 (P3.3): A1 b l1 = = 6.46 m, 0.5 P1 0.5(b + 2 l 1 ) A2 b l2 B2 = = = 12.00 m. 0.5 P2 0.5(2 l 2 ) B1 =

Both houses have moderately insulated floor, df < B, so the thermal transmittances of the floor (P3.6) are 2 λg π B1 W ln ( + 1) = 0.274 2 , π B 1 + df df m K 2 λg π B2 W Uf2 = ln ( + 1) = 0.211 2 . π B 2 + df df m K Uf1 =

i

W , K

Problems in Building Physics

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P3.22 Abw , Ubw

z

Abf , Ubf a

b

The calculation of heat losses through the ground for heated (conditioned) basement requires (a) the structure of the basement floor: the layer thicknesses d 1 = 0.2 m, d 2 = 0.05 m and d 3 = 0.01 m; and the layer thermal conductivities λ 1 = 1 W/(m K), λ 2 = 0.033 W/(m K) and λ 3 = 0.19 W/(m K) and (b) the structure of the basement wall: the layer thicknesses d 4 = 0.08 m, d 5 = 0.19 m and d 6 = 0.01 m; and the layer thermal conductivities λ 4 = 0.04 W/(m K), λ 5 = 0.49 W/(m K) and λ 6 = 0.51 W/(m K). In a heating season, the heat flows downwards through the basement floor, and horizontally through the basement wall, so the external surface resistances are Rsi,f = 0.17 m2 K/W and Rsi,bw = 0.13 m2 K/W for the floor and basement wall, respectively (Table 3.1). The internal surface resistance is Rse = 0.04 m2 K/W for both. The thickness of the wall w and the total equivalent thicknesses of the floor df (P3.4) and the basement wall dbw (P3.5) are then w = d 4 + d 5 + d 6 = 0.28 m, d1 d2 d3 + + + Rse ) = 4.24 m, λ1 λ2 λ3 d4 d5 d6 = λg (Rsi,bw + + + + Rse ) = 5.15 m, λ4 λ5 λ6

df = w + λg (Rsi,f + dbw

where λg = 2 W/(m K) is the thermal conductivity of the sandy ground (Table P3.3). The geometry of the basement, the floor plan dimensions a = 10 m and b = 7.5 m, and the depth of the basement floor below ground level z = 1.5 m are also required to complete the calculation. For the characteristic dimension of the house (P3.3) we get: B=

A ab = = 4.29 m. 0.5 P 0.5(2 a + 2 b)

The house has a well-insulated floor, df + 0.5z > B, so the thermal transmittance of the basement floor (P3.10) is Ubf =

λg W = 0.288 2 . 0.457 B + df + 0.5 z m K

On the other hand, the thermal transmittance of the basement wall (P3.11) is Ubw =

2 λg 0.5 df z W + 1) = 0.297 2 . (1 + ) ln ( πz df + z dbw m K

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To calculate the ground heat transfer coefficient (P3.12), we must first calculate the area of the basement floor Abf and the area of the basement wall below the ground level Abw : Abf = a b = 75 m2 , Abw = (2 a + 2 b)z = 52.5 m2 Ô⇒ Hg = Abf Ubf + Abw Ubw = 37.2

W . K

P3.23 The calculation of heat losses through the ground for slab-on-ground floors requires the structure of the floor: the layer thicknesses d 1 = 0.2 m, d 2 = 0.05 m and d 3 = 0.01 m; the layer thermal conductivities λ 1 = 1 W/(m K), λ 2 = 1.2 W/(m K) and λ 3 = 0.16 W/(m K); and the thickness of the external wall w = 0.4 m. The internal and external temperatures are θ i = 20 ○ C and θ e = −2 ○ C, respectively. This implies that the heat flows downwards through the floor, and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.17 m2 K/W, respectively (Table 3.1). The total equivalent thickness of the floor (P3.4) is then df = w + λg (Rsi +

d1 d2 d3 + + + Rse ) = 1.17 m, λ1 λ2 λ3

where λg = 1.5 W/(m K) is the thermal conductivity of the clay ground (Table P3.3). The geometry of the floor is used to calculate the area A, the exposed perimeter P and the characteristic dimension of the house B (P3.3): A = a b − a ′ b′ = 100 m2 , P = 2 a + 2 b = 44 m, A B= = 4.55 m. 0.5 P The house has a moderately insulated floor, df < B, so the thermal transmittance of the floor (P3.6) is 2 λg πB W Uf = ln ( + 1) = 0.501 2 . π B + df df m K Heat losses through the ground are mitigated by horizontal edge insulation in the form of extruded polystyrene of thickness dn = 0.03 m, thermal conductivity λn = 0.033 W/(m K) and width D = 1 m. This is quantified by the additional equivalent thickness d ′ (P3.15)

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and the resulting negative linear thermal transmittance Ψg,e (P3.16): d ′ = dn ( Ψg,e = −

λg − 1) = 1.33 m, λn

λg D D W + 1)] = −0.134 . [ln ( + 1) − ln ( π df df + d ′ mK

The ground heat transfer coefficient (P3.18) and the heat losses through the ground are W , K Φg = Hg (θ i − θ e ) = 970 W.

Hg = A Uf + P Ψg,e = 44.2

P3.24

z

Abw , Ubw

Abf , Ubf l

b

The calculation of heat losses through the ground for heated (conditioned) basement requires (a) the structure of the basement floor: the layer thicknesses d 1 = 0.2 m, d 2 = 0.03 m, d 3 = 0.065 m and d 4 = 0.01 m; and the layer thermal conductivities λ 1 = 1 W/(m K), λ 2 = 0.035 W/(m K), λ 3 = 1.2 W/(m K) and λ 4 = 1.3 W/(m K) and (b) the structure of the basement wall: the layer thicknesses d 5 = 0.1 m, d 6 = 0.19 m and d 7 = 0.01 m; and the layer thermal conductivities λ 5 = 0.04 W/(m K), λ 6 = 0.49 W/(m K) and λ 7 = 0.51 W/(m K). The internal and external temperatures are θ i = 21 ○ C and θ e = −6 ○ C, respectively. This implies that the heat flows downwards through the floor and horizontally through the basement wall, so the external surface resistances are Rsi,f = 0.17 m2 K/W and Rsi,bw = 0.13 m2 K/W for the floor and basement wall, respectively (Table 3.1). The internal surface resistance is Rse = 0.04 m2 K/W for both. The thickness of the wall w and the total equivalent thicknesses of the floor df (P3.4) and the basement wall dbw (P3.5) are then w = d 5 + d 6 + d 7 = 0.3 m, d1 d2 d3 d4 + + + + Rse ) = 2.29 m, λ1 λ2 λ3 λ4 d5 d6 d7 = λg (Rsi,bw + + + + Rse ) = 4.62 m, λ5 λ6 λ7

df = w + λg (Rsi,f + dbw

where λg = 1.5 W/(m K) is the thermal conductivity of the clay ground (Table P3.3).

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The geometry of the floor is used to calculate the characteristic dimension (P3.3) as B=

A bl = = 6 m. 0.5 P 0.5(b + 2 l)

The depth of the basement floor below ground level is z = 1.5 m. The flat therefore has a moderately insulated floor, df + 0.5z < B and the thermal transmittance of the basement floor (P3.9) is Ubf =

2 λg πB W ln ( + 1) = 0.270 2 . π B + df + 0.5 z df + 0.5 z m K

On the other hand, the thermal transmittance of the basement wall (P3.11) is Ubw =

2 λg 0.5 df z W + 1) = 0.233 2 . (1 + ) ln ( πz df + z dbw m K

To calculate the ground heat transfer coefficient (P3.12) and the heat losses through the ground, we must first calculate the area of the basement floor Abf and the area of the basement wall below the ground level Abw : Abf = b l = 96 m2 , Abw = (b + 2 l)z = 48 m2 Ô⇒ Hg = Abf Ubf + Abw Ubw = 37.2

W , K

Φg = Hg (θ i − θ e ) = 1000 W.

P3.25 The (natural) air change rate nlk is calculated from the air change rate at 50 Pa, n 50 = 1.67 × 10−4 1/s, and the leakage-infiltration ratio N = 20 (3.47). Combining it with the volume of the house V = 300 m3 , we obtain the (natural) air leakage rate (3.43) as q V,lk = nlk V =

n 50 m3 V = 2.5 × 10−3 . N s

To calculate the heat flow rate due to the air leakage (3.41), we also need the density of air ρ = 1.2 kg/m3 , the specific heat capacity of air c p = 1000 J/(kg K) and the internal and external temperatures, θ i = 20 ○ C and θ e = 4 ○ C, respectively: Φlk = q V,lk ρ c p (θ i − θ e ) = 48 W.

Problems in Building Physics

104

The heat flow rate due to the forced ventilation without recuperator is calculated with the same expression from the air flow rate of forced ventilation q V,fv = 3.3 × 10−2 m3/s as Φfv = q V,fv ρ c p (θ i − θ e ) = 640 W. However, most of the energy that would otherwise be lost in this process is retained by the recuperator of efficiency η = 0.75 . The heat flow rate due to the forced ventilation with recuperator is therefore Φrp = (1 − η) Φfv = 160 W.

i

Note that all values given in this problem correspond to the typical minimum conditions for passive houses. Therefore, the calculated values give a good estimate of the actual heat losses for the building with four occupants.

P3.26 The direct heat transfer coefficient includes the heat losses through the elements that directly separate the conditioned space from the external environment, the roof of area Arf = 60 m2 and thermal transmittance Urf = 0.26 W/(m2 K); the windows of area Awd = 46.5 m2 and thermal transmittance Uwd = 1.2 W/(m2 K); the walls of area Awl = 130.6 m2 and thermal transmittance Uwl = 0.21 W/(m2 K); the doors of area Adr = 2.1 m2 and thermal transmittance Udr = 1 W/(m2 K); and the wall-balcony thermal bridge of length lwb = 6 m and linear thermal transmittance Ψwb = 0.4 W/(m K). Direct heat transfer coefficient (3.37) is therefore Hd = Arf Urf + Awd Uwd + Awl Uwl + Adr Udr + lwb Ψwb = 103.3

W . K

The ground heat transfer coefficient (P3.8) includes the heat losses through the slab-onground of area Asg = 60 m2 and thermal transmittance Usg = 0.28 W/(m2 K) as Hg = Asg Usg = 16.8

W . K

The air change rate of n = 5.56 × 10−5 1/s and the volume of the house V = 336 m3 give the air leakage rate q V (3.43), which is then combined with the density of air ρ = 1.2 kg/m3 and the specific heat capacity of air c p = 1000 J/(kg K) to obtain the ventilation heat transfer coefficient (3.44) as Hve = ρ c p q V = ρ c p n V = 22.4

W . K

P3 Heat transfer in building components

105

The transmission heat transfer coefficient is the sum of the direct and the ground heat transfer coefficients (3.48) as in Htr = Hd + Hg = 120.1

W . K

The total heat losses are the sum of transmission and ventilation heat losses as in Φ = Φtr + Φve = Htr (θ i − θ e ) + Hve (θ i − θ e ) = 5700 W, where θ e = −20 ○ C and θ i = 20 ○ C are the external and internal temperatures, respectively.

P3.27 The maximum heat flow rate Φhp that can be transferred by a heat pump depends on the provided electrical power P = 4000 W and the coefficient of performance (1.34), as COP =

Qhp Φhp = . A P

On the other hand, the coefficient of performance depends on the temperatures of the external air Te and water Tw = 328 K (θ w = 55 ○ C). Taking into account that the actual coefficient of performance is f = 0.5 times the ideal coefficient of performance (1.44), the maximum heat flow rate is Φhp = P COP = P f

Tw . Tw − Te

The heat flow rate of heat losses, quantified by the transmission heat transfer coefficient Htr = 250 W/K and the ventilation heat transfer coefficient Hve = 40 W/K, amounts to Φhl = (Htr + Hve )(Ti − Te ), where Ti = 294 K (θ i = 21 ○ C) is the internal temperature. Φ/kW Φ hp

15 10

Φhl

5 θ′ −20

−15

−10

−5

0

5

10

θ e /○ C

Note that as the external temperature Te decreases, the maximum heat flow rate of a heat pump decreases and the heat flow rate of the heat losses increases. If we decrease the

Problems in Building Physics

106

temperature sufficiently, the heat pump will no longer be able to compensate for all the heat losses and an auxiliary heat source will have to provide some of the required heat (the shaded area in the preceding plot). The lowest external temperature T ′ at which the heat pump is still sufficient is obtained by equating the heat flow rates: Φhp = Φhl , Tw Pf = (Htr + Hve )(Ti − T ′ ). Tw − T ′ The temperature T ′ is therefore ¿ Á Tw + Ti 2 T + T P f Tw w i À( T′ = ±Á = 260.5 K. ) − Tw Ti + 2 2 Htr + Hve Here, we have chosen the minus sign because the plus sign would have represented a cooling situation if the expression (1.44) had also been valid for cooling.

i

A heating system that combines a heat pump with an auxiliary heating source is called the bivalent system and the temperature T ′ (θ ′ ) is called the bivalent point.

P4 Moisture in building components P4.1 The initial water vapour pressure p 1 is obtained from the ideal gas law (1.12), where the volume and temperature of humid air are V = 1 m3 and T = 293 K (θ = 20 ○ C), respectively; and the mass and molar mass of water are m = 9 × 10−3 kg and M = 0.018 kg/mol, respectively: p1 V = n R T = Ô⇒ p 1 =

m R T, M

mRT = 1218 Pa. MV

As Dalton’s law implies (Section 1.4.1), when the pressure of humid air increases, the pressure of all its components increases proportionally. The ratio of the final pressure pa,2 to the initial pressure pa,1 = 1 bar of humid air is therefore equal to the ratio of the final p 2 to the initial water vapour pressure as in pa,2 p 2 = . pa,1 p 1 The air is saturated when the relative humidity reaches 100 %, that is, when the water vapour pressure reaches the pressure at saturation psat (4.7). We obtain the water vapour pressure at saturation at 20 ○ C, psat = 2337 Pa, from the quasi-empirical equation (PA.1). For the final pressure we finally get: pa,2 =

p2 psat pa,1 = pa,1 = 1.9 bar. p1 p1

Condensation in compressed air systems is a common problem, and such systems should be drained regularly. If this is not taken care of, a mixture of air and moisture can discharge from an air outlet, such as an air hose of tyre inflators at filling stations.

i 107

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108

P4.2 Heat is released from liquid water of mass mw = 3 kg and specific heat capacity cw = 4200 J/(kg K) and from dry air of mass ma and specific heat capacity ca = 1000 kJ/(kg K). The volume of the room and the mass of the dry air are calculated from the dimensions of the room a = 4 m, b = 5 m, h = 2.5 m and the density of dry air ρa = 1.20 kg/m3 as V = a b h = 50 m3 , ma = ρa V = 60 kg. The released heat (1.20) is Qr = mw cw (θ w − θ ′ ) + ma ca (θ a − θ ′ ), where θ w = 41 ○ C and θ a = 21 ○ C are the initial temperatures of water and air, respectively, and θ ′ = 16 ○ C is the final temperature of all substances. On the other hand, the heat is absorbed by part of the water of mass me that evaporated. The amount of absorbed heat (1.30) is Qa = me qv , where qv = 2.46 × 106 J/kg is the specific heat of evaporation of water. If we equate the released and the absorbed heat, Qr = Qa , we get: me =

1 [mw cw (θ w − θ ′ ) + ma ca (θ a − θ ′ )] = 0.25 kg. qv

Now we calculate the water vapour pressure from ideal gas law (1.12): pV = nRT = Ô⇒ p =

me R T′ M

me R T ′ = 667 Pa. MV

Here, M = 0.018 kg/mol is the molar mass of water and T ′ = 289 K is the final temperature. After calculating the water vapour pressure at saturation (PA.1) at θ ′ = 16 ○ C, psat = 1817 Pa, we get for the relative humidity (4.7): φ=

i

p = 37 %. psat

Note that evaporation and cooling processes occur simultaneously, the former over the entire temperature range. Fortunately, only the initial and final states matter, so we can rationalise both processes as if they occurred in separate, sequential steps.

P4 Moisture in building components

109

The problem assumes that the room is not only perfectly thermally insulated but also perfectly airtight. In practice, air leakage is an important mechanism of the water vapour transfer out of or into the room and has a significant impact on the actual relative humidity.

P4.3 The initial water vapour pressure p 1 , calculated from the initial relative humidity φ 1 = 60 % (4.7) and the water vapour pressure at saturation (PA.1) for the initial temperature θ 1 = 21 ○ C, psat,1 = 2486 Pa, is p 1 = φ 1 psat,1 = 1.5 kPa. The initial pressure of dry air pd,1 is obtained by subtracting the initial water vapour pressure from the initial pressure of humid air pa,1 = 100 kPa (1.13) as pd,1 = pa,1 − p 1 = 98.5 kPa. The initial mass of water vapour m 1 and the mass of dry air md are now easily calculated using ideal gas law from the volume V = 0.3 m3 and the initial temperature T1 = 294 K (1.12): pV = nRT =

m RT M

p1 V M = 3.3 × 10−3 kg, R T1 pd,1 V Md md = = 350.6 × 10−3 kg. R T1 Ô⇒ m 1 =

Here, M = 0.018 kg/mol and Md = 0.029 kg/mol are the molar masses of water vapour and dry air, respectively. The humid air is then cooled in the freezer. The water vapour pressure at saturation (PA.1) for the final temperature θ 2 = −18 ○ C is psat,2 = 124 Pa. Because this pressure is obviously smaller than the water vapour pressure of the cooled air and the relative humidity is consequently larger than 100 %, the deposition of water must occur. The deposition continues until the relative humidity drops to 100 %. The final water vapour pressure p 2 must therefore be equal to psat,2 . The final mass of remaining water vapour m 2 (1.12) and the mass of deposited water ∆m are then m2 =

p 2 V M psat,2 V M = = 0.3 × 10−3 kg, R T2 R T2 ∆m = m 1 − m 2 = 3.0 × 10−3 kg,

where T2 = 255 K is the final temperature in the freezer.

i

Problems in Building Physics

110

The final pressures of the dry air pd,2 and the humid air pa,2 in the freezer are pd,2 =

md R T2 = 85.4 kPa, Md V

pa,2 = pd,2 + p 2 = pd,2 + psat,2 = 85.6 kPa. Note that the mass of dry air and volume did not change during the process.

i

The water deposition causes the formation of a layer of ice on the walls of the freezer. In addition, the water deposition and the cooling in the freezer create a relative vacuum, which ensures that the freezer doors are well sealed even without the locks.

P4.4 At the beginning, the temperature is θ 1 = 2 ○ C and the air is saturated, φ 1 = 100 % = 1, so the water vapour pressure is (4.7) p 1 = φ 1 psat,1 = 705 Pa, where the initial water vapour pressure at saturation at temperature θ 1 , psat,1 = 705 Pa, is obtained from the quasi-empirical equation (PA.1). As Dalton’s law implies (Section 1.4.1), when the pressure of humid air decreases, the pressure of all its components decreases proportionally. The ratio of the intermediate pressure pa,2 = 70 kPa to the initial pressure pa,1 = 100 kPa of humid air is therefore equal to the ratio of the intermediate water vapour pressure p 2 to the initial water vapour pressure p 1 as in pa,2 p 2 = , pa,1 p 1 so the intermediate water vapour pressure is p2 =

pa,2 p 1 = 494 Pa. pa,1

On the other hand, due to the drop in temperature, the water vapour pressure at saturation also drops. The water vapour pressure at saturation at temperature θ 2 = −17 ○ C is psat,2 = 137 Pa. Because p 2 is larger than psat,2 and the relative humidity is larger than 100 %, the deposition of water must occur. The deposition continues until the relative humidity drops to 100 %. The final water vapour pressure p 3 must therefore be equal to psat,2 . The masses of water vapour before and after deposition calculated with the ideal gas law (1.12) are p2 V M = 4.2 × 10−3 kg, R T2 p 3 V M psat,2 V M m3 = = = 1.2 × 10−3 kg, R T2 R T2 m2 =

P4 Moisture in building components

111

where T2 = 256 K is the temperature of the cooled air, V = 1 m3 the volume of air and M = 0.018 kg/mol the molar mass of water. The change in the mass of water vapour ∆m corresponds to the mass of water vapour deposited in the form of snowflakes. The number of snowflakes is therefore N=

∆m m 2 − m 3 = = 1000, m′ m′

where m′ = 3 × 10−6 kg is the mass of one snowflake.

P4.5 Dry air of mass md and specific heat capacity cd = 1000 kJ/(kg K) releases heat when cooled from initial temperature θ to final temperature θ 0 = 14 ○ C. The amount of released heat (1.20) is Qr = md cd (θ − θ 0 ). On the other hand, liquid water of mass m 0 and specific heat of vaporisation qv = 2.47 × 106 J/kg at θ 0 absorbs heat and vaporises. The amount of absorbed heat (1.30) is Qa = m 0 qv . If we equate the released and the absorbed heat, Qr = Qa , divide by the mass of dry air md and use the definition of the mass ratio x (4.11), we get: md cd (θ − θ 0 ) = m 0 qv , cd (θ − θ 0 ) = x 0 qv . The highest temperature θ is achieved for the maximum value of the final mass ratio x 0 , that is, for the maximum amount of water vapour. This is achieved when the air is saturated, φ = 100 % (4.7), that is, when the water vapour pressure is equal to the water vapour pressure at saturation at temperature θ 0 , psat = 1598 Pa (PA.1). The corresponding mass ratio (Section 4.2) is x 0 = 0.622

psat g = 0.010 = 10 , patm − psat kg

where patm = 1.013 × 105 Pa is the atmospheric pressure. The highest temperature of dry air is thus θ=

qv x 0 + θ 0 = 38.6 ○ C. cd

We can generalise the above derivation for the initial air that is already humid, that is, already contains water of mass m, and whose relative humidity is φ and mass ratio is x.

Problems in Building Physics

112

We equate the heat released by cooling the dry air of mass md and the water vapour of mass m from θ to θ 0 with the heat required to evaporate additional water vapour of mass m 0 − m, which we then divide by the mass of dry air to get: (m c + md cd )(θ − θ 0 ) = (m 0 − m)qv , (x c + cd )(θ − θ 0 ) = (x 0 − x)qv . Here c = 1900 J/(kg K) is the specific heat capacity of water vapour. To obtain the solution, we must combine the preceding equation with the expression for the mass ratio of the initial humid air, φpsat x = 0.622 patm − φpsat and the quasi-empirical equation (PA.1) relating the water vapour pressure at saturation psat to the temperature θ. This set of three complicated equations can only be solved numerically. Fortunately, the process of evaporative cooling is adiabatic, and in psychrometric diagrams all adiabatic processes are straight lines. g

0

5 20

20

15 % 30

%

10 %

40

x / kg 10

40

%

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 100

1.16

30 1.18

25

kg m3

70

ρ/

1.20 60

50

1.22

15 40

θ/○ C

20

h/

k kg J

1.24 30

10 1.26

20

5 1.28

10

0 1.30

−5 1.32 0

1.34

−10

%

p = 101.3 kPa

P4 Moisture in building components

% φ

%

20

40

%

50

60

70 %

%

% 80

%

p = 101.3 kPa

90

70

100

113

60

15

g

40

x / kg

30

%

50

h/

k kg J

10

30

20

%

20

5 10

10 %

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

35

0 40

The preceding figures contain a red line connecting the point representing saturated air at (dry bulb) temperature θ 0 = 14 ○ C with the point representing dry air at (dry bulb) temperature θ = 38.6 ○ C. Each point on this line represents the highest temperature of humid air that can be cooled to 14 ○ C by evaporation of liquid water alone. For example, humid air of temperature 20.3 ○ C and relative humidity 50 % can also be cooled to 14 ○ C. The line represents the so-called wet bulb temperature for 14 ○ C. As can be seen in the preceding figures, the line is almost parallel to the specific enthalpy lines. This is because the slope of an adiabatic (h = constant) line depends on the reference point. All specific enthalpy lines have a common reference point, namely liquid water and dry air at temperature 0 ○ C (Section 4.3), so they are not only parallel but also equidistant to each other. On the other hand, the reference point of a particular wet bulb temperature line is liquid water and saturated air at particular wet bulb temperature, so that the wet bulb temperature lines are neither parallel nor equidistant to each other. The reference points of the wet bulb temperature lines are generally closer to the conditions of practical adiabatic processes than the reference point of the specific enthalpy lines. However, in addition to the problems already mentioned, wet bulb temperature lines are less suitable for energy calculations. Because of their similarity, both types of lines can usually be used interchangeably. In this book we will only use specific enthalpy lines, even if the wet bulb temperature has to be determined. This problem clearly shows the advantages of using psychrometric charts for psychrometric calculations.

i

i

i

Problems in Building Physics

114 g

0

5 20

2

20

15 % 30

%

10 %

40

x / kg 10

40

%

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 100

1.16

30

1

P4.8

P4.7

1.18

25

kg m3

70

ρ/

%

P4.6

1.20

50

1.22

2

15 40

θ/○ C

60

2

1

20

h/

k kg J

1.24 30

10 1.26

20

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34 % φ

%

20

40

%

50

60

70 %

%

% 80

%

p = 101.3 kPa

90

70

100

−10

60

2

15

50

1

g

2

k kg J

40

10

h/

P4.6

30

20

1

20

%

2

P4.8

5 10

10 %

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

x / kg

30

%

P4.7

35

0 40

P4 Moisture in building components

P4.6 Mollier diagram Condensation on the glass pane is the result of the drop in temperature of humid air. However, as long as the relative humidity is below 100 % during cooling, condensation does not occur and the mass of water vapour m is constant. And because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the cooling process, starting from point 1 at θ 1 = 30 ○ C and φ 1 = 40 %, proceeds vertically downwards, x = constant. The conditions for condensation are fulfilled only when the relative humidity reaches φ 2 = 100 % at point 2. The dew point, the temperature at point 2 read from the psychrometric chart, is θ 2 = 15 ○ C.

ashrae-style chart Condensation on the glass pane is the result of the drop in (dry bulb) temperature of humid air. However, as long as the relative humidity is below 100 % during cooling, condensation does not occur and the mass of water vapour m is constant. And because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the cooling process, starting from point 1 at θ 1 = 30 ○ C and φ 1 = 40 %, proceeds horizontally left, x = constant. The conditions for condensation are fulfilled only when the relative humidity reaches φ 2 = 100 % at point 2. The dew point, the (dry bulb) temperature at point 2 read from the psychrometric chart, is θ 2 = 15 ○ C.

P4.7 Mollier diagram If the air is cooled by evaporation of liquid water alone, that is without energy being removed from the system by other colder objects, the process is adiabatic and the enthalpy H is constant. And because the mass of dry air md is always constant, the specific enthalpy h (4.12) is also constant. This implies that the cooling process, starting from point 1 at θ 1 = 30 ○ C and φ 1 = 40 %, proceeds diagonally downwards, h = constant. The lowest possible temperature is achieved when the relative humidity reaches φ 2 = 100 % at point 2. The wet bulb temperature, the temperature at point 2 read from the psychrometric chart, is θ 2 = 20 ○ C. ashrae-style chart If the air is cooled by evaporation of liquid water alone, that is without energy being removed from the system by other colder objects, the process is adiabatic and the enthalpy H is constant. And because the mass of dry air md is always constant, the specific enthalpy h (4.12) is also constant. This implies that the cooling process, starting from point 1 at θ 1 = 30 ○ C and φ 1 = 40 %, proceeds diagonally upwards, h = constant. The lowest possible (dry bulb) temperature is achieved when the relative humidity reaches φ 2 = 100 % at point 2. The wet bulb temperature, the (dry bulb) temperature at point 2 read from the psychrometric chart, is θ 2 = 20 ○ C.

115

Problems in Building Physics

116

P4.8 Mollier diagram While the air is heated, no condensation occurs and the mass of

water vapour m is constant. Because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the heating process, starting from point 1 at θ 1 = 20 ○ C and φ 1 = 50 %, proceeds vertically upwards, x = constant. The process is finished when the temperature has reached 38 ○ C at the final point 2. At this point, the relative humidity and specific enthalpy read from the psychrometric chart are 18 % and 57 kJ/kg, respectively.

ashrae-style chart While the air is heated, no condensation occurs and the mass

of water vapour m is constant. Because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the heating process, starting from point 1 at θ 1 = 20 ○ C and φ 1 = 50 %, proceeds horizontally right, x = constant. The process is finished when the (dry bulb) temperature has reached 38 ○ C at the final point 2. At this point, the relative humidity and specific enthalpy read from the psychrometric chart are 18 % and 57 kJ/kg, respectively. g

x / kg 10

5

20

%

15 3

20

40

0 10 %

P4.9

0%

% 40

φ

50

%

60

%

1.14

35

% 70 % 80 % 90 % 2 100

1.16

1

30 1.18

25

kg m3

70

ρ/

1.20 60

50

1.22

15

30

10

h/

3

1.24

k kg J

40

θ/○ C

20

1.26

20

5 1.28

10

0 1.30

−5 1.32 0

1.34

−10

p = 101.3 kPa

P4 Moisture in building components

117

Mollier diagram On cooling from the initial point 1 at θ 1 = 30 ○ C and φ1 = 70 %,

the air is initially unsaturated, φ < 100 %. Thus the mass of water vapour m does not change and because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the process must proceed vertically downwards, x = constant. After the air is saturated at point 2, condensation begins. Because the relative humidity cannot be larger than 100 %, the process proceeds along the curve φ = 100 % until the temperature reaches 11 ○ C at the final point 3. We see that in the second leg of the cooling process mass ratio x decreases, which implies that the mass of water vapour decreases due to the water condensation. The mass ratios at three points read from the psychrometric chart are x 1 = x 2 = 18.8 g/kg = 0.0188 and x 3 = 8.2 g/kg = 0.0082. We also need the initial mass of humid air ma,1 , which is calculated from its initial volume V1 , the dimensions of the room a = 8 m, b = 6 m and c = 3 m and the initial density read from the psychrometric chart ρ 1 = 1.152 kg/m3 as ma,1 = ρ 1 V1 = ρ 1 a b c = 165.9 kg. The important intermediate step is to calculate the mass of dry air md (P4.2) as md =

ma,1 = 162.8 kg. 1 + x1

The mass of the eliminated, that is, condensed water is equal to the decrease of the mass of water vapour. The change in the mass of water vapour, calculated in terms of the mass ratio, is therefore ∆m = m 3 − m 1 = md (x 3 − x 1 ) = −1.7 kg. As already mentioned in Section 4.3, the mass of dry air is always constant and we do not have to calculate it multiple times. However, for pedagogical purposes, we will do so in this one case. First, we note that whereas the mass of dry air is constant, the mass of humid air must have changed because the mass of water vapour has changed. To obtain the final mass of humid air ma,3 we must take this into account by subtracting the mass of eliminated water vapour, as in ma,3 = ma,1 + ∆m = 164.2 kg. Now we apply the same expression as before to calculate the mass of dry air at the final point 3 as ma,3 md = = 162.8 kg. 1 + x3 As predicted, the values are equal.

!

% φ

%

20 50

60

70 %

1

40

%

%

2

80

%

90

p = 101.3 kPa

100

70

%

Problems in Building Physics

118

60

15

g

40 k kg J

10

h/ 30

x / kg

30

%

50

20

3

%

20

5 10

10 %

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

35

0 40

ashrae-style chart On cooling from the initial point 1 at θ 1 = 30 ○ C and φ1 = 70 %,

the air is initially unsaturated, φ < 100 %. Thus the mass of water vapour m does not change and because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the process must proceed horizontally left, x = constant. After the air is saturated at point 2, condensation begins. Because the relative humidity cannot be larger than 100 %, the process proceeds along the curve φ = 100 % until the (dry bulb) temperature reaches 11 ○ C at the final point 3. We see that in the second leg of the cooling process mass ratio x decreases, which implies that the mass of water vapour decreases due to the water condensation. The mass ratios at three points read from the psychrometric chart are x 1 = x 2 = 18.8 g/kg = 0.0188 and x 3 = 8.2 g/kg = 0.0082. We also need the mass of dry air, which is calculated from the initial volume of humid air V1 , the dimensions of the room a = 8 m, b = 6 m and c = 3 m, and the initial specific volume read from the psychrometric chart vd,1 = 0.885 m3/kg (P4.1) as md =

V1 abc = = 162.7 kg. vd,1 vd,1

The mass of the eliminated, that is, condensed water is equal to the decrease of the mass of water vapour. The change in the mass of water vapour, calculated in terms of the mass ratio, is therefore ∆m = m 3 − m 1 = md (x 3 − x 1 ) = −1.7 kg. As already mentioned in Section 4.3, the mass of dry air is always constant and we do not have to calculate it multiple times. On the one hand, we observe the decrease in specific volume. On the other hand, the whole process is isobaric and the volume of

P4 Moisture in building components

119

humid air also decreases due to the decrease in temperature T and the decrease in the amount of substance n due to the water condensation. If we perform a rather simple but lengthy calculation of the volume reduction and use the final specific volume read from the psychrometric chart, we will obtain the same value for the mass of dry air. g

x / kg 10

5

20

%

15 3

20

40

0 10 %

P4.10

0%

4

0%

φ

50

%

60

%

1.14

35 1.16

1

30 1.18

25

kg m3

70

ρ/

2

1.20 60

4

20

50

1.22

15 40

θ/○ C

% 70 % 80 % 90 % 100

3 30

10

kg

h/

kJ

1.24

1.26

20

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram We easily locate the initial point 1 at θ 1 = 30 ○ C and φ1 = 60 %

and the final point 4 at θ 4 = 20 ○ C and φ 4 = 65 % in the psychrometric chart. On cooling from point 1, the air is initially unsaturated, φ < 100 %, so the mass of water vapour and mass ratio do not change and the process proceeds vertically downwards, x = constant. After the air is saturated at point 2, condensation begins and the process proceeds along the curve φ = 100 %. The crucial step is to determine at what location the cooling process stops. We can find this information only by going back in time from point 4. The air is brought to point 4 by heating. While the air is being heated, the mass of water vapour and the mass ratio do not change, and the process proceeds vertically upwards, x = constant. So if we go back in time, we go vertically downwards from point 4 until we reach the curve φ = 100 %. The

Problems in Building Physics

120

obtained point 3 is obviously the location at which the air passes from the cooling to the heating process. The plot of the entire procedure of air manipulation in the psychrometric chart is complete. The mass ratios of all points, x 1 = x 2 = 16.0 g/kg = 0.0160 and x 3 = x 4 = 9.5 g/kg = 0.0095, and the specific enthalpies of all points, h 1 = 71 kJ/kg, h 2 = 62 kJ/kg, h 3 = 37 kJ/kg and h 4 = 44 kJ/kg, are now read from the chart. The important intermediate step is to calculate the mass of dry air md from the mass of humid air ma,1 = 100 kg (P4.2), as md =

ma,1 = 98.4 kg. 1 + x1

The mass of condensed water is equal to the decrease in the mass of water vapour. We calculate the change in the mass of water vapour using the mass ratio (4.11) as ∆m = m 4 − m 1 = md (x 4 − x 1 ) = −0.64 kg. The energies removed in cooling from point 1 to point 3 and added in heating from point 3 to point 4 are equal to the decrease and increase in the enthalpy of air, respectively. We calculate the changes in the enthalpy using the specific enthalpy (4.12): Q 13 = ∆H 13 = H 3 − H 1 = md (h 3 − h 1 ) = −3.35 × 106 J, Q 34 = ∆H 34 = H 4 − H 3 = md (h 4 − h 3 ) = 0.69 × 106 J.

% φ

%

20

60

2

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

Note that point 2 is irrelevant to the calculation.

1

15

g

40

10

k kg J

h/ 30

3

4

20

%

20

5 10

10 %

0

25

30

0 . 88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

x / kg

30

%

50

35

0 40

P4 Moisture in building components

121

ashrae-style chart We easily locate the initial point 1 at θ 1 = 30 ○ C and φ1 = 60 %

and the final point 4 at θ 4 = 20 ○ C and φ 4 = 65 % in the psychrometric chart. On cooling from point 1, the air is initially unsaturated, φ < 100 %, so the mass of water vapour and mass ratio do not change and the process proceeds horizontally left, x = constant. After the air is saturated at point 2, condensation begins and the process proceeds along the curve φ = 100 %. The crucial step is to determine at what location the cooling process stops. We can find this information only by going back in time from point 4. The air is brought to point 4 by heating. While the air is being heated, the mass of water vapour and the mass ratio do not change, and the process proceeds horizontally right, x = constant. So if we go back in time, we go horizontally left from point 4 until we reach the curve φ = 100 %. The obtained point 3 is obviously the location at which the air passes from the cooling to the heating process. The plot of the entire procedure of air manipulation in the psychrometric chart is complete. The mass ratios of all points, x 1 = x 2 = 16.0 g/kg = 0.0160 and x 3 = x 4 = 9.5 g/kg = 0.0095, and the specific enthalpies of all points, h 1 = 71 kJ/kg, h 2 = 62 kJ/kg, h 3 = 37 kJ/kg and h 4 = 44 kJ/kg, are now read from the chart. The important intermediate step is to calculate the mass of dry air md (P4.2) as md =

ma,1 = 98.4 kg. 1 + x1

The mass of condensed water is equal to the decrease in the mass of water vapour. We calculate the change in the mass of water vapour using the mass ratio (4.11) as ∆m = m 4 − m 1 = md (x 4 − x 1 ) = −0.64 kg. The energies removed in cooling from point 1 to point 3 and added in heating from point 3 to point 4 are equal to the decrease and increase in the enthalpy of air, respectively. We calculate the changes in the enthalpy using the specific enthalpy (4.12): Q 13 = ∆H 13 = H 3 − H 1 = md (h 3 − h 1 ) = −3.35 × 106 J, Q 34 = ∆H 34 = H 4 − H 3 = md (h 4 − h 3 ) = 0.69 × 106 J. Note that point 2 is irrelevant to the calculation.

!

Problems in Building Physics

122 g

0

5 %

20

15 % 30

20

40

x / kg 10

10 %

P4.11

% 40

φ 5

0%

1.14

35

60

1.16

1

30 1.18

25

kg m3

% 70 % 80 % 90 % 0 10 70

ρ/

%

1.20 60

20 50

1.22

15

2

h/

1.24

30

10

k kg J

40

θ/○ C

3

1.26

20

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram From the initial point 1 at θ 1 = 30 ○ C and φ1 = 65 % we read the

initial mass ratio, x 1 = 17.4 g/kg = 0.0174, the initial specific enthalpy h 1 = 74.5 kJ/kg and the initial density ρ 1 = 1.152 kg/m3 . We also need the mass of humid air ma,1 , which is calculated from its initial volume V1 and the dimensions of a room a = 6.2 m, b = 4 m and c = 2.85 m as ma,1 = ρ 1 V1 = ρ 1 a b c = 81.4 kg. The important intermediate step is to calculate the mass of dry air md from the mass of humid air (P4.2) as ma,1 md = = 80 kg. 1 + x1 The energies removed in cooling and added in heating, Q 12 = −3.4 × 106 J and Q 23 = 0.64 × 106 J, respectively, are equal to the change in the enthalpy of air. We write the change in the enthalpy using the specific enthalpy (4.12): Q 12 = ∆H 12 = H 2 − H 1 = md (h 2 − h 1 ), Q 23 = ∆H 23 = H 3 − H 2 = md (h 3 − h 2 ).

P4 Moisture in building components

123

From this we then obtain the specific enthalpies after cooling and after heating: Q 12 kJ = 32 , md kg Q 23 kJ h3 = h2 + = 40 . md kg h2 = h1 +

Now we have to draw both processes in the chart. On cooling from point 1, the air is initially unsaturated, φ < 100 %, so the mass of water vapour and mass ratio do not change and the process proceeds vertically downwards, x = constant. After the air is saturated, condensation begins. The cooling proceeds along the curve φ = 100 % until the specific enthalpy reaches 32 kJ/kg at point 2. In heating, the mass of water vapour and mass ratio do not change, and heating proceeds vertically upwards, x = constant, until the specific enthalpy reaches 40 kJ/kg at point 3. From the final point 3 we read the final relative humidity φ 3 = 60 %, the final temperature θ 3 = 19 ○ C and the final mass ratio x 3 = 8.3 g/kg = 0.0083. The mass of condensed water is equal to the decrease in the mass of water vapour. We calculate the change in the mass of water vapour using the mass ratio (4.11) as

% φ

%

20 50

60

70 %

%

% 80

%

p = 101.3 kPa

90

70

100

∆m = m 3 − m 1 = md (x 3 − x 1 ) = −0.73 kg.

40

%

1 60

15

g

40

h/

k kg J

10

30

2

20

3

%

20

5 10

10 %

0

25

30

0 . 88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

x / kg

30

%

50

35

ashrae-style chart From the initial point 1 at θ 1 = 30 ○ C and φ1 = 65 % we read

0 40

the initial mass ratio, x 1 = 17.4 g/kg = 0.0174, the initial specific enthalpy h 1 = 74.5 kJ/kg and the initial specific volume vd,1 = 0.883 m3/kg (P4.1). We also need the mass of dry air,

Problems in Building Physics

124

which is calculated from the initial volume of humid air V1 and the dimensions of a room a = 6.2 m, b = 4 m and c = 2.85 m as md =

V1 abc = = 80 kg. vd,1 vd,1

The energies removed in cooling and added in heating, Q 12 = −3.4 × 106 J and Q 23 = 0.64 × 106 J, respectively, are equal to the change in the enthalpy of air. We write the change in the enthalpy using the specific enthalpy (4.12): Q 12 = ∆H 12 = H 2 − H 1 = md (h 2 − h 1 ), Q 23 = ∆H 23 = H 3 − H 2 = md (h 3 − h 2 ). From this we then obtain the specific enthalpies after cooling and after heating: Q 12 kJ = 32 , md kg Q 23 kJ h3 = h2 + = 40 . md kg h2 = h1 +

Now we have to draw both processes in the chart. On cooling from point 1, the air is initially unsaturated, φ < 100 %, so the mass of water vapour and mass ratio do not change and the process proceeds horizontally left, x = constant. After the air is saturated, condensation begins. The cooling proceeds along the curve φ = 100 % until the specific enthalpy reaches 32 kJ/kg at point 2. In heating, the mass of water vapour and mass ratio do not change, and heating proceeds horizontally right, x = constant, until the specific enthalpy reaches 40 kJ/kg at point 3. From the final point 3 we read the final relative humidity φ 3 = 60 %, the final (dry bulb) temperature θ 3 = 19 ○ C and the final mass ratio x 3 = 8.3 g/kg = 0.0083. The mass of condensed water is equal to the decrease in the mass of water vapour. We calculate the change in the mass of water vapour using the mass ratio (4.11) as ∆m = m 3 − m 1 = md (x 3 − x 1 ) = −0.73 kg.

P4 Moisture in building components

125 g

0

5 %

20

15 % 30

20

40

x / kg 10

10 %

P4.12

% 40

φ 5

0%

1.14

35

60

1.16

% 70 % 80 % 90 % 0 10

2

30 1.18

25

kg m3

70

ρ/

%

1.20 60

50

1.22

15

3

40

h/

kJ

1.24 30

10

kg

θ/○ C

20

1.26

5

20

1 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram While heating the humid air from point 1 at θ 1 = 5 ○ C and φ1 =

75 %, the mass of water vapour m and the mass ratio x (4.11) are constant. This implies that the process proceeds vertically upwards, x = constant, until the temperature reaches 30 ○ C at point 2. During the adiabatic humidification, no energy is added or removed, the enthalpy H and the specific enthalpy h (4.12) are constant. This implies that the process proceeds diagonally downwards, h = constant. Because the air is saturated at the end of the process, the final point 3 is at relative humidity φ 3 = 100 %. All mass ratios, x 1 = x 2 = 4.0 g/kg = 0.0040 and x 3 = 10.3 g/kg = 0.0103, all specific enthalpies h 1 = 15 kJ/kg and h 2 = h 3 = 40.5 kJ/kg and the initial density ρ 1 = 1.266 kg/m3 are read from the psychrometric chart. Now we calculate the mass of dry air md entering the room (P4.2) from the inlet mass ma,1 and the volume of humid air V1 , and the corresponding inlet mass flow rate of dry air q m,d

Problems in Building Physics

126

from the inlet air flow rate q V,1 = 2.78 × 10−2 m3/s by dividing by time, (4.2) and (3.42): md = q m,d =

ma,1 ρ 1 V1 = , 1 + x1 1 + x1

ρ 1 q V,1 kg = 3.50 × 10−2 . 1 + x1 s

We calculate the mass of the removed water ∆m by simply taking the difference between the masses of water vapour leaving and water vapour entering the room (4.11) and the corresponding mass flow rate q m by dividing by time: ∆m = m 3 − m 1 = (x 3 − x 1 )md , q m = (x 3 − x 1 ) q m,d = 2.2 × 10−4

kg . s

Similarly, we calculate the energy consumed for heating E by simply taking the difference between the enthalpies of the air before and after heating (4.12), and corresponding power P by dividing by time: E = ∆H = H 2 − H 1 = (h 2 − h 1 )md ,

% φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

P = (h 2 − h 1 ) q m,d = 890 W.

60

15

g

40 3

x / kg

30

%

50

h/

k kg J

10

30

20

%

20

5 10

1

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

10 %

2

35

0 40

ashrae-style chart While heating the air from point 1 at θ 1 = 5 ○ C and φ1 = 75 %, the mass of water vapour m and the mass ratio x (4.11) are constant. This implies that the process proceeds horizontally right, x = constant, until the temperature reaches 30 ○ C at point 2. During the adiabatic humidification, no energy is added or removed, the

P4 Moisture in building components

127

enthalpy H and the specific enthalpy h (4.12) are constant. This implies that the process proceeds diagonally upwards, h = constant. Because the air is saturated at the end of the process, the final point 3 is at relative humidity φ 3 = 100 %. All mass ratios, x 1 = x 2 = 4.0 g/kg = 0.0040 and x 3 = 10.3 g/kg = 0.0103, all specific enthalpies h 1 = 15 kJ/kg and h 2 = h 3 = 40.5 kJ/kg and the initial specific volume vd,1 = 0.793 m3/kg are read from the psychrometric chart. Now we calculate the mass of dry air md entering the room (P4.1) from the inlet volume of humid air V1 and the corresponding inlet mass flow rate of dry air q m,d from the inlet air flow rate q V,1 = 2.78 × 10−2 m3/s by dividing by time, (4.2) and (3.42): md = q m,d =

V1 , vd,1

q V,1 kg = 3.50 × 10−2 . vd,1 s

We calculate the mass of the removed water ∆m by simply taking the difference between the masses of water vapour leaving and water vapour entering the room (4.11) and the corresponding mass flow rate q m by dividing by time: ∆m = m 3 − m 1 = (x 3 − x 1 )md , q m = (x 3 − x 1 ) q m,d = 2.2 × 10−4

kg . s

Similarly, we calculate the energy consumed for heating E by simply taking the difference between the enthalpies of the air before and after heating (4.12), and corresponding power P by dividing by time: E = ∆H = H 2 − H 1 = (h 2 − h 1 )md , P = (h 2 − h 1 ) q m,d = 890 W.

Problems in Building Physics

128 g

0

5 %

20

15 % 30

20

40

x / kg 10

10 %

P4.13

% 40

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 0 10

1.16

30

2

1.18

25

kg m3

70

ρ/

%

1.20

50

1.22

15 40

θ/○ C

60

4

20

3 h/

k kg J

1.24 30

10 1.26

20

5

1

1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram The entire procedure of air manipulation consists of three processes, the first heating from point 1 to point 2, followed by an adiabatic humidification to saturation from point 2 to point 3 and the second heating from point 3 to point 4. We easily locate the initial point 1 at θ 1 = 4 ○ C and φ 1 = 80 % and the final point 4 at θ 4 = 20 ○ C and φ 4 = 65 % in the psychrometric chart. While heating the air from point 1, the mass of water vapour m and the mass ratio x (4.11) are constant, so the process proceeds vertically upwards, x = constant. We must determine at what location the heating process stops. We can find this information only by going back in time from point 4. The initially saturated air is brought to point 4 by the second heating process, the process that proceeds vertically upwards. So if we go back in time, we go vertically downwards from point 4 until we reach saturation at point 3, that is, φ 3 = 100 %. The air is brought to saturation by an adiabatic humidification. In an adiabatic process, no energy is added or removed, the enthalpy H and the specific enthalpy h (4.12) are constant, so we go back in time diagonally upwards, h = constant. Point 2 is the location where the vertical line from point 1 and the diagonal line from point 3 meet.

P4 Moisture in building components

129

We immediately read the temperature after the first heating at point 2, θ 2 = 27 ○ C. All the specific enthalpies h 1 = 14 kJ/kg, h 2 = h 3 = 37 kJ/kg and h 4 = 44 kJ/kg, the initial mass ratio x 1 = 4 g/kg = 0.004 and the initial density ρ 1 = 1.27 kg/m3 are also read from the psychrometric chart. The important step is to calculate the mass of dry air md from the mass of humid air (P4.2) as ma,1 ρ 1 V1 md = = = 100 kg, 1 + x1 1 + x1 where V1 = 79 m3 and ma,1 are the initial volume and mass of humid air, respectively. The energy consumed for heating is equal to the increase in the enthalpy of air. There are two heating processes, but because we are interested only in the total energy, we can simply consider the change between the initial state 1 and the final state 4. We calculate the change in enthalpy using the specific enthalpy (4.12) as

% φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

Q = ∆H = H 4 − H 1 = md (h 4 − h 1 ) = 3.0 × 106 J.

60

15

g

3

10

4

h/

k kg J

40

30

20

%

20

5 10

1

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

10 %

2

0.78

0.76

−10

x / kg

30

%

50

35

0 40

ashrae-style chart The entire procedure of air manipulation consists of three processes, the first heating from point 1 to point 2, followed by an adiabatic humidification to saturation from point 2 to point 3 and the second heating from point 3 to point 4. We easily locate the initial point 1 at θ 1 = 4 ○ C and φ 1 = 80 % and the final point 4 at θ 4 = 20 ○ C and φ 4 = 65 % in the psychrometric chart. While heating the air from point 1, the mass of water vapour m and the mass ratio x (4.11) are constant, so the process proceeds horizontally right, x = constant. We must determine at what location the heating process stops.

Problems in Building Physics

130

We can find this information only by going back in time from point 4. The initially saturated air is brought to point 4 by the second heating process, the process that proceeds horizontally right. So if we go back in time, we go horizontally left from point 4 until we reach saturation at point 3, that is, φ 3 = 100 %. The air is brought to saturation by an adiabatic humidification. In an adiabatic process, no energy is added or removed, the enthalpy H and the specific enthalpy h (4.12) are constant, so we go back in time diagonally downwards, h = constant. Point 2 is the location where the horizontal line from point 1 and the diagonal line from point 3 meet. We immediately read the (dry bulb) temperature after the first heating at point 2, θ 2 = 27 ○ C. All the specific enthalpies h 1 = 14 kJ/kg, h 2 = h 3 = 37 kJ/kg and h 4 = 44 kJ/kg and the initial specific volume vd,1 = 0.790 m3/kg are also read from the psychrometric chart. The important step is to calculate the mass of dry air md (P4.1) as md =

V1 = 100 kg, vd,1

where V1 = 79 m3 is the initial volume. The energy consumed for heating is equal to the increase in the enthalpy of air. There are two heating processes, but because we are interested only in the total energy, we can simply consider the change between the initial state 1 and the final state 4. We calculate the change in the enthalpy using the specific enthalpy (4.12) as Q = ∆H = H 4 − H 1 = md (h 4 − h 1 ) = 3.0 × 106 J.

P4 Moisture in building components

131 g

0

5

20

15 % 40

%

% 30

20

40

x / kg 10

10 %

P4.14

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 0 10

1.16

30

25

kg m3

70

ρ/

1

1

1.18

%

(b)

1.20

60

20 50

15

2

2

30

10

kg

h/

kJ

1.24

40

θ/○ C

(a) 1.22

1.26

20

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram In this problem we will indicate the location of the inlet air with point 1 and the properties of the outlet air with point 2. The heat that the air conditioner of the coefficient of performance COP = 3.8 removes from the room air (1.33) is Q = COP A, where A is the electrical work supplied to the air conditioner. This heat is equal to the decrease in the enthalpy of the air (4.12) as in Q = −∆H = H 1 − H 2 = md (h 1 − h 2 ). The important step is to calculate the mass of dry air md (P4.2) from the initial volume V1 and density ρ 1 of humid air as md =

ma,1 ρ 1 V1 = , 1 + x1 1 + x1

where ma,1 is the initial mass of humid air, and x 1 is the initial mass ratio (4.11).

Problems in Building Physics

132

Putting the three preceding equations together and dividing by the time, we get: ρ 1 V1 (h 1 − h 2 ), 1 + x1 ρ 1 q V,1 COP P = (h 1 − h 2 ). 1 + x1 COP A =

Here, q V,1 = 0.2 m3/s is the inlet air flow rate (3.42) and P = 1.1 × 103 W is the electrical power of the air conditioner.

Situation (a) From the initial point 1 at θ 1 = 27 ○ C and φ1 = 30 % we read the initial

mass ratio x 1 = 6.6 g/kg = 0.0066, the density ρ 1 = 1.172 kg/m3 and the specific enthalpy h 1 = 44 kJ/kg. The final specific enthalpy is then h 2 = h 1 − COP P

1 + x1 kJ = 26 . ρ 1 q V,1 kg

As long as the relative humidity is below 100 %, no condensation occurs and the mass of water vapour m is constant. And because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the cooling process, starting from point 1, proceeds vertically downwards, x = constant, until the specific enthalpy reaches 26 kJ/kg at point 2. The final temperature read from the final point 2 is θ 2 = 9.2 ○ C.

Situation (b) From the initial point 1 at θ 1 = 27 ○ C and φ1 = 50 % we read the initial

mass ratio x 1 = 11.1 g/kg = 0.0111, the density ρ 1 = 1.168 kg/m3 and the specific enthalpy h 1 = 55.5 kJ/kg. The final specific enthalpy is then h 2 = h 1 − COP P

1 + x1 kJ = 37.4 . ρ 1 q V,1 kg

As in the previous situation, the cooling process, starting from point 1, proceeds vertically downwards, x = constant. After the air is saturated, condensation begins. Because the relative humidity cannot be larger than 100 %, the process proceeds along the curve φ = 100 % until the specific enthalpy reaches 37.4 kJ/kg at point 2. The final temperature read from the final point 2 is θ 2 = 13.3 ○ C.

i

The main reason why the air is cooled to a lower temperature in the situation (a) is that there is no condensation of water. In the situation (b), only part of the removed energy is sensible heat, corresponding to a reduction in temperature, whereas the rest is latent heat, corresponding to condensation. Condensation can therefore significantly reduce the actual efficiency of the air conditioner.

P4 Moisture in building components

% φ

%

20

40

%

50

60

70 %

%

% 80

%

p = 101.3 kPa

90

70

100

133

60

15

30

%

50

(b)

g

10

k kg J

h/ 30 20

2

20

(a)

2

%

1 5

10

10 %

0

25

30

0.88

20

0 . 86

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

x / kg

1

40

35

ashrae-style chart In this problem we will indicate the location of the inlet air with point 1 and the properties of the outlet air with point 2. The heat that the air conditioner of the coefficient of performance COP = 3.8 removes from the room air (1.33) is Q = COP A, where A is the electrical work supplied to the air conditioner. This heat is equal to the decrease in the enthalpy of the air (4.12) as in Q = −∆H = H 1 − H 2 = md (h 1 − h 2 ). The important step is to calculate the mass of dry air md from the initial volume V1 and specific volume vd,1 (P4.1) as V1 . md = vd,1 Putting the three preceding equations together and dividing by the time, we get: V1 (h 1 − h 2 ), vd,1 q V,1 COP P = (h 1 − h 2 ). vd,1 COP A =

Here, q V,1 = 0.2 m3/s is the inlet air flow rate (3.42) and P = 1.1 × 103 W is the electrical power of the air conditioner.

0 40

Problems in Building Physics

134

Situation (a) From the initial point 1 at θ 1 = 27 ○ C and φ1 = 30 % we read the initial specific volume vd,1 = 0.859 m3/kg and the specific enthalpy h 1 = 44 kJ/kg. The final specific enthalpy is then h 2 = h 1 − COP P

vd,1 kJ = 26 . q V,1 kg

As long as the relative humidity is below 100 %, no condensation occurs and the mass of water vapour m is constant. And because the mass of dry air md is always constant, the mass ratio x (4.11) is also constant. This implies that the cooling process, starting from point 1, proceeds horizontally left, x = constant, until the specific enthalpy reaches 26 kJ/kg at point 2. The final (dry bulb) temperature read from the final point 2 is θ 2 = 9.2 ○ C.

Situation (b) From the initial point 1 at θ 1 = 27 ○ C and φ1 = 50 % we read the initial specific volume vd,1 = 0.866 m3/kg and the specific enthalpy h 1 = 55.5 kJ/kg. The final specific enthalpy is then h 2 = h 1 − COP P

vd,1 kJ = 37.4 . q V,1 kg

As in the previous situation, the cooling process, starting from point 1, proceeds horizontally left, x = constant. After the air is saturated, condensation begins. Because the relative humidity cannot be larger than 100 %, the process proceeds along the curve φ = 100 % until the specific enthalpy reaches 37.4 kJ/kg at point 2. The final (dry bulb) temperature read from the final point 2 is θ 2 = 13.3 ○ C.

i

The main reason why the air is cooled to a lower (dry bulb) temperature in the situation (a) is that there is no condensation of water. In the situation (b), only part of the removed energy is sensible heat, corresponding to a reduction in (dry bulb) temperature, whereas the rest is latent heat, corresponding to condensation. Condensation can therefore significantly reduce the actual efficiency of the air conditioner.

P4 Moisture in building components

135 g

0

5

20

15 % 40

%

% 30

20

40

x / kg 10

10 %

P4.15

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 0 10

1.16

30 1.18

25

2

kg m3

70

ρ/

m

1.20

60

1

20 50

1.22

15 40

h/

kJ

1.24 30

10

kg

θ/○ C

%

1.26

20

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34 % φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

−10

60

15

1

g

m

40

h/

k kg J

10

30

20

20

%

2 5

10

10 %

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

x / kg

30

%

50

35

0 40

Problems in Building Physics

136

We determine point 2, the location of the second humid air introduced in the problem on the psychrometric chart, directly from (dry bulb) temperature θ 2 = 25 ○ C and relative humidity φ 2 = 30 %. The mass ratio x 2 = 5.9 g/kg = 0.0059 is then read from this point. The important intermediate step is to calculate the mass of dry air md,2 from the mass of humid air ma,2 = 33 kg (P4.2), as md,2 =

ma,2 = 32.8 kg. 1 + x2

On the other hand, from the given data, that is, the mass of humid air ma,1 = 50 kg, the mass of water vapour m 1 = 0.7 kg and the (dry bulb) temperature θ 1 = 20 ○ C, we cannot directly determine the location of the first humid air introduced in the problem on the psychrometric chart. We must first calculate the mass of dry air md,1 and the mass ratio x 1 (4.11) by writing the mass of humid air as the sum of the mass of dry air md,1 and the mass of water vapour m 1 : ma,1 = md,1 + m 1 Ô⇒ md,1 = ma,1 − m 1 = 49.3 kg, m1 g x1 = = 0.0142 = 14.2 . md,1 kg Now we determine point 1, the location of the first humid air, from the mass ratio x 1 and the (dry bulb) temperature θ 1 . The humid air mixture is located somewhere on the line connecting the locations representing its components. The exact point is determined by the fraction of one component in the mixture in terms of the mass of dry air. For the first component, the mass fraction is md,1 = 0.6 = 60 %. md,1 + md,2 The mixture is therefore located at 60 % (3/5) of the length of the line. It is obvious that point m, the location of the humid air mixture, must be closer to point 1 than to point 2 because the first component makes up a larger fraction of the mixture. The (dry bulb) temperature of the humid air mixture θ m = 22 ○ C and its relative humidity φm = 66 % are now read from the obtained point. To obtain the correct result, the mass fraction must be calculated in terms of the mass of dry air and not in terms of the mass of humid air.

P4 Moisture in building components

137 g

0

5

20

15 % 40

%

% 30

20

40

x / kg 10

10 %

P4.16

φ 5

0%

1.14

35

60

1.16

30

2

1.18

25

kg m3

% 70 % 80 % 90 % 0 10 70

ρ/

%

1.20

m 50

1.22

15 40

h/

kJ

1.24 30

10

kg

θ/○ C

60

20

1.26

20

1

5 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram We directly determine the locations for the first and second humid air introduced in the problem using the (dry bulb) temperatures and relative humidities θ 1 = 5 ○ C, φ 1 = 75 % and θ 2 = 25 ○ C, φ 2 = 80 %, respectively. The mass ratios x 1 = 4.0 g/kg = 0.0040 and x 2 = 16.0 g/kg = 0.0160, and the densities ρ 1 = 1.266 kg/m3 and ρ 2 = 1.173 kg/m3 are now read from the psychrometric chart. The point m representing the humid air mixture is somewhere on the line connecting points 1 and 2 representing its components. We determine its exact location by finding the intersection with the value xm = 12 g/kg. In the chart, the distance between points 1 and m, d 1m , is approximately twice the distance between points 2 and m, d 2m . Thus, the ratio of the first air to the second air in terms of the mass of dry air is md,1 d 2m 1 = ≈ . md,2 d 1m 2 Because point m is closer to point 2 than to point 1, the fraction of the second humid air must obviously be larger than the fraction of the first humid air.

Problems in Building Physics

138

The next step is to calculate the masses of dry air md,1 and md,2 from the volumes of humid air V1 and V2 (P4.2) as ma,1 ρ 1 V1 = , 1 + x1 1 + x1 ma,2 ρ 2 V2 = = , 1 + x2 1 + x2

md,1 = md,2

where ma,1 and ma,2 are the masses of humid air. Combining the expressions and writing the volumes in terms of the air flow rates q V,1 and q V,2 = 11.0 m3/h (3.42) we get:

% φ

%

20

40

%

50

60

%

70 %

80

%

90

p = 101.3 kPa

100

70

%

md,1 ρ 1 V1 1 + x 2 ρ 1 q V,1 1 + x 2 1 = = = md,2 ρ 2 V2 1 + x 1 ρ 2 q V,2 1 + x 1 2 1 ρ2 1 + x1 m3 Ô⇒ q V,1 = q V,2 = 5.0 . 2 ρ1 1 + x2 h

60

2

15

30

%

50

g

x / kg

m 40

h/

k kg J

10

30

20

%

20

5 10

1

0

25

30

0 . 88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

10 %

35

0 40

ashrae-style chart We directly determine the locations for the first and second humid air introduced in the problem using the (dry bulb) temperatures and relative humidities θ 1 = 5 ○ C, φ 1 = 75 % and θ 2 = 25 ○ C, φ 2 = 80 %, respectively. The specific volumes vd,1 = 0.793 m3/kg and vd,2 = 0.866 m3/kg are now read from the psychrometric chart. The point m representing the humid air mixture is somewhere on the line connecting points 1 and 2 representing its components. We determine its exact location by finding the intersection with the value xm = 12 g/kg.

P4 Moisture in building components

139

In the chart, the distance between points 1 and m, d 1m , is approximately twice the distance between points 2 and m, d 2m . Thus, the ratio of the first air to the second air in terms of the mass of dry air is md,1 d 2m 1 = ≈ . md,2 d 1m 2 Because point m is closer to point 2 than to point 1, the fraction of the second humid air must obviously be larger than the fraction of the first humid air. The next step is to calculate the masses of dry air md,1 and md,2 from the volumes of humid air V1 and V2 (P4.1) as V1 , vd,1 V2 = . vd,2

md,1 = md,2

Combining the expressions and writing the volumes in terms of the air flow rates q V,1 and q V,2 = 11.0 m3/h (3.42) we get: md,1 V1 vd,2 q V,1 vd,2 1 = = = md,2 V2 vd,1 q V,2 vd,1 2 1 vd,1 m3 Ô⇒ q V,1 = q V,2 = 5.0 . 2 vd,2 h

Problems in Building Physics

140 g

0

5 %

20

15 % 40

% 30

20

40

x / kg 10

10 %

P4.17

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 0 10

1.16

30 1.18

25

kg m3

70

ρ/

2

1.20

60

m

20

50

1.22

15

1

40

θ/○ C

%

h/

k kg J

1.24 30

10 1.26

20

5 1.28

0 10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram We label the initial state of the first humid air introduced in the problem as point 0 on the psychrometric chart. Its location is determined by the temperature θ 0 = 1 ○ C and the relative humidity φ 0 = 100 % (saturation). The mass ratio x 0 = 4.0 g/kg = 0.0040 and the density ρ 0 = 1.285 kg/m3 are then read from the psychrometric chart. While heating, the mass of water vapour m and the mass ratio x (4.11) are constant. This implies that the process, starting from point 0 proceeds vertically upwards, x = constant, until the temperature reaches θ 1 = 13 ○ C at point 1. On the other hand, we label the initial state of the second humid air introduced in the problem as point 2 on the psychrometric chart. Its location is determined by the temperature θ 2 = 21 ○ C and the relative humidity φ 2 = 90 %. The mass ratio x 2 = 14.0 g/kg = 0.0140 and the density ρ 2 = 1.190 kg/m3 are then read from the psychrometric chart. The important intermediate step is to calculate the masses of dry air md,0 = md,1 and md,2 from the masses of humid air ma,0 and ma,2 (P4.2), and the volumes of humid air V0 =

P4 Moisture in building components

141

32 m3 and V2 = 104 m3 : ma,0 ρ 0 V0 = = 41 kg, 1 + x0 1 + x0 ma,2 ρ 2 V2 = = = 122 kg. 1 + x2 1 + x2

md,1 = md,2

The humid air mixture is located somewhere on the line connecting the locations representing its components. The exact point is determined by the fraction of one component in the mixture in terms of the mass of dry air. For the first component, the mass fraction is md,1 = 0.25 = 25 %. md,1 + md,2

% φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

The mixture is therefore located at 25 % (1/4) of the length of the line. It is obvious that point m, the location of the humid air mixture, must be closer to point 2 than to point 1 because the second component makes up a larger fraction of the mixture. The temperature of the humid air mixture θ m = 19 ○ C and its relative humidity φm = 84 % are now read from the obtained point.

60

15

2 30

%

50

g

h/

k kg J

10

30

20

%

20

5 10

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

10 %

1

0

0

x / kg

m 40

35

0 40

ashrae-style chart We label the initial state of the first humid air introduced in the problem as point 0 on the psychrometric chart. Its location is determined by the temperature θ 0 = 1 ○ C and the relative humidity φ 0 = 100 % (saturation). The specific volume vd,0 = 0.782 m3/kg is then read from the psychrometric chart. While heating, the mass of water vapour m and the mass ratio x (4.11) are constant. This implies that heating process, starting from point 0 proceeds horizontally right, x = constant, until the temperature reaches θ 1 = 13 ○ C at point 1.

Problems in Building Physics

142

On the other hand, we label the initial state of the second humid air introduced in the problem as point 2 on the psychrometric chart. Its location is determined by the temperature θ 2 = 21 ○ C and the relative humidity φ 2 = 90 %. The specific volume vd,2 = 0.852 m3/kg is then read from the psychrometric chart. The important intermediate step is to calculate the masses of dry air md,0 = md,1 and md,2 from the volumes of humid air V0 = 32 m3 and V2 = 104 m3 (P4.1): V0 = 41 kg, vd,0 V2 = = 122 kg. vd,2

md,1 = md,2

The humid air mixture is located somewhere on the line connecting the locations representing its components. The exact point is determined by the fraction of one component in the mixture in terms of the mass of dry air. For the first component, the mass fraction is md,1 = 0.25 = 25 %. md,1 + md,2 The mixture is therefore located at 25 % (1/4) of the length of the line. It is obvious that point m, the location of the humid air mixture, must be closer to point 2 than to point 1 because the second component makes up a larger fraction of the mixture. The (dry bulb) temperature of the humid air mixture θ m = 19 ○ C and its relative humidity φm = 84 % are now read from the obtained point.

P4 Moisture in building components

143 g

0

5

20

15 % 40

%

% 30

20

40

x / kg 10

10 %

P4.18

1.14

φ 5

i

35

60

1.16

30 1.18

25

0%

o

kg m3

% 70 % 80 % 90 % 0 10 70

ρ/

%

1.20 60

50

m

1.22

15 40

h/

kJ

1.24 30

10

kg

θ/○ C

20

1.26

20

5

f 1.28

10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

Mollier diagram We label the state of the inlet humid air at θ i = 35 ○ C and φi = 30 % as point i in our psychrometric chart. During the adiabatic humidification, no energy is added or removed, so the enthalpy H and specific enthalpy h (4.12) are constant. This implies that the process proceeds diagonally downwards, h = constant, until the relative humidity reaches φo = 80 %, whereupon the air leaves through the outlet. We label the state of the outlet humid air as point o in our psychrometric chart. We label the state of the fresh humid air at θ f = 4 ○ C and φf = 80 % as point f in our psychrometric chart. The fresh and the outlet humid air flows are then mixed adiabatically. The humid air mixture is located somewhere on the line connecting points f and o. To determine the exact location, we must go back in time from point i. The air is brought to point i by heating. While the air is being heated, the mass of water vapour and the mass ratio do not change, and the process proceeds vertically upwards. So, we go back in time from point i vertically downwards until we reach the humid air mixture line. The obtained location, labelled as point m, represents the state of the humid air mixture. From the chart, we read the distance between points o and m, dom , and the distance between points f and m, dfm . Using these two distances, we write the ratio of the fresh air

Problems in Building Physics

144 to the outlet air in terms of the mass of dry air as md,f dfm = = 0.68. md,o dom

Because point m is closer to point o than to point f, the mass of the outlet dry air must obviously be larger than the mass of the fresh dry air. We must relate the masses of dry air, md,f and md,o , to the corresponding masses of humid air, ma,f and ma,o . To do this, we first read the mass ratios from the psychrometric chart, xf = 4.0 g/kg = 0.0040 and xo = 15.0 g/kg = 0.0150 and then write (P4.2): ma,f , 1 + xf ma,o = , 1 + xo

md,f = md,o

where ma,f and ma,o are the masses of the fresh and outlet humid air, respectively. Finally, combining the expressions, we get ma,f md,f 1 + xf 2 40 = ≈ 0.67 ≈ = . ma,o md,o 1 + xo 3 60

% φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

Thus, the humid mass ratio of fresh air to outlet air is 40 %:60 %.

60

o

15

g

i m

10

h/

k kg J

40

x / kg

30

%

50

30

20

%

20

5 10

f

0

25

30

0.88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

0.78

0.76

−10

10 %

35

0 40

ashrae-style chart We label the state of the inlet humid air at θ i = 35 ○ C and

φi = 30 % as point i in our psychrometric chart. During the adiabatic humidification, no energy is added or removed, so the enthalpy H and specific enthalpy h (4.12) are constant.

P4 Moisture in building components

145

This implies that the process proceeds diagonally upwards, h = constant, until the relative humidity reaches φo = 80 %, whereupon the air leaves through the outlet. We label the state of the outlet humid air as point o in our psychrometric chart. We label the state of the fresh humid air at θ f = 4 ○ C and φf = 80 % as point f in our psychrometric chart. The fresh and the outlet humid air flows are then mixed adiabatically. The humid air mixture is located somewhere on the line connecting points f and o. To determine the exact location, we must go back in time from point i. The air is brought to point i by heating. While the air is being heated, the mass of water vapour and the mass ratio do not change, and the process proceeds horizontally right. So, we go back in time from point i horizontally left until we reach the humid air mixture line. The obtained location, labelled as point m, represents the state of the humid air mixture. From the chart, we read the distance between points o and m, dom , and the distance between points f and m, dfm . Using these two distances, we write the ratio of the fresh air to the outlet air in terms of the mass of dry air as md,f dfm = = 0.68. md,o dom Because point m is closer to point o than to point f, the mass of the outlet dry air must obviously be larger than the mass of the fresh dry air. We must relate the masses of dry air, md,f and md,o , to the corresponding masses of humid air, ma,f and ma,o . To do this, we first read the mass ratios from the psychrometric chart, xf = 4.0 g/kg = 0.0040 and xo = 15.0 g/kg = 0.0150 and then write (P4.2): ma,f , 1 + xf ma,o = , 1 + xo

md,f = md,o

where ma,f and ma,o are the masses of the fresh and outlet humid air, respectively. Finally, combining the expressions, we get ma,f md,f 1 + xf 2 40 = ≈ 0.67 ≈ = . ma,o md,o 1 + xo 3 60 Thus, the humid mass ratio of fresh air to outlet air is 40 %:60 %.

Problems in Building Physics

146

P4.19 The density of damp timber ρt = 450 kg/m3 and the mass ratio of water to dry matter u = 20 % = 0.2 are sufficient to fully describe the properties of damp timber. To relate the mass concentration of water w (4.15) and the density of dry timber ρd to the density of damp timber, we first note that the mass of dry timber md and the mass of water mw add up to the mass of damp timber mt (4.14) as in mt = md + mw . If we divide the expression by the volume V , we get: mt md mw = + V V V Ô⇒ ρt = ρd + w. Note that the mass concentration of water and the density of dry timber add up to the density of damp timber. Now we relate the mass concentration of water and the density of the dry timber to the mass ratio: u=

mw mw V w = = md V md ρd w Ô⇒ u = . ρd

We obtained two equations with two unknowns whose solution is: u kg ρt = 75 3 , 1+u m 1 kg ρd = ρt = 375 3 . 1+u m w=

P4.20 We must first determine the temperature at the internal surface θ si from the thermal transmittance U = 2 W/(m2 K), and the internal and external temperatures, θ i = 20 ○ C and θ e = −5 ○ C, respectively. For that we use the expression for the density of heat flow rate q (3.5) and a form of Fourier’s law for the internal air layer (3.6): q = U(θ i − θ e ) = 50 q=

W , m2

θ i − θ si Ô⇒ θ si = θ i − q Rsi = 13.5 ○ C. Rsi

Here, we have taken into account that for a vertical building element the heat flows horizontally and the internal surface resistance is Rsi = 0.13 m2 K/W (Table 3.1).

P4 Moisture in building components

147

Because the temperature at the wall surface is lower than that in the room, the relative humidity at the wall surface φsi is larger than that in the room φi = 65 %. To determine its value, we assume that the water vapour pressures at the wall surface and in the room are equal (Section 4.5.3). From the definition of the relative humidity (4.7) we get: psi = pi , φsi psat (θ si ) = φi psat (θ i ) Ô⇒ φsi =

psat (θ i ) φi = 98 %. psat (θ si )

Here, psat (θ i ) = 2337 Pa and psat (θ si ) = 1547 Pa are obtained from the quasi-empirical equation (PA.1).

P4.21 We must first determine the minimum temperature at the internal surface θ si that corresponds to the maximum relative humidity at the internal surface φsi = 80 %. Their values are related to the internal relative humidity and temperature, φi = 70 % and θ i = 20 ○ C, respectively. To obtain the relationship, we assume that the water vapour pressures at the wall surface and in the room are equal, as we argued in Section 4.5.3. From the definition of the relative humidity (4.7), we get: psi = pi , φsi psat (θ si ) = φi psat (θ i ) φi Ô⇒ psat (θ si ) = psat (θ i ) = 2045 Pa. φsi Here, psat (θ i ) = 2337 Pa is obtained from the quasi-empirical equation (PA.1). The minimum temperature θ si = 17.9 ○ C is then obtained from the inverse form of the quasiempirical equation (PA.2). There are two ways to determine the minimum total thermal resistance Rtot corresponding to the external temperature θ e = 0 ○ C. In the first method, we first calculate the density of heat flow rate using a version of Fourier’s law for the internal air layer (3.6), as q=

θ i − θ si W = 16.5 2 . Rsi m

Here, we have taken into account that for a vertical building element, the heat flows horizontally and the internal surface resistance is Rsi = 0.13 m2 K/W (Table 3.1). Similarly, we connect the temperatures θ i and θ e to obtain the total thermal resistance: q=

θi − θe θi − θe m2 K Ô⇒ Rtot = = 1.22 . Rtot q W

Problems in Building Physics

148

In the second method, we write the definition of the temperature factor of the internal surface (3.8) and its expression for the walls (3.9) to obtain the total thermal resistance directly: f Rsi = Ô⇒ Rtot

θ si − θ e Rtot − Rsi = θi − θe Rtot Rsi m2 K = = 1.22 . θ si −θ e W 1 − θ i −θ e

P4.22 We must first determine the temperature at the internal surface θ si that corresponds to the relative humidity at the internal surface φsi = 80 %, if the internal relative humidity and temperature are φi = 55 % and θ i = 21 ○ C, respectively. As we argued in Section 4.5.3, we can assume that the water vapour pressures at the wall surface and in the room are equal. From the definition of the relative humidity (4.7) we get: psi = pi , φsi psat (θ si ) = φi psat (θ i ) φi Ô⇒ psat (θ si ) = psat (θ i ) = 1709 Pa. φsi Here, psat (θ i ) = 2486 Pa is obtained from the quasi-empirical equation (PA.1). The temperature θ si = 15.0 ○ C is then obtained from the inverse form of the quasi-empirical equation (PA.2). In the case of thermal bridges, the characteristic temperatures cannot be determined using the method described in Section 3.1.2. But if we know the value of the temperature factor of the internal surface, f Rsi = 0.71, we can relate temperatures θ i and θ si to the external temperature θ e (3.8): θ si − θ e θi − θe θ si − f Rsi θ i Ô⇒ θ e = = 0.4 ○ C. 1 − f Rsi f Rsi =

P4.23 We must first determine the minimum temperature at the internal surface θ si that corresponds to the maximum relative humidity at the internal surface φsi = 80 %. Their values are related to the internal relative humidity and temperature, φi = 70 % and θ i = 22 ○ C, respectively. To obtain the relationship, we assume that the water vapour pressures at the

P4 Moisture in building components

149

wall surface and in the room are equal, as we argued in Section 4.5.3. From the definition of the relative humidity (4.7), we get: psi = pi , φsi psat (θ si ) = φi psat (θ i ) φi Ô⇒ psat (θ si ) = psat (θ i ) = 2312 Pa. φsi Here, psat (θ i ) = 2642 Pa is obtained from the quasi-empirical equation (PA.1). The minimum temperature θ si = 19.8 ○ C is then obtained from the inverse form of the quasiempirical equation (PA.2). Next we determine the minimum total thermal resistance Rtot , which corresponds to the minimum temperature at the internal surface. We can do this in two ways. In the first method, we first calculate the density of heat flow rate using a version of Fourier’s law (3.6) that relates the temperatures θ si and θ i as q=

θ i − θ si W = 16.7 2 . Rsi m

Here, we have taken into account that for a vertical building element the heat flows horizontally and the internal surface resistance is Rsi = 0.13 m2 K/W (Table 3.1). Similarly, we connect the temperatures θ i and θ e = −5 ○ C to obtain the total thermal resistance: q=

θi − θe θi − θe m2 K Ô⇒ Rtot = = 1.62 . Rtot q W

In the second method we write the definition of the temperature factor of the internal surface (3.8) and its expression for the walls (3.9) to obtain the total thermal resistance directly: f Rsi = Ô⇒ Rtot

θ si − θ e Rtot − Rsi = θi − θe Rtot Rsi m2 K = = 1.62 . −θ e W 1 − θθsii −θ e

Finally, we determine the minimum thickness of the second (eps) layer d 2 . For this step we also need the structure of the building component: the thickness d 1 = 0.012 m and thermal conductivity λ 1 = 0.6 W/(m K) for the first layer; the thermal conductivity λ 2 = 0.04 W/(m K) for the second layer; the thickness d 3 = 0.18 m and thermal conductivity λ 3 = 0.8 W/(m K) for the third layer; and the thickness d 4 = 0.01 m and thermal conductivity λ 4 = 0.5 W/(m K) for the fourth layer. The total thermal resistance (3.2) is Rtot = Rse +

d1 d2 d3 d4 + + + + Rsi λ1 λ2 λ3 λ4

Problems in Building Physics

150 and the thickness of the second layer is d 2 = λ 2 (Rtot − Rse −

d1 d3 d4 − − − Rsi ) = 0.047 m. λ1 λ3 λ4

Note that the external relative humidity φe = 90 % is irrelevant to the solution.

P4.24 We begin the solution by obtaining the external and internal water vapour pressures at saturation, psat,e = 813 Pa (θ e = 4 ○ C) and psat,i = 2337 Pa (θ i = 20 ○ C), respectively, from the quasi-empirical equation (PA.1). These are then used to calculate the external and internal water vapour pressures from the external and internal relative humidities, φe = 80 % and φi = 65 %, respectively (4.7): pe = φe psat,e = 650 Pa, pi = φi psat,i = 1519 Pa.

ve′

ve q′V

pe Te

vi qV

qV

pe Ti ve q′V

The external air must be heated to the internal temperature before it enters the room, as shown in the preceding figure. In the heating process, the temperature and volume of humid air increase, whereas the relative humidity and mass concentration of water vapour decrease. However, because the process is isobaric, the water vapour pressure pe does not change. Therefore, we can use the water vapour pressure calculated for the external air at 4 ○ C also for the external air heated to 20 ○ C. Now we calculate the mass concentration of water vapour (4.9) for the air flows entering and leaving the room by taking the internal temperature, Ti = 293 K: pe = 4.8 × 10−3 Rv Ti pi vi = = 11.2 × 10−3 Rv Ti ve =

kg , m3 kg . m3

On the other hand, N = 10 persons each emit me = 5 × 10−2 kg of water vapour per t = 1 h. The total mass flow rate of water vapour (4.2) emitted by the persons is q m,e = N

me . t

P4 Moisture in building components

If we want to keep the amount of water vapour constant, the water vapour emitted by persons must be equal to the water vapour removed by ventilation (P4.3), as in q m,e = q m,ve = q(vi − ve ), from which we finally get the air flow rate as qV =

q m,e Nme m3 = = 78 . vi − ve (vi − ve )t h

P4.25 We begin the solution by obtaining the water vapour pressures at saturation for water, internal and external air, psat,w = 3563 Pa (θ w = 27 ○ C), psat,i = 4003 Pa (θ i = 29 ○ C) and psat,e = 872 Pa (θ e = 5 ○ C), respectively, from the quasi-empirical equation (PA.1). The former is then used to calculate the mass concentration of water vapour at saturation at water temperature (4.9) psat,w kg vs = = 25.7 × 10−3 3 , Rv Tw m where Tw = 300 K is the temperature of the water. From the relative humidity of external air (4.7) φe = 90 %, we can calculate its water vapour pressure as pe = φe psat,e = 785 Pa and the mass concentration of water vapour of the heated external air entering the swimming pool hall as pe kg ve = = 5.6 × 10−3 3 , Rv Ti m where Ti = 302 K is the internal temperature. Using the same reasoning as in the solution to problem P4.24 , we combined the water vapour pressure of external air pe with the internal temperature Ti . The mass flow rate of the water vapour emitted by the pool of dimensions l = 50 m and w = 25 m, and area A = l w = 1250 m2 must be equal to the mass flow rate of water vapour removed by ventilation of air flow rate q V = 7 m3/s (P4.3): q m,ve = q m,ev , q V (vi − ve ) = A h m,ev (vs − vi ). The internal mass concentrations of water vapour is therefore vi =

q V ve + A h m,ev vs kg = 17.3 × 10−3 3 . q V + A h m,ev m

151

Problems in Building Physics

152

Now we calculate the water vapour pressure pi (4.9) and relative humidity φ i (4.7) of internal air as pi = vi Rv Ti , pi vi Rv Ti φi = = = 60 %. psat,i psat,i

i

Note that the expression for the evaporation mass flow rate is similar to the expressions for the mass convection of water vapour described in Section 4.5.2 and Newton’s law of cooling (2.30). That suggests that the mass transfer is triggered by the difference between the mass concentration of saturated water vapour at the water surface vs and the mass concentration of water vapour in the adjacent environment vi . Indeed, the mass transfer of water vapour is facilitated by convection, but the actual coefficient h m,ev is smaller than expected for such assumptions, because the generation of water vapour at the surface is limited for energetic reasons. vi cv

vs

The value of evaporative surface coefficient of mass transfer given in the problem was inspired by those given by standard vdi 2089 [6].

% φ

%

20

40

%

50

60

70 %

%

% 80

%

90

70

p = 101.3 kPa

100

P4.26

60

15

g

i

40

h/

k kg J

10

30

20

%

20

5 10

10 %

e

0

25

30

0 . 88

0 . 86

20

0 . 84

15 θ/○ C

3

10

m kg

5

v d/

0

0.82

0 . 80

−5

h

0.78

0.76

−10

x / kg

30

%

50

35

0 40

P4 Moisture in building components

153 g

0

5

20

15 % 30

20

%

10 %

40

x / kg 10

40

%

φ 5

0%

1.14

35

60

% 70 % 80 % 90 % 100

1.16

30 1.18

25

kg m3

70

ρ/

%

i

h

1.20 60

50

1.22

15 40

h/

kJ

1.24 30

10

kg

θ/○ C

20

1.26

20

5 1.28

e 10

0 1.30

−5 1.32 0

p = 101.3 kPa

1.34

−10

The internal humid air at (dry bulb) temperature θ i = 23 ○ C and relative humidity φi = 60 % leaves the room at outlet air flow rate q V,i = 0.03 m3/s. We label the internal air as point i in the psychrometric chart and read the mass ratio xi = 10.5 g/kg = 0.0105, the density ρi = 1.184 kg/m3 and the specific enthalpy hi = 49.9 kJ/kg. Now we calculate the mass of dry air md leaving the room (P4.2) and the corresponding outlet mass flow rate of dry air q m,d by dividing by time, (4.2) and (3.42): ρi Vi ma,i = , 1 + xi 1 + xi ρi q V,i kg = = 3.5 × 10−2 . 1 + xi s

md = q m,d

The external air at (dry bulb) temperature θ e = 2 ○ C and relative humidity φe = 80 % enters the room at inlet air flow rate that is not equal to the outlet air flow rate. That is because the inlet air is colder and denser and expands when it warms up to the internal (dry bulb) temperature. However, we can assume that the ventilation process does not

!

Problems in Building Physics

154

change the amount of dry air in the room, so the mass flow rates of dry air at the outlet and the inlet should be equal. We label the external air as point e in the psychrometric chart and read the mass ratio xe = 3.5 g/kg = 0.0035 and the specific enthalpy he = 10.7 kJ/kg. We calculate the mass of the removed water vapour ∆m by simply taking the difference between the masses of water vapour leaving and water vapour entering the room (4.11) and the corresponding mass flow rate q m by dividing by time: ∆m = mi − me = (xi − xe )md , q m = (xi − xe ) q m,d = 2.5 × 10−4

kg . s

Similarly, we calculate the heat of the total (enthalpic) heat losses Q by simply taking the difference between the enthalpies of the air leaving and the air entering the room (4.12), and the corresponding heat flow rate Φ by dividing by time (2.2): Q = ∆H = Hi − He = (hi − he )md , Φ = (hi − he ) q m,d = 1380 W. To calculate the heating power required for ventilation, we must first determine the specific enthalpy of the heated external air. We label the heated external air as point h in the psychrometric chart and read the specific enthalpy hh = 32.0 kJ/kg. Similarly to the preceding heat flow rate of the total heat losses, we write: Φ = (hh − he ) q m,d = 750 W.

i

We obtain very similar (though less accurate) value if we use the expression for ventilation heat losses (3.41) as Φ = ρ c p q V (θ i − θ e ) = 760 W, where ρ = 1.2 kg/m3 and c p = 1.0 × 103 J/(kg K) are the density and specific heat capacity at constant pressure.

i

The difference between the total (enthalpic) heat flow rate and the heating power required for the ventilation is the power needed to convert the liquid water into vapour that has left the room through ventilation (dashed line in the psychrometric chart).

P4.27 We first determine the mass concentration of carbon monoxide from the molar fraction yi = 20 ppm = 2 × 10−5 (P4.4) as vi =

M′ ρ kg yi = 2.32 × 10−5 3 , M m

P4 Moisture in building components

155

where ρ = 1.2 kg/m3 is the density of air and M = 0.029 kg/mol and M ′ = 0.028 kg/mol are the molar masses of air and carbon monoxide, respectively. On the other hand, the air leakage rate, obtained from the air change rate n = 0.8 1/h and the volume of the building V = 150 m3 (3.43), is q V = n V = 120

m3 . h

At steady state, the generated mass flow rate of carbon monoxide is equal to the mass flow rate removed by ventilation, q m = q m,ve , so that the generated mass flow rate (P4.3) is q m = q V (vi − ve ) = 2.8 × 10−3

kg , h

where we have taken into account that the external mass concentration of carbon monoxide is negligible, ve = 0.

P4.28 First, we convert the given activity concentrations of radon 222, 450 Bq/m3 , 30 Bq/m3 and 150 Bq/m3 , into mass concentrations, 79.1 × 10−18 kg/m3 , 5.3 × 10−18 kg/m3 and 26.4 × 10−18 kg/m3 (P4.6). Next we calculate the (natural) air leakage rate from the air leakage rate at 50 Pa, q 50 = 1500 m3/h (3.47) as q 50 m3 q V,l = = 75 , N h where N = 20 is the leakage-infiltration ratio. We calculate the mass flow rate of radon removed by the air leakage using expression (P4.3) as q m,l = q V,l (vi1 − ve ) = 5.5 × 10−15

kg , h

where vi1 = 79.1 × 10−18 kg/m3 and ve = 5.3 × 10−18 kg/m3 are the initial internal and external mass concentrations of radon. In an equilibrium situation the radon mass flow rate entering from the ground q m,g is equal to the radon mass flow rate removed by the air leakage, q m,g = q m,l = 5.5×10−15 kg/h. We can reduce the internal mass concentration of radon to the final value vi2 = 26.4 × 10−18 kg/m3 by replacing an air leakage of air leakage rate q V,l with a forced ventilation of air flow rate q V,f . The required air flow rate is q V,f =

q m,g m3 = 260 . vi2 − ve h

Problems in Building Physics

156

P4.29 The characteristic temperatures were already determined in the solution to problem P3.3 . First, we calculate the characteristic water vapour pressures at saturation, such as psat (θ e ) = 401 Pa and psat (θ i ) = 2337 Pa, with the quasi-empirical equation (PA.1). Second, we calculate the external and internal water vapour pressures from the external and internal relative humidities (4.7), φe = 65 % and φi = 65 %, respectively: pe = φe psat (θ e ) = 261 Pa, pi = φe psat (θ i ) = 1519 Pa. Third, we calculate the equivalent air layer thicknesses (4.29) for stone, concrete and eps, sd,s , sd,c and sd,e from the given water vapour resistance factors µs = 200, µc = 120 and µe = 60, respectively: sd,s = µs ds = 4 m, sd,c = µc dc = 24 m, sd,e = µe de = 9 m. For the external and internal surface equivalent air layer thicknesses we use sd,se = sd,si = 0.01 m. And finally, we calculate the water vapour pressures using the adapted Fick’s first law (4.38). We calculate the density of water vapour flow rate д by writing down Fick’s first law between the internal and external environments, and then we calculate the characteristic water vapour pressures by writing down Fick’s first law for individual layers.

Situation ∣stone∣concrete∣ e se 1

si i q д

The total equivalent air layer thickness is sd,tot = sd,se + sd,s + sd,c + sd,si = 28 m.

P4 Moisture in building components

157

To obtain the characteristic water vapour pressures, we write: pi − pe kg = 8.98 × 10−9 2 , sd,tot m s pse − pe д д = δ0 Ô⇒ pse = pe + sd,se = 261 Pa, sd,se δ0 p′ − pse д д = δ0 1 Ô⇒ p′1 = pse + µs ds = 441 Pa, µs ds δ0 psi − p′1 д д = δ0 Ô⇒ psi = p′1 + µc dc = 1519 Pa. µc dc δ0 д = δ0

All results are shown in the following table.

e se 1 si i

θ/○ C −5.00 −2.37 −1.71 11.45 20.00

psat /Pa 401 501 530 1352 2337

p/Pa 261 261 441 1519 1519

This low-insulated wall is characterised by a low temperature at the internal surface θ si = 11.45 ○ C. It follows that at the internal surface psi > psat (θ si ) and φsi > 100 %, revealing condensation at this location. Nevertheless, the calculated water vapour pressures are incorrect because in practice it is impossible for the relative humidity to exceed 100 %. The correct pressures can be determined only with the Glaser method.

Situation ∣stone∣concrete∣eps∣ e se 1

2 q д

The total equivalent air layer thickness is sd,tot = sd,se + sd,s + sd,c + sd,e + sd,si = 37 m.

si i

i

Problems in Building Physics

158

To obtain the characteristic water vapour pressures, we write: pi − pe kg = 6.80 × 10−9 2 , sd,tot m s pse − pe д д = δ0 Ô⇒ pse = pe + sd,se = 261 Pa, sd,se δ0 p′ − pse д д = δ0 1 Ô⇒ p′1 = pse + µs ds = 397 Pa, µs ds δ0 p′ − p′1 д д = δ0 2 Ô⇒ p′2 = p′1 + µc dc = 1213 Pa, µc dc δ0 psi − p′2 д д = δ0 Ô⇒ psi = p′2 + µe de = 1519 Pa. µe de δ0 д = δ0

All results are shown in the following table.

e se 1 2 si i

i

θ/○ C −5.00 −4.79 −4.73 −3.66 19.30 20.00

psat /Pa 401 409 410 450 2238 2337

p/Pa 261 261 397 1213 1519 1519

This well-insulated wall has a high temperature at the internal surface θ si = 19.30 ○ C. It follows that at the internal surface psi < psat (θ si ) and φsi < 100 %, implying that condensation has been eliminated at this location. However, because thermal insulation is on the inner side of the wall, the interface temperature θ ′2 = −3.66 ○ C is low. It follows that at this interface p′2 > psat (θ ′2 ) and φ′2 > 100 %, revealing interstitial condensation at this location. Nevertheless, the calculated water vapour pressures are incorrect because in practice it is impossible for the relative humidity to exceed 100 %. The correct pressures can be determined only with the Glaser method.

Situation ∣stone∣eps∣concrete∣ e se 1

si i

2 q д

The total equivalent air layer thickness is sd,tot = sd,se + sd,s + sd,e + sd,c + sd,si = 37 m.

P4 Moisture in building components

159

To obtain the characteristic water vapour pressures, we write: pi − pe kg = 6.80 × 10−9 2 , sd,tot m s pse − pe д д = δ0 Ô⇒ pse = pe + sd,se = 261 Pa, sd,se δ0 p′ − pse д д = δ0 1 Ô⇒ p′1 = pse + µs ds = 397 Pa, µs ds δ0 p′ − p′1 д д = δ0 2 Ô⇒ p′2 = p′1 + µe de = 703 Pa, µe de δ0 psi − p′2 д д = δ0 Ô⇒ psi = p′2 + µc dc = 1519 Pa. µc dc δ0 д = δ0

All results are shown in the following table.

e se 1 2 si i

θ/○ C −5.00 −4.79 −4.73 18.23 19.30 20.00

psat /Pa 401 409 410 2093 2238 2337

p/Pa 261 261 397 703 1519 1519

In this well-insulated wall, thermal insulation is on the outer side of the wall, so the interface temperature θ ′2 = 18.23 ○ C is high. It follows that at this interface p′2 < psat (θ ′2 ) and φ′2 < 100 %, implying that condensation has been eliminated at this location. In all three cases, we observe that the surface water vapour pressures are practically equal to those in the corresponding environments, pse ≈ pe and psi ≈ pi . This backs up the common assumption in almost all practical situations that they are simply equal and that we can neglect the external and internal surface equivalent air layer thicknesses sd,se = sd,si = 0 m (Section 4.5.3).

P4.30 e se 1

si i

2 q дout

дin дc

The input data for the Glaser diagram are (a) the structure of the building component, the thicknesses d 1 = 0.02 m, d 2 = 0.2 m and d 3 = 0.08 m; the thermal conductivities λ 1 = 1.5 W/(m K), λ 2 = 1 W/(m K) and λ 3 = 0.035 W/(m K); and the water vapour resistance

i

Problems in Building Physics

160

factors µ 1 = 50, µ 2 = 120 and µ 3 = 60 and (b) the environmental data, the temperatures θ i = 20 ○ C and θ e = 0 ○ C and the relative humidities φi = 45 % and φe = 80 %. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). First, we calculate the thermal transmittance U = 0.375 W/(m2 K) (3.3), the density of heat flow rate q = 7.49 W/m2 (3.5) and the characteristic temperatures with (3.6), or we calculate the total thermal resistance Rtot = 2.67 m2 K/W (3.2) and the characteristic temperatures with (3.7). Second, we calculate the characteristic water vapour pressures at saturation (PA.1). The results are shown in the following table.

e se 1 2 si i

θ/○ C 0.00 0.30 0.40 1.90 19.03 20.00

psat /Pa 611 624 628 700 2200 2337

p/Pa 488 488

1052 1052

Third, we calculate the equivalent air layer thicknesses (4.29), (4.35) and the internal and external water vapour pressures (4.7): sd,1 = µ 1 d 1 = 1 m, sd,2 = µ 2 d 2 = 24 m, sd,3 = µ 3 d 3 = 4.8 m, sd,tot = sd,1 + sd,2 + sd,3 = 29.8 m, pe = φe psat (θ e ) = 488 Pa, pi = φi psat (θ i ) = 1052 Pa. We assume that pse = pe and psi = pi . Finally, we plot the functions psat (sd ) and p(sd ) from the data in the shaded part of the table, as shown in the following figure. As we explained in Section 4.6.1, we first try to plot p(sd ) by connecting the points pe and pi by a straight line, which is shown as a dotted line in the plot. For the dotted line, p > psat and φ > 100 % in a certain section of the plot, so that condensation must occur there.

P4 Moisture in building components

161

p/kPa дin

дout

2.5

psa

t

2.0 1.5 1.0

p 0.5

pi

pc pe sd,1

sd,2 5

10

sd,3 15

20

25

30 sd /m

However, in practice, the relative humidity cannot be larger than 100 % and the p profile cannot go above the psat profile. Therefore, we must redraw the p profile as a series of straight lines that touch but never go above the psat profile, which is shown as a solid line in the plot. We see that interstitial condensation is expected at the interface between layers 2 and 3 and that the water vapour pressure at condensation is p = psat = pc = 700 Pa. The amount of condensed water at the interface between two layers is estimated by calculating the difference in density of water vapour flow rate (4.38) between the water vapour entering дin and leaving дout : pi − pc , sd,3 pc − pe дout = δ 0 , sd,1 + sd,2 pi − pc pc − pe kg = δ0 ( − ) = 12.9 × 10−9 2 . sd,3 sd,1 + sd,2 m s дin = δ 0

дc = дin − дout

The amount of water condensed in t = 30 d = 2.59 × 106 s is then ∆ρA = дc t = 0.034

kg . m2

In this problem, we must also estimate the thickness of the polyethylene sheet of water vapour resistance factor µ 4 = 100 000 that would prevent condensation. As shown in the following figure, we extend the first (outer) section of the p profile and find the intersection with a horizontal line starting at the former pi point. Then, we read the equivalent air layer thickness sd,4 = 36.7 m. Finally, the sheet thickness (4.29) is d4 =

sd,4 = 3.7 × 10−4 m. µ4

Problems in Building Physics

162 p/kPa 2.5

psat 2.0 1.5

pi

1.0

p 0.5

pe sd,1

sd,2 10

i

sd,3 20

sd,4

30

40

50

60

70

sd /m

Note that water vapour barriers are usually sold in terms of rounded equivalent air layer thicknesses rather than physical thicknesses. So in this case, we would have to install a water vapour barrier of sd = 50 m.

P4.31 e se

1

si i

2 q

дout

дin дev

The input data for the Glaser diagram are (a) the structure of the building component: the thicknesses d 1 = 0.1 m, d 2 = 0.1 m and d 3 = 0.1 m; the thermal conductivities λ 1 = 0.16 W/(m K), λ 2 = 0.035 W/(m K) and λ 3 = 0.16 W/(m K); and the water vapour resistance factors µ 1 = 16, µ 2 = 1 and µ 3 = 16 and (b) the environmental data, the temperatures θ i = 20 ○ C and θ e = 5 ○ C and the relative humidities φi = 50 % and φe = 80 %. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). First, we calculate the thermal transmittance U = 0.234 W/(m2 K) (3.3), the density of heat flow rate q = 3.51 W/m2 (3.5) and the characteristic temperatures with (3.6), or we calculate the total thermal resistance Rtot = 4.28 m2 K/W (3.2) and the characteristic temperatures with (3.7). Second, we calculate the characteristic water vapour pressures at saturation (PA.1). The results are shown in the following table.

P4 Moisture in building components

163

θ/○ C 5.00 5.14 7.33 17.35 19.54 20.00

e se 1 2 si i

psat /Pa 872 880 1024 1980 2272 2337

p/Pa 697 697

1168 1168

Third, we calculate the equivalent air layer thicknesses (4.29), (4.35) and the internal and external water vapour pressures (4.7): sd,1 = µ 1 d 1 = 1.6 m, sd,2 = µ 2 d 2 = 0.1 m, sd,3 = µ 3 d 3 = 1.6 m, sd,tot = sd,1 + sd,2 + sd,3 = 3.3 m, pe = φe psat (θ e ) = 697 Pa, pi = φi psat (θ i ) = 1168 Pa. We assume that pse = pe and psi = pi . Finally, we plot the functions psat (sd ) and p(sd ) from the data in the shaded portion of the table, as shown in the following figure. As we explained in Section 4.6.1, we first try to plot p(sd ) by connecting the points pe and pi by a straight line, which is shown as a dotted line in the plot. For the dotted line, p < psat , φ < 100 % everywhere, so that no condensation occurs. p/kPa дout

2.5

дin

p sat 2.0 1.5

p

pev

pi

1.0 0.5

pe sd,1 0.5

1.0

sd,2 1.5

sd,3 2.0

2.5

3.0

3.5 sd /m

However, condensed water is present in the building component, and it is evident from the plot that the interstitial condensation must have occurred at the interface between layers 1 and 2. Because water evaporates there, the relative humidity must be 100 % and the water vapour pressure at evaporation must be p = psat = pev = 1024 Pa. Therefore, we must redraw the p profile as a series of straight lines that touch the psat profile at the indicated location, which is shown as a solid line in the plot.

Problems in Building Physics

164

The amount of evaporated water at the interface between two layers is estimated by calculating the difference in density of water vapour flow rate (4.38) between the water vapour leaving дout and entering дin : pi − pev , sd,2 + sd,3 pev − pe дout = δ 0 , sd,1 pev − pe pi − pev kg дev = дout − дin = δ 0 ( − ) = 23.9 × 10−9 2 . sd,1 sd,2 + sd,3 m s дin = δ 0

The amount of water evaporated in t = 30 d = 2.59 × 106 s is then ∆ρA = дev t = 0.062

kg . m2

P4.32 e se 1

si i

2 q дin

дout дc

The input data for the Glaser diagram are (a) the structure of the building component: the thicknesses d 1 = 0.015 m, d 2 = 0.08 m and d 3 = 0.2 m; the thermal conductivities λ 1 = 2.8 W/(m K), λ 2 = 0.035 W/(m K) and λ 3 = 2.3 W/(m K); and the water vapour resistance factors µ 1 = 10 000, µ 2 = 150 and µ 3 = 130 and (b) the environmental data, the temperatures θ i = 20 ○ C and θ e = 0 ○ C and the relative humidities φi = 45 % and φe = 70 %. Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). First, we calculate the thermal transmittance U = 0.392 W/(m2 K) (3.3), the density of heat flow rate q = 7.85 W/m2 (3.5) and the characteristic temperatures with (3.6), or we calculate the total thermal resistance Rtot = 2.55 m2 K/W (3.2) and the characteristic temperatures with (3.7). Second, we calculate the characteristic water vapour pressures at saturation (PA.1). The results are shown in the following table.

P4 Moisture in building components

e se 1 2 si i

165

θ/○ C 0.00 0.31 0.36 18.30 18.98 20.00

psat /Pa 611 625 626 2102 2193 2337

p/Pa 427 427

1052 1052

Third, we calculate the equivalent air layer thicknesses (4.29), (4.35) and the internal and external water vapour pressures (4.7): sd,1 = µ 1 d 1 = 150 m, sd,2 = µ 2 d 2 = 12 m, sd,3 = µ 3 d 3 = 26 m, sd,tot = sd,1 + sd,2 + sd,3 = 188 m, pe = φe psat (θ e ) = 427 Pa, pi = φi psat (θ i ) = 1052 Pa. We assume that pse = pe and psi = pi . Finally, we plot the functions psat (sd ) and p(sd ) from the data in the shaded portion of the table, as shown in the following figure. As we explained in Section 4.6.1, we first try to plot p(sd ) by connecting the points pe and pi by a straight line, which is shown as a dotted line in the plot. For the dotted line, p > psat and φ > 100 % in a certain section of the plot, so that condensation must occur there. p/kPa дin

дout

2.5

p sat 2.0 1.5 1.0

p 0.5

pi

pc

pe sd,1 20

40

60

80

sd,2 100

120

140

160

sd,3 180

200 sd /m

However, in practice, the relative humidity cannot be larger than 100 % and the p profile cannot go above the psat profile. Therefore, we must redraw the p profile as a series of straight lines that touch but never go above the psat profile, which is shown as a solid line in the plot. We see that interstitial condensation is expected at the interface between layers 1 and 2 and that the water vapour pressure at condensation is p = psat = pc = 626 Pa. The amount of condensed water at the interface between two layers is estimated by calculating the difference in density of water vapour flow rate (4.38) between the water

Problems in Building Physics

166 vapour entering дin and leaving дout :

pi − pc , sd,2 + sd,3 pc − pe дout = δ 0 , sd,1 pi − pc pc − pe kg = δ0 ( − ) = 1.97 × 10−9 2 . sd,2 + sd,3 sd,1 m s дin = δ 0

дc = дin − дout

The amount of water condensed in t = 30 d = 2.59 × 106 s is then kg . m2

∆ρA = дc t = 0.0051

In this problem, we must also estimate the thickness of the polyethylene sheet of water vapour resistance factor µ 4 = 100 000 that would prevent condensation. As shown in the following figure, we extend the first (outer) section of the p profile and find the intersection with a horizontal line starting at the former pi point. Then, we read the equivalent air layer thickness sd,4 = 282 m. Finally, the sheet thickness (4.29) is sd,4 = 2.8 × 10−3 m. µ4

d4 =

p/kPa 2.5

psat 2.0 1.5

pi

1.0

p 0.5

pe sd,1 100

i

sd,2 sd,3 200

sd,4 300

400

500 sd /m

Note that water vapour barriers are usually sold in terms of rounded equivalent air layer thicknesses rather than physical thicknesses. So in this case, we would have to install a water vapour barrier of sd = 300 m.

P4 Moisture in building components

167

P4.33 e se

3

1 2

4 si i

q дout

дin дc

The input data for the Glaser diagram are (a) the structure of the building component: the thicknesses d 1 = 0.12 m, d 2 = 0.02 m, d 3 = 0.05 m, d 4 = 0.2 m and d 5 = 0.0125 m; the thermal conductivities λ 1 = 0.76 W/(m K), λ 3 = 0.04 W/(m K), λ 4 = 0.13 W/(m K) and λ 5 = 0.21 W/(m K); the water vapour resistance factors µ 1 = 16, µ 3 = 60, µ 4 = 8 and µ 5 = 10; and the air layer emittances ε 1 = ε 2 = 0.9 and (b) the environmental data, the temperatures θ i = 20 ○ C and θ e = −1 ○ C and the relative humidities φi = 60 % and φe = 80 %. The average temperature and the temperature difference in the air layer are assumed to be T = 274 K (θ = 1 ○ C) and ∆T < 5 K (∆θ < 5 ○ C), respectively. Because the airspace is unventilated, and ∆T < 5 K, we get hca = 1.25 W/(m2 K) and hra = 3.82 W/(m2 K) for the convective (Table P3.1) and radiative (P3.1) surface coefficients, respectively. This gives the thermal resistance of airspace Ra = 0.197 m2 K/W (3.10). Because the building component is vertical, the heat flows horizontally and the surface resistances are Rse = 0.04 m2 K/W and Rsi = 0.13 m2 K/W (Table 3.1). First, we calculate the thermal transmittance U = 0.296 W/(m2 K) (3.13), the density of heat flow rate q = 6.23 W/m2 (3.5) and the characteristic temperatures with (3.6), or we calculate the total thermal resistance Rtot = 3.37 m2 K/W (3.12) and the characteristic temperatures with (3.7). Second, we calculate the characteristic water vapour pressures at saturation (PA.1). The results are shown in the following table.

e se 1 2 3 4 si i

θ/○ C −1.00 −0.75 0.23 1.46 9.24 18.82 19.19 20.00

psat /Pa 562 574 621 679 1166 2172 2222 2337

p/Pa 450 450

1402 1402

Third, we calculate the equivalent air layer thicknesses (4.29), (4.35): sd,1 = µ 1 d 1 = 1.92 m, sd,3 = µ 3 d 3 = 3 m, sd,4 = µ 4 d 4 = 1.6 m, sd,5 = µ 5 d 5 = 0.13 m, sd,tot = sd,1 + sd,2 + sd,3 + sd,4 + sd,5 = 6.65 m. The given equivalent air layer thickness of the unventilated airspace is sd,2 = 0 m, which is a reasonable assumption because we already assumed the same for the other air layers,

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sd,se = sd,si = 0 m. We also calculate the internal and external water vapour pressures (4.7): pe = φe psat (θ e ) = 450 Pa, pi = φi psat (θ i ) = 1402 Pa. We assume that pse = pe and psi = pi . Finally, we plot the functions psat (sd ) and p(sd ) from the data in the shaded portion of the table, as shown in the following figure. As we explained in Section 4.6.1, we first try to plot p(sd ) by connecting the points pe and pi by a straight line, which is shown as a dotted line in the plot. For the dotted line, p > psat and φ > 100 % in a certain section of the plot, so that condensation must occur there. p/kPa дin

дout

2.5

p sa

2.0

t

1.5

p

pi

1.0 0.5

pc

pe sd,1 1

sd,3 2

3

sd,4 4

5

sd,5 6

7 sd /m

However, in practice, the relative humidity cannot be larger than 100 % and the p profile cannot go above the psat profile. Therefore, we must redraw the p profile as a series of straight lines that touch but never go above the psat profile, which is shown as a solid line in the plot. We see that interstitial condensation is expected at the interface between layers 1 and 2 and that the water vapour pressure at condensation is p = psat = pc = 621 Pa. The amount of condensed water at the interface between two layers is estimated by calculating the difference in density of water vapour flow rate (4.38) between the water vapour entering дin and leaving дout : pi − pc , sd,2 + sd,3 + sd,4 + sd,5 pc − pe , дout = δ 0 sd,1 pi − pc pc − pe kg = δ0 ( − ) = 1.52 × 10−8 2 . sd,2 + sd,3 + sd,4 + sd,5 sd,1 m s дin = δ 0

дc = дin − дout

The amount of water condensed in t = 30 d = 2.59 × 106 s is then ∆ρA = дc t = 0.0395

kg . m2

P5 Basics of waves P5.1 To characterise the properties of waves, either the frequency or the wavelength is sufficient, because they are connected by the dispersion relation (5.10). Thus, the frequencies for the wavelengths λ 1 = 5 × 10−7 m (visible light) and λ 2 = 1 × 10−5 m (infrared light) are c0 = 6.0 × 1014 Hz, λ1 c0 f2 = = 3.0 × 1013 Hz, λ2 f1 =

where c 0 , the speed of light, is the speed of all electromagnetic waves.

P5.2 With two parallel walls, we are confronted with the situation of two fixed boundaries (Fig. 5.12, left), and the frequencies of the normal modes (5.19) are fn = n

c . 2L

The shortest distance is obtained for the first normal mode, n = 1. Thus the distance for the frequency f = 110 Hz is c L= = 1.56 m, 2f where c is the speed of sound.

P5.3 In the case of the ear canal, we are confronted with the situation of one fixed boundary (tympanic membrane) and one free boundary (ear canal opening), as shown in Fig. 5.12 on the right. The frequencies of the normal modes (5.20) are thus f n = (2n − 1)

c , 4L

169

170

Problems in Building Physics where L = 2.5 × 10−2 m is the length of the ear canal and c is the speed of sound. The equation gives an infinite harmonic series, however we must take into account that human hearing is capable of processing only frequency range from 20 Hz to 20 kHz (Section 6.3.2). The three values smaller than 20 kHz are 3.4 kHz, 10.3 kHz and 17.2 kHz.

P6 Sound propagation P6.1

I 2 , L p,2

I 1 , L p,1

S

Generally, we can solve acoustic problems with fundamental quantities (power, intensity) or with auxiliary quantities (level). However, in this problem we cannot avoid using both because the level gain ∆L p = 20 dB is an auxiliary quantity, whereas the area of the tympanic membrane S = 4.3 × 10−5 m2 and the sound power P are fundamental quantities. First, we use the auxiliary quantities. The quietest detectable sound corresponds to the sound pressure level L p,1 = 0 dB (Section 6.3.2). The amplified sound pressure level at the tympanic membrane is therefore L p,2 = L p,1 + ∆L p = 20 dB. We must convert the sound pressure level at the tympanic membrane into corresponding sound intensity I 2 (6.24): L p,2 = 10 lg

I2 I2 I2 Ô⇒ lg = 0.1 L p,2 Ô⇒ = 100.1 L p,2 Ô⇒ I 2 = I 0 100.1 L p,2 . I0 I0 I0

Then we multiply it by the area of the tympanic membrane to get the sound power (6.17) as P = S I 2 = S I 0 100.1 L p,2 = 4.3 × 10−15 W. Second, we use the fundamental quantities. The quietest detectable sound corresponds to the sound intensity I 1 = 1 × 10−12 W/m2 (Section 6.3.2).

171

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We must convert the level gain into the sound intensity amplification factor f I (6.24): ∆L p = L p,2 − L p,1 = 10 lg

I2 I1 I 2 I 1 −1 I2 I2 − 10 lg = 10 lg [ ( ) ] = 10 lg Ô⇒ lg = 0.1 ∆L p I0 I0 I0 I0 I1 I1 I2 Ô⇒ f I = = 100.1 ∆L p = 100. I1

The sound intensity at the tympanic membrane must be multiplied by the area of the tympanic membrane to get the sound power (6.17), as P = S I 2 = S f I I 1 = 4.3 × 10−15 W.

i

Note that the multiplication by a fundamental quantity f I is converted into the addition of an auxiliary quantity ∆L p . This is typical for all calculations with auxiliary (level) quantities.

P6.2 The sound power level is calculated from the sound power P = 2 × 10−6 W from its definition (6.23) as P L W = 10 lg = 63 dB. P0 To obtain the sound intensity and the sound pressure level at distance r = 10 m from the sound source, we use the expressions for a point sound source, (6.30) and (6.31): P W = 1.6 × 10−9 2 , 4π r 2 m r L p = L W − 20 lg − 11 dB = 32 dB. r0 I=

Alternatively, the sound pressure level can also be calculated from the sound intensity from its definition (6.24) as I L p = 10 lg = 32 dB. I0

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173

P6.3 From the sound pressure level L p,1 = 70 dB and the distance from the sound source r 1 = 3 m we calculate the linear sound power level using the expression for a linear sound source (6.35): L p,1 = L′W − 10 lg

r1 r1 − 8 dB Ô⇒ L′W = L p,1 + 10 lg + 8 dB = 82.8 dB. r0 r0

Using the same expression, we calculate the distance for the sound pressure level L p,2 = 58 dB: L p,2 = L′W − 10 lg

′ r2 r2 r2 − 8 dB Ô⇒ lg = 0.1(L′W − L p,2 − 8 dB) Ô⇒ = 100.1(L W −L p,2 −8 dB) r0 r0 r0

Ô⇒ r 2 = r 0 100.1(L W −L p,2 −8 dB) = 48 m. ′

P6.4 The sound power level is calculated from the sound pressure level at distance r = 0.5 m from the loudspeaker, L p = 105 dB, with the expression for a point sound source (6.31): L p = L W − 20 lg Ô⇒ L W = L p + 20 lg

r − 11 dB r0

r + 11 dB = 110 dB. r0

The electrical power of the loudspeaker Pe = 10 W is already given, so to obtain the efficiency, we have to calculate the sound power of the loudspeaker P using the definition of the sound power level (6.23): L W = 10 lg

P P P Ô⇒ lg = 0.1 L W Ô⇒ = 100.1 L W P0 P0 P0 Ô⇒ P = P0 100.1 L W = 0.1 W.

The efficiency η is therefore η=

P = 0.01. Pe

At first glance, the calculated efficiency seems too small, but its value is actually very typical: loudspeakers are very inefficient in terms of usable energy conversion.

i

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174

P6.5 To add three sound power levels L W,1 = 55 dB, L W,2 = 50 dB and L W,3 = 45 dB, we have to use the special expression for the level addition (6.27): L W = 10 lg (100.1L W,1 + 100.1L W,2 + 100.1L W,3 ) = 56.5 dB. Multiple adjacent point sound sources can be treated as a single point sound source if the distance from the sources is large enough, that is, much larger than the distance between the sound sources themselves. We can therefore use the expression for a point sound source (6.31). At distance r, the sound is no longer audible, so the sound pressure level must be L p = 0 dB (Section 6.3.2). Taking these two facts together, we get: L p = L W − 20 lg

r r r − 11 dB Ô⇒ lg = 0.05(L W − L p − 11 dB) Ô⇒ = 100.05(L W −L p −11 dB) r0 r0 r0 Ô⇒ r = r 0 100.05(L W −L p −11 dB) = 189 m.

P6.6 The electric motor is obviously a point sound source, so we calculate the sound pressure level caused by an electric motor of sound power level L W,1 = 90 dB at distance r 1 = 15 m (6.31) as r1 L p,1 = L W,1 − 20 lg − 11 dB = 55.5 dB. r0 Similarly, we conclude that the road is a linear sound source, so we calculate the sound pressure level caused by a road of linear sound power level L′W,2 = 75 dB at distance r 2 = 25 m (6.35) as r2 L p,2 = L′W,2 − 10 lg − 8 dB = 53.0 dB. r0 The total sound pressure level is simply the sum of the sound pressure levels (6.27) caused by the electric motor and road: L p = 10 lg (100.1L p,1 + 100.1L p,2 ) = 57.4 dB.

P6.7 Generally, we can solve acoustic problems with fundamental quantities (power, intensity) or with auxiliary quantities (level). However, in this problem we cannot avoid using both because the maximum sound pressure level L p = 55 dB is an auxiliary quantity, whereas the linear sound power of the road P1′ = 2.5 × 10−5 W/m and the sound power of the device P2 are fundamental quantities. Note that the distances to the road and to the device,

P6 Sound propagation

175

r 1 = 20 m and r 2 = 10 m, respectively, are actually fundamental quantities, but are not converted even when auxiliary quantities are used. First, we use the auxiliary quantities. For the road, which is a linear sound source, we calculate the linear sound power level using its definition (6.33) and then its contribution to the sound pressure level (6.35): P1′ = 74.0 dB, P0′ r1 = L′W,1 − 10 lg − 8 dB = 53.0 dB. r0 L′W,1 = 10 lg

L p,1

Because we know the maximum total sound pressure level and the contribution of the road, we calculate the maximum acceptable contribution of the device using the expression for the level addition (6.27): L p = 10 lg (100.1L p,1 + 100.1L p,2 ) Ô⇒ L p,2 = 10 lg (100.1L p − 100.1L p,1 ) = 50.7 dB. For the device, which is a point sound source, we calculate its maximum sound power level (6.31) from its maximum contribution: L p,2 = L W,2 − 20 lg Ô⇒ L W,2 = L p,2 + 20 lg

r2 − 11 dB r0

r2 + 11 dB = 81.7 dB. r0

Finally, we convert the result into the sound power using the definition of sound power level (6.23): L W,2 = 10 lg

P2 P2 P2 Ô⇒ lg = 0.1 L W,2 Ô⇒ = 100.1 L W,2 P0 P0 P0

Ô⇒ P2 = P0 100.1 L W,2 = 1.49 × 10−4 W.

Second, we use the fundamental quantities. Using the definition of sound pressure level (6.24), we first calculate the maximum total sound intensity: I I I Ô⇒ lg = 0.1 L p Ô⇒ = 100.1 L p I0 I0 I0 W Ô⇒ I = I 0 100.1 L p = 3.16 × 10−7 2 . m

L p = 10 lg

For the road, which is a linear sound source, we calculate its contribution to the sound intensity (6.34) as P′ W I 1 = 1 = 1.99 × 10−7 2 . 2π r 1 m

Problems in Building Physics

176

Then we subtract the result from the maximum total sound intensity to get the maximum acceptable contribution of the device as in I 2 = I − I 1 = 1.17 × 10−7

W . m2

Finally, we calculate the maximum sound power of the device, which is a point sound source, from its maximum contribution (6.30): I2 =

i

P2 Ô⇒ P2 = 4π r 22 I 2 = 1.47 × 10−4 W. 4π r 22

Note that there is a slight difference in the result. This is because in the level expressions (6.31) and (6.35) the constants 11 dB and 8 dB are slightly approximative. But even with these approximations, the result is sufficiently precise.

P6.8 r1

r1

LW

LW

L p,A L p,B

LW

LW r3

r2

If the distances to both loudspeakers of the sound power level L W are equal, r 1 = 2.5 m, the sound pressure level is L p,A = 40 dB. The loudspeakers are point sound sources, so to obtain the expression for L p,A , we first write down the contribution of one loudspeaker L p,1 (6.31) and multiply it by 2 using the expression for the level multiplication (6.29): r1 − 11 dB, r0 r1 − 20 lg − 11 dB + 10 lg 2. r0

L p,1 = L W − 20 lg L p,A = L p,1 + 10 lg 2 = L W

The sound power level of each loudspeaker is therefore L W = L p,A + 20 lg

r1 + 11 dB − 10 lg 2 = 55.9 dB. r0

To get the sound pressure level of the loudspeakers L p,B at different distances, r 2 = 4 m and r 3 = 1 m, we must write down the contributions of each loudspeaker separately (6.31)

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177

and sum them using the expression for the level addition (6.27): r2 − 11 dB = 32.9 dB, r0 r3 L p,3 = L W − 20 lg − 11 dB = 44.9 dB, r0 0.1L p,2 (10 L p,B = 10 lg + 100.1L p,3 ) = 45.2 dB. L p,2 = L W − 20 lg

P6.9 The total sound pressure level L p = 80 dB is related to the contributions of the first and second sound source, L p,1 and L p,2 = 75 dB, respectively, by the expression for the level addition (6.27). The contribution of the first sound source is obtained by rearranging the equation: L p = 10 lg (100.1L p,1 + 100.1L p,2 ) Ô⇒ L p,1 = 10 lg (100.1L p − 100.1L p,2 ) = 78.3 dB. Determining the noise of the sound source of interest by switching it off is a common practice when a significant (nonnegligible) amount of noise originates from other sound sources that cannot be muted. Measuring the noise with the sound source of interest switched off gives the so-called background noise, which is then subtracted from the total noise.

P6.10 To determine the average of two sound pressure levels, L p,1 = 60 dB and L p,2 = 50 dB, we must use the level averaging expression (6.28). However, the weights of the two values are not equal, because the first and second sound pressure levels apply to 3 h and 7 h periods, respectively. To obtain the correct average value, we take the average of the total 10 values, of which n 1 = 3 are equal to the first sound pressure level and n 2 = 7 to the second sound pressure level as L p = 10 lg [

1 (n 1 100.1L p,1 + n 2 100.1L p,2 )] = 55.7 dB. n1 + n2

P6.11 To determine the average of the daytime, evening and night sound pressure levels, Ld = 61 dB, Le = 58 dB and Ln = 52 dB, respectively, we must use the level averaging expression (6.28). However, we must take into account that the daytime, evening and night periods are 12 h, 4 h, and 8 h long, respectively (Section 6.3.3). Therefore, we take

i

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178

the average of the total 24 values, of which 12 values are equal to the daytime sound pressure level, 4 values to the evening sound pressure level and 8 values to the night sound pressure level, as in L = 10 lg [

1 (12 ⋅ 100.1L d + 4 ⋅ 100.1L e + 8 ⋅ 100.1L n )] = 59.0 dB. 24

P6.12 The multiplication of a fundamental quantity (power, intensity) by a factor N is converted into the addition of a value ∆L to an auxiliary quantity (level). This is described by the expression for the level multiplication (6.29), L N = L 1 + 10 lg N , where the multiplication factor is N = 5. The increase in level is thus ∆L = L N − L 1 = 10 lg N = 7.0 dB.

P6.13 First, we calculate the linear sound power levels of each vehicle category by multiplying the linear sound power levels of one light, one medium heavy and one heavy vehicle per hour, L′W,1 = 55.4 dB, L′W,2 = 59.8 dB, and L′W,3 = 62.7 dB, respectively, by the respective number of vehicles N 1 = 285, N 2 = 27 and N 3 = 13 (6.29): L′W,1,tot = L′W,1 + 10 lg N 1 = 79.9 dB,

L′W,2,tot = L′W,2 + 10 lg N 2 = 74.1 dB,

L′W,3,tot = L′W,3 + 10 lg N 3 = 73.8 dB.

In the next step, we add up the determined linear sound power levels (6.27) as L′W = 10 lg (100.1L W,1,tot + 100.1L W,2,tot + 100.1L W,3,tot ) = 81.7 dB. ′

′

′

We could have combined addition and multiplication without calculating the intermediate values to get: L′W = 10 lg (N 1 100.1L W,1 + N 2 100.1L W,2 + N 3 100.1L W,3 ) = 81.7 dB. ′

′

′

P7 Building acoustics

15 m

P7.1

6m

31 m

d3 1.5 m

d1

4.5 m 4.5 m

10 m

15 m

d2

d1

The preceding figure shows the floor plan (left) and the vertical cross section of the room (right). The cross section is taken along the dash-dotted line on the floor plan. First we calculate the path length of the direct sound shown in the figure on the left as d1 =

√ (31 m)2 + (10 m)2 = 32.6 m.

The easiest way to determine the path length of the reflected sound is to construct the image source by mirroring the real source over the reflecting surface. The path length of the sound reflected from the right wall shown in the figure on the left is d2 =

√ (31 m)2 + (40 m)2 = 50.6 m

and the path length of the sound reflected from the ceiling shown in the figure on the right is √ d3 =

d 12 + (9 m)2 = 33.8 m.

179

Problems in Building Physics

180 The time delays (Section 7.2.1) are therefore d2 − d1 = 53 ms, c d3 − d1 ∆t 3 = = 3.6 ms, c ∆t 2 =

where c is the speed of sound.

P7.2 I 1 , L p,1 I 2 , L p,2

In terms of fundamental quantities, the reflected sound energy is obtained by multiplying the incident sound energy by the reflectance ρ (7.1), a number smaller than one. The multiplication by a fundamental quantity smaller than one is converted into the addition of an auxiliary quantity smaller than zero or into the subtraction of a positive auxiliary quantity ∆L p = 7.0 dB. Because the initial sound pressure level L p,1 is larger than the final L p,2 , we write (6.24): ∆L p = L p,1 − L p,2 = 10 lg

I1 I2 I 1 −1 I 2 I2 − 10 lg = −10 lg [( ) ] = −10 lg = −10 lg ρ I0 I0 I0 I0 I1

Ô⇒ lg ρ = −0.1 ∆L p Ô⇒ ρ = 10−0.1 ∆L p . Absorbance is therefore α = 1 − ρ = 1 − 10−0.1 ∆L p = 0.80.

P7.3 Each of two reflections from the wall corresponds to the multiplication of the sound intensity by the reflectance ρ (7.1), as in I2 = ρ ρ I1 = ρ2 I1 . Note that, as explained in the solution to problem P7.2 , the multiplication of a fundamental quantity, such as sound intensity, by a value ρ smaller than one is converted into the subtraction of a positive auxiliary quantity ∆L p from an auxiliary quantity, such as sound pressure level. In the case of two reflections we get: L p,2 = L p,1 − ∆L p − ∆L p = L p,1 − 2∆L p .

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181

The total sound pressure level reduction (7.9) is therefore 2∆L p = −2 ⋅ 10 lg ρ = −2 ⋅ 10 lg(1 − α) = 1.9 dB, where α = 0.2 is the absorbance.

P7.4 To solve the problem, we use the sound path lengths d 1 = 32.6 m, d 2 = 50.6 m and d 3 = 33.8 m of the direct sound, the sound reflected from the wall and the sound reflected from the ceiling, respectively, which were calculated in the solution to problem P7.1 . The attenuation due to the distance from the point sound source (geometrical divergence) is already part of the expression for a point sound source (6.31), so the sound pressure level of the direct sound L p,1 is L p,1 = L W − 20 lg

d1 − 11 dB = L W − 41.26 dB. r0

The reflection additionally reduces the sound pressure level of the sounds. Because the sound pressure level is an auxiliary quantity, this is taken into account by a simple subtraction of the attenuation due to the reflection ∆L p (7.9): ∆L p = −10 lg ρ = −10 lg(1 − α) = 0.46 dB. The sound pressure levels of the sound reflected from the wall L p,2 and the sound reflected from the ceiling L p,3 are therefore written as d2 − 11 dB − ∆L p = L W − 45.54 dB, r0 d3 − 20 lg − 11 dB − ∆L p = L W − 42.03 dB, r0

L p,2 = L W − 20 lg L p,3 = L W

where α = 0.1 is the absorbance. We easily determine the sound pressure level differences between the reflected sounds and the direct sound by the subtraction as ∆L p,2 = L p,1 − L p,2 = 4.28 dB, ∆L p,3 = L p,1 − L p,3 = 0.78 dB. The sum of the sound pressure levels (6.27) of the two reflected sounds is L p,23 = 10 lg (100.1L p,2 + 100.1L p,3 ) = 10 lg [100.1L W ⋅ (10−0.1⋅45.54 + 10−0.1⋅42.03 )] = L W + 10 lg (10−0.1⋅45.54 + 10−0.1⋅42.03 ) = L W − 40.43 dB.

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182

We see that two reflected sounds together have a larger sound pressure level than the direct sound and that the difference is ∆L p,23 = L p,23 − L p,1 = 0.82 dB.

7.5 m

P7.5

7.5 m

d2

d1

4.5 m

i

Although the sound pressure level of each reflected sound is necessarily smaller than the sound pressure level of the direct sound, typically two reflected sounds are already sufficient to prevail. In fact, the number of reflections is much larger and, as we will see in the solution to problem P7.8 , the contribution of the direct sound can generally be neglected.

13 m

First we calculate the path length of the direct sound as d1 =

√ (13 m)2 + (4.5 m)2 = 13.8 m.

The easiest way to determine the path length of the reflected sound is to construct the image source by mirroring the real source over the reflecting surface. The path length of the sound reflected from the right wall shown in the preceding figure is d2 =

√ (13 m)2 + (19.5 m)2 = 23.4 m.

The time delay (Section 7.2.1) is then ∆t = where c is the speed of sound.

d2 − d1 = 0.028 s, c

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183

The attenuation due to the distance from the point sound source (geometrical divergence) is already part of the expression for a point sound source (6.31), so the sound pressure level of the direct sound is L p,1 = L W − 20 lg

d1 − 11 dB = 26.2 dB, r0

where L W = 60 dB is the sound power level. The reflection additionally reduces the sound pressure level of the sound, which is taken into account by a simple subtraction of the attenuation due to the reflection (7.9), which amounts to ∆L p = −10 lg ρ = −10 lg(1 − α). The sound pressure level of the reflected sound is therefore L p,2 = L W − 20 lg

d2 − 11 dB + 10 lg(1 − α) = 21.1 dB, r0

where α = 0.1 is the absorbance of the wall.

3.95 m

P7.6

d2 d1 5.3 m

2.9 m

2.9 m

First, we calculate the path length of the direct sound as d1 =

√ (3.95 m)2 + (5.3 m)2 = 6.61 m.

The easiest way to determine the path length of the reflected sound is to construct the image source by mirroring the real source over the reflecting surface. The path length of the sound reflected from the barrier, shown in the preceding figure, is d2 =

√ (3.95 m)2 + (11.1 m)2 = 11.78 m.

The sound pressure level of the direct sound is calculated using the expression for a linear sound source (6.35) as L p,1 = L′W − 10 lg

d1 − 8 dB = 53.8 dB, r0

Problems in Building Physics

184 where L′W = 70 dB is the linear sound power level.

To obtain the sound pressure level of the reflected sound, we must additionally subtract the attenuation due to the reflection (7.9), as in the solution to problem P7.5 : L p,2 = L′W − 10 lg

d2 − 8 dB + 10 lg(1 − α) = 50.3 dB, r0

where α = 0.2 is the absorbance of the barrier. The total sound pressure level is simply the sum of the sound pressure levels (6.27) of the direct and reflected sound: L p = 10 lg (100.1L p,1 + 100.1L p,2 ) = 55.4 dB.

i

Note that according to cnossos-eu the reflection from the road is already taken into account in the linear sound power level. Here the calculation provides a very reliable result.

P7.7 If we know the acoustic properties of the room, we can calculate the reverberation time from the Sabine’s equation (7.18). Therefore, we need the equivalent absorption area and the volume of the room. The latter is obtained from the dimensions of the room a = 10 m, b = 20 m and h = 4 m as V = a b h = 800 m3 . To obtain the equivalent absorption area (7.12), we have to combine the absorbances of glass αg = 0.03, timber αt = 0.06 and concrete αc = 0.02 with the respective areas. Because the room has Nw = 6 windows, each of area Sw = 10 m2 , the areas of glass Sg , timber St and concrete Sc are Sg = Nw Sw = 60 m2 , St = a b = 200 m2 , Sc = a b + 2 a h + 2 b h − Sg = 380 m2 . On the other hand, the equivalent absorption area of each of Ns = 70 seats is given to be As = 0.5 m2 and must be simply added to the remaining absorption area as in A = αg Sg + αt St + αc Sc + Ns As = 56.4 m2 . The reverberation time is therefore T60 = (0.163

s V ) ⋅ = 2.3 s. m A

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185

In acoustic calculations, it is common to use predefined values of equivalent absorption area for common objects. Those come from various sources, such as design manuals. In addition, the calculations are usually performed separately for different frequency bands, as the acoustic properties of materials depend on the frequency.

P7.8

3m

d 9m

For the sound pressure level of the direct sound, we must first calculate the sound path length as √ d = (9 m)2 + (3 m)2 = 9.5 m and insert it into the expression for a point sound source (6.31) as in L p,dir = L W − 20 lg

d − 11 dB = 39.5 dB, r0

where L W = 70 dB is the sound power level of speech. For the sound pressure level of the diffuse sound, we must first calculate the equivalent absorption area of the room of dimensions a = 10 m, b = 15 m and h = 4 m. The absorbances of the walls/ceiling and floor are αwc = 0.04 and αf = 0.06, respectively, whereas the areas of the walls/ceiling and floor are Swc = a b + 2 a h + 2 b h = 350 m2 , Sf = a b = 150 m2 , respectively. The equivalent absorption area of the room (7.12) also includes Np = 60 people, each of equivalent absorption area Ap = 1 m2 , so the total equivalent absorption area is A = αwc Swc + αf Sf + Np Ap = 83 m2 . The sound pressure level of the diffuse sound (7.15) is L p,dif = L W − 10 lg

A = 56.8 dB. 4S 0

i

Problems in Building Physics

186

The total sound pressure level is simply the sum of the sound pressure levels of the direct and all reflected sounds (6.27): L p = 10 lg (100.1L p,dir + 100.1L p,dif ) = 56.9 dB.

i

As already mentioned, the diffuse sound, which corresponds to the sound reflected from the surfaces in the room, is significantly larger than the direct sound in most of the room. In most practical situations, direct sound can therefore be neglected. The notable exceptions are the situations where the receiver and the source are very close to each other, or the rooms where a proper diffuse sound field cannot be established due to large dimensions and large absorbance.

P7.9 Adding absorptive surfaces or changing their acoustic properties directly changes the equivalent absorption area; the change in reverberation time is only a consequence of the change in the equivalent absorption area. So, we start by calculating the initial equivalent absorption area from the initial reverberation time T60,1 = 2 s. To do this, we calculate the volume of the room from its dimensions a = 20 m, b = 10 m and h = 4 m, as V = a b h = 800 m3 , and use Sabine’s equation (7.18): T60,1 = (0.163

s V s V Ô⇒ A 1 = (0.163 ) ⋅ = 65.2 m2 . )⋅ m A1 m T60,1

If we include Np = 30 people, each of equivalent absorption area Ap = 0.7 m2 , the intermediate equivalent absorption area is A 2 = A 1 + Np Ap = 86.2 m2 , which is used to calculate the intermediate reverberation time T60,2 as T60,2 = (0.163

s V = 1.51 s. )⋅ m A2

In the last step, we cover the concrete ceiling of area S = a b = 200 m2 and absorbance αc = 0.05 with an absorber of absorbance αa = 0.5. The concrete no longer absorbs sound, so we have effectively replaced the concrete surface with the absorber

P7 Building acoustics

187

surface or, in other words, changed the absorbance of the ceiling. The final equivalent absorption area (7.12) is therefore A 3 = A 2 + αa S − αc S = A 2 + (αa − αc )S = 176.2 m2 , which is used to calculate the final reverberation time T60,3 as T60,3 = (0.163

s V = 0.74 s. )⋅ m A3

P7.10 Adding absorptive surfaces or changing their acoustic properties directly changes the equivalent absorption area; the change in reverberation time is only a consequence of the change in the equivalent absorption area. So we start by calculating the initial equivalent absorption area from the initial reverberation time T60,1 = 2 s. To do this we calculate the volume of the auditorium from its dimensions a = 10.8 m, b = 7.1 m and h = 3.8 m as V = a b h = 291.4 m3 , and use Sabine’s equation (7.18): T60,1 = (0.163

s V s V Ô⇒ A 1 = (0.163 ) ⋅ = 23.7 m2 . )⋅ m A1 m T60,1

If we include Np = 34 people of the equivalent absorption area Ap = 0.44 m2 each, the intermediate equivalent absorption area is A 2 = A 1 + Np Ap = 38.7 m2 , which is used to calculate the intermediate reverberation time T60,2 as T60,2 = (0.163

s V = 1.23 s. )⋅ m A2

The final equivalent absorption area is determined from the target reverberation time T60,3 = 0.74 s as s V A 3 = (0.163 ) ⋅ = 64.2 m2 . m T60,3 The change in the equivalent absorption area is achieved by covering the concrete wall of area S and absorbance αc = 0.06 with an absorber of absorbance αa = 0.5. The concrete no longer absorbs sound, so we have effectively replaced the concrete surface with the absorber surface or, in other words, changed the absorbance of part of the wall. We

Problems in Building Physics

188

therefore write a relationship between the two equivalent absorption areas (7.12) and obtain the area of the covered concrete: A 3 = A 2 + αa S − αc S = A 2 + (αa − αc )S A3 − A2 Ô⇒ S = = 58 m2 . αa − αc

P7.11 r1

r2

In the initial situation, the sound pressure level is written with the expression for a point sound source (6.31) as r1 L p,1 = L W − 20 lg − 11 dB, r0 where r 1 = 40 m is the initial distance between the person and the sound source. The barrier additionally reduces the sound pressure level. Because the sound pressure level is an auxiliary quantity, this is taken into account in the situation of the free sound field by a simple subtraction of the sound reduction index R = 5 dB (7.8) as in L p,2 = L W − 20 lg

r2 − 11 dB − R, r0

where r 2 is the final distance between the person and the sound source. Because the loudness, that is, the sound pressure level, does not change, L p,1 = L p,2 , we obtain: r1 r2 r 1 r 2 −1 r1 − 20 lg = 20 lg [ ( ) ] = 20 lg r0 r0 r0 r0 r2 r1 r1 = 100.05 R Ô⇒ lg = 0.05 R Ô⇒ r2 r2 Ô⇒ r 2 = r 1 10−0.05 R = 22.5 m.

R = 20 lg

i

In this problem we have neglected the effect of diffraction (Fig. 6.17). In typical situations, especially for barriers of large sound reduction index, the contribution of the diffracted sound is larger than the contribution of the transmitted sound behind the barrier. Note that the attenuation due to the diffraction depends on the geometry of the barrier and the frequency of sound [2].

P7 Building acoustics

189

P7.12 The sound pressure level without the barrier is written with the expression for a linear sound source (6.35) as r L p = L′W − 10 lg − 8 dB, r0 where r = 32 m is the distance between the road and the building. The barrier additionally reduces the sound pressure level, which is taken into account in the situation of the free sound field by a simple subtraction of the sound reduction index R = 10 dB (7.8), as in L p = L′W − 10 lg

r − 8 dB − R. r0

We therefore calculate the linear sound power level from the sound pressure level at the building, L p = 57 dB, as L′W = L p + 10 lg

r + 8 dB + R = 90 dB. r0

d2 d1 5.3 m

5.3 m

2.95 m

P7.13

6.1 m

First, we calculate the path length of the sound transmitted through the barrier as d1 =

√ (2.95 m)2 + (6.1 m)2 = 6.8 m.

The geometrical divergence of the transmitted and reflected sound is already taken into account in the expression for a linear sound source (6.35). As in the solution to problem P7.12 , to calculate the sound pressure level of the transmitted sound we must additionally subtract the attenuation due to the transmission, that is, the sound reduction index (7.8), as in L p,1 = L′W − 10 lg

d1 − 8 dB − R = 43.7 dB, r0

where L′W = 75 dB and R = 15 dB are the linear sound power level and the sound reduction index of the barrier, respectively.

Problems in Building Physics

190

The easiest way to determine the path length of the sound reflected from the left fac¸ade is to construct the image source by mirroring the real source over the reflecting surface. The path length of the sound reflected from the wall shown in the preceding figure is d2 =

√ (2.95 m)2 + (16.7 m)2 = 17.0 m.

As in the solution to problem P7.5 , to calculate the sound pressure level of the transmitted sound we must additionally subtract the attenuation due to the reflection (7.9), as in L p,2 = L′W − 10 lg

d2 − 8 dB + 10 lg(1 − α) = 53.7 dB, r0

where α = 0.2 is the absorbance of the fac¸ade. The total sound pressure level is simply the sum of the sound pressure levels of the transmitted and reflected sound (6.27): L p = 10 lg (100.1L p,1 + 100.1L p,2 ) = 54.1 dB.

i

i

Because we know the geometry of the barrier, we can also calculate the contribution of the diffracted sound (Fig. 6.17), using the standard iso 9613-2 [2]. This calculation is relatively complex and frequency dependent. For this problem, the attenuation due to the diffraction is 5 dB at 20 Hz and 20 dB at 20 kHz and is thus comparable to the attenuation due to the transmission. Nevertheless, the reflected sound remains by far the most important contribution to the total sound pressure level. Note that according to cnossos-eu the reflection from the road is already taken into account in the linear sound power level. Here the calculation provides a very reliable result.

P7.14 The sound pressure level in the source room depends on the properties of the sound source, the sound power level L W = 80 dB, and on the acoustic properties of the source room, the equivalent absorption area A 1 = 12 m2 (7.15), as in L p,1 = L W − 10 lg

A1 = 75 dB. 4S 0

The sound pressure level in the receiving room also depends on the properties of the wall between the rooms, the area S = 8 m2 and the apparent sound reduction index R′ = 50 dB, and on the acoustic properties of the receiving room, the equivalent absorption area A 2 = 20 m2 (7.28), as in L p,2 = L p,1 − R′ + 10 lg (

S ) = 21 dB. A2

P7 Building acoustics

191

S, R ′

A1

L p,1

LW (7.15)

A2

L p,2 (7.28)

In this type of problem, it is important to distinguish between the functions of each equivalent absorption area. As shown in the preceding figure, the sound pressure level in the source room L p,1 depends on the acoustic properties of the source room, the equivalent absorption area A 1 . On the other hand, the sound pressure level in the receiving room L p,2 depends directly only on the acoustic properties of the receiving room, the equivalent absorption area A 2 ; only indirectly (via L p,1 ) does it also depend on A 1 . So, if we already know the value of L p,1 , the value of A 1 is redundant.

P7.15 To calculate the apparent sound reduction index, we need the sound pressure levels in both rooms, the area of the wall between them and the equivalent absorption area of the receiving room. To calculate the latter, we also need the reverberation time and the volume of the receiving room. We calculate the area of the wall between the rooms and the volume of the receiving room from the given floor plan dimensions and the height of the rooms h = 2.5 m as S = l 2 h = 7.5 m2 , V2 = w 2 l 2 h = 18.8 m3 . The equivalent absorption area of the receiving room is obtained from Sabine’s equation (7.18): T60,2 = (0.163

s V2 s V2 Ô⇒ A 2 = (0.163 ) ⋅ = 12.2 m2 . )⋅ m A2 m T60,2

The apparent sound reduction index (7.28) is R ′ = L p,1 − L p,2 + 10 lg (

S ) = 40.9 dB. A2

This problem represents a typical situation of the field measurements of the apparent sound reduction index. Because the sound pressure level in the source room is determined by the measurement, other information such as the volume, reverberation time and equivalent absorption area of the source room are redundant.

i

Problems in Building Physics

192

P7.16 To relate the sound pressure levels in the two rooms, we need the apparent sound reduction index, the area of the wall between the rooms and the equivalent absorption area of the receiving room. To calculate the latter, we also need the reverberation time and the volume of the receiving room. We calculate the area of the wall between the rooms and the volume of the living (receiving) room from the given floor plan dimensions and the height of the rooms h = 2.6 m as S = l 1 h = 5.2 m2 , V2 = w 2 l 2 h = 39 m3 . The equivalent absorption area of the living (receiving) room is obtained from the reverberation time T60,2 = 0.5 s (7.18) as T60,2 = (0.163

s V2 s V2 Ô⇒ A 2 = (0.163 ) ⋅ = 12.7 m2 . )⋅ m A2 m T60,2

Knowing the maximum sound pressure level in the living (receiving) room L p,2 = 35 dB and the apparent sound reduction index R′ = 57 dB, we calculate the maximum sound pressure level in the engine (source) room L p,1 (7.28) as L p,1 = L p,2 + R ′ − 10 lg (

S ) = 95.9 dB. A2

The sound field in the engine (source) room is diffuse, so knowing the sound power level of the device L W = 100 dB, we estimate the minimum equivalent absorption area of the engine (source) room (7.15) to be: L p,1 = L W − 10 lg

A1 A1 A1 Ô⇒ lg = 0.1(L W − L p,1 ) Ô⇒ = 100.1(L W −L p,1 ) 4S 0 4S 0 4S 0 Ô⇒ A 1 = 4 S 0 100.1(L W −L p,1 ) = 10.3 m2 .

To achieve the minimum equivalent absorption area, we place absorbers on all surfaces of the engine (source) room, whose total area is S 1 = 2 w 1 l 1 + 2 w 1 h + 2 l 1 h = 28.8 m2 . The minimum absorbance is then obtained from the definition of the equivalent absorption area (7.12) as A1 A 1 = α S 1 Ô⇒ α = = 0.36. S1

P8 Illumination P8.1 θ1 θ 1 −θ 2

θ2 d

l

θ2

x θ1

As shown in the preceding figure, the ray changes direction as it enters the glass pane due to the diffraction. The relationship between the angle of incidence θ 1 = 60○ and the angle of refraction θ 2 (8.8) is sin θ 1 c 1 = = n, sin θ 2 c 2 where n = 1.5 is the ratio of the speeds of light in air and glass. When the ray emerges on the other side of the pane, it changes its direction back to the initial, but is displaced by the distance x. We write the relationships between the thickness of the glass pane d = 4 mm and the path length of the diffracted ray in the glass pane l, and between x and l as cos θ 2 =

d , l

sin(θ 1 − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 =

x . l

The ray displacement is then x=

d (sin θ 1 cos θ 2 − cos θ 1 sin θ 2 ). cos θ 2

193

Problems in Building Physics

194 Using the expressions cos θ 1 = preceding expression as

√ √ 1 − sin2 θ 1 and cos θ 2 = 1 − sin2 θ 2 we can rewrite the

¿ ⎛ Á 1 − sin2 θ 1 ⎞ À ⎟ = 2.0 mm. x = d sin θ 1 ⎜1 − Á n 2 − sin2 θ 1 ⎠ ⎝

i

The ray displacement is small and always smaller than the thickness of the glass pane. Because it is negligible compared to the typical total length of the ray path between the light source and the receiver, we can safely assume that the ray through the glass pane between these two objects is simply straight.

P8.2

Ω

2R r

The preceding figure shows the Sun of radius R = 6.96 × 108 m on the left and the observer on the right. If we draw a sphere centred in the observer and passing through the Sun (blue line), whose radius is equal to the distance between the Sun and the observer r = 1.50 × 1011 m, we see that the area of the sphere segment within the Sun is A = π R2 . The solid angle, calculated from its definition (8.16), is Ω=

A πR 2 = 2 = 6.76 × 10−5 sr. r2 r

Because the solid angle is very small, the illuminance can be calculated from the luminance Lv = 1.44 × 109 cd/m2 (8.24) without integration as Ev = Lv Ω cos θ i = 9.74 × 104 lx. To obtain the maximum illuminance, we assumed perpendicular incident rays, θ i = 0○ .

P8 Illumination

195

P8.3 a θi a

0

h

r θi √

2a 2

0

√

2a 2

The preceding figure shows the ground plan (left) and the vertical cross section (right). The cross section is taken along the dash-dotted line on the ground plan. The distance between the point 0 and the light source r and the angle of incidence θ i are √ √ 2 r = ( 22a ) + h 2 = 8.7 m, √

⎛ 22a ⎞ θ i = arctan = 23.8○ . ⎝ h ⎠ Because of symmetry, all four lamps have equal contribution to the illuminance at point 0. The total illuminance is the sum of the contributions of all four lamps (8.19): Ev = 4

Iv cos θ i . r2

If the target illumination at the point 0 is Ev = 50 lx, the luminous intensity must be Iv =

r 2 Ev = 1050 cd. 4 cos θ i

Note that the angle of incidence θ i is always determined with respect to the perpendicular line.

Problems in Building Physics

196

P8.4 l

l

x2

x1

d

l

x2

x1

θ i,1

r1

θ i,1

h

h

0

0

x1

θ i,2

r2

θ i,2

x2

0

The preceding figure shows the ground plan (top) and two vertical cross sections (bottom). The cross sections are taken along the dashdotted lines on the ground plan. The distances between the point 0 and the lamps r, their horizontal components x and the angles of incidence θ i are √ 2 x 1 = ( 32l ) + d 2 = 16.2 m, √ 2 x 2 = ( 2l ) + d 2 = 7.8 m, √ r 1 = x 12 + h 2 = 16.6 m, √ r 2 = x 22 + h 2 = 8.8 m, x1 ) = 76○ , h x2 = arctan ( ) = 63○ . h

θ i,1 = arctan ( θ i,2

Because the light is directed only to the lower half of the space and the solid angle of the full space is 4π, the solid angle of the emitted light is Ω = 2π. The luminous intensity, calculated from the luminous flux Φv = 10 000 lm (8.17), is Iv =

Φv = 1590 cd. Ω

Because of symmetry, the outer lamps have equal contribution Ev1 and the inner lamps have equal contribution Ev2 . The illuminance at point 0 is the sum of the contributions of all four lamps (8.19): Ev = 2 Ev1 + 2 Ev2 = 2

Iv Iv cos θ i,1 + 2 2 cos θ i,2 = 21.6 lx. 2 r1 r2

P8 Illumination

197

For more complex nonisotropic light sources, the luminous intensity Iv is given as a function of the angle of emittance θ e .

i

P8.5

2.2 m

1m

r1

1m

θ i,1 θ i,2

2m

r2

First, we calculate the path length and angle of incidence of the direct light as r1 =

√ (1 m)2 + (2.2 m)2 = 2.4 m, 1m θ i,1 = arctan = 24○ . 2.2 m

The easiest way to determine the path length of the reflected light is to construct the image source by mirroring the real source over the reflecting surface. As shown in the preceding figure, the path length and angle of incidence of the light reflected from the mirror are r2 =

√ (3 m)2 + (2.2 m)2 = 3.7 m, 3m θ i,2 = arctan = 54○ . 2.2 m

Because the light source of luminous flux Φv = 1500 lm is isotropic, that is, the light is directed into the entire space, the solid angle of the emitted light is Ω = 4π and the luminous intensity (8.17) is Φv Iv = = 119 cd. Ω The illuminance of the floor is the sum of the contributions of the direct and reflected light (8.19): Iv Iv Ev = Ev1 + Ev2 = 2 cos θ i,1 + 2 cos θ i,2 = 23.7 lx. r1 r2 In most cases, the contribution of the light reflections from the walls cannot be neglected, but because these reflections are diffuse, their calculation requires much more complex mathematical methods.

i

Problems in Building Physics

198

P8.6

1.4 m

0.75 m

0.75 m

0.75 m

r1 θ i,1

r2 θ i,2

First we calculate the path length and the angle of incidence of the direct light as r1 =

√ (1.4 m)2 + (0.75 m)2 = 1.59 m, 1.4 m θ i,1 = arctan = 62○ . 0.75 m

The easiest way to determine the path length of the reflected light is to construct the image source by mirroring the real source over the reflecting surface. As shown in the preceding figure, the path length and angle of incidence of the light reflected from the mirror are r2 =

√ (1.4 m)2 + (2.25 m)2 = 2.65 m, 1.4 m θ i,2 = arctan = 32○ . 2.25 m

Because the light source of luminous flux Φ = 750 lm is isotropic, that is, the light is directed into the entire space, the solid angle of the emitted light is Ω = 4π and the luminous intensity (8.17) is Φv Iv = = 60 cd. Ω The illuminance of the floor is the sum of the contributions of the direct and reflected light (8.19): Iv Iv Ev = Ev1 + Ev2 = 2 cos θ i,1 + 2 cos θ i,2 = 18.4 lx. r1 r2

P8 Illumination

199

P8.7 3

R

∆h

∆h

2

γ2

15 m

0

1 10 m

d3

d2 β1

d2 2 10 m

α0

0

β3

β4

L 4

d3 3 10 m

d4

∆h

d1

γ3

0

4 γ4

0 d4

The preceding figure shows the ground plan of the buildings on the left and three cross sections on the right. The cross sections are taken along the dash-dotted lines on the ground plan. To determine the azimuths, we first calculate the azimuth of the direction perpendicular to the fac¸ade α 0 . For a nonrotated building, the fac¸ade faces south and the azimuth is 180○ . Because the building is rotated clockwise, the angle of rotation must be added, as in α 0 = 180○ + θ = 210○ . The angle deflections of points 1 to 4 relative to the preceding direction are β 2 = 0○ , 10 m = 33.7○ , 15 m 20 m β 4 = arctan = 53.1○ . 15 m

β 1 = β 3 = arctan

To get the azimuths of points 1 to 4, we must add and subtract the deflection angle β i for the clockwise and anticlockwise rotated points, respectively: α 1 = α 0 + β 1 = 243.7○ , α 2 = α 0 = 210○ , α 3 = α 0 − β 3 = 176.3○ , α 4 = α 0 − β 4 = 156.9○ .

Problems in Building Physics

200 The azimuths for the left and right sides of the fac¸ade are αL = α 0 − 90○ = 120○ , αR = α 0 + 90○ = 300○ .

To determine the angles of elevation, we first calculate the horizontal distances from point 0 to points 1 to 4: d 2 = 15.0 m, √ d 1 = d 3 = (15 m)2 + (10 m)2 = 18.0 m, √ d 4 = (15 m)2 + (20 m)2 = 25.0 m. The angles of elevation for points 1 to 4, calculated using the horizontal distances and the vertical distance ∆h = 22.5 m − 1.5 m = 21.0 m, are ∆h = 54.5○ , d2 ∆h γ 1 = γ 3 = arctan = 49.4○ , d3 ∆h γ 4 = arctan = 40.0○ . d4 γ 2 = arctan

70

L 60

12:00

R

13:00

11:00

φ = 47○

14:00

10:00

2 3

9:00

50

1

15:00 2 1j un y al 2 12m1j u

40

4

8:00

16:00

γ/○

r ap g 2 02 2au

30

7:00

17:00 r m ae p 2 02 2s

20 6:00

18:00 b fe t 1 7 3o c 2

0 60

n ja v 1 9 2n o e c 2 1d 2

10

80

100

120

140

160

180

200

220

240

260

280

300

α/○

Having drawn the building and both sides of the fac¸ade in the sun path diagram for geographical latitude 47○ , we identify two periods during which there is a direct sunlight exposure at an angle of elevation larger than 15○ , namely 08:40–11:10 and 15:50–16:40, which is a total of 3 h 20 min.

P8 Illumination

201

P8.8

0 R

15 m d5

β5

β1 β2 β3

L

d1 d2 1

β4 d 4

2

4

5

d3

20 m

3

20 m 10 m

10 m

15 m

α0

We must take into account the exposed points of the two buildings south of the northernmost building, that is, the edges of these buildings visible from point 0. The exposed points are marked 1 to 5 in the preceding figure. To determine the azimuths, we first calculate the azimuth of the direction perpendicular to the fac¸ade α 0 . For a nonrotated building, the fac¸ade faces south and the azimuth is 180○ . Because the building is rotated anticlockwise, the angle of rotation must be subtracted as in α 0 = 180○ − θ = 170○ . To get the azimuths of points 1 to 5, we have to add and subtract the deflection angle β i for the clockwise and anticlockwise rotated points, respectively: 45 m = 98.4○ , 15 m 25 m = α 0 − β 2 = α 0 − arctan = 111.0○ , 15 m 25 m = α 0 − β 3 = α 0 − arctan = 134.5○ , 35 m 10 m = α 0 − β 4 = α 0 − arctan = 136.3○ , 15 m 10 m = α 0 + β 5 = α 0 + arctan = 203.7○ . 15 m

α 1 = α 0 − β 1 = α 0 − arctan α2 α3 α4 α5

The azimuths for the left and right sides of the fac¸ade are αL = α 0 − 90○ = 80○ , αR = α 0 + 90○ = 260○ .

Problems in Building Physics

202

To determine the angles of elevation, we first calculate the horizontal distances from point 0 to points 1 to 5: √ (45 m)2 + (15 m)2 = 47.4 m, √ d 2 = (25 m)2 + (15 m)2 = 29.2 m, √ d 3 = (25 m)2 + (35 m)2 = 43.0 m, √ d 4 = d 5 = (10 m)2 + (15 m)2 = 18.0 m. d1 =

As in the solution to problem P8.7 , the angles of elevation for points 1 to 5 are calculated using the horizontal distances and the vertical distance ∆h = 20 m − 1.5 m = 18.5 m: ∆h d1 ∆h γ 2 = arctan d2 ∆h γ 3 = arctan d3 ∆h γ 4 = γ 5 = arctan d4 γ 1 = arctan

70

= 21.3○ , = 32.4○ , = 23.3○ , = 45.7○ .

12:00

L 60

R

13:00

11:00

φ = 47○

14:00

10:00 9:00

50

15:00 un y al 2 12m1j u

40

2 1j

5

4

2 30

16:00

r ap g 2 02 2au

γ/○

8:00

7:00

17:00 r m ae p 2 02 2s

3

1 20 6:00

18:00 b fe t 1 7 3o c 2

0 60

n ja v 1 9 2n o e c 2 1d 2

10

80

100

120

140

160

180

200

220

240

260

280

300

○

α/

Having drawn the buildings and both sides of the fac¸ade in the sun path diagram for geographical latitude 47○ , we identify two periods during which there is a direct sunlight exposure at an angle of elevation larger than 15○ , namely 08:55–09:50 and 13:20–16:40, which is a total of 4 h 15 min.

P8 Illumination

203

P8.9 θi γ

The illuminance depends on the angle of incidence. We are looking for the smallest possible angle of incidence of the Sun’s rays on a winter solstice day. Note, however, that the angle of incidence θ i is determined with respect to the perpendicular (vertical) line, whereas the angle of elevation γ is determined with respect to the horizontal plane, so their relationship is θ i + γ = 90○ . The largest possible angle of elevation on a winter solstice day can be read from the sun path diagram for geographical latitude 47○ , γ = 19.6○ . Because the Sun is an isotropic light source, that is, the light is directed into the entire space, the solid angle of the emitted light is Ω = 4π. The effective luminous intensity of the Sun at the Earth’s surface is then obtained from the luminous flux of the Sun Φv = 3.58 × 1028 lm and the optical transmittance of the atmosphere τ = 0.77 (8.17) as Iv = τ

Φv = 2.2 × 1027 cd. Ω

The illuminance of a horizontal surface (8.19), calculated from the distance between the Sun and the Earth r = 1.50 × 1011 m, is Ev =

Iv cos(90○ − γ) = 3.3 × 104 lx. r2

!

PA Appendix The expressions for calculating the water vapour pressure at saturation (4.4) specified by iso 13788 [4] and used in this book are 17.269 θ ) , θ ≥ 0 ○ C, 237.3 + θ 21.875 θ = 610.5 exp ( ) , θ < 0 ○ C. 265.5 + θ

psat = 610.5 exp ( psat

(PA.1)

The inverted expressions (4.5) are p

θ=

sat ) 237.3 ln ( 610.5

p

sat ) 17.269 − ln ( 610.5

, psat ≥ 610.5 Pa, (PA.2)

p

θ=

sat ) 265.5 ln ( 610.5

p

sat ) 21.875 − ln ( 610.5

, psat < 610.5 Pa.

Table PA.1 gives the water vapour pressure at saturation that is determined by a more precise and complex expression [8].

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Pinterić, Problems in Building Physics, https://doi.org/10.1007/978-3-031-47668-6

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Table PA.1: Water vapour pressure at saturation in Pa. °C −20 −19 −18 −17 −16 −15 −14 −13 −12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 −0 +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12 +13 +14 +15 +16 +17 +18 +19 +20 +21 +22 +23 °C

.0 103.2 113.6 124.9 137.2 150.7 165.3 181.2 198.5 217.3 237.7 259.9 283.9 310.0 338.2 368.7 401.8 437.5 476.1 517.7 562.7 611.2 611.2 657.1 706.0 758.1 813.5 872.5 935.3 1002 1073 1148 1228 1313 1403 1498 1599 1706 1819 1938 2065 2198 2339 2488 2645 2811 .0

.1 102.3 112.5 123.7 135.9 149.3 163.8 179.5 196.7 215.4 235.6 257.6 281.4 307.3 335.3 365.6 398.3 433.8 472.1 513.4 558.0 606.1 615.7 661.8 711.0 763.5 819.3 878.6 941.8 1009 1080 1156 1236 1322 1412 1508 1609 1717 1830 1951 2078 2212 2354 2503 2661 2828 .1

.2 101.3 111.5 122.6 134.7 147.9 162.3 177.9 194.9 213.4 233.5 255.3 279.0 304.6 332.4 362.4 394.9 430.1 468.1 509.1 553.4 601.2 620.2 666.6 716.1 768.9 825.0 884.8 948.3 1016 1088 1164 1245 1330 1421 1518 1620 1728 1842 1963 2091 2226 2368 2519 2678 2845 .2

.3 100.3 110.4 121.4 133.4 146.5 160.8 176.3 193.2 211.5 231.4 253.1 276.5 301.9 329.5 359.3 391.6 426.5 464.2 504.9 548.8 596.2 624.7 671.4 721.3 774.3 830.8 891.0 954.9 1023 1095 1172 1253 1339 1431 1528 1630 1739 1854 1975 2104 2240 2383 2534 2694 2862 .3

.4 99.34 109.3 120.3 132.2 145.1 159.3 174.7 191.4 209.6 229.4 250.8 274.1 299.3 326.6 356.2 388.2 422.9 460.3 500.7 544.3 591.3 629.2 676.3 726.4 779.8 836.7 897.2 961.5 1030 1103 1180 1261 1348 1440 1538 1641 1750 1866 1988 2117 2254 2398 2550 2711 2880 .4

.5 98.39 108.3 119.1 130.9 143.8 157.8 173.1 189.7 207.7 227.3 248.6 271.7 296.7 323.8 353.2 384.9 419.3 456.4 496.5 539.8 586.5 633.8 681.1 731.6 785.4 842.6 903.4 968.2 1037 1110 1188 1270 1357 1450 1548 1652 1761 1878 2001 2131 2268 2413 2566 2727 2897 .5

.6 97.45 107.3 118.0 129.7 142.5 156.4 171.5 187.9 205.8 225.3 246.4 269.3 294.1 321.0 350.1 381.6 415.7 452.6 492.3 535.3 581.6 638.4 686.0 736.8 790.9 848.5 909.7 974.9 1044 1118 1196 1278 1366 1459 1558 1662 1773 1890 2013 2144 2282 2428 2581 2744 2915 .6

.7 96.51 106.3 116.9 128.5 141.1 154.9 169.9 186.2 204.0 223.3 244.2 266.9 291.5 318.2 347.1 378.4 412.2 448.7 488.2 530.9 576.8 643.0 691.0 742.1 796.5 854.4 916.1 981.6 1051 1125 1204 1287 1375 1469 1568 1673 1784 1902 2026 2157 2296 2443 2597 2760 2932 .7

.8 95.58 105.2 115.8 127.3 139.8 153.5 168.4 184.5 202.1 221.3 242.0 264.5 289.0 315.4 344.1 375.1 408.7 445.0 484.1 526.4 572.1 647.7 695.9 747.4 802.1 860.4 922.4 988.4 1058 1133 1212 1296 1384 1479 1578 1684 1796 1914 2039 2171 2310 2458 2613 2777 2950 .8

.9 94.67 104.2 114.7 126.1 138.5 152.1 166.8 182.9 200.3 219.3 239.9 262.2 286.4 312.7 341.1 371.9 405.2 441.2 480.1 522.1 567.4 652.4 700.9 752.7 807.8 866.5 928.9 995.2 1066 1140 1220 1304 1394 1488 1589 1695 1807 1926 2052 2185 2325 2473 2629 2794 2968 .9

PA Appendix

207

Table PA.1: Water vapour pressure at saturation in Pa (continued). °C +24 +25 +26 +27 +28 +29 +30 +31 +32 +33 +34 +35 +36 +37 +38 +39 +40 +41 +42 +43 +44 +45 +46 +47 +48 +49 +50 °C

.0 2986 3170 3364 3568 3783 4009 4247 4497 4760 5036 5325 5629 5948 6283 6633 7000 7385 7788 8210 8651 9113 9596 10 100 10 630 11 180 11 750 12 350 .0

.1 3004 3189 3384 3589 3805 4032 4271 4523 4787 5064 5355 5660 5981 6317 6669 7038 7425 7830 8253 8697 9160 9645 10 150 10 680 11 230 11 810 12 410 .1

.2 3022 3208 3404 3610 3827 4056 4296 4549 4814 5092 5385 5692 6014 6351 6705 7076 7464 7871 8297 8742 9208 9695 10 200 10 740 11 290 11 870 12 480 .2

.3 3040 3227 3424 3631 3850 4079 4321 4574 4841 5121 5415 5723 6047 6386 6742 7114 7504 7913 8340 8788 9255 9745 10 260 10 790 11 350 11 930 12 540 .3

.4 3058 3246 3444 3653 3872 4103 4346 4601 4868 5150 5445 5755 6080 6421 6778 7152 7544 7955 8384 8833 9303 9795 10 310 10 840 11 400 11 990 12 600 .4

.5 3077 3266 3465 3674 3895 4127 4370 4627 4896 5179 5475 5787 6113 6456 6815 7191 7584 7997 8428 8880 9352 9845 10 360 10 900 11 460 12 050 12 660 .5

.6 3095 3285 3485 3696 3917 4150 4396 4653 4924 5208 5506 5819 6147 6491 6851 7229 7625 8039 8472 8926 9400 9896 10 410 10 950 11 520 12 110 12 730 .6

.7 3114 3305 3506 3717 3940 4174 4421 4680 4951 5237 5536 5851 6181 6526 6888 7268 7665 8081 8517 8972 9449 9946 10 470 11 010 11 580 12 170 12 790 .7

.8 3132 3324 3526 3739 3963 4199 4446 4706 4979 5266 5567 5883 6214 6562 6926 7307 7706 8124 8561 9019 9497 9997 10 520 11 070 11 640 12 230 12 850 .8

Table PA.2: Table of acronyms.

Acronym ashrae cnossos-eu eps iso osb vdi xps

Meaning American Society of Heating, Refrigerating and Air-Conditioning Engineers Common Noise Assessment Methods in Europe Expanded Polystyrene International Organization for Standardization Oriented Strand Board Verein Deutscher Ingenieure Association of German Engineers Extruded Polystyrene

.9 3151 3344 3547 3761 3986 4223 4471 4733 5007 5296 5598 5916 6248 6597 6963 7346 7747 8167 8606 9066 9546 10 050 10 570 11 120 11 690 12 290 12 920 .9

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Table PA.3: Symbols and names of building physics quantities.

Symbol A A A A B c c c0 COP d dbw df E Ev f f f Rsi д H H h hc hr I Iv Lp Lv LW L′W M m m ma md N NA n n n pr P P

Unit J m2 Bq m2 m m/s J/(kg K) m/s 1 m m m J lx Hz 1 1 kg/(m2 s) J W/K J/kg W/(m2 K) W/(m2 K) W/m2 cd dB cd/m2 dB dB kg/mol kg kg kg kg 1 1/mol mol 1/s 1/s W W

Name work area radioactive activity equivalent absorption area characteristic dimension of the floor speed of sound in air at 20 ○ C specific heat capacity speed of light in vacuum coefficient of performance thickness total equivalent thickness of the basement wall total equivalent thickness of the floor mechanical energy illuminance frequency fractional area temperature factor of the internal surface density of water vapour flow rate enthalpy heat transfer coefficient specific enthalpy convective surface coefficient radiative surface coefficient time-averaged sound intensity luminous intensity sound pressure level luminance sound power level linear sound power level molar mass mass mass of water vapour mass of (humid) air mass of dry air leakage-infiltration ratio Avogadro’s constant amount of substance air change rate air change rate at the reference pressure difference power sound power

PA Appendix

209

Table PA.3: Symbols and names of building physical quantities (continued)

Symbol P′ p p patm pa pd psat Q q qf qm q pr qV qv P R R R Ra Rse Rsi Rtot Rv S sd

Unit W/m Pa Pa Pa Pa Pa Pa J W/m2 J/kg kg/s m3/s m3/s J/kg m dB m2 K/W J/(mol K) m2 K/W m2 K/W m2 K/W m2 K/W J/(kg K) m2 m

T Tn t U u V v v vd w x y α α α αl

K s s W/(m2 K) m3 m3 kg/m3 m/s m3/kg kg/m3 1 ppm 1 1 1 1/K

Name linear sound power pressure water vapour pressure atmospheric pressure pressure of (humid) air pressure of dry air water vapour pressure at saturation heat density of heat flow rate specific heat of fusion mass flow rate air leakage rate at the reference pressure difference air flow rate specific heat of vaporisation exposed perimeter sound reduction index thermal resistance molar gas constant thermal resistance of airspace external surface resistance internal surface resistance total thermal resistance gas constant for water vapour area water vapour diffusion equivalent air layer thickness temperature reverberation time time thermal transmittance mass ratio of water to dry matter volume mass concentration of water vapour wind speed specific volume mass concentration of water mass ratio of water vapour to dry gas molar fraction absorptance absorbance azimuth linear expansion coefficient

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Table PA.3: Symbols and names of building physical quantities (continued)

Symbol αV γ δ0

Unit 1/K 1 kg/(m s Pa)

ε η θ θi λ λ µ ρ ρ ρA σ τ Φ Φv φ φ Ψ Ω

1 1 ○

C 1 m W/(m K) 1 kg/m3 1 kg/m2 W/(m2 K4 ) 1 W lm 1 1 W/(m K) sr

Name cubic expansion coefficient angle of elevation or altitude water vapour permeability with respect to vapour pressure emittance or emissivity efficiency temperature angle of incidence wavelength thermal conductivity water vapour resistance factor density reflectance surface density Stefan-Boltzmann constant transmittance heat flow rate luminous flux relative humidity geographical latitude linear thermal transmittance solid angle

Table PA.4: Physical constants.

Quantity speed of sound in air at 20 ○ C speed of light in vacuum sound intensity reference value sound power reference value linear sound power reference value standard atmospheric pressure molar gas constant Avogadro constant gas constant for water vapour water vapour permeability with respect to vapour pressure of air standard density of liquid water Stefan Boltzmann constant

Symbol c c0 I0 P0 P0′ patm R NA Rv δ0

Value 343.2 m/s 2.998 × 108 m/s 1.0 × 10−12 W/m2 1.0 × 10−12 W 1.0 × 10−12 W/m 1.013 × 105 Pa 8.314 J/(mol K) 6.022 × 1023 1/mol 461.5 J/(kg K) 2 × 10−10 kg/(m s Pa)

ρ0 σ

1.00 × 103 kg/m3 5.670 × 10−8 W/(m2 K4 )

Bibliography [1] iso 6946:2017, Building components and building elements – Thermal resistance and thermal transmittance – Calculation method. [2] iso 9613-2:1996 Acoustics – Attenuation of sound during propagation outdoors – Part 2: General method of calculation. [3] iso 13370:2017, Thermal performance of buildings – Heat transfer via the ground – Calculation methods. [4] iso 13788:2012, Hygrothermal performance of building components and building elements – Internal surface temperature to avoid critical surface humidity and interstitial condensation – Calculation methods. [5] iso 13789:2017, Thermal performance of buildings – Transmission and ventilation heat transfer coefficients – Calculation method. [6] vdi 2089:2010, Building Services in swimming baths – Indoor pools. [7] W.D. Keidel, W.D. Neff, eds. Handbook of Sensory Physiology (Vol. 1). Springer-Verlag, 1974. [8] D. Sonntag, Meteorol. Zeitschrift 3 (2) 51–66 (1994).

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 M. Pinterić, Problems in Building Physics, https://doi.org/10.1007/978-3-031-47668-6

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