el-wakil nuclear heat transport's solution manual [1 ed.]

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1 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

Chapter 4 Reactor Heat Generation

2 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-1 Known; 𝑈𝑂2 𝑆𝑂4 ; 10% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑, 𝑖𝑛 𝑎 50% 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑎𝑞𝑢𝑒𝑜𝑢𝑠 𝑤𝑎𝑡𝑒𝑟𝑦 ; @500℉; 𝜌𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝜌𝑤𝑎𝑡𝑒𝑟 Find; 𝑁𝑓𝑓 =? Analysis; 𝑁𝑓𝑓 =

𝐴𝑣 𝑀𝑓𝑓

× 𝜌𝑓𝑓 × 𝑖

𝜌𝑓𝑓 = 𝑟𝑓𝜌𝑓𝑚 ; 𝜌𝑓𝑚 = 0.5 𝜌𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 0.5 𝜌𝑤𝑎𝑡𝑒𝑟 → 𝜌𝑓𝑓 = 0.5 𝑟𝑓𝜌𝑤𝑎𝑡𝑒𝑟 (𝐴𝑠𝑠𝑢𝑚𝑒 2000𝑝𝑠𝑖𝑎) 𝑇𝑎𝑏𝑙𝑒 𝐸 − 1 → 𝜌𝑤𝑎𝑡𝑒𝑟 @500℉ = 49.618

𝑙𝑏𝑚

𝑓𝑡 3

𝑇𝑎𝑏𝑙𝑒 𝐻 − 5 → 𝜌𝑤𝑎𝑡𝑒𝑟 ≡ 49.618 × 0.01602 = 0.795 𝑓=

𝑟𝑀𝑓𝑓 +(1−𝑟)𝑀𝑛𝑓 𝑟𝑀𝑓𝑓 + 1−𝑟 𝑀𝑛𝑓 +𝑀(𝑂2 𝑆𝑂4 )

→ 𝑁𝑓𝑓 =

6.022×10 23 235.0439

→ 𝑓 = 0.65

× 0.5 × 0.1 × 0.65 × 0.795 × 1

→ 𝑁𝑓𝑓 = 6.620 × 1019 # 𝑐𝑚3 [𝑎𝑛𝑠]

𝑔𝑟

𝑐𝑚3

3 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-2 Known; 𝐹𝑎𝑠𝑡 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 20%𝑃𝑢239 𝑂2 + 80%𝑈 238 𝑂2 ; 𝜌𝑓𝑚 = 10

𝑔𝑟

𝑐𝑚3

Find; 𝑎) 𝑁𝑃𝑢 239 , 𝑁𝑈 238 =? ; 𝑏) 𝑇𝑕𝑒 % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠 𝑏𝑦 𝑃𝑢239 , 𝑈 238 =? (𝑖𝑓 𝐸𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 = 1𝑀𝑒𝑉) 𝑐) 𝑅𝑒𝑝𝑒𝑎𝑡 𝑏 𝑓𝑜𝑟 𝐸𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 = 2𝑀𝑒𝑉 Analysis; a) 𝑁𝑃𝑢 239 =

𝐴𝑣 𝑀𝑃𝑢 239

× 𝜌𝑃𝑢 239 × 𝑖

𝜌𝑃𝑢 239 𝑂2 = 0.2 × 𝜌𝑃𝑢 239 𝑂2 𝑈𝑂2 = 0.2 × 10 = 2 𝜌𝑃𝑢 239 = 𝑓𝜌𝑃𝑢 239 𝑂2 ; 𝑓 =

𝑀𝑃𝑢 239 𝑀𝑃𝑢 239 +𝑀𝑂 2

→ 𝜌𝑃𝑢 239 = 0.88 × 2 = 1.7638 → 𝑁𝑃𝑢 239 =

6.022×10 23 239

𝑔𝑟

=

239 239+32

𝑐𝑚3

𝑔𝑟

𝑐𝑚3 = 0.88

;𝑖 = 1

× 1.7638 × 1

→ 𝑁𝑃𝑢 239 = 4.44 × 1021 # 𝑐𝑚3 [𝑎𝑛𝑠] 𝑁𝑈 238 = 𝑁𝑈 =

𝐴𝑣 𝑀𝑈 238

× 𝜌𝑈 238 × 𝑖

𝜌𝑈𝑂2 = 0.8 × 𝜌𝑃𝑢 239 𝑂2 𝑈𝑂2 = 0.8 × 10 = 8

𝑔𝑟

𝑐𝑚3

4 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝜌𝑈 238 = 𝑓𝜌𝑈𝑂2 ; 𝑓 =

𝑀𝑈 238 𝑀𝑈 238 +𝑀𝑂 2

=

→ 𝜌𝑈 238 = 0.8814 × 8 = 7.052 → 𝑁𝑈 238 =

6.022×10 23 238

238

= 0.8814

238+32

𝑔𝑟

𝑐𝑚3 ; 𝑖 = 1

× 7.052 × 1

→ 𝑁𝑈 238 = 17.84 × 1021 # 𝑐𝑚3 [𝑎𝑛𝑠] b) % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 = → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 =

𝛴𝑓 𝛴𝑓

𝑃𝑢 239

𝑃𝑢 239

+𝛴𝑓 238 𝑈

𝑁𝑃𝑢 239 𝜍 𝑓 𝑁𝑃𝑢 239 𝜍 𝑓

𝑃𝑢 239

+𝑁𝑈 238 𝜍 𝑓 238 𝑃𝑢 239 𝑈

𝐹𝑖𝑔. 4 − 5 (@ 1 𝑀𝑒𝑉 ≡ 106 𝑒𝑉) → 𝜍𝑓 𝑃𝑢 239 = 1.6 𝑏 𝐹𝑖𝑔. 4 − 6 (@ 1 𝑀𝑒𝑉 ≡ 106 𝑒𝑉) → 𝜍𝑓 𝑈 238 = 0.02 𝑏 → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 = 4.44×10 21 ×1.6×10 −24 4.44×10 21 ×1.6×10 −24 +17.84×10 21 ×0.02×10 −24

= 0.952

→ % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 = 95.2% [𝑎𝑛𝑠] → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑈 238 = 1 − 0.952 = 0.0478 → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑈 238 = 4.8% [𝑎𝑛𝑠] c) 𝐹𝑖𝑔. 4 − 5 (@ 2 𝑀𝑒𝑉 ≡ 2 × 106 𝑒𝑉) → 𝜍𝑓 𝑃𝑢 239 = 2 𝑏 𝐹𝑖𝑔. 4 − 6 (@ 2 𝑀𝑒𝑉 ≡ 2 × 106 𝑒𝑉) → 𝜍𝑓 𝑈 238 = 0.56 𝑏

5 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

→ % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 = 4.44×10 21 ×2×10 −24 4.44×10 21 ×2×10 −24 +17.84×10 21 ×0.56×10 −24

= 0.468

→ % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑃𝑢239 = 46.8% [𝑎𝑛𝑠] → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑈 238 = 1 − 0.468 = 0.532 → % 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 𝑏𝑦 𝑈 238 = 53.2% [𝑎𝑛𝑠]

6 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-3 Known; 𝑚𝐻𝑒 = 2.07 × 107

𝑙𝑏𝑚

𝑕𝑟 ; 𝑇𝑖 = 800 ℉ ≡ 1260 𝑅;

𝐺𝑟𝑎𝑝𝑕𝑖𝑡𝑒 − 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑; 𝑃 = 2000 𝑀𝑊(𝑡) Find; 𝜍𝑓 =? Analysis; 𝑃 = 𝑚𝐶𝑝 ∆𝑇 𝑇𝑎𝑏𝑙𝑒 𝐸 − 2 → 𝐶𝑝 𝐻𝑒 = 1.248 𝐵𝑡𝑢 𝑙𝑏 ℉ 𝑚 𝑇𝑎𝑏𝑙𝑒 𝐻 − 11 → 2000 𝑀𝑊 𝑡 ≡ 6.83 × 109 𝐵𝑡𝑢 𝑕𝑟 → ∆𝑇 =

6.83×10 9

2.07×10 7 ×1.248

= 264.38 ℉

∆𝑇 = 264.38 ℉ ; 𝑇1 = 800 ℉ → 𝑇2 = 800 + 264.38 = 1064.38 ℉ → 𝑇𝑎𝑣 . =

𝑇1 +𝑇2 2

= 932.19 ℉ ≡ 500.1 ℃ 𝑇

𝜍𝑓 = 0.8862 𝑓 𝑇 𝜍𝑓 0 ( 0 )0.5 𝑇

𝐹𝑖𝑔. 4 − 3 → 𝑓 𝑇 @𝑇𝑎𝑣 . =500 ℃ = 0.91 (𝑓𝑜𝑟 𝑈 235 − 𝑓𝑖𝑠𝑠𝑖𝑜𝑛) 𝐴𝑝𝑝. 𝐵 → 𝜍𝑓 0 235 = 577.1 𝑏 ; 𝑇0 = 293 𝐾 ; 𝑇 = 773 𝐾 𝑈

→ 𝜍𝑓 = 0.8862 × 0.91 × 577.1 × (

293 0.5 ) 773

= 286.53 𝑏 [𝑎𝑛𝑠]

7 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-4 Known; 𝑈𝑂2 3% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝑇 = 500 ℉ ; 𝑃 = 2000 𝑝𝑠𝑖𝑎 ; 𝑂𝑟𝑑𝑖𝑛𝑎𝑟𝑦 𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑 ; 𝐹𝑢𝑒𝑙/𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 𝑟𝑎𝑡𝑖𝑜 = 1: 1.9 ; 2 𝑏𝑎𝑟𝑛𝑠 𝑜𝑓 1/𝑉 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑎𝑡𝑜𝑚 𝑜𝑓 𝑕𝑦𝑑𝑟𝑜𝑔𝑒𝑛 Find; 𝜍𝑓 =? Analysis; 𝑉 𝐹𝑢𝑒𝑙 𝑉𝑀𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 𝑉𝐹𝑢𝑒𝑙



𝑉𝐶𝑜𝑟𝑒

= =

1 1.9 1

1+1.9

= 34.48% ;

𝑉𝑊𝑎𝑡𝑒𝑟 𝑉𝐶𝑜𝑟𝑒

=

1.9 1+1.9

= 65.52%

235 𝑁𝑈 235 = 7.115 × 1020 𝑈 𝑐𝑚3 𝑓𝑢𝑒𝑙 (𝐸𝑥𝑎𝑚𝑝𝑙𝑒 4 − 1)

→ 𝑁𝑈′ 235 = 7.115 × 1020 × 0.3448 = 235 2.4532 × 1020 𝑈 𝑐𝑚3 𝑐𝑜𝑟𝑒

𝑁𝐻 = 2 𝑁𝐻2 𝑂 ; 𝑁𝐻2 𝑂 = 𝐴𝑝𝑝. 𝐸 → 𝜌𝐻2 𝑂

𝐴𝑣 𝑀𝐻 2 𝑂

× 𝜌𝐻2 𝑂

@500 ℉, 2000 𝑝𝑠𝑖𝑎

= 49.618

𝑙𝑏𝑚

𝑓𝑡 3

8 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

→ 𝜌𝐻2 𝑂 ≡ 0.7948 → 𝑁𝐻 = 2 ×

𝑔𝑟

6.022×10 23 18

𝑐𝑚3 × 0.7948 = 5.32 × 1022 𝐻 𝑐𝑚3 𝑤𝑎𝑡𝑒𝑟

→ 𝑁𝐻′ = 5.32 × 1022 × 0.6552 = 3.485 × 1022 𝐻 𝑐𝑚3 𝑐𝑜𝑟𝑒 𝐴𝑡𝑜𝑚𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 = 𝑈

235

𝐻=

𝑁𝑈′ 235

𝑁𝐻′

=

2.4532×10 20 3.485×10 22

= 0.007

→ 𝐹𝑖𝑔. 4 − 4 → 𝜍𝑓 0 = 420 𝑏 𝑇

𝜍𝑓 = 0.851 𝜍𝑓 0 ( 0 )0.5 ; 𝑇 = 500℉ ≡ 260℃ ≡ 533 𝐾 𝑇

→ 𝜍𝑓 = 0.851 × 420 × → 𝜍𝑓 = 265 𝑏 [𝑎𝑛𝑠]

293 0.5 533

= 265.0019 𝑏

9 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-5 Known; 𝐻𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑏𝑎𝑟𝑒 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 (𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙) ; 𝑓𝑢𝑒𝑙: 𝑈𝑂2 𝑆𝑂4 ; 10% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐻2 𝑂 @500 ℉ 𝜑𝑐𝑜 = 1014 ; 𝜌𝑓 = 0.255

𝑔𝑟

𝑐𝑚3 ;

Find; 𝐻

𝑞 ′′′ (0.499𝑅 , ) =? 4

Analysis; 𝑞 ′′′ = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 → 𝑞 ′′′ = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜 𝑐𝑜𝑠

𝜋𝑧 𝐻

𝐽0

2.405 𝑟 𝑅

(𝐵𝑦 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑡𝑕𝑒 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕) 𝐺𝑓 = 190

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

𝑁𝑓𝑓 = 𝑁𝑈 235 =

𝐴𝑣 𝑀𝑈 235

× 𝜌𝑈 235

𝜌𝑈 235 = 𝑟 𝜌𝑈 ; 𝜌𝑈 = 𝑓𝜌𝑈𝑂2 𝑆𝑂4 ; 𝜌𝑈 235 = 𝑟𝑓𝜌𝑈𝑂2 𝑆𝑂4 ; 𝑟 = 0.1 ; 𝑔𝑟 𝜌𝑈𝑂2 𝑆𝑂4 = 0.255 𝑐𝑚3 𝑓=

𝑟𝑀𝑈 235 + 1−𝑟 𝑀𝑈 238 𝑟𝑀𝑈 235 + 1−𝑟 𝑀𝑈 238 +𝑀 𝑂2 𝑆𝑂4

10 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

→𝑓=

0.1×235.0439+0.9×238.0508 0.1×235.0439+0.9×238.0508+32+6×16

→ 𝑓 = 0.65 𝑁𝑓𝑓 =

6.022×10 23 235.0439

× 0.1 × 0.65 × 0.255

→ 𝑁𝑓𝑓 = 4.25 × 1019 # 𝑐𝑚3 𝑇

𝜍𝑓 = 0.8862 𝜍𝑓 0 ( 0 )0.5 𝑇

𝑇 = 500 ℉ ≡ 260 ℃ ≡ 533 𝐾 𝐴𝑝𝑝. 𝐵 → 𝜍𝑓 0 235 = 577.1 𝑏 ; 𝑇0 = 293 𝐾 𝑈

→ 𝜍𝑓 = 0.8862 × 577.1 × → 𝜍𝑓 = 379.2 × 10

−24

𝑐𝑚

𝜋𝑧

2.405 𝑟

𝐻

𝑅

𝜑 = 𝜑𝑐𝑜 𝑐𝑜𝑠( ) 𝐽0 ( 𝐻

𝜋

= 379.2 𝑏

) 𝐽0 (

2.405 × 0.499𝑅

)

𝑅 2 𝐴𝑝𝑝 .𝐶→𝐽 0 1.2 =0.6711

𝑐𝑜𝑠 4 = 2 14

→ 𝜑 = 0.475 × 10

533

2

)

𝐻

𝜋× 4

→ 𝜑 = 1014 𝑐𝑜𝑠(

293 0.5

#

𝑠. 𝑐𝑚2

→ 𝑞 ′′′ = 190 × 4.25 × 1019 × 379.2 × 10−24 × 0.475 × 1014 → 𝑞 ′′′ = 1.45447 × 1014 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 𝑞 ′′′ = 2.251 × 106 𝐵𝑡𝑢 𝑕𝑟. 𝑓𝑡 3 [𝑎𝑛𝑠]

11 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-6 Known; 𝑈𝑂2 3% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝑛 = 10000 ; 𝑇 = 500 ℉ ; 𝐻 = 20 𝑓𝑡 ; 𝐷𝐹𝑢𝑒𝑙 = 0.6 𝑖𝑛 ; 𝐷𝐶𝑜𝑟𝑒 = 8 𝑓𝑡 ; 𝜑𝑐𝑜 = 1013 𝐻𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑 ; 𝐻𝑒𝑡𝑒𝑟𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕𝑠 Find; 𝑄𝑡 =? Analysis; ′′′ 𝐸𝑥𝑎𝑚𝑝𝑙𝑒 4 − 4 → 𝑄𝑡 = 0.289 𝑛 𝐴𝑠 𝐻 𝑞𝑐𝑜 𝐷𝐹𝑢𝑒𝑙 = 0.6 𝑖𝑛 ≡ 1.524 𝑐𝑚 → 𝐴𝑠 =

𝜋 (1.524)2 4

= 1.82 𝑐𝑚2

𝐻 = 20 𝑓𝑡 ≡ 609.6 𝑐𝑚 ′′′ 𝑞𝑐𝑜 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜

𝐺𝑓 = 180

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

𝑁𝑓𝑓 = 7.115 × 1020 # 𝑐𝑚3 (𝐸𝑥𝑎𝑚𝑝𝑙𝑒 4 − 1) 𝑇

𝜍𝑓 = 0.8862 𝑓 𝑇 𝜍𝑓 0 ( 0 )0.5 𝑇

12 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝐹𝑖𝑔. 4 − 3 𝐷2 𝑂 , 𝑇𝑎𝑣 . = 500 ℉ ≡ 260 ℃ ≡ 533 𝐾 → 𝑓 𝑇 = 0.93 𝐴𝑝𝑝. 𝐵 → 𝜍𝑓 0 = 577.1 𝑏 293 0.5

→ 𝜍𝑓 = 0.8862 × 0.93 × 577.1 × (

533

)

→ 𝜍𝑓 = 352.64 𝑏 ′′′ → 𝑞𝑐𝑜 = 180 × 7.115 × 1020 × 352.64 × 10−24 × 1013 ′′′ → 𝑞𝑐𝑜 = 4.5163 × 1014 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 ′′′ → 𝑞𝑐𝑜 = 72.35 𝑊𝑎𝑡𝑡 𝑐𝑚3

→ 𝑄𝑡 = 0.289 × 10000 × 1.82 × 609.6 × 72.35 → 𝑄𝑡 = 231.98 × 103 𝑘𝑊 𝑡 [𝑎𝑛𝑠]

13 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-7 Known; 𝑇𝑕𝑒𝑟𝑚𝑎𝑙 𝑕𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 ; 𝐻 = 4 𝑓𝑡 ; 𝐷 = 4 𝑓𝑡 ; 𝜑𝑐𝑜 = 1013 ; 𝜆𝑒 = 0.5 𝑓𝑡 ; 𝑁𝑓𝑓 = 6 × 1020 # 𝑐𝑚3 ; 𝑇 = 500 ℉ ≡ 260 ℃ ≡ 533 𝐾 ; Find; 𝑞 ′′′ (𝑟 = 0.832 𝑓𝑡 , 𝑧 = 1.667 𝑓𝑡) =? Analysis; 𝜋𝑧

2.405 𝑟

𝐻𝑒

𝑅𝑒

𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 → 𝜑 = 𝜑𝑐𝑜 𝑐𝑜𝑠( ) 𝐽0 (

)

𝐻𝑒 = 𝐻 + 2𝜆𝑒 = 4 + 2 × 0.5 = 5 𝑓𝑡 𝑅𝑒 = 𝑅 + 𝜆𝑒 = 2 + 0.5 = 2.5 𝑓𝑡 𝐺𝑓 = 190

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

(𝑖𝑛 𝑕𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑎𝑠𝑒𝑠) 𝑇

𝜍𝑓 = 0.8862 𝑓(𝑇)𝜍𝑓 0 ( 0 )0.5 𝑇

𝐹𝑖𝑔. 4 − 3 → 𝑓 𝑇 = 0.93 𝐴𝑝𝑝. 𝐵 → 𝜍𝑓 0 = 577.1 𝑏 293 0.5

→ 𝜍𝑓 = 0.8862 × 0.93 × 577.1 × (

533

→ 𝜍𝑓 = 352.64 𝑏

)

14 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL 𝜋×1.667

→ 𝜑 = 1013 × 𝑐𝑜𝑠(

5 0.4998

2.405×0.832

) × 𝐽0 (

2.5 𝐴𝑝𝑝 .𝐶→0.8463

)

→ 𝜑 = 4.23 × 1012 # 𝑠. 𝑐𝑚2 𝑞 ′′′ = 190 × 6 × 1020 × 352.64 × 10−24 × 4.23 × 1012 → 𝑞 ′′′ = 1.6977 × 1014 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 𝑞 ′′′ = 2.6276 × 106 𝐵𝑡𝑢 𝑕𝑟. 𝑓𝑡 3 [𝑎𝑛𝑠]

15 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-8 Known; 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑟𝑒 ; 𝐻 = 4.8 𝑓𝑡 ; 𝐷𝐶𝑜𝑟𝑒 = 4 𝑓𝑡 ; 𝜑𝑐𝑜 = 1013 ; 𝜍𝑓 = 500 𝑏 ; 𝑈𝑂2 20% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 𝜆𝑒 = 0.186 𝑓𝑡 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 ; 𝜆𝑒 = 0.3 𝑓𝑡 (𝑖𝑛 𝑎𝑥𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛) Find; ′′′ 𝜑 @ 𝑡𝑕𝑒 𝑟𝑖𝑚𝑠 =? ; 𝑞𝑀𝑎𝑥 =?

Analysis; 𝑎) 𝜑 = 𝜑𝑐𝑜 𝑐𝑜𝑠

𝜋𝑧 𝐻𝑒

𝐽0

2.405 𝑟 𝑅𝑒 𝐻

@(𝑟 = 𝑅 = 2 𝑓𝑡 , 𝑧 = ± = 2.4 𝑓𝑡) 2

𝐻𝑒 = 𝐻 + 2𝜆𝑒 = 4.8 + 2 × 0.3 = 5.4 𝑓𝑡 𝑅𝑒 = 𝑅 + 𝜆𝑒 = 2 + 0.186 = 2.186 𝑓𝑡 𝜋×2.4

→ 𝜑 = 1013 × 𝑐𝑜𝑠(

5.4 0.174

2.405×2

) × 𝐽0 (

→ 𝜑 = 1.921 × 1011 # 𝑠. 𝑐𝑚2 [𝑎𝑛𝑠] ′′′ 𝑞𝑀𝑎𝑥 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜

𝐺𝑓 = 180

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

(𝑖𝑛 𝑓𝑢𝑒𝑙)

)

2.186 𝐴𝑝𝑝 .𝐶→0.1104

16 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑁𝑓𝑓 = 𝑁𝑈 235 = 𝜌𝑈 235 = 10.5 𝑓=

𝐴𝑣 𝑀𝑈 235

𝑔𝑟

× 𝜌𝑈 235 =

𝐴𝑣 𝑀𝑈 235

× 𝑟 × 𝑓 × 𝜌𝑈 235

𝑐𝑚3 ; 𝑟 = 0.2

𝑟𝑀𝑈 235 + 1−𝑟 𝑀𝑈 238 𝑟𝑀𝑈 235 + 1−𝑟 𝑀𝑈 238 +𝑀 𝑂2

→𝑓= 𝑁𝑓𝑓 =

0.2×235.0439+0.8×238.0508 0.2×235.0439+0.8×238.0508+2×16 6.022×10 23 235.0439

= 0.8812

× 0.2 × 0.8812 × 10.5

→ 𝑁𝑓𝑓 = 4.74 × 1021 # 𝑐𝑚3 ′′′ → 𝑞𝑀𝑎𝑥 = 180 × 4.74 × 1021 × 500 × 10−24 × 1013 ′′′ → 𝑞𝑀𝑎𝑥 = 4.266 × 1015 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 ′′′ 𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 𝑞𝑀𝑎𝑥 = 6.6025 × 107 𝐵𝑡𝑢 𝑕𝑟. 𝑓𝑡 3 [𝑎𝑛𝑠]

17 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-9 Known; 𝑇𝑕𝑒𝑟𝑚𝑎𝑙 𝑕𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 − 𝑓𝑢𝑒𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 ; 𝑈 𝑠𝑎𝑙𝑡 𝑖𝑛 𝑕𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 ; 𝐺𝑟𝑎𝑝𝑕𝑖𝑡𝑒 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 ; 𝑆𝑝𝑕𝑒𝑟𝑖𝑐𝑎𝑙 ; 𝐷 = 200 𝑐𝑚 ; 𝑁𝑓𝑓 = 1020 # 𝑐𝑚3 ; 𝜑𝑐𝑜 = 1013 ; 𝑇𝐵𝑢𝑙𝑘 = 550 ℉ ; 𝑅𝑒 = 104.3 𝑐𝑚 ; Find; 𝑄𝑡 =? Analysis; 𝑅

𝜋𝑟

𝜋

𝑅𝑒

𝑄𝑡 = 1.175 × 10−5 𝑁𝑓𝑓 𝜍𝑓 𝑅𝑒 𝜑𝑐𝑜 [( 𝑒 )2 𝑠𝑖𝑛



(𝐸𝑞. 4 − 23 ; 𝑓𝑜𝑟 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑) 𝐷 = 200 𝑐𝑚 → 𝑅 = 100 𝑐𝑚 𝑇

𝜍𝑓 = 0.8862 𝑓 𝑇 𝜍𝑓 0 ( 0 )0.5 𝑇

𝑇𝐵𝑢𝑙𝑘 = 550 ℉ ≡ 287.78 ℃ ≡ 561 𝐾 , 𝐷2 𝑂 ; 𝐹𝑖𝑔. 4 − 3 → 𝑓 𝑇 = 0.93 𝐴𝑝𝑝. 𝐵 → 𝜍𝑓 0 = 577.1 𝑏 293 0.5

→ 𝜍𝑓 = 0.8862 × 0.93 × 577.1 × (

561

→ 𝜍𝑓 = 343.73 𝑏

)

𝑅𝑒 𝜋

𝑟 𝑐𝑜𝑠

𝜋𝑟 𝑅 ] 𝑅𝑒 0

18 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑄𝑡 = 1.175 × 10−5 × 1020 × 343.73 × 10−24 × 104.3 × 1013 × [

104.3 2 𝜋

𝑠𝑖𝑛

𝜋×100



104.3

104.3 𝜋 12 𝐵𝑡𝑢

→ 𝑄𝑡 = 1.4468 × 10

𝑟 𝑐𝑜𝑠

𝜋×100

𝑕𝑟 [𝑎𝑛𝑠]

104.3

]

19 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-10 Known; 𝐻𝑒𝑡𝑒𝑟𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜𝑟𝑒 𝑃𝑊𝑅 ; 𝑈𝑂2 3% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝐷𝐶𝑜𝑟𝑒 = 6.25 𝑓𝑡 ; 𝐻𝐶𝑜𝑟𝑒 = 8.5 𝑓𝑡 ; 𝐷𝑃𝑒𝑙𝑙𝑒𝑡𝑠 = 0.3 𝑖𝑛 ; 𝑄𝑡 = 53 𝑀𝑊(𝑡) ; 𝑚 = 4 × 106

𝑙𝑏𝑚

𝑕𝑟 ; 𝑛 = 20000 ;

𝑇𝑖 = 460 ℉ ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕𝑠 Find; 𝜑𝑐𝑜 =? Analysis; ′′′ 𝐸𝑞. 4 − 30𝑏 → 𝑄𝑡 = 0.289 𝑛 𝐴𝑠 𝐻 𝑞𝑐𝑜 𝐷𝑃𝑒𝑙𝑙𝑒𝑡𝑠 = 0.3 𝑖𝑛 ≡ 0.762 𝑐𝑚 → 𝐴𝑠 =

𝜋(0.762)2 4

= 0.456 𝑐𝑚2

𝐻𝐶𝑜𝑟𝑒 = 8.5 𝑓𝑡 ≡ 259.08 𝑐𝑚 ′′′ → 𝑞𝑐𝑜 =

53 ×10 6 0.289×20000 ×0.456×259.08

= 77.615

𝑊(𝑡)

𝑐𝑚3

′′′ 𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 𝑞𝑐𝑜 = 4.845 × 1014 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 ′′′ 𝑞𝑐𝑜 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜

𝐺𝑓 = 180

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

‫‪20 Nuclear Heat Transport‬‬ ‫‪Problem’s Solutions‬‬ ‫‪M. M. EL-WAKIL‬‬

‫)‪𝑁𝑓𝑓 = 7.115 × 1020 # 𝑐𝑚3 (𝐸𝑥𝑎𝑚𝑝𝑙𝑒 4 − 1‬‬

‫باید از منودارآب معمولی استفاده کنیم ولی بو دلیل کمبود اطالعات؛‬ ‫𝑇‬

‫‪𝜍𝑓 = 0.8862 𝑓(𝑇)𝜍𝑓 0 ( 0 )0.5‬‬ ‫𝑇‬

‫;)𝑟𝑜𝑟𝑟𝐸 𝑑𝑛𝑎 𝑦𝑟𝑇( ‪𝐹𝑜𝑟 𝑇0‬‬ ‫𝑇∆ 𝑝𝐶𝑚 = 𝑄‬ ‫℉ ‪𝑇𝑖 = 460‬‬ ‫𝑟𝑕‬ ‫℉ ‪= 470‬‬

‫‪460+480‬‬

‫𝑚𝑏𝑙‬

‫‪𝑚 = 4 × 106‬‬

‫= ‪𝐴𝑠𝑠𝑢𝑚𝑒 𝑇0 = 480 ℉ → 𝑇𝑎𝑣 .‬‬

‫‪2‬‬

‫℉ 𝑏𝑙 𝑢𝑡𝐵 ‪𝑇𝑎𝑏𝑙𝑒 𝐸 − 1 𝑃𝑊𝑅 @ 2000 𝑝𝑠𝑖𝑎 → 𝐶𝑝 = 1.12‬‬ ‫𝑚‬ ‫𝑟𝑕 𝑢𝑡𝐵 ‪𝑇𝑎𝑏𝑙𝑒 𝐻 − 11 → 𝑄𝑡 = 1.80836 × 108‬‬ ‫℉ ‪= 40.365‬‬

‫‪1.80836 ×10 8‬‬ ‫‪4×10 6 ×1.12‬‬

‫= ‪→ ∆𝑇 = 𝑇0 − 460‬‬ ‫℉ ‪→ 𝑇0 = 500.365‬‬

‫℉ ‪= 480‬‬

‫‪460+500‬‬ ‫‪2‬‬

‫= ‪𝐴𝑠𝑠𝑢𝑚𝑒 𝑇0 = 500℉ → 𝑇𝑎𝑣 .‬‬

‫℉ 𝑏𝑙 𝑢𝑡𝐵 ‪𝑇𝑎𝑏𝑙𝑒 𝐸 − 1 𝑃𝑊𝑅 @ 2000 𝑝𝑠𝑖𝑎 → 𝐶𝑝 = 1.13‬‬ ‫𝑚‬ ‫℉ ‪= 40‬‬

‫‪1.80836 ×10 8‬‬ ‫‪4×10 6 ×1.13‬‬

‫= ‪→ ∆𝑇 = 𝑇0 − 460‬‬

‫)𝑡𝑝𝑒𝑐𝑐𝑎( ℉ ‪→ 𝑇0 = 500‬‬ ‫𝐾 ‪= 480℉ ≡ 248.89℃ ≡ 522‬‬

‫‪460+500‬‬ ‫‪2‬‬

‫= ‪𝑇0 = 500 ℉ → 𝑇𝑎𝑣 .‬‬

‫‪𝐹𝑖𝑔. 4 − 3 → 𝑓 𝑇 = 0.93‬‬

21 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝐴𝑝𝑝. 𝐵 → 𝜍𝑓0 = 577.1 𝑏 𝜍𝑓 = 0.8862 × 0.93 × 577.1 × (

293 0.5 ) 522

→ 𝜍𝑓 = 356.34 𝑏 → 𝜑𝑐𝑜 =

′ ′′ 𝑞 𝑐𝑜

𝐺𝑓 𝑁𝑓𝑓 𝜍 𝑓

=

4.845×10 14 180×7.115×10 20 ×356.34×10 −24

→ 𝜑𝑐𝑜 = 10.62 × 1012 # 𝑠. 𝑐𝑚2 [𝑎𝑛𝑠]

22 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-11 Known; 𝑇𝑕𝑒𝑟𝑚𝑎𝑙 𝑕𝑒𝑡𝑒𝑟𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 ; 𝐶𝑢𝑏𝑒 𝑓𝑜𝑟𝑚 ; 𝑈𝑟𝑎𝑛𝑖𝑢𝑚 𝑚𝑒𝑡𝑎𝑙 ; 20% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝑎 = 4 𝑓𝑡 ; 𝑛 = 4900 ; 𝐷𝐹𝑢𝑒𝑙 = 0.45 𝑖𝑛 ; 𝑄𝑡 = 5 × 108 𝐵𝑡𝑢 𝑕𝑟 ; 𝑇 = 500 ℉ ; 𝐺𝑟𝑎𝑝𝑕𝑖𝑡𝑒 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 ; 𝑁𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑈𝑛𝑑𝑖𝑠𝑡𝑜𝑟𝑡𝑒𝑑 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕𝑠 Find; 𝜑𝑎𝑣 . =? ; 𝜑𝑀𝑎𝑥 =? Analysis; 𝑎) 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝑕𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑝𝑙𝑎𝑛𝑒 → 𝑧 = 0 𝜑𝑎𝑣 . =

1 𝐴

𝜑 𝑑𝐴 → 𝜑𝑎𝑣 . =

𝑧 = 0 → 𝜑 = 𝜑𝑐𝑜 𝑐𝑜𝑠 → 𝜑𝑎𝑣 . = → 𝜑𝑎𝑣 . =

1 𝑎2 𝜑 𝑐𝑜 𝑎2

𝜋𝑥 𝑎

𝑐𝑜𝑠

𝑎 𝑎 2 2 𝜑 −𝑎 2 −𝑎 2 𝑐𝑜 2𝑎 2 𝜋

𝑎 𝑎 2 2 𝑎 𝑎 𝜑 − − 𝐴 2 2 𝜋𝑦 1

𝑎

𝑐𝑜𝑠

→ 𝜑𝑎𝑣 . =

𝑑𝑥𝑑𝑦

𝜋𝑥

4 𝜋2

𝑎

𝑐𝑜𝑠

𝜋𝑦 𝑎

𝑑𝑥𝑑𝑦

𝜑𝑐𝑜 [𝑎𝑛𝑠]

𝑏) 𝐻𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑏𝑦 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑓𝑢𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 2

𝜋𝐻

𝜋

2𝐻𝑒

→ 𝑞𝑡 = 𝑞𝑐′′′ 𝐴𝑠 𝐻𝑒 𝑠𝑖𝑛

23 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL 2

𝐻𝑒 = 𝐻 → 𝑞𝑡 = 𝑞𝑐′′′ 𝐴𝑠 𝐻 𝜋

′′′ 𝐶𝑢𝑏𝑖𝑐 𝑐𝑜𝑟𝑒 → 𝑞𝑐′′′ = 𝑞𝑐𝑜 𝑐𝑜𝑠

𝐴′ =

𝑎2

→ 𝑞𝑡′ =

𝑛

→ 𝑞𝑡′ = → 𝑞𝑡′ =

𝑛 2 𝑎2 𝜋 2𝑛 𝜋𝑎 2

𝑞𝑡 𝐴′

=

𝑛 𝑎2

𝜋𝑥 𝑎

𝑐𝑜𝑠

𝜋𝑦 𝑎

𝑞𝑡

𝐴𝑠 𝐻𝑞𝑐′′′ ′′′ 𝐴𝑠 𝐻𝑞𝑐𝑜 𝑐𝑜𝑠

𝜋𝑥 𝑎

𝑐𝑜𝑠

𝜋𝑦 𝑎 2𝑎 2 𝜋

→ 𝑄𝑡 = → 𝑄𝑡 =

𝑎 𝑎 2 2 𝑐𝑜𝑠 𝜋𝑥 𝑎 − 2 −𝑎 2 𝜋𝑎 𝑎 2 2𝑛 4𝑎 8𝑛 ′′′ ′′′ 𝐴 𝐻𝑞𝑐𝑜 = 3 𝐴𝑠 𝐻𝑞𝑐𝑜 𝜋𝑎 2 𝜋 2 𝑠 𝜋 𝑄𝑡 𝜋 3 7 𝐵𝑡𝑢 2𝑛

′′′ 2 𝐴𝑠 𝐻𝑞𝑐𝑜

′′′ → 𝑞𝑐𝑜 =

8𝑛𝐴𝑠 𝐻

𝑐𝑜𝑠

𝜋𝑦 𝑎

= 8.952 × 10

′′′ 𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 𝑞𝑐𝑜

𝑕𝑟 𝑓𝑡 3 = 5.78 × 1015 𝑀𝑒𝑉

𝑠. 𝑐𝑚3

′′′ 𝑞𝑐𝑜 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜

𝐺𝑓 = 180 𝑁𝑓𝑓 =

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

𝐴𝑣 𝑀235

× 𝜌235 =

𝐴𝑣 𝑀235

× 𝑟 × 𝜌𝑈

𝑇𝑎𝑏𝑙𝑒 4 − 2 @500 ℉ → 𝜌𝑈 = 18.815

𝑔𝑟

𝑐𝑚3 ; 𝑟 = 0.2 ;

24 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

→ 𝑁𝑓𝑓 = 9.64 × 1021 # 𝑐𝑚3 𝐹𝑖𝑔. 4 − 3 𝐺𝑟𝑎𝑝𝑕𝑖𝑡𝑒 , 𝑇 = 500 ℉ ≡ 260 ℃ ≡ 533 𝐾 → 𝑓 𝑇 = 0.93 293 0.5

→ 𝜍𝑓 = 0.8862 × 0.93 × 577.1 × (

533

)

→ 𝜍𝑓 = 352.64 𝑏 → 𝜑𝑐𝑜 =

′′′ 𝑞 𝑐𝑜

𝐺𝑓 𝑁𝑓𝑓 𝜍 𝑓

=

5.78×10 15 180×9.64×10 21 ×352.64×10 −24

→ 𝜑𝑐𝑜 = 9.452 × 10

21 #

𝑐𝑚 3

≅ 1013 [𝑎𝑛𝑠]

25 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-12 Known; 𝐹𝑎𝑠𝑡 𝑕𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 ; 𝐹𝑙𝑢𝑖𝑑 − 𝑓𝑢𝑒𝑙𝑒𝑑 ; 𝐻 = 8 𝑓𝑡 ; 𝐻𝑒 = 12 𝑓𝑡 ; 𝐷 = 5 𝑓𝑡 ; 𝑁𝑓𝑓 = 1020 # 𝑐𝑚3 ; 𝜍𝑓 = 5 𝑏 ; 𝜑

𝐻

𝑧= 2

= 1015 ; 𝜑 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑖𝑛 𝑡𝑕𝑒 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

Find; 𝑄𝑡 =? ; Analysis; 𝑞 ′′′ = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑 𝑄𝑡 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜

𝐻 2 𝐻 −2

𝜋𝑧

𝑐𝑜𝑠( ) 𝜋𝑟 2 𝑑𝑧 𝐻𝑒

𝜋𝑧

𝜑 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑖𝑛 𝑡𝑕𝑒 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 → 𝜑 = 𝜑𝑐𝑜 𝑐𝑜𝑠( ) 𝐻𝑒

𝜑

𝐻

𝑧= 2

15

= 10

15

→ 10

= 𝜑𝑐𝑜 𝑐𝑜𝑠(

𝜋×4 12

)

→ 𝜑𝑐𝑜 = 2 × 1015 𝐺𝑓 = 190 𝐻 2 𝐻 −2

𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛

𝜋𝑧 𝜋𝐷 2

𝑐𝑜𝑠( ) 𝐻𝑒

4

𝑑𝑧 =

𝜋𝐷 2 𝐻𝑒 4

𝜋

𝐻 2 𝐻 −2

𝜋 𝐻𝑒

𝑐𝑜𝑠

𝜋𝑧 𝐻𝑒

𝑑𝑧 =

𝜋𝐷 2 𝐻𝑒 4

𝐻

𝜋𝑧 2 𝑠𝑖𝑛 𝜋 𝐻𝑒 −𝐻 2

26 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL (𝑇𝑎𝑏𝑙𝑒 𝐻 −12)

1.9×10 14 𝑀𝑒𝑉

𝑠.𝑐𝑚 3

≡2.94×10 6 𝐵𝑡𝑢

→ 𝑄𝑡 = 190 × 1020 × 5 × 10−24 × 2 × 1015 × 6

12

→ 𝑄𝑡 = 2.94 × 10 × × 1.732 × 𝜋 8 𝐵𝑡𝑢 → 𝑄𝑡 = 3.82 × 10 𝑕𝑟 [𝑎𝑛𝑠]

25𝜋 4

1.732

𝑕𝑟.𝑓𝑡 3 12 𝜋52 𝜋

4

𝑠𝑖𝑛

𝜋𝑧 4 12 −4

27 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-13 Known; 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 ; 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 − 𝑡𝑦𝑝𝑒 𝑓𝑢𝑒𝑙: 3.5 𝑖𝑛 𝑤𝑖𝑑𝑒, 0.2 𝑖𝑛 𝑡𝑕𝑖𝑐𝑘, 0.005 𝑖𝑛 𝑐𝑙𝑎𝑑 (304𝐿 𝑠𝑡𝑎𝑖𝑛𝑙𝑒𝑠𝑠 𝑠𝑡𝑒𝑒𝑙) ; 𝐻 = 4 𝑓𝑡 ; 𝐷 = 4 𝑓𝑡 ; 𝑛 = 1000 ; 𝜑𝑐 𝑞𝑐′′′

𝑟=0.5 𝑓𝑡 ,𝑧=0

𝑟=0.5 𝑓𝑡 ,𝑧=0

= 3.057 × 1013 ; = 2.221 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 ;

𝐴𝑥𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑏𝑦 2 𝑓𝑡 𝑙𝑖𝑔𝑕𝑡 𝑤𝑎𝑡𝑒𝑟 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 Find; 𝜑𝑀𝑎𝑥 = 𝜑𝑐𝑜 =? ; 𝑄𝑡 =? ; Analysis; 𝑎) 𝐸𝑞. 4 − 25𝑎 → 𝜑𝑐 = 𝜑𝑐𝑜 𝐽0

2.405 𝑟

→ 𝜑𝑐𝑜 =

𝑅𝑒

𝑅𝑒 = 𝑅 = 2 𝑓𝑡 , 𝑟 = 0.5 𝑓𝑡 → 𝜑𝑐𝑜 =

𝐽0

𝜑 𝑐 𝑟=0.5 𝑓𝑡 ,𝑧=0 2.405 ×0.5 𝐽0 2

𝜑𝑐 2.405 𝑟 𝑅𝑒

=

3.057×10 13 𝐽 0 0.6 𝐴𝑝𝑝 . 𝐶 →0.912

→ 𝜑𝑐𝑜 = 3.352 × 1013 # 𝑠. 𝑐𝑚2 𝑎𝑛𝑠 𝑏) 𝐸𝑞. 4 − 29𝑎 → 𝑄𝑡 =

4𝑛 𝜋𝑅 2

𝐴𝑠 𝐻𝑒 sin

𝜋𝐻 2𝐻𝑒

′′′ 𝑞𝑐𝑜

𝑅 𝑟 𝐽0 0

2.405 𝑟 𝑅𝑒

𝑑𝑟

28 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑅𝑒 = 𝑅 → 𝑄𝑡 = 0.275𝑛𝐴𝑠 𝐻𝑒 sin 𝐴𝑠 =

0.2 12

×

3.5 12

𝜋𝐻 2𝐻𝑒

′′′ 𝑞𝑐𝑜

= 0.00486 𝑓𝑡 2

𝑇𝑎𝑏𝑙𝑒 3 − 1 𝑓𝑜𝑟 𝑙𝑖𝑔𝑕𝑡 𝑤𝑎𝑡𝑒𝑟 → 𝐿 = 2.88 𝑐𝑚 ≡ 0.095 𝑓𝑡 2 𝑓𝑡 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 → 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 > 2𝐿 → 𝜆𝑒 = 𝜓𝑒 = 𝐿 → 𝜆𝑒 = 0.095 𝑓𝑡 → 𝐻𝑒 = 𝐻 + 2𝜆𝑒 = 4 + 2 × 0.095 = 4.19 𝑓𝑡 ′′′ 𝐸𝑞. 4 − 25𝑏 → 𝑞𝑐′′′ = 𝑞𝑐𝑜 𝐽0 ′′′ 𝑅 = 𝑅𝑒 → 𝑞𝑐𝑜 =

2.405 𝑟

𝑞 𝑐′′′ 𝑟=0.5 𝑓𝑡 ,𝑧=0 2.405 ×0.5 𝐽0 2

𝑅𝑒

=

2.221×10 7 0.912

′′′ → 𝑞𝑐𝑜 = 2.435 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 0.997

→ 𝑄𝑡 = 0.275 × 1000 × 0.00486 × 4.19 × sin 2.435 × 107 → 𝑄𝑡 = 13.6 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑎𝑛𝑠

𝜋×4 2×4.19

×

29 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-14 Known; 𝑃0 = 2000 𝑀𝑊 ; 𝜃0 = 1 𝑌𝑒𝑎𝑟 ; 𝑎) 𝜃𝑠 = 1 𝑕𝑜𝑢𝑟 ; 𝑏) 𝜃𝑠 = 2 𝑕𝑜𝑢𝑟𝑠 ; Find; 𝑄𝑠 =? ; Analysis; 𝐸𝑞. 4 − 35 → 𝐸𝑠 = 0.128 𝜃𝑠 0.74 𝑇𝑎𝑏𝑙𝑒 𝐻 − 11 → 𝑃0 = 2000 𝑀𝑊 ≡ 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 𝑎) 𝜃𝑠 = 1 𝑕𝑜𝑢𝑟 ≡ 3600 𝑠𝑒𝑐 → 𝐸𝑠 = 0.128 (3600)0.74 → 𝐸𝑠 = 54.81

𝐵𝑡𝑢 𝑠𝑒𝑐 𝑓𝑡 3 𝑕 𝑟 𝐵𝑡𝑢 𝑓𝑡 3 𝑕 𝑟



54.81 𝐵𝑡𝑢 𝑕𝑟 3600

𝐵𝑡𝑢

54.81 𝐵𝑡𝑢 𝑕𝑟

× 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝑄𝑠 = 103.89 × 106 𝐵𝑡𝑢 [𝑎𝑛𝑠] 𝑄𝑠 =

3600

𝐵𝑡𝑢

𝑏) 𝜃𝑠 = 2 𝑕𝑜𝑢𝑟 ≡ 7200 𝑠𝑒𝑐 → 𝐸𝑠 = 0.128 (7200)0.74 → 𝐸𝑠 = 91.54

𝐵𝑡𝑢 𝑠𝑒𝑐 𝑓𝑡 3 𝑕 𝑟 𝐵𝑡𝑢 𝑓𝑡 3 𝑕 𝑟



91.54 𝐵𝑡𝑢 𝑕𝑟 3600

𝐵𝑡𝑢

30 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL 91.54 𝐵𝑡𝑢 𝑕𝑟

× 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝑄𝑠 = 173.52 × 106 𝐵𝑡𝑢 [𝑎𝑛𝑠] 𝑄𝑠 =

3600

𝐵𝑡𝑢

31 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-15 Known; 𝑇𝑕𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 ; 𝐻𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑 ; ′′′ 𝑀𝑜𝑟𝑒 𝑡𝑕𝑎𝑛 1 𝑦𝑒𝑎𝑟 ; 𝑞𝑐𝑜 = 5 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 ;

𝐷 = 10 𝑓𝑡 ; 𝐻 = 10 𝑓𝑡 ; 𝑅𝑒 = 1.5 𝑅 ; 𝐻𝑒 = 1.5 𝐻 ; 𝑉𝐹𝑢𝑒𝑙 = 0.3 𝑉𝐶𝑜𝑟𝑒 Find; 𝑄𝑠 =? , 𝐶𝑜𝑚𝑝𝑎𝑟𝑒 𝑡𝑜 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛; Analysis; 𝐹𝑢𝑒𝑙 𝑣𝑜𝑙𝑢𝑚𝑒

𝐸𝑞. 4 − 24 → 𝑞𝑕′′′ = 𝑞 ′′′

𝐶𝑜𝑟𝑒 𝑣𝑜𝑙𝑢𝑚𝑒

𝑉𝐹𝑢𝑒𝑙 = 0.3 𝑉𝐶𝑜𝑟𝑒 → 𝑞𝑕′′′ = 0.3 𝑞 ′′′ 𝑞 ′′′ = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜 𝑐𝑜𝑠

𝑄𝑡 =

𝐻 2 𝐻 −2

𝑅 ′′′ 𝑞 0

→ →

15 𝜋

𝑐𝑜𝑠 sin

𝜋𝑧 𝐻𝑒

𝜋𝑧 5 15 −5

𝐻𝑒

𝐽0

2.405 𝑟 𝑅𝑒

2𝜋𝑟𝑑𝑟𝑑𝑧

′′′ → 𝑄𝑡 = 0.3 𝑞𝑐𝑜 2𝜋

𝐻 2 𝐻 −2

𝜋𝑧

𝐻 2 𝐻 −2

𝑑𝑧 = =

30 𝜋

𝐻𝑒 𝜋

𝑅 𝜋𝑧 𝑐𝑜𝑠 0 𝐻𝑒

𝐻 2 𝐻 −2

sin

𝐽0

2.405 𝑟

𝑟𝑑𝑟𝑑𝑧

𝑅𝑒

𝐻

𝜋 𝐻𝑒

𝜋𝑧 5 15 0

𝑐𝑜𝑠

𝜋𝑧 𝐻𝑒

= 8.27

𝑑𝑧 =

𝐻𝑒 𝜋

𝜋𝑧 2 sin 𝐻𝑒 −𝐻 2

32 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

→ →

𝑅 0

𝑟𝐽0

7.5 2.405

2.405 𝑟 𝑅𝑒

× 5 × 𝐽1

𝑑𝑟 =

𝑅𝑒 2.405

2.405×5

7.5 1.6 𝐴𝑝𝑝 .𝐶→0.5699

𝑟𝐽1

2.405 𝑟

𝑅

𝑅𝑒

0

= 8.886

→ 𝑄𝑡 = 0.3 × 5 × 107 × 2𝜋 × 8.27 × 8.886 → 𝑄𝑡 = 6.926 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝐸𝑠 = 0.128 (24 × 3600)0.74 → 𝐸𝑠 = 575.75

𝐵𝑡𝑢 𝑠𝑒𝑐 𝑓𝑡 3 𝑕 𝑟 𝐵𝑡𝑢 𝑓𝑡 3 𝑕 𝑟



575.75 𝐵𝑡𝑢 𝑕𝑟 3600

𝐵𝑡𝑢

575.75 𝐵𝑡𝑢 𝑕𝑟

× 6.926 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝑄𝑠 = 1.108 × 109 𝐵𝑡𝑢 [𝑎𝑛𝑠] (𝐷𝑢𝑟𝑖𝑛𝑔 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 𝑑𝑎𝑦 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑠𝑕𝑢𝑡𝑑𝑜𝑤𝑛) 𝑄𝑠 =

3600

𝐵𝑡𝑢

𝑄 = 𝑄𝑡 × 𝑡 = 6.926 × 109 𝐵𝑡𝑢 𝑕𝑟 × 24 𝑕𝑟 → 𝑄 = 166.22 × 109 𝐵𝑡𝑢 [𝑎𝑛𝑠] (𝐷𝑢𝑟𝑖𝑛𝑔 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛)

33 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-16 Known; 𝑃0 = 2000 𝑀𝑊 𝑡 ; 𝜃0 = 1 𝑚𝑜𝑛𝑡𝑕 ; 𝑎) 𝜃𝑠 = 1 𝑕𝑜𝑢𝑟 ; 𝑏) 𝜃𝑠 = 2 𝑕𝑜𝑢𝑟𝑠 ; Find; 𝑄𝑠 =? ; Analysis; 𝑇𝑎𝑏𝑙𝑒 𝐻 − 11 → 𝑃0 = 2000 𝑀𝑊 ≡ 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 𝑎) 𝑓𝑖𝑛𝑖𝑡𝑒 𝜃0 ; 𝑓 𝜃0 = 1 − (1 +

2.592×10 6 −0.2 ) 3600

= 0.732

→ 𝐸𝑠′ = 𝑓 𝜃0 𝐸𝑠 → 𝐸𝑠′ = 0.732 × 0.128 (3600)0.74 → 𝐸𝑠′ = 40.113

𝐵𝑡𝑢 𝑠𝑒𝑐 𝑓𝑡 3 𝑕 𝑟 𝐵𝑡𝑢 𝑓𝑡 3 𝑕 𝑟



40.113 𝐵𝑡𝑢 𝑕𝑟 3600

𝐵𝑡𝑢

40.113 𝐵𝑡𝑢 𝑕𝑟

× 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝑄𝑠 = 76 × 106 𝐵𝑡𝑢 [𝑎𝑛𝑠] 𝑄𝑠 =

3600

𝐵𝑡𝑢

𝑏) 𝑓𝑖𝑛𝑖𝑡𝑒 𝜃0 ; 𝑓 𝜃0 = 1 − (1 +

2.592×10 6 −0.2 ) 2×3600

= 0.692

→ 𝐸𝑠′ = 𝑓 𝜃0 𝐸𝑠 → 𝐸𝑠′ = 0.692 × 0.128 (2 × 3600)0.74 →

𝐸𝑠′

= 63.35

𝐵𝑡𝑢 𝑠𝑒𝑐 𝑓𝑡 3 𝑕 𝑟 𝐵𝑡𝑢 𝑓𝑡 3 𝑕 𝑟



63.35 𝐵𝑡𝑢 𝑕𝑟 3600

𝐵𝑡𝑢

34 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL 63.35 𝐵𝑡𝑢 𝑕𝑟

× 6.824 × 109 𝐵𝑡𝑢 𝑕𝑟 → 𝑄𝑠 = 120.1 × 106 𝐵𝑡𝑢 [𝑎𝑛𝑠] 𝑄𝑠 =

3600

𝐵𝑡𝑢

(𝐶𝑜𝑚𝑝𝑎𝑟𝑒 𝑎𝑛𝑠𝑤𝑒𝑟𝑠 𝑡𝑜 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 4 − 14)

35 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

4-17 Known; 𝑇𝑎𝑏𝑙𝑒 4 − 3, 100% 𝑒𝑛𝑟𝑖𝑐𝑕𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 ; Find; 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑃𝑜𝑤𝑒𝑟 (𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟) =? ; 𝑚 𝑎) 𝐷𝑢𝑟𝑖𝑛𝑔 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑏) 𝑂𝑛𝑒 𝑦𝑒𝑎𝑟 𝑎𝑓𝑡𝑒𝑟𝑤𝑎𝑟𝑑 Analysis; 𝑎) 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑝𝑜𝑤𝑒𝑟 =

𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐵𝑡𝑢 𝐹𝑢𝑒𝑙 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

𝑙𝑏 𝑚

𝑕𝑟 𝑓𝑡 3

𝑓𝑡 3

𝑇𝑎𝑏𝑙𝑒 𝐻 − 12 → 1 𝑊 𝑐𝑚3 ≡ 9.662 × 104 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 𝑔𝑟 𝑙𝑏𝑚 𝑇𝑎𝑏𝑙𝑒 𝐻 − 5 → 1 𝑐𝑚3 ≡ 62.43 𝑓𝑡 3 𝑆. 𝑃. (𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟) = 𝑚

𝑃( 𝑊

)×9.662×10 4 𝑐𝑚 3 𝑔𝑟 𝜌( )×62.43 𝑐𝑚 3

= 1547.65

𝑃( 𝑊 ) 𝑐𝑚 3 𝑔𝑟 𝜌( ) 𝑐𝑚 3

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝑆𝑟 90 = 174.1 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚 137 𝐵𝑡𝑢 𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝐶𝑠 = 503.98 𝑙𝑏𝑚 𝑕𝑟 144 𝐵𝑡𝑢 𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝐶𝑒 = 3022.76 𝑙𝑏𝑚 𝑕𝑟 147 𝐵𝑡𝑢 𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝑃𝑚 = 257.94 𝑙𝑏𝑚 𝑕𝑟 𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝑃𝑜 210 = 2.197 × 105 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚

36 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝑃𝑢238 = 854.3 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚 242 5 𝐵𝑡𝑢 𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃. 𝐶𝑚 = 1.54 × 10 𝑙𝑏𝑚 𝑕𝑟 𝑏) 𝑂𝑛𝑒 𝑦𝑒𝑎𝑟 𝑎𝑓𝑡𝑒𝑟𝑤𝑎𝑟𝑑 → 𝑒𝑛𝑟𝑖𝑐𝑕𝑚𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑕𝑎𝑛𝑔𝑒 → 𝑁 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑕𝑎𝑛𝑔𝑒 𝑃 = 𝑅∆𝐸 = 𝜆𝑁∆𝐸 ; 𝑁 = 𝑁0 𝑒 −𝜆𝜃 → →

𝑆.𝑃.′ 𝑆.𝑃.

=

𝑃′ 𝑃

=

𝑁′

𝑁′ 𝑁

𝑁

= 𝑒 −𝜆𝜃

= 𝑒 −𝜆𝜃 ; 𝜆 =

0.693 𝜃1

2

−0.693 ×1

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝑆𝑟 90 = 𝑒 28 × 174.1 → 𝑆. 𝑃.′ 𝑆𝑟 90 = 169.84 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚 −0.693 ×1

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝐶𝑠137 = 𝑒 33 → 𝑆. 𝑃.′ 𝐶𝑠137 = 493.51 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚

× 503.98

−0.693 ×365 (𝑑 )

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝐶𝑒144 = 𝑒 285 (𝑑 ) → 𝑆. 𝑃.′ 𝐶𝑒144 = 1244.4 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚

× 3022.76

−0.693 ×1

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝑃𝑚147 = 𝑒 2.5 × 257.94 → 𝑆. 𝑃.′ 𝑃𝑚147 = 195.49 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚 −0.693 ×365 (𝑑 ) 138 .4(𝑑)

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝑃𝑜 210 = 𝑒 → 𝑆. 𝑃.′ 𝑃𝑜 210 = 3.53 × 104 𝐵𝑡𝑢

𝑙𝑏𝑚 𝑕𝑟

× 2.197 × 105

37 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL −0.693 ×1

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝑃𝑢238 = 𝑒 86 → 𝑆. 𝑃.′ 𝑃𝑢238 = 847.4 𝐵𝑡𝑢 𝑙𝑏 𝑕𝑟 𝑚

× 854.3

−0.693 ×365 (𝑑) 163 (𝑑)

𝑇𝑎𝑏𝑙𝑒 4 − 3 → 𝑆. 𝑃.′ 𝐶𝑚242 = 𝑒 → 𝑆. 𝑃.′ 𝐶𝑚242 = 3.26 × 104 𝐵𝑡𝑢

𝑙𝑏𝑚 𝑕𝑟

× 1.54 × 105

1 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

Chapter 5 Heat Conduction in Reactor Elements

2 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-1 Known; 𝐻𝑒𝑎𝑣𝑦 − 𝑤𝑎𝑡𝑒𝑟 𝑐𝑜𝑜𝑙𝑒𝑑 ; 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 − 𝑢𝑟𝑎𝑛𝑖𝑢𝑚 𝑚𝑒𝑡𝑎𝑙 ; 𝐷𝑅𝑜𝑑 = 0.9 𝑖𝑛 ; 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑐𝑙𝑎𝑑 ; 𝑡𝐶𝑙𝑎𝑑 = 0.05 𝑖𝑛 ; 𝑇𝑀𝑎𝑥 = 700℉ ; 𝑇𝐵𝑢𝑙𝑘 = 180℉ ; 𝑕 = 5000 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; Find; 𝜑 =? ; 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑝𝑜𝑤𝑒𝑟 𝑘𝑊 𝑘𝑔 =? ; 𝑡𝑠 =? ; 𝑇𝑐 =? ; 𝑃𝑀𝑖𝑛 =? ; 𝜑𝑀𝑎𝑥 =? (𝑡𝑚 = 𝑡𝑚𝑒𝑙𝑡𝑖𝑛𝑔 ) Analysis; 𝑇𝑚 −𝑇𝑓

𝐸𝑞. 5 − 51𝑏 → 𝑞𝑠 = → 𝑞 ′′′ =

→𝑅=

𝑞𝑠 𝜋𝑅 2 𝐿

0.9 2×12

=

𝑐

𝐶 2𝜋𝑐𝐿 𝑅+𝑐 𝐿𝑛 𝑅

4𝑘 𝑓

1

+𝑕 2𝜋 𝑅+𝑐 𝐿

𝑇𝑚 −𝑇𝑓 𝑅2 + 4𝑘 𝑓

𝑅2 2𝑘 𝑐 𝑅+𝑐 𝐿𝑛 𝑅

𝑅2

+2𝑕 𝑅+𝑐

𝑓𝑡 = 0.0375 𝑓𝑡 ; 𝑐 =

→ 𝑞 ′′′ = 0.00141

→ 𝑁𝑓𝑓 =

𝑅 + 2𝑘 𝑓 2𝜋𝑅𝐿 𝑘

0.05 12

𝑓𝑡 = 0.0042 𝑓𝑡

700−180 +

0.00141 2𝑘 𝑐 0.0417 𝐿𝑛 0.0375

6.022×10 23 235.0439

0.00141 2×5000 0.0417

+

× 0.5 × 0.1 × 0.65 × 0.795 × 1

→ 𝑁𝑓𝑓 = 6.620 × 1019 # 𝑐𝑚3 [𝑎𝑛𝑠]

3 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-2 Known; 𝐷𝐹𝑢𝑒𝑙 = 0.5 𝑖𝑛 ; 𝑈𝑂2 3% 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝑡𝐻𝑒𝑙𝑖𝑢𝑚 = 0.003 𝑖𝑛 ; 𝑍𝑖𝑟𝑐𝑎𝑙𝑜𝑦 2 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 ; 𝑡𝐶𝑙𝑎𝑑 = 0.03 𝑖𝑛 ; 𝑃 = 1000 𝑝𝑠𝑖𝑎 ; 𝑕 = 10000 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; 𝑇𝑐 − 𝑇𝑓 = 30.4 ℉ ; 𝑘𝐻𝑒 = 0.16 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ; 𝑘𝐶𝑙𝑎𝑑 = 8 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ Find; 𝑇𝑀𝑎𝑥 =? ; Analysis; 𝑇𝑎𝑏𝑙𝑒 𝐷 − 1𝑎 → 𝑇𝑓 = 544.6 ℉ 𝑇𝑐 − 𝑇𝑓 = 30.4 ℉ → 𝑇𝑐 = 575 ℉ 𝑞 ′′ 𝐴𝑅𝑜𝑑 = 𝑞 ′′′ 𝑉𝐹𝑢𝑒𝑙 → 𝑞 ′′ × 2𝜋(𝑅 + 𝑎 + 𝑐)𝐿 = 𝑞 ′′′ 𝜋𝑅2 𝐿 𝑞 ′′ =

𝑅 2 𝑞 ′′′ 2 𝑅+𝑎+𝑐

→ 𝑞 ′′′ = → 𝑞 ′′′ =

; 𝑞 ′′ = 𝑕 𝑇𝑐 − 𝑇𝑓

𝑕 𝑇𝑐 −𝑇𝑓 2 𝑅+𝑎+𝑐 𝑅2 0.5 +0.003 +0.03 10000 30.4 2 2 12 0.5 2 2×12

= 3.3 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3

4 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑇𝑠1 − 𝑇𝑐 =

𝑞 ′′′ 𝑅 2

𝐿𝑛(

2

𝑅+𝑎 ) 𝑅

𝑘 𝐻𝑒

+

𝐿𝑛(

𝑘𝑐

0.5

→ 𝑇𝑠1 = 575 +

𝑅+𝑎 +𝑐 ) 𝑅+𝑎

3.3×10 7 × 2×12 2

2

𝐿𝑛(

0.25+0.003 ) 0.25

0.16

→ 𝑇𝑠1 = 1220 ℉

𝐴𝑠𝑠𝑢𝑚𝑒 𝑇𝑀𝑎𝑥 = 4500 ℉ → 𝑇𝑓 = 𝑘𝐹𝑢𝑒𝑙

4500+1220 2

(@2860 ℉)

𝑇𝑀𝑎𝑥 − 𝑇𝑠1 =

= 2860

= 1.1 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉

𝑞 ′′′ 𝑅 2 4𝑘 𝐹𝑢𝑒𝑙

→ 𝑇𝑀𝑎𝑥 = 1220 +

3.3×10 7 ×

→ 𝑇𝑀𝑎𝑥 ≅ 4475 [𝑎𝑛𝑠]

0.45 2 2×12

4×1.1

= 4475

+

𝐿𝑛(

0.25+0.003 +0.03 ) 0.25+0.003

8

5 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-3 Known; 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 − 𝑠𝑕𝑒𝑙𝑙 𝑓𝑢𝑒𝑙 ; 𝑡𝐹𝑢𝑒𝑙 = 0.2 𝑖𝑛 ; 𝐷𝑖𝑛𝑠𝑖𝑑𝑒 = 4 𝑖𝑛 ; 𝑞 ′′′ = 50 × 106 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 ; 𝑘𝑓 = 10 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ; 𝑡𝑆𝑡𝑒𝑎𝑚 = 700 ℉ ; 𝑕1 = 400 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; 𝑕2 = 280 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 ; (𝐻𝑖𝑛𝑡: 𝑀𝑎𝑦 𝑏𝑒 𝑡𝑟𝑒𝑎𝑡𝑒𝑑 𝑎𝑠 𝑎 𝑓𝑙𝑎𝑡 𝑠𝑙𝑎𝑏) Find; 𝑥𝑀𝑎𝑥 =? ; 𝑥1 =? ; 𝑥2 =? Analysis; 𝜋𝑅𝑜2 −𝜋𝑅𝑖2 𝐿

𝑉

𝑞 ′′ = 𝑞 ′′′ × = 𝑞 ′′′ × 𝐴

2𝜋𝑅𝑜 𝐿+2𝜋𝑅𝑖 𝐿

2.2 2 − 12 12

→ 𝑞 ′′ = 50 × 106

2

= 𝑞 ′′′

𝑅𝑜 −𝑅𝑖 2

= 4.167 × 105 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2

𝑎) 𝐸𝑞. 5 − 63 → 𝑇 𝑟 = 𝑇𝑖 −

𝑞 ′′′ (𝑟 2 −𝑅𝑖2 ) 4 𝑘𝑓

𝑇𝑀𝑎𝑥 @𝑟𝑀𝑎𝑥 → →−

2𝑟𝑞 ′′′ 4 𝑘𝑓



𝑑𝑇 𝑑𝑟

− [ 𝑇𝑖 − 𝑇𝑜 −

𝑞 ′′′

(𝑅𝑜2 4 𝑘𝑓



𝑅𝑖2 )]

=0;

𝑇𝑖 − 𝑇𝑜 −

𝑞 ′′′ 4 𝑘𝑓

𝑅𝑜2 − 𝑅𝑖2

1

𝑟 𝑅

𝐿𝑛 𝑅𝑜 𝑖

=0

𝑟 𝑅𝑖 𝑅 𝐿𝑛( 𝑅𝑜 ) 𝑖

𝐿𝑛( )

6 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

2 → 𝑟𝑀𝑎𝑥 =

−𝑘 𝑓 𝑞 ′′′

𝑇𝑖 −𝑇𝑜 +

2 𝑅2 𝑜 −𝑅 𝑖 4

𝑅

𝐿𝑛 𝑅𝑜 𝑖

→ 𝑟𝑀𝑎𝑥 = 2.16 𝑖𝑛 [𝑎𝑛𝑠] 𝑏)

7 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-4 Known; 𝑈𝑂2 ; 𝐷𝑃𝑒𝑙𝑙𝑒𝑡 = 0.5 𝑖𝑛 ; 𝐿𝑃𝑒𝑙𝑙𝑒𝑡 = 0.6 𝑖𝑛 ; 𝐶𝑙𝑎𝑑 𝑧𝑖𝑟𝑐𝑎𝑙𝑜𝑦2 𝑐𝑎𝑛 ; 𝑡𝐶𝑙𝑎𝑑 = 0.032 𝑖𝑛 ; 𝐷𝑂𝑢𝑡𝑠𝑖𝑑𝑒

𝑐𝑙𝑎𝑑

= 0.57 𝑖𝑛 ; 𝐻𝑒𝑙𝑖𝑢𝑚 𝑔𝑎𝑝 ;

′′′ 𝐵𝑡𝑢 1.5 % 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 ; 𝑃 = 50 𝑀𝑊 𝑡 ; 𝑞𝑎𝑣 . = 74444 𝑕𝑟𝑓𝑡 2 ;

4 % 𝑜𝑓 𝑡𝑕𝑒 𝑡𝑕𝑒𝑟𝑚𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑡𝑕𝑒 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 ; 𝑡𝑓 = 545 ℉ ; 𝑘𝐶 = 8 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡℉ ; 𝑘𝑓 = 1.15 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡℉ ; 𝑘𝐻𝑒 = 0.135 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡℉ ; 𝑕 = 10000 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡 2 ℉ Find; ′′′ 𝑞𝑛𝑒𝑤 =? 𝑖𝑓 𝐷 → 𝐷 − 0.12 ;

Analysis; ′′ 𝑞𝐹𝑢𝑒𝑙 = 0.96 × 74444 = 71466.24 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2

𝑞 ′′′ 𝑉𝐹𝑢𝑒𝑙 = 𝑞 ′′ 𝐴𝐹𝑢𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 → 𝑞 ′′′ 𝜋𝑅2 𝐿 = 2𝜋(𝑅 + 𝑎 + 𝑐)𝐿𝑞 ′′ → 𝑞 ′′′ = → 𝑞 ′′′ =

2𝜋(𝑅+𝑎+𝑐)𝐿𝑞 ′′ 𝜋𝑅 2 𝐿

=

0.2855 71466 .24×2× 12 0.25 ( 12 )2

2𝑞 ′′ (𝑅+𝑎+𝑐) 𝑅2

= 7821265 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3 ℉

8 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑇𝑚 = 𝑇𝑓 +

𝑞 ′′′ 𝑅 2

1

2

2𝑘 𝑓

→ 𝑇𝑚 = 545 +

+

𝑅+𝑎 𝑅

𝐿𝑛

+

𝑘 𝐻𝑒

𝑅+𝑎 +𝑐 𝑅+𝑎

𝐿𝑛

+

𝑘𝑐

0.25 7821265 ×( 12 )2

1

2

2×1.15

+

𝐿𝑛

1 𝑕 𝑅+𝑎+𝑐

0.2535 0.25

0.135

11040.285512 → 𝑇𝑚 = 1490.12 ℉ 𝐷 → 𝐷 − 0.12 → 𝑅′ = 0.19 ′′′ → 𝑞𝑛𝑒𝑤 =

2(𝑇𝑚 −𝑇𝑓 ) 𝑅 ′ +𝑎 𝐿𝑛 𝑅′ 1 2 ′ 𝑅 + 2𝑘 𝑓 𝑘 𝐻𝑒

′′′ → 𝑞𝑛𝑒𝑤 =

𝐿𝑛

+

𝑅 ′ +𝑎 +𝑐 𝑅 ′ +𝑎 𝑘𝑐

1 𝑕 𝑅 ′ +𝑎 +𝑐

+

2(1490.12−545) 0.193 𝐿𝑛 0.19 0.19 2 1 + 0.135 12 2×1.15

0.225 𝐿𝑛 0.193 8

+

+

10 4

1 0.225 12

′′′ → 𝑞𝑛𝑒𝑤 = 1.31 × 107 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 3



′′′ 𝑞𝑛𝑒𝑤 𝐹𝑢𝑒𝑙

2

=

′′′ 𝑅 ′ 𝑞 𝑛𝑒𝑤

2(𝑅 ′ +𝑎+𝑐)

=

0.19 2 ×1.31×10 7 12 0.19+0.035 2× 12

′′′ → 𝑞𝑛𝑒𝑤 = 87575.9 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 𝐹𝑢𝑒𝑙 ′′′ → 𝑞𝑛𝑒𝑤 = 𝑇𝑜𝑡𝑎𝑙

′′′ 𝑞 𝑛𝑒𝑤 𝐹𝑢𝑒𝑙 0.96

=

87576 0.96

′′′ → 𝑞𝑛𝑒𝑤 = 91224.9 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 [𝑎𝑛𝑠] 𝑇𝑜𝑡𝑎𝑙

0.2855

+

𝐿𝑛 0.2535 8

+

9 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-5 Known; 𝑆𝑝𝑕𝑒𝑟𝑖𝑐𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑐𝑜𝑟𝑒 ; 20 % 𝑒𝑛𝑟𝑖𝑐𝑕𝑒𝑑 𝑜𝑓 𝑈𝑂2 ; 𝐷𝑐𝑜𝑟𝑒 = 5 𝑓𝑡 ; 𝐷𝑃𝑒𝑙𝑙𝑒𝑡 = 1 𝑖𝑛 ; 𝑡𝑚 − 𝑡𝑠 = 3000 ℉ ; 𝑔𝑟 𝜑𝑐𝑜 = 1013 ; 𝑘𝑓 = 1.1 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡 ℉ ; 𝜌𝑓 = 11 𝑐𝑚3 ; 𝜍𝑓 = 500 𝑏 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑕𝑠 ; Find; 𝑟 =? Analysis; 𝐸𝑞. 5 − 66 → 𝑇𝑚 − 𝑇𝑠 = → 𝑅𝐹𝑢𝑒𝑙 = → 𝑞 ′′′ =

𝐷𝑃𝑒𝑙𝑙𝑒𝑡

=

2

6 𝑘 𝑓 (𝑇𝑚 − 𝑇𝑠 ) 𝑅2

1 2

=

𝑖𝑛 ≡

𝑞 ′′′ 𝑅 2 6 𝑘𝑓 0.5 12

𝑓𝑡

6×1.1×3000 0.5

( 12 )2

→ 𝑞 ′′′ = 1.14 × 107 𝐵𝑡𝑢 𝑕𝑟𝑓𝑡 3 𝑇𝑎𝑏𝑙𝑒𝐻 − 12 → 𝑞 ′′′ ≡ 7.36554 × 1014 𝑀𝑒𝑉 𝑠. 𝑐𝑚3 𝜋𝑟

𝐸𝑞. (4 − 3) → 𝑞

→ 𝑅𝐶𝑜𝑟𝑒 =

𝐷𝐶𝑜𝑟𝑒 2

′′′

= 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑 = 𝐺𝑓 𝑁𝑓𝑓 𝜍𝑓 𝜑𝑐𝑜 5

= = 2.5 𝑓𝑡 2

sin 𝑅 𝑒 𝜋𝑟 𝑅𝑒

10 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

𝑁𝑓𝑓 =

𝐴𝑣 𝑀𝑓𝑓

𝜌𝑓 𝑟𝑖 =

6.023×10 23 ×11×0.2×1 235.0439

= 5.637 × 1021 # 𝑐𝑚3

→ 𝐺𝑓 = 180 𝑀𝑒𝑉 𝑓𝑖𝑠𝑠𝑖𝑜𝑛 → 7.36554 × 1014 = 180 × 5.637 × 1021 × 500 × 10−24 × 13

10

×

sin

𝜋𝑟 𝑅 𝐶𝑜𝑟𝑒 𝜋𝑟

𝑅 𝐶𝑜𝑟𝑒

→ sin

𝜋𝑟 𝑅𝐶𝑜𝑟𝑒

= 0.1452 ×

𝜋𝑟 𝑅𝐶𝑜𝑟𝑒

→ sin (1.257𝑟) = 0.1825𝑟

11 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-6 Known; 𝑆𝑁𝐴𝑃 ; 560 𝑔𝑟 𝑜𝑓 𝑃𝑢238 𝐶 ; 𝑆𝑝𝑕𝑒𝑟𝑒 𝑓𝑢𝑒𝑙 ; 𝐷𝐹𝑢𝑒𝑙 = 0.6072 𝑓𝑡 ; 𝜌𝐹𝑢𝑒𝑙 = 12.5

𝑔𝑟

𝑐𝑚3 ;

𝑘𝐹𝑢𝑒𝑙 = 12 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ; 𝑇𝑆𝑕𝑒𝑙𝑙 = 235 ℉ ; 𝑕 = 10 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 Find; 𝑞 ′′′ =? ; 𝑡𝑚 =? ; Analysis;

12 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-7 Known; 𝑃𝑙𝑎𝑡𝑒 − 𝑡𝑦𝑝𝑒 ; 𝑡𝐹𝑢𝑒𝑙 = 0.2 𝑖𝑛 ; 𝑡1 = 0.1 𝑖𝑛 ; 𝑘1 = 1 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ; 𝑁𝑈 235 1 = 1021 # 𝑐𝑚3 ; 𝑡2 = 0.1 𝑖𝑛 ; 𝑘2 = 0.9 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ; 𝑁𝑈 235 2 = 8 × 1021 # 𝑐𝑚3 ; 𝜑 = 1014 ; 𝑇𝑀𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 = 500 ℉ ; 𝑕 = 5000 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 2 ℉ ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 Find; 𝑡𝑀𝑎𝑥 =? Analysis;

13 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-8 Known; 𝑃𝑒𝑏𝑏𝑙𝑒 − 𝑏𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 ; 𝑃𝑊𝑅 ; 𝑆𝑝𝑕𝑒𝑟𝑖𝑐𝑎𝑙 𝑓𝑢𝑒𝑙 ; 𝐷𝐹𝑢𝑒𝑙 = 1 𝑖𝑛 ; 𝑃 = 2500 𝑝𝑠𝑖𝑎 ; 𝐶𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦 𝑠𝑢𝑏𝑐𝑜𝑜𝑙𝑒𝑑 ; ′′′ 𝑞𝑐𝑜 = 107 𝐵𝑡𝑢 𝑕𝑟. 𝑓𝑡 3 ; 𝑘𝑓 = 10 𝐵𝑡𝑢 𝑕𝑟 𝑓𝑡 ℉ ;

𝐼𝑔𝑛𝑜𝑟𝑒 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 Find; 𝑡𝑀𝑎𝑥 𝑖𝑛 𝑡𝑕𝑒 𝑐𝑜𝑟𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 =? ; Analysis;

14 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-9 Known; 𝑆𝑝𝑕𝑒𝑟𝑖𝑐𝑎𝑙 𝑠𝑕𝑒𝑙𝑙 𝑜𝑓 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 ∆𝑟 𝑎𝑡 𝑟 ; 𝐻𝑒𝑎𝑡 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 ; Find; 𝐷𝑒𝑟𝑖𝑣𝑒 𝐸𝑞. 5 − 64 ; Analysis;

15 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL

5-10 Known; 𝑈𝑛𝑐𝑙𝑎𝑑 𝑕𝑜𝑙𝑙𝑜𝑤 𝑡𝑕𝑖𝑐𝑘 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑓𝑢𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ; 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑕𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 ; 𝐾𝑛𝑜𝑤𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 ; 𝐻𝑒𝑎𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑡𝑜 𝑓𝑙𝑜𝑤 𝑡𝑕𝑟𝑜𝑢𝑔𝑕 𝑏𝑜𝑡𝑕 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 ; Find; ′′′ 𝑟𝑚 𝑤𝑕𝑒𝑟𝑒 𝑡 = 𝑡𝑀𝑎𝑥 =? ; 𝑞𝑖𝑛𝑛𝑒𝑟

Analysis;

𝑠𝑢𝑟𝑓𝑎𝑐𝑒

′′′ , 𝑞𝑜𝑢𝑡𝑒𝑟

𝑠𝑢𝑟𝑓𝑎𝑐𝑒

=? ;

Chapter 6 Heat Conduction in Reactor Elements

6 1 Known : thickness  3 in ;   1013 ; h  3 MeV

#

; Ti  To ; k  28

Btu hrft F

Analysis: a) Eq.( 6-18 )  xm 

  1  μk 1 Ln  (ti  t o )  ( 1  e  μL )  μ μL  qoL 

  0.282cm -1 Table (6 - 1)      7.86 1   0.282   8.595 ft 1 0.03281 Mev cm3 . sec Btu  qo  8.46  1012  1.5477  10 8  13 .09  10 4 hrft3 L  3in  7.62 cm qo  h  0.282  1013  3  8.46  1012

1

Mev Btu  1.5477  10 8 3 cm . sec hrft3

1  1  e ( 0.2827.62)   xm  Ln    xm  3.15cm  1.24in  0.1034 ft[ans] 0.282  0.282  7.62  q k (ti  t o ) qo e  L  1 b) Eq.6  20      (1  ) A L  L ( x o )

q 1.309  10 5   A 8.595

 e ( 0.2827.62)  1  1  0.282  7.62    q Btu   8968 .87 [ans] A hrft3

x l  Eq.6  21  

q k (ti  to ) qo L e  L  1   (e  ) A L L L

q 1.309  10 5  ( 0.2827.62) e ( 0.2827.62)  1   e  A 8.595  0.282  7.62  q Btu   4484 .82 [ans] A hrft3 

 x q  x c) Eq.( 6-17 )  tm  ti  (to  ti )  2o  (e  μL  1 )  (e μx  1) L μkL  5 1.309  10 1.24in ( 0.2827.62)   t m  ti  (e  1)  (e 0.2823.15  1) 2  (8.595 )  28  3in  m

 tm  ti  14 .14478  F [ans]

62 Known : h  3

MeV ;   1014 ; iron ; total atenuation  90% ;   0.282cm -1 #

Analysis : total attenuatio n 90%  2  0.1o qt 1 qt 2 q q   o (1  e  L )  1 (1  e  L ) A A   1

2

 0 h (1  e  L )  1 h (1  e  L ) 1

2

 0 (1  e  L )  0 e  L (1  e  L ) 1

1

2

2  0.10 ; 2  1e  L  o e  L e  L  0 e   ( L  L )  0.10  0 e   ( L  L ) 2

1

1

2

1

2

2

1  e  L  e  L  e   ( L  L ) Ln (0.55)  1  e  L  e  L  0.1  1.1  2e  L  L1    ( L L )  0.282 0.1  e  L1  2.21cm  0.07 ft[ans] 1

1

1

2

1

1

2

 0.1  e   ( L  L )  L2  1

2

 L2  0.2 ft[ans]

Ln (0.1)  L1  0.282

1

1

63 Known :   1012 ; h  2

MeV Btu ; k  131 # hrft F

Analysis : q1 k (ti  to ) qo L e L  1 Eq.(6  21)    (e  ) A L  L L  6in  0.5 ft 2   ) Al  0.0431 cm @ 2 Mev gr 1     Al  0.0431  2.7  0.116 cm  Table(6  2)   Al  2.7 gr 3

Table(6  1) 

cm

  Al 



0.116 ft 1  3.547 ft 1 0.03281

qo  I  0.116  2 1012  0.233 1012 MeV

sec cm

3

 3.6 10 3 Btu

q1 131  (400  335 ) 3.6 10 3  3.5470.5 e 3.5470.5  1     e  A 0.5 0.547  3.547  0.5  q Btu  1  17332 .87 A hrft2 q2 K (ti  t0 ) qo e L  1 Eq.(6  20 )    (1  ) A L  L qo  2 h  2  1e  L   L 1  0 e  photons 0  1012 cm2 . sec  w w

AL AL

hrft3

Lw  1 ft 2   ) w  0.0493 cm @ 2Mev gr 1     w  0.0493 1  0.0493 cm   w  1 gr 3

cm  0.0493 1  w  ft  1.503 ft 1 0.03281 1  1012 e 3.5360.5  1.78 1011   12 1.50313.5470.5 10    10  e  3 . 787  10  2 

 qo  3.787 1010  0.116  2  8.814 10 9

Mev Btu  136 .4 3 sec.cm hrft3

q2 131  (282  250 ) 136 .4  e 3.5470.5  1  Btu    1    9127 .86 A 0.5 3.547  3.547  0.5  hrft2 q q q Btu  total  1  2  17332 .87  9127 .86  26460 .73 [ans] A A A hrft2

64 Known : 0  1013 ; h  3

MeV #

Analysis:

  0.282cm -1 q0  0 h  1013  0.282  3  1.5488  10 8 q0  130935 .42  13 .09  10 4

Btu hrft3

KA qoA  L e L  1  qx L  o(isulation)  (ti  to )  e   L   L    8.595  8.595 2 28 130935 .42 e  1   (500  to )   e 2  8 . 595 0.5 8.595    2   to  563 .3 F [ans] qo K qo e L 1   (ti  to )  1   A L  L  qo 28 130935 .42  e 8.5950.5  1    (500  563 .3)  1   A 0.5 8.595  8.595  0.5  qo Btu  15265 .3 [ans] A hrft2

65 Known : E  h  5 Find : o Analysis : dE qt  st dt Eq.6  19   qo  E

MeV q t ;  2 10 5 Btu hrft2 # A



qoA T (1  e L )  Vc 2200 m  0.0253 ev s  t neutron absorption only 



App. B) For nickel 60 (From   σ a  2.6 b  2.6 10 - 24 cm2

6.022 10 23 N  8.9  8.943 10 22 # 3  Σ  0.2325 cm1 cm 59 .9308   1011 # 2 ; E  0.0253 eV cm .sec qo  0.2325  0.0253 1011  5.88 10 8 eV 3  5.88 10 2 Mev 3 cm . sec cm . sec 1 MeV   1.5477 10 8 Btu 3 3  cm . sec hrft    q  9.110 6 Btu hrft3 q V T  o (1  e L )   c ; L  3in  0.25ft  7.62cm  A t T q   o (1  e L ) t Lc lbm   8.9 gr 3  555 .627 3 cm ft

9.110 6 Btu

T hrft3   (1  e 0.23257.62 ) t 0.2325 cm1  7.62cm  555 .627 lbm  0.103 Btu lbm F ft3  T   0.08975 10 6  0.83  7.45 10 8 F [ans] hr t

66 Known :

67 Known: t  4in ; iron ; h  5MeV ;

qt  2  10 5 Btu hrft2 A

Analysis: qt q0  1  e L  A  L  4in  0.333 ft  10 .16 cm Eq.(6  19 ) 

h  5MeV  7.595  10 16 Btu Table(6  1) 

2     0.0314 cm g  m  iron    0.247 cm1  7.522 ft 1

Table(6  2)  iron  7.86

gm

cm3

   

qt 7.522  2  10 5 A q0   q0   1.638  10 6 Btu  L 7.5220.333 hrft3 1  e  1  e 



1.638  10 6 Photon q0  h     2.867  10 20 16 7.522  7.595  10 hrft2 Photon    8.57233  1013 [ans] sec cm2

68 Known :  0  4  1012 ; h  3

Mev ; B  1.5 #

Analysis: 1  0.0953 cm 1   1   2  0.279 cm  1  o (1  B )e   L    L  2  1 (1  B )e  1  4  1012  2.5e 0.095382.54  1.44  1012  1 1 1

2 2

2  1.44  1012  2.5e 0.2792.54  0.873  1012  2 qo  0 1E  4  1012  0.0953  3  1.1436  1012  1.5477  10 8 q00  1.77  10 4

Btu hrft3

q01  1  2 E  1.44  1012  3  0.279  1.5477  10 8 q1 1.8654  10 4

Btu hrft3

qt q1  (1  e   L )  h(t o  400 ) A 2 2 2

1.8654  10 4 (1  e 1.417 )  200 (t o  400 )  t o  408 .31 F [ans] 0.279  30 .48 8  0.095330.48 q1t q00 1.77  10 4  L 12  (1  e )  (1  e ) A 1 0.0953  30 .48 2  0.27930.48 q2 t q01 1.8654  10 4  L 12  (1  e )  (1  e ) A 2 0.279  30 .48 1 1

2 2

q1t Btu  5214 .75 A hrft2

q 2t Btu  1661 .94 [ans] A hrft2

69 Known : L  3in ; hυ  3

MeV q Btu ; t  10 6 # A hrft2

Find :  Analysis : qt q0  E  (1  e L )  0 (1  e L ) A   h  3MeV  4.56 10 16 Btu Table(6  2)    0.282 cm1  8.6 ft 1 L  3in  0.25 ft qt 10 6 A  o   E (1  e L ) 4.56 10 16  (1  e 8.60.25 ) photon o  7.4233 1014 [ans] sec cm2

6  10 Known : h  1

MeV ; B  3 ;   0.10 ;   0.797 cm1 #

Analysis :

  0 (1  B)e L  0.10 e L 

0.1  0.025 4

L   Ln (0.025 )  L  L  1.822 in[ans]

 Ln (0.025 )  4.628 cm 0.797

6  11 Known:L  6in ;   1014

photon MeV ; hυ  5 ; cm2 .sec photon

Btu ; t f  300  F 2 hrft F Find : a )The two surface temprature b) Max tempa ture within the plate

h  1000



Analysis: @ x  0  h(ti  t )   q x0

k q0 e L  1   (ti  t0 )  1   (1) L  L 

@ x  L  h(t0  t )  q x L

k q0 L e L  1   (ti  t0 )   e   ( 2) L  L 

 2k (ti  t0 ) q0 e L  1   L (1) - (2)  h( ti  t0 )   1  e  2  L  L  2k  q0 e L  1    L   h  (ti  t0 )  1  e  2  (3) L  L   Table(6  2) Steel    0.248 cm1  7.559 ft 1 k  63 .9W

mK

 36 .92 Btu

hrftF

q0  0 E  0.248 1014  5  1.24 1014 MeV

3

cm sec

 1.919 10 6 Btu

2  36 .92  1.919 10 6  e 7.60.5  1   7.60.5 (3)  1000   2 1  e  (ti  t0 )  0.5  7.559  7. 6  0. 5    (ti  t0 )  111 .73 F

ft 3 hr

a)  36 .92 111 .73 1.919 10 6  e 7.60.5  1  (1)  1000( ti  300 )    1   0.5 7.559 7 . 6  0 . 5    ti  300  179 .3  ti  479 .3 F [ans]  t0  ti  111 .73  t0  367 .56  F [ans] b) 1 1  7.559  36 .92 1  e 7.60.5  Ln  (111 .73)  6 7.559 1.919 10  0.5 7.559  0.5   xm  0.163 ft  1.955 in Eq.(6  18)  xm 

tm  t ( xm )  Eq.(6  17 ) 0.163 1.919 10 6  0.163 7.60.5  7.60.5 tm  479 .3  (111 .73)   ( e  1 )  ( e  1 ) 2  0.5 (7.559 )  36 .92  0.5 tm  1035 .83 F  557 .7 C[ans] !‫ اصالح شده و حماسبات جمدداً تکرار شوند‬k ‫برای پاسخ دقیق تر باید مقدار‬

6  12 Known : L1  L2  L ; h  5

MeV ; 0  1014 ; 2  0.10 #

Analysis :

1  0e L 2  1e L  (0e L )e L 2  0e 2 L  e 2 L  0.1  e 20.245L  0.1  2  0.10 L  4.7cm  0.154 ft[ans] q1 0 h  1014  5  0.245  1.225  1014 q1 1895932 .5

MeV sec.cm3

Btu [ans] hrft3

q2  1h  3.162  1013  5  0.245  1.5477  10 8 Btu [ans] hrft3 q1 q1 1895932 .5  (1  e L )  (1  e 0.2454.7 ) A  0.245  0.03281 q1 Btu  32724837 [ans] A hrft2 q2 q2 Btu  (1  e L )  10346356 [ans] A  hrft2

q2  599422 .13

6  13 Known : E 

5MeV photon 1 ; 0  1014 ;  0 .1 # sec.cm2 0

Analysis : Table(6  1) 

2    0.0314 cm gm      0.247 cm1  7.523 ft 1

Table(6  2)    7.86

gm

cm3

  

a)  1  0e x  0.10  e 7.523L  0.1  7.53 L  Ln (0.1)  L  0.306 ft  3.67 in b) qt q0  (1  e L )  0 E (1  e L ) A  qt  1014  5  (1  e 7.5230.306 )  1.5477  10 8 A qt Btu  228652 .85 [ans] A hrft2

6  14 known: Analysis :  k (1  e L )  xm  Ln (t  t )    q0L i o L   1  1  e L  ti  to  xm  Ln     L  MeV  14   10 h   3 0 1  photon  MeV   1013 h 2  1  photon 1

1  0.282 cm1  8.6 ft 1  1 1  2  0.47 cm  14 .33 ft  1  (1  e 8.61 )  x1m  Ln  0.25 ft 8.6  8.6  1  8 .6 MeV q0  0 1h1  1014   3  8.46  1013 30 .48 sec.cm3 Btu q0  13 .103  10 5 hrft3 MeV q  2 h 2  1013  0.47  1  4.7  1012 sec.cm3 MeV q  7.274  10 4  0.47  1  4.7  1012 sec.cm3 4 Btu q  7.274  10 hrft3 q  x m  t1m  ti  20  1 (e  L  1)  (e  x m  1)  k L  1

1 1

13 .103  10 5  0.25 8.1  t1m  500 F  (e  1)  (e 8.60.25  1)   900 .83 F  2 (8.6)  28  1  

 (1  e 14.33 )  Ln  1  1 14 .33  1   L   x2 m  Ln (1  e )   2   2 L 14 .33  x2 m  0.186 ft 2

7.274  10 4  0.186 14.331  14.330.186 x2 m  500 F  ( e  1 )  ( e  1 ) 2  (14 .33)  28  1 

t2 m  509 .41 F qt q0 13 .103  10 5  L  (1  e )   (1  e 8.6 )  1.52335  10 5 A  8.6 q1t Btu  1.52332  10 5 [ans] A hrft2 q2 t q 7.274  10 4  L  2 (1  e )  (1  e 14.33 )  0.51  10 4 A  14 .33 q2 t Btu  0.51  10 4 [ans] A hrft2 1

2

Chapter 7 Heat Conduction in Reactor Elements

7 1 known: Analysis : q 

T T x y x y T T  1  k  1 ny n  k  1 nx n 2 2 2 y 2 x

y   x  h  1  1(T f  Tn )  0 2  2  x 2 1 1 1 1 (x  y )  q  kTny  kTn  kTnx  kTn  hxTf  hxTn  0 4 2 2 2 2 2 x 2hx 2hx  q  Tny  Tnx  T f  Tn  Tn  Tn 2k k k x 2 q hx   , Bi   t g    t g  Tny  Tnx  2 BiT f  (2  2 Bi )Tn 2k k   T  T  2 BiT f t g  Tn  ny nx  [ans] 2  2 Bi 2  2 Bi

72 known: Analysis : T  Tn x T T  x y  q   3    1  k   1  ny  k  y  1  nx n 2 2  2 y x  T  Tn y T T y   x  k  x  1  ny k  1  nx n  h  1   1(T f  Tn )  0 y 2 x 2  2  3qx 2 1 1 1 (x  y )   kTny  kTn  kTnx  kTn  kTny  kTn  kTnx 4 2 2 2 1  kTn  hxTf  hxTn  0 2 3qx 2 2hx 2hx   Tny  2Tnx  2Tny  Tnx  T f  Tn  2Tn  2Tn  Tn  Tn 2k k k x 2 q hx   , Bi   t g    3t g  Tny  Tnx  2Tnx  Tny   2 BiT f  (6  2 Bi )Tn 2 k k   T  Tnx  2Tnx  Tny   2 BiT f 3t g  Tn  ny  [ans] 6  2 Bi 6  2 Bi

73 known: k  1.085

Btu Btu ; t f  1000  F ; q  1.5  10 7  hrft F hrft3

Analysis : temprature s at m - s related by one dimentiona l heat flow  tempratur e at m is maximum  t m  4500  F 0.25 2 ( ) (1.5  10 7 ) 0.5 qx 2 x   0.25in ; t g   12  t g  3000 .19  F 2 2k 2  1.058 Heat balance for point m; t t t m  s s  t g  4500  t s  3000  t s  1500  F 2 Heat balance for point ; q g  q yi  q yo  0  q  x 

t t x  L  x  L  k  m s  h  x  L  (t f  t s )  0 2 x

x 2  k (t m  t s )  hx(t f  t s )  0 2 x 2 q  (t m  t s ) hx 2k    Bi ts  t f k  q

2

 0.25    12   7 1.5  10   1.085  (4500  1500 ) 2  Bi   12 1500  1000 Bik h x 12  1.085 Btu h  624 .96  0.25 hrft2 F 12 for point 1; q g  q xi  q xo  q yi  q yo  0 q  x  x  1  0  k  x 

t t t t t 2  t1  k  x  3 1  k  x  3 1  0 x x x

 q x 2  k (t 2  t1 )  2k (t 3  t1 )  0  t1 

2t 3  t 2 2  t g 3 3

for point 2; (t  t ) x 2 x (t 4 t 2 ) )  k  x  1 2  k    h  x  (t f  t 2 ) 2 x 2 x x t 4  t 2 k  0 2 x x  q ( ) 2  k (t1  t 2 )  hx(t f  t 2 )  k (t 4  t 2 )  0 2 t1  t 4  Bit f t g  t2   2  Bi 2  Bi q   (

for point 3; t t x 2 x t 4  t 3 k ( )  k  x  1 3  h  x  (t f  t 3 )  0 2 2 x x x 2 k    q  (t 4  t 3 )  k (t1  t 3 )  hx(t f  t 3 )  0 2 2 2t1  t 4  2 Bit f 2t g  t3   3  2 Bi 3  2 Bi q  

for point 4; (t  t ) x 2 x (t 3  t 4 ) ) k   k  x  2 4  h  x  (t f  t 4 )  0 2 2 x x t 2  t 3  2 Bit f t g  t4   2  2 Bh 2  2 Bi q   (

t 2  2t3   t1  3  2000    3t1  t 2  2t3  6000  t  t1  t 4  1071 .4   t  14t  t  15000  2    1  14 2 4      t3  2t1  t 4  1111 .1  2t1  27 t3  t 4  3000     t 2  t3  26t 4  27000  27   t t t 4  2 3  1038 .5  26  

t1  3402 .51  F   t 2  1396 .33 F

t 3  1405 .6  F   t 4  1146 .22  F 

74 known: Analysis : (x  y ) qs  hxTs  T f  q3  hxT3  T f 

q4  hxT4  T f  q2  hxT2  T f  Method1;   2qs  2q3  2q4  q2 qtotal   2hxTs  T f   2hxT3  T f   2hxT4  T f   hxT2  T f   qtotal   hx2Ts  2T3  2T4  T2  7T f   qtotal

0.25  3000  2811 .2  2292 .44  1396 .33  7000  12    32549 .61 Btu  qtotal hrft     624 .96   qtotal

Method2; 0.25    q  As  q  5  x   1.5  10  5   qtotal   12  2

   32552 .1 Btu  qtotal hrft  

7

2

75 known: Analysis : 2

 0.25  10    2 qΔx 12   Δt g    20 2k 2  10 .85 6

2t 2  2t s 1  t g 4 2 t t t t 1 t 3  2 8 4 s  t g 4 2 t t t t 1 t 5  4 6 10 s  t g 4 2 2t  2t8 1 t7  2  t g 4 2 t t t t 1 t 9  8 10 12 4  t g 4 2 2t  2t12 1 t11  8  t g 4 2

n 1 ) t1  n3 ) n5 ) n7 ) n9 ) n 11 )

t1  t3  t5  t 7 1  t g 4 2 t t t t 1 n 4 ) t 4  3 9 5 s  t g 4 2 t t t t 1 n 6 ) t 6  s s s 5  t g 4 2 t t t t 1 n 8 ) t 8  7 9 11 3  t g 4 2 t  t  2t s 1 n 10 ) t10  9 5  t g 4 2 t  t  2t s 1 n 12 ) t12  9 11  t g 4 2 n 2 ) t2 

After 8 steps; t1  435 .06 ; t 2  450 .51 ; t 3  452 .85 ; t 4  446 .74 ; t 5  434 .75 t 6  418 .69 ; t 7  472 .14 ; t 8  473 .04 ; t 9  458 .77 ; t10  433 .38 t11  467 .41 ; t12  441 .55

76 known: 0.48in  0.24in ; q  2  10 6 h  100

Btu ; hrft3

Btu Btu ; k  10 ; Δ  Δy  0.12" hrft2  F hrft F

Analysis:  . qgen.  qconv  q  a  b  h  2  (a  b)  (Ts  T f ) 0.24 0.48  0.24 0.48    100  2      (Ts  600 ) 12 12 12 12     Ts  733 .33 F  2  10 6 

2

 0.12  2  10    qx 2 12   t g    10  F 2k 2  10  2t 2  2t s t g 2t 2  2  733 .33 10  t   t    1  1 4 2 4 2   t  t1  3t3  t g t  t1  3  733 .33  10 2  2  4 2 4 2  t1  741 .9  F [ans] ; t 2  740 .5 F [ans] 6

77 known: Analysis : t t x y y  k  1  n  x n 2 2 2 x t n  y  t n x x y k 1  q   1  q  1  0 2 x 2 2 x  y qx 2 x   t n  x  t n  t n  y  t n  2q 0 2k k t n  x  t n  y qx 1 tn    t g [ans] 2 k 2 a ) q 

t t x y y  k  1  n  x n 2 2 2 x t n  y  t n x y k 1  q  1  0 2 x 2 x  y qx 2 qx   t n  x  t n  t n  y  t n  0 2k k t n  x  t n  y qx 1 tn    t g [ans] 2 2k 2 b) q 

78 known: k  2

Btu 6 Btu    ; q  10 ; x  y  0.24in hrft F hrft3

Analysis : q  a  b  q  (2a  b) 10 6 

0.48 0.96 0.48 0.96   4 Btu   q   2     q  2  10 12 12 12 12  hrft2 

t 6  500  F 2

 0.24  10    qx 2 12   t g    100  F 2k 22 0.24 4 2  10  qx 12  200  F  k 2 2t 2  t 7  t 4 1  t   t g 1  4 2  t  t1  t3  t5  t8  1 t g 2 4 2  t3  2t 2  t 6  t9  qx  1 t g  4 k 2  2t  2t5 qx 1 t 4  1   t g  4 k 2  t  2t 2  t 4  t 6  qx  1 t g 5 4 k 2  2t  2t8 1 t 7  1  t g 4 2   2t  t  t 1 t8  2 7 9  t g 4 2   t3  t8 qx 1 t    t g 9 4 k 2  6

79 known : Analysis : t n   y  t n  t t x y y x   1  k   1 n   x n   k   1 2 2 2 x 2 x  x y   h  t f  t n    0 2   2 q x 2 2hx hx     t n   x  t n   y  t f  tn  2  2  2k k k   t n   x  t n   y  2 Bix t f t g  tn    [ans] 2  2 Bi 2  2 Bi a ) q  

x y  2 2  1  k  x  1 t n  y  t n  h y  x 2  y 2 t  t   0 b) q   f n 2 2 y  2  q x 2 2hx hx     t n   x  t n   y  t f  tn  2  2  2k k k   tn 

t n  y  (1  2 2 ) Bit f 1  (1  2 ) Bi



t g 2  2(1  2 ) Bi

[ans]

7  10 known: Analysis : ‫; وتر‬ q 



t n  y  t n t t x  y  1  k  y  1  n  x n  k  x  1  2 x y



 h x 2  y 2  1 t f  t n   0

qx 2 2hx x  y    t n  x  t n  y  t f  2t n  2k k t n  x  t n  y  2 Bit f t g  tn   2  2 Bi 2  2 Bi

2hx tn k

t 1  518 .525 ; t 2  542 .623 ; t 3  540 .631 ; t 4  525 .549 ; t 5  504 .945   t 6  626 .160 ; t 7  609 .697 ; t 8  530 .506 ; t 9  541 .491

7  11 known: Analysis : a  b  fx  fy 37 s; x  a   x 2  2x   a  ( 5  1)x a fx   f x  f y  5 1 x t g  160 F 2

2

Ts  800 F 2t2  2t5  320  t  1 4  t  2t6  t1  t3  320 2 4 t1  1238 .7 F  t  t  2t7  320   t3  2 4 t  1195 . 5 F 2  4  t3  1094 .8 F  2t8  t3  1120  t4   4  t4  1224 .3 F    2t6  t1  1120 t5  t5  1121 .8 F 4  t  1064 .3 F  6 t2  t5  t7  1120  t6  t7  819 .8 F 4  t  1341 .1 F 8 t6  t8  t3  1120  t  7 4  5  1 t7  2 5  1 t4  1600  160 5  Eq .( 7  26 )  t  8  2 5









7  12 known: Analysis : 

sss ; 3x   x  2x  a   f  2 2  2 2

2

2

1



y

sss ; 3x   2x   2x  a  f   f   5  2 2

2

2

2

x

y

qx 2 t g   200  F 2k t s  600 F 2t2  t4  1000  t  1  4 t1  916 .796  F    t  t1  t3  t5  1000 t  883 . 414 F 2 2   4 t3  774 .545  F  Eq.(7  27 )  t  405 .71  0.226 t  0.248 t  3 6 2  t4  900 .357 F   t1  2t5  1000 t4  t  842 .315  F 4  5 t6  662 .602  F  Eq.(7  27 )  t5  270  0.226 t6  0.226 t4  0.248 t2   Eq.(7  26 )  t6  509  0.095 t3  0.095 t5

7  13 known: slab 1  2in ; q  10 6

Btu ; t s  1000  F 3 hrft

Analysis : 2

 0.25  10    qx 2 12   a ) t g    t g  200  F 2k 2  11 .085  t s  1000 F 6

t1  1717 .5 F 4t1  2t 2  2t5  400  4t  t  t  2t  400 t 2  1693 .4  F 2 1 3 6   t3  1607 .4  F 4t3  t 2  t 4  2t 7  400    4 t  t  2 t  1400  4 3 t 4  1411 F 8    4 t  2 t  t  1400 6 1  5 t5  1541 .6 F 4t 6  t5  t 7  t 2  1400 t  1524 .4  F  6  4t 7  t 6  t8  t3  1400 t 7  1462 .5 F 4t  t  t  2400 t  1318 .4  F  8 4 7 8

b)   t  1000 3 q   ( x, y )  (a 2  x 2 )(b 2  y 2 ) 2 2 4 k (a  b ) 3 q  t ( x, y )  (a 2  x 2 )(b 2  y 2 )  1000  F 2 2 4 k (a  b )  0.25    t 5  t  0,   1729 F  12   0.75   t 4  t ,0   1420  F  12  0.25 10 6  1.085 720   12  47914 .27  Btu  q1   2   hrft  0.25  2 12 0.25 10 6  1.085 420   12  32290 .27  Btu  q 2   2   hrft  0.25  2 12

Chapter 8 Heat Conduction in Reactor Elements

8 1 Known : t 0  850  F ; t f  200  F ; h  2

Btu hrft2 F

Find : necessary time for the ball to cool down to 300  F Analysis:  t f  t1   CV  t f  t1 ( )    Ln     Ln  t t  hA t  t  f 0   f 0 CV C V    1 1 1  1 1   1  hA 1 h  A1  Eq.(8  2a )   

t f  t1 850  300   575  F 2 2 Cp  0.121 for steel :      501 .2  4 3 R  0.121  501 .2 0.121  501 .2  3 D     12    0.83533 2 2 2 6   4R     200  300     0.8353 Ln    1.5635 hr  200  850     1.5635 hr[ans] t

82 Btu ;   700 lbm 3 ; ft lbm Btu t0  t f  100  F ; t2  500  F ; h  200 hrft2 F  Known : D  1in ; C  0.04

Find : a )time constant of the fuel b)time it take the fuel to reach 499  F Analysis :



C1 1V1 C1 1  V     hA1 h A

pipe   

C1 1  D    h 4

0.04  700  1  3    2.9  10 hr 200  12  4   t  t ( )   1    Ln f 1   2.917  10 3 Ln  t  t 400   f 0     0.017 hr  63 sec[ans]



83 Known : Analysis : qin  qout  q gen 

dEs dt

Ax    t t n x  t n t n x  t n  t n n  kA  kA  qAx  C V x x       kt n kt n  x kt n  x kt n      qx  Cxtn    Cxtn x x x x 2k   k   t n  x  tn x   q  Cxtn     Cx  t n  x  x C   k 2k   k  q        Fo    t  1  t  t  t    n n 2 x 2 Cx 2 Cx 2 n  x n  x C  Cx  q  t n    1  2 Fo t n  Fot n x  t n x   [ans] C

84 Known : 1.25in  0.25in ; t 0  1000  F ; q  1.5  10 7

Btu ; t s  1000  F ; 3 hrft

Btu lbm Btu ;   740 3 ; C  0.06  hrft F ft lbm Analysis: 7 0.125 1 . 5  10 ( ) 2 t t t t qx tg 12 x  0.125 ; t g    750  F ; tn  ny nx ny nx  2k 2  1.085 4 2 k  1.085

t s  t s  t s  t2 750  t   1  4 2  t  t s  t s  t1  t s  750 2 4 2  t  t  t  t 750  t3  s s 2 4  4 2  t s  t s  t3  t5 750   t4  4 2  t  t s  t s  t4  t4  750  5 4 2

t2   t   1125 F 1  4  t1  1281  F t  t1  t3  875  F   2 4 t2  1624 .2 F  t t   t3  2 4  875  F  t 3  1716 .1 F 4  t  1240  F t3  t5  4  t   875 F 4 t5  1745 .2 F 4  t  t4  875   5 4

99 .5 t5  1736 .275  F 100 1  F0 x 2 x 2 C F0   F0      4 x 2  4k 2  0.125     740  0.06 12       1.11  10 3 hr 4  1.085    1.11  10 3  3600  4 sec

t5 

85 Known : Analysis : t ny  t n t nx  t n y x  q  x   1  k  y  1   k  1 2 x 2 y t ny  t n t n   t n x x   k  1  h  y  1  t f  t n     C  y   1  2 y 2  q 2k  2k  k  k    t  t  t  t C Cx 2 nx Cx 2 n Cx 2 ny Cx 2 n k  k  2h 2h      t  t  t  t t t n   y n f Cx 2 Cx 2 Cx Cx n n n qx 2  q   Fo  2 Fot g    C k    2h 2hkx  hx k   2  2 BiFo   2 k Cx 2  Cx Ckx 

 t n   1  4 Fo  2 BiFo t n  2 Fotnx  Fot ny  t ny   2 BiFot f  2 Fot g [ans]

86 Known :

8  7) Known : Analysis : t n y  t n t n x  t n x x  q   y  1  k   1   k  y  1 2 2 y x t n y  t n t n    t n x x  k   1  q  y  1    C   y  1  2 y 2  q k  k  2k  2k  k    t  t  t  t  t 2 n  y 2 n 2 n  x 2 n C Cx Cx Cx Cx Cx 2 n  y k  2q   t   t n  t n   2 n Cx Cx qx   2q 2qkx   2 Fo   2  C  x  Ck  x k    t n    1  4 Fo t n  2 Fotn x  Fot n y  t n y   2 Fo

qx  2 Fot g [ans] k

88 Known :

89 Known : Analysis : t ny  t n t nx  t n 3 x  q   x  y  1  k   1   k  y  1  4 2 y x t ny  t n t nx  t n y  x y   k  x  1   k  1  h    t f  t n  y 2 x 2   2 t n   t n 3    C   x  y  1  4  q 2k  2k  4k  4k    t  t  t  tn 2 ny 2 n 2 n x C 3Cx 3Cx 3Cx 3Cx 2 4k  4k  2k  2k   t  t  t  tn 2 n y 2 n 2 n  x 3Cx 3Cx 3Cx 3Cx 2 4hxk 4hxk      t  tn  tn  tn f 3kCx 2 3kCx 2 4 2    t n   1  4 Fo  BiFo t n  Fot ny  t nx  2t nx  2t ny  2 Bit f   2 Fot g [ans] 3 3  

8  10 Known :

8  11 Known :

8  12 Known :

8  13 Graphical techniques.

8  14 Known :

8  15) 0.6 Btu ft ; k  1.25 ; t f  1000 F ;  12 hrft F lbm Btu ; ρ  676 3 ; C  0.06 ft lbm  F

Known : r0  0.6in  q  6  10 6

Btu hrft 3

Analysis: 

SteadyStat e  t  0,     n 1

n

2

2 Bi 1  2 2 J 0 ( n ) n  Bi





0.6 1  2.1795  hr   12 Bi    10  at Bi  10  2  5.0332 k 1.25   7.9569   3  2  10 1 2  10 1  t  0,       2 2 2 2 2 2 2.1795 2.1795  10 J 0 (2.1795) 5.0332 5.0332  10 J 0 (5.0332) 250 











2  10 1   0.3384  0.0560  0.0191  0.413 2 2 7.9569 7.9569  10 J 0 (7.9569) 2



t 0,    t  0,  



qr0 tf k 2

2

 0.6  6  10    12    t 0,    0.413   1000  5961.94 F 1.25   2 Bi 1 n 2   t 0 ,    1  e  2  2 2 J 0 ( n ) n 1 n n  Bi  k 1.25       4500  1000  0.29166 t 0 ,   t 0 ,   t  f 2 2  qr0 0 . 6    6  10 6      12  6























 0.29166  0.3384(1  e 2.1795  )  0.056(1  e 5.0332  )  0.0191(1  e 7.9569  ) 2

        0

2

2

Chapter 9 Heat Conduction in Reactor Elements

9  1) Known : cubical 20 ft ; Natural Uranium ; T  80 F ; T  190 F i

Analysis : D  D m  m  m   V A   V     4   4 Kg  at 80 F   300 K   1.1614   i m3 i

o

i

  1.1614 i

i

c

i

2

2

hole

e

i

Kg lbm  0.06243  0.0725 m3 ft 3

   3  2  1  2  lb m  0.0725  15        0.04745 4   12  

 12  

sec

lb lb  170.83 sec hr  4A  V   p VD   Re   m  0.04745

c

e





4m  D  D  hole

f

q  m C (T  T ) P

0

i

C (at 135 F )  1.008 kJ 

P



kg K

q  170.83  1.008  0.23886(190  80)  4525.3

Btu hr

o

  4 D  D  2 4 D    D D D   D  D  12 2

2

hole

f

e

hole

hole

f

e

f

4  170.83 Re   13178.781 1 3      198.76  2419.1  12 12   flow is turbulant Pr at 135 F  0.7031 Ditus Bulter  Nu  0.023 Re Pr W k(at 135 )  28.52  10 m.K hD kNu Nu  h k D 0.8

0.4

3



e

e

 Nu  0.023(13178.781) (0.703)  39.48 28.52  0.57782  39.48 Btu h  3. 9 [ans ] hrft F 1    12  q 452.3 Btu q    41484.9 1  V hrft     20   2  12  0.8

0.4

2

2

fuel

3

0.602  10 # N   0.0072  18.5  3.413  10 235.04 cm 528 σ  0.8862  577.1 10   481.9  10 cm 595 q q  G  N     G N 24

20

ff

3

 24

 24

f

max

max

f

f

ff

max

max

f

f

ff

41484.9 180  481.9  10  3.413  10  1.5477  10 #   9  10 [ans ] cm .sec

 

 24

max

10

max

2

20

8

2

92 known:

9  3) Btu ; t  5000 F ; hrft Btu V  375 fps ; P  6atm ; t  1000 F ; k  35 hrft F Known : UO ; D  0.5in ; q  2  10 2

f



6

m

3



Bulk

bulk

g



Analysis: a) 5000  2000  3500 F 2 Btu at 3500 F  K  1.1 hrft F

t  2000 assumed  t 



-

s1

f





 0. 5  2  10     qR 2  12   t t   t  5000   4803 F 4k 4  1.1 2

6

2



s

m

s

f

D D D e

2

1

 RC qR  1 1 t t  Ln    2 k R h( R  C )     0.5  2  10   1  1 2  12   0.25  0.1   4803  1000  Ln  Ln   0 . 25  0 . 1 2 35 0 . 25     h     12    Btu  h  3.9181 [ans ] hrft F 2

g

s

Bulk

g

g

2

6

2

b)  D  4  V m 4  G    V D D 4 Fluid mass - flow rate note : C  cross  sec tional area of channel D  0.0211  2.3 L  h    Pr    CG  D G      D D D 2

e

2

2

e

e

e

0.14

2 3

w

0.2

m

e

p

m

e

2

1

9  4) Known : t  400 F , m  100000 

f

lbm hr

Analysis : h  3MeV  

( iron )

 0.282cm

1

q h  0.282  3  10  8.46  10 13

12

0

MeV Btu  1.309  10 sec .cm hrft 5

3

3

 k  t  t   1 1  e   x  3.152cm  0.10342 ft Ln    qL L  t  425    t  415 t  t  10 KA(t  t ) qA  e  1  q   1   L   L  1

x 

 L

m

i

0

m

0

max

i

max

i

 L

i

o

o

x o

Btu hrft q Btu q  h(t  t )  h   5979 t t hrft F

t  t  q  8968.5 i

x o

o

2



w

2



w

D  e

4A 1  D  2a  ft P 3 c

e

hD  D V  Nu  0.023 Re Pr   0.023  Pr k    0.8

0.8

0.4

    hD V     Pr     0.023k 0 .8

e

0.2

e

 0.4

e

0.4

   

lbm lbm  μ  0 . 335 , ρ  54 . 218  hrft ft if P  2000 psia   C  1.062 Btu , k  0.388 Btu  hrft F hrft F 3

P

 V  5284.92

ft [ans ] hr

2



9  5) Known : m  100000 lbm

; t  400 F ; t  1049 F 

hr



f1

f2

4 in  0.135in ; 3%Uo ; clad  0.005 ; ρ  10.5 gr 2

UO2

cm

2

D  8 ft ; H  8 ft core

core

Analysis:

  0.289A Hq q  G N   t

s

co

N

f

ff

ff

f

co

co

A  rfi M

for 3% Uo enriched : N  2

v

ff

fm

ff

f 

rM  ( 1  r)M 0.03  235.493  0.97  238.05  rM  ( 1  r)M  M 0.03  235.0493  0.97  238.05  2  15.99 ff

nf

ff

nf

O2

6.023  10 f  0.8815  N   10.5  0.03  0.8815  1 235.0439 # N  7.115  10 cm ( 0.135  2  0.005 )( 4  2  0.005 ) A   3.464  10 ft 12  12 t t 400  1049 t    724.5 F 2 2 at t  724 F  358 C  f(t)  0.92 23

ff

20

ff

3

3

2

s

f1



f2

fm





fm

t 528  577.1  10  0.92  0.8862 t 1182   3.141 10 cm σ  σ f(t)0.8826 f

o

fo

 22

f

2

 24

Q  m C Δt t

p

C  1.248 p

f

Btu lbm.F

at 10 atm

Btu 1 hr Q  0.289nA Hq  0.2389nA H G N σ φ  1.5477  10 2 Q  100000  1.248( 1049  400 )  80995200 t

8

t

1&2  Q

t

s

co

co

f

ff

f

 80995200  1.995  10 φ

 φ  φ  4.06  10 max

s

6

co

13

# [ans] sec .cm 2

co

96 known:

9  7) Known: UO ; d  0.5 in ; gap  0.003 in ; S  0.7 in 2

Particuler sec tion: 520 F and 15 fps ; q  5  10 Btu 

7

Analysis: 4A 4  ( S  R ) D   P 4 S  8 R  2R 2

2

c

t

e

t

t

  0.5  2 4   0. 7     0.03  0.003     2   144  D  0.283  0.7   0.283  4  8  2       12 12 2     D  0.034 ft 2

e

e

V  15 fps  15  3600 ft

hr

 0.7  s  ft   S S 12   1 . 24  C  0 . 042  0.024   D  D  0.566 ft  D   12 C  0.042  1.24 - 0.024  0.02808

hrft

2

hD k  D V  Nu  C Re Pr   C Re Pr  h  C   Pr k D    t  520 F  k  0.3475 ; Pr  0.851   P  2000psi    43.527 ;   0.249 0.028  0.3475  0.0343  15  3600  43.527  h   0.351 0.0343 0 . 249   h  7452.5 Btu hrft F q  (R L)  2R  L  h  (t  t ) 0.8

1 3

0.8

1 3

0.8

e

e

e



0.8

1 3

2

2

t

w

f

 0.25  5  10   qR 12    581.7 F t t   t  520  0.283 2R h 2  7452.5  12 t  t 520  581.7 t    550.85 F 2 2 TableD  1a  1000 psi  P  1100 psi 2

7

2



w

f



w

t

f



w

film

sat

at average  1055psi [ans]

1 3

9  8) Known : k  25.4 k  45 f

c

Analysis: a) (t  t )4k qR (750  600)  4  25.4 t t   q   4k R  0.45     2 12  Btu  q  4.335 10 hrft 2

m

m

s

s

f

2

2

f

7

3

qR t t  2 s

2

f

 1 RC  1  Ln    k R h ( R  C )     c

  0.45    0.45   4.335 10     0 . 025   1 1  2 12    Ln 2   600  500  0.45  45  2  h 0.025   0.45           2 12  2      Btu h  4452 [ans ] hrft F 2

7

2

b) qR 1 t t   2 h( R  C ) 2

c

f

 0.45  4.335 10    2 12   t  500   2 2

7

1  0.45 0.025  4452   12   2 12 t  582  P )  1347.5 psia[ans ] c

c

w

min

9  9) Known : P  2000psia ; d  0.55in ; d fuel

q  4  10 Btu 7

hrft

3

; t

overall

 0.6 in ; S  0.66 in ;

 540 F 

coolant

Analysis : 4A 4 s  R 1 D    P 4 s  8 R  2R 12 2

2

c

t

e

t

t

D  0.024 ft e

S  0.66 S     1.1  D  0. 6  D S C  0.042  0.024  0.0222 D Nu  C Re Pr 0 .8

1 3

hD  D V   c    C    k     k  0.8

e

1 3

p

e

 P  47.181lbm ;   0.239 lbm ft hrft  Water at 200 psi & 540 F   k  0.3371 Btu hrft F ; C  1.225 Btu lbm R  3





k D c   h  C  V   D     k  0.8

e

0.8



p

1 .3

p

e

1 3

0.0222  0.3371  0.0222  47.181   1.225  0.239  h    V 0.024 0.239    0.3371  0.8

0.8

h  1.0332V

0 .8

2π(R  C)  L  h  (t  t )  πR Lq 2

w

f

πR Lq R q h  2π(R  C)L(t  t ) 2(R  C)(t  t ) 2

2

w

f

w

f

 0.275     4  10 12  1.0332  V    8134.3  0.6  1.0332     50  12  V  77250.8 ft  21.46 fps[ans] hr 2

7

0 .8

9  10) Known: 300 F ; 10atm ; V  350 fps ; D  1 in ; T (real)  280 

fs

Analysis; P  15 atm k  0.11    0.686  T  300 F     0.0605 P  0.0718 1 350  3600     0.0718 VD  12  Re    124611.6  0.0605 k Btu h  0.023 Re Pr  h  311.371 D hrft F r



0.8

0.4

2

q h(t  t )  311.371  (280  300)  6227.42 1

w

f

Btu hrft

2

In the second step :   1  T  T 1  M  2   2

fs

f

  1.4    (Rg T )  1.4 

1545.08   4.17  10  760  4   V 350  3600   13091346.9  M    0.09625  13091346.9 1 2

c

f

8

 1.4  1  T  7601   0.09655   761.408 R  301.408 F 2   T T F   0.89  T  T  0.81T  T   T  301.253 F T T 2





fs

fa



f

R

fa

fs

f

fs

f

fa

f

q  h(t  t )  311.371(280  301.263)  6617.64 2

w

fa

q  q  q  6227.42  (6617.46)  390.04 1

2

Btu hrft

2

Btu [ans ] hrft 2

Chapter 10 Liquid Metal Coolants

10  1 Known : Liquid metal ; Pr  o ; k  50 Btu d tube  1in ; t w  1000  F ; t(at y 

ro

2

hrft  F

; Re  1 10 6 ;

)  900  F

Analysis : (a) Base on Fig 9 - 3b for 

t t y 1   w  0.5 2 ro t w  t fc

1000  900 1   t fc  800  F [ans ] 1000  t fc 2

(b) t  t w  ( tf 



ro

o

t (r )dv



ro

dv

o

tf 

r  1)(t w  t fc ) for Pr  o ro



ro

o

r (t w  (( )  1)(t w  t fc )2rdr ro



ro

o

2 [(t w  t fc )(( 

2rdr

2 2 r3 r )  ) 2  t w r 2 ]oro 2(t w  t fc )( ro  ro )  t w ro2 3ro 2 3 2  2 ro ro2

 (t w  t fc )

2 1  t w  t w  t fc 3 3 3 2 1  t f   1000   800  933.3 F [ans ] 3 3 

(c) Base on equation 10 - 3  Nu  5  0.025 Pe 0.8 Pr  o  Pe  Re Pr  Nu  h

hD 5 k

5  50  3000 Btu hrft 2 F 1 12

q w  h(t f  t w )  3000(933.33  1000)  200010 Btu

hrft 2

[ans ]

10  2 Known : t i  932 F ; D  1 in ; h  5000 Btu



hrft 2 F

Analysis :    52.07 lbm 3   0.588 lbm fthr  ft   ti  932 F    Pr  0.0046 k  38.61 Btu   hrft F   hD Nu   7  0.025 Pe0.8 k Pe  Re Pr 1 5000  12  7  0.025  Pe0.8  Pe  532  38.61 VD 4m  Re Pr  532  Re  115652.1    D  m  4450 lbm [ans ] hr 

10  3 known : D fm  0.4in ; H  6ft ; D pin  0.45in ; S  0.54in ; V  20 fps ; T fc  700 c ; (t c  t f ) max  25 F Analysis : S 0.54   1.2 D 0.45 c p  0.3003 Btu  lbm F      48.88 lbm  3   ft    from Table 10 - 2  for Na at 700 c   0.45 lbm  fthr   k  34.1 Btu  hrft F    P  0.004   r  4A De  c p 2

2 1 3 0.45 1 2 D 1 ( (0.542  ) s sin 60   1 2 2 2 4 2 4 De  4  4 D  0.45 3 0.45 12 12 3s  6  3 D (3  0.54  6   ) 2 6 2 6 De  0.01595 ft

Re 

VDe 48.88  20  3600  0.01595   124743.5124  0.45

Pe  Re Pr  0.004  124743.81  500 from figure 10 - 4  for Pe  500 & S

 1.2  Nu  10 D Nu k 34.1 Btu h  10   21379.3 De 0.01595 hrft 2 F   2 ( R  C )h(tc  t f ) L  q  R 2 Lqco    qco

2(

2( R  C )(tc  t f )h R2

0.45 )  25  21379.3 Btu 2  12  72.155  106 0 .4 2 hrft 3 ( ) 2  12 2

 0.4  Btu   2 2 12     Qt   Qt  As Hqco  6  72.155  240517.2375 hr [ans ] fuel pin   4

10  4 Known : 0.3 in gap ; fuel material thickness  0.25 in ; V  20 fps ; t f  1110  F ; t cmax  1160  F Analysis : b  a  De  2a  2  Re 

VD 

0.3  0.05 ft 12

   50.51 lbm 3  ft     lbm   0.503  fthr table  k  36.24 Btu  hrft F   Pr  0.0042    50.51 20  3600  0.05 0.505  Re  357944.8819  Re 

Pe  Re Pr  1503.37 Equation (10  6)  N u  5.8  0.02Pe 0.8  12.76 hDe Nuk 12.761 36.24 h   9249.3 k D 0.05 q ( A  S )  hA(t c  t f ) Nu 

h 9249.3  q   (t c  t f )  (1160  1110) 1 0.25 s  2 12 Btu  q   44.39664 10 6 [ans ] hrft 3

Chapter 11 Heat Transfer with Change in Phase

11  1 Known : t sat  212  F ;   4.03

1bf ; D  4.68  10 -3 in ft

Analysis :   ft 3 v  26 . 8 g    lbm t  212 F    h fg  970.3 Btu  lbm     h fg  J ; J  778.16 ftlbf Btu 2t sat t  t sat   g rc  t  t sat 

4v g t sat

 Dc

4  4.03  26.8  212 4.68  10 3 778.16  970.3  12   t  311.03 F [ans ]  t  t sat 

11  2 Known : Saturated pool boiling ; P  300psia ; h radiation  200

Btu hrft 2 F

Analysis:  ft 3  t  t  417 . 33 F v  0 . 0189 sat f f  lbm  3 P  300psia  v g  1.5433 ft h f  393.82 Btu lbm lbm  h fg  809 Btu h g  1202.8 Btu lbm lbm   f   g 0.6 g 0.25 Eq.(11  5)  qc  143 h fg  g ( ) ( ) g gc 1   1    1  qc  143  809    0.0189 1.5433  1 1.5433     1.5433   Btu  qc  1.044  10 6 hrft 2 qc  hr (t surface  t sat. )  t surface 

qc 1.044  10 6  t sat   417.33 hr 200

t surface  5638.72 F [ans ]

0.6

1

      

11  3 Known : P  100 psia ; g 

1 g ; S.F.  2 ; A  4in  0.25in 6

Analysis:   ft 3  t  t  327 . 28 F v  0 . 01774 f sat f  lbm  P  100psia    3 ft h fg  888.8 Btu  v g  4.432 lbm lbm    f   g 0.6 g 0.25 Eq.(11  5)  qc  143h fg  g ( ) ( ) g gc 1 1      1  qc  143  888.8    0.01774 4.432  1 4.432     4.432   Btu  qc  501817 hrft 2 q 501817 Btu  qallowable  c   250908 S.F. 2 hrft 2 A 2(a  b)  L    qallowable  qallowable  qallowable V ab L 4.25 2( ) 12   qallowable  250908  4 0.25  12 12 Btu   qallowable  2.56  10 7 [ans ] hrft 3

0.6

1  ( ) 0.25 6

11  4 known : P  2000 psia ; t b  500 F ; D in  1in ; D o  1.2in ; t c  0.03in ; V  30fps Analysis : P  2000 psia  t sat  635.82 F ; t b  500 F  Subcooled flow boiling G Eq.(11  11)  q  C ( 6 ) m (t sat  tb ) 0.22 10 t sat  635.82 F       1 lbm 3  ft  P  2000 psia   f 0.0257 c  0.445  10 6    m  0.5  G   V 

1 1bm  30  3600  4.202  10 6 0.0257 hrft 2

4.202  10 6 0.5 ) (635.82  50) 0.22 6 10 Btu  qc  2.68  10 6 hrft 2 qc  Di L qc   (( Do  2Cc ) 2 ) 4 Btu   qBurnout  353.6  10 6 [ans ] hrft 3  qc  0.445  10 6 (

11  5 known : BWR ; P  600psia ; Ti  470 F ; Vi  4 fps ; A fuel  0.0218 ft 2 ; d  2in ; Burnout may occur at center plane Analysis: 2  As H q t  qcenter



   qcenter

qt  2  10 6  Btu   2.882  10 7 2 As H 2  0.018  5 hrft 3

P  600psia  Tsat  486.21 F ; Ti  470 F  Saturatd flow boiling Eq.(11 - 15)  qc(600 psia )  qc(at 1000 psia)  440(1000 - P) Eq.(11 - 14)  qc(at 1000 psia) 1 1 lbm        50 . 5 i f  v f 0.0198 ft 3 Ti  470 F   h  452.7 Btu lbm  i lbm G    V  50.5  (4  3600)  727200 hrft 2 x1  0.197  0.108  10 6 G  x1  0.1184 x2  0.254  0.026  10 6 G  x2  0.235 2 2 ) 1bm 12 m  GA  727200    15865 4 hr P  600 psia  h f  471.6 Btu ; h fg  731.6 Btu lbm lbm q 1 q  m [h f  x h fg  hi ]  x  [  hi  h f ] m h fg (

1  10 6 1  x [  452.7  471.6]  15865 731.6  x  0.0603 x  x1  qc  0.705  10 6  0.237G qc  0.705  10 6  0.237  727200  877346.4

Btu hrft 2

 qc (600 psia )  877346.4  440  (1000  600)  1053346 A DL Btu  q  2.528 10 7 2 V D hrft 3  L 4   Reactor is not safe [ans]  qcritcal

qc  q  qcenter

Btu hrft 2

11  7 Known : Liquid Sodium ; V  20 fps ; T  700 C ;   0 ; S .F .  2 d i  0.5in ; d o  1in ; qout Analysis: lbm hrft 2 liquid metals in flow boiling  Eq.(11 - 19), Eq.(11 - 20) Na at 700 C Table 10 - 2      48.88 

G 48.88  20  3600   381.875  150 2 (L )2   D  4   0.5   12  L  Eq.(11  20)  qc  140 G 0.5 D 0.2 ( ) 0.15 D  0.5   qc  140  (3519360) 0.5     12 

0.2

 4     0.5   12 

qc q Btu  c  125034.96 S .F . 2 hrft 2 A q  q V  di L 4d  q  q 2 i 2  2 do  di (d o  d i2 ) L 4 0.5 4 Btu 12  q   4001118.635 2 2 hrft 3  1   0.5       12   12  Btu   4  10 6  qmax [ans ] hrft 3 q 

0.15

 250069.92

Btu hrft 2

11  8 known : PWR ; P  2000 psia ; Ti  580 F ; q t  2.33  10 6

Btu ; At  2.7 ft 2 ; Af  0.01 ft 2 hr

Analysis : P  2000 psia  Tsat  635.82 F ; Ti  580 F  Subcooled flow boiling c  0.445  10 6  G Eq.(11  11)  qc  C ( 6 ) m (t sat  tb ) 0.22 ;   10 m  0.5  lbm Ti  580 F    44.493 3 ft q tb  TH ; m C p (TH  Ti )  t 2 2 2 qt  TH  Ti  2 2m C p C p (2000 psia,590 F )  1.3945

Btu T  H  600.9 F  lbm F 2

4  10 4 1bm m hr  4  10 6 1bm G  2 Af 0.01 ft hrft 2  4  10 4  6  qc  0.445  10   0.01  10 6   Btu  qc  1.945  10 6 hrft 2 q q ave  t At  q ave 

0.5

    (635.82  600.9) 0.22   

2.33  10 6 Btu  8.63  105 2.7 hrft 2

 qmax q 1.945  10 6  c   2.25[ans ]  qave qave 8.63  105

11  9 known : Vertical tube ; P  1000psia ; V  10fps ; D  1in ; H  12ft ; Tw  530 F Analysis : We can assume vertical furface  Eq.(11 - 27)   t sat  544.61 F  P  1000 psia    l  46.30 lbm 3 ft     2.244 lbm  g ft 3

h fg  649.4 Btu

lbm

k l  0.3273 Btu

hrft F

 l  0.233 lbm hrft

g  84.616  10 6 1

 46.30  (46.30  2.244)  84.616  10 6  649.4  0.3273  4  h  1.6   4  0.233  12  (544.61  530)   Btu  h  630 hrft 2 F 1 Btu q  hAT  630  (   12)  (544.61  530)  28919.3 12 hr 2

1   1 lbm 12 m  1V1 A1   10  3600       440.641 0.4456 4 hr q 28919.3 lbm m f    44.53 h fg 649.4 hr m g  m  m f  440.641  44.53  396.11

lbm ans hr

11  10 Known : P  1psia ; N  400 tubes (40rows,10deep) ; D  1in ; L  10ft ; h w  312

Analysis : a ) n  hw  At (Ts  Tw )  m h fg  m 

nAhw (Tsat  Tw ) h fg

  Tsat  101.74 F  3 P  1 psia  v f  0.01614 ft lbm   l  1.624 lbm hrft 

3 vg  333.6 ft

k l  0.365 Btu

lbm

h fg  1036.3

hrft F

1

  l (  l   g ) gh fg K l3  4 Eq.(11  32)  h c  0.725   ND l (t s  t w )    1 1 1 (  )  32.17  3600 1036.3  (0.365) 3     h c  0.725 0.01614 0.01614 333.6  2.4    400    10 1.624  (101.74  98)    12    Btu  h c  1163.8  2 hr Fft  2.4  2 A  Dl    20  4 ft  12  1163.76  400 12.57  4  (101.74  98)  m  1036.3 lbm ans  m  21111.51 hr b) h c . At (Tsat  Tw )  hw At (Tw  Twater )  Twater  Tw 

1163.8  (101.74 - 98) 312  84  F [ans]

 Twater  98  Twater

hc (Tsat  Tw ) hw

1 4

Btu lbm

btu ; t w  98 F 2 hrft F

Chapter 12 Two-Phase Flow

12  1 Known : L  6 ft ; D  1in ; uniformly heated ; P  600 psia ; Ti  460 F ; Vi  2.5 fps ;  e  0.6 Analysis: h  (h f  xh fg )  hi Ti  460 F  hi  441.4 Btu

lbm

, i 

1 1 lbm  ft 3 vi 0.0196

Tsat  486.21 F  3 v  0.0201 ft lbm  f 3  P  600psia  v g  0.7698 ft lbm  h f  471.6 Btu lbm  h fg  731.6 Btu lbm  0. 1  vg     0.67     vg    1  v f   1 Eq.(12  12)   1    1     v  xe   f    e    1   13.496  xe  0.0741 xe  h  (471.6  0.0741  731.6)  441.4  84.41 Btu 2

1  1 12    m  Vi  i Ai  2.5  3600   0.0196 4  m  2504.46 lbm hr m h qt  m h  qt  L 2504.46  84.41  qt   35233.6 Btu [ans ] hrft 6



lbm

12  2 Known : P  600 psig ; Vi  1.4 fps ; Q t  5  10 6 Analysis : Tsat  488.76 F  3 v  0.7526 ft g lbm  3  P  600  14.7  615psia  v f  0.02015 ft lbm  h f  474.6 Btu lbm  h fg  728.35 Btu lbm  Tin  Tsat  Tsub  477.76 F Tin  477.76 F  h i  461.8 Btu

lbm

q t 5  10   65.79 Btu lbm m 76000 1 h  h f  hi  h  h f  xe h fg  hi  xe  h fg

q t  m h  h 

6

1 65.79  474.6  461.8  0.07275 728.34  xe  7.3%[ans ]  xe 

Vi  1.4 fps -7 P  600 psia Fig 12   S  2.66  xe  0.0728 1 Eq.(12  9)    1 x 1 ( ) x v 0.02015   f S   2.66  0.0712 vg 0.7526 1  0.52987 1  0.07275 1 ( )  0.0712 0.07275   53%[ans]  

Btu lbm ; m  76000 ; Tsub  11 F hr hr

12  3 Known : Boiling sodium ; P  15.4psia ; S  4 ; Ti  1140 F ;  e  0.8 Analysis: for sodium Tsat  2100 R  3 v  2.1656  10 2 ft lbm  f 3  P  15.4  v g  56.185 ft lbm  hg  2312.1 Btu lbm  h f  649.7 Btu lbm  1 Eq.(12  10)  xe  1 1 1 ( )



 

vf vg

S 

 h fg  h f  hg  1662.4 Btu



2.1656  10 2  4  1.542  10 3 56.185

 xe 

1

1  0.8 1 1 ( ) 0.8 1.542  10 3 h  h f  xe h fg  hi

 0.00613

 h  649.7  0.00613  1662.4  498.5  h  161.4

Btu [ans ] 1bm

lbm

12  4 Known : BWR ; A  1 in 2 ; P  9.153psia ; Vi  20 fps ;

 e  0.96 ; Ti  1900 R ; q  2.37  10 6 Btu hr Analysis : Tsat  2000 R  3 v  90.914 ft lbm  g 3  P  9.1533 psia  v f  2.1340  10 2 ft lbm  Btu h f  619.1 lbm  hg  2304.1 Btu lbm  Ti  1900 R  hi  588.8 Btu m   iVi Ai 

lbm

3 ; vi  2.1024  10 2 ft

lbm

Vi Ai vi 2

1 20  3600     12   23782.344 lbm  m  2.1024  10 2 hr q  m h ; h  h f  xe h fg  hi  xe 

 2.37  10 6  1 qt 1  [  hi  h f ]   619.1  588.8  h fg m 2304.1  619.1  23782.34 

 xe  0.0412 1 1 x  1    x  v V (1   ) x    ;   f  S  S   g  (1  x) vg Vf Eq.(12  8)   

(1  0.96) 0.0412 90.914   0.96 (1  0.0412) 2.134  10 2  S  7.62[ans ] S

12  5 Known : BWR ; H  12 ft ; P  1200psia ; sinusoidal heat generation ; H 0  4 ft ; Tsubinlet  27.22 F Analysis : qs 1   H    1  cos 0   qt 2   H  q 1    4  1  s  1  cos    qt 2   12   4 h f  hi qs  q t h f  xe h fg  hi 

h f  hi h f  xe h fg  hi

 xe 



1 4



1 4  h f  hi   hi  h f h fg



Tsat  567.22 F  3 v  0.3619 ft g lbm  3  P  1200 psia  v f  0.0223 ft lbm  h f  571.7 Btu lbm  h fg  611.7 Btu lbm  Tsub  27.22 F  Ti  Tsat  Tsub  Ti  540 F  hi  536.6 Btu

lbm

1 3  571.7  536.6 611.7  xe  0.17214  xe  17.2%[ans ]  xe 

12  6 Known : BWR ; H  6 ft ; P  700psia ; Tsubinlet  23.1 F ; xe  0.06 Analysis : h f  hi qs  q t h f  xe h fg  hi Tsat  503.1 F  3 v  0.6554 ft lbm  g 3  P  700 psia  v f  0.0205 ft lbm  Btu h f  491.5 lbm  h fg  709.7 Btu lbm  Ti  Tsat  Tsub  Ti  503.1  23.1  480  hi  464.4 Btu 

qs 491.5  464.4   0.386 q t 491.5  0.06  709.7  464.4

a )Uniformly; qs H 0   H 0  0.386  6  2.32 ft [ans ] qt H  H B  H  H 0  6  2.32  3.68 ft [ans ] b)Sinusoid ally; qs 1   H 0  1  cos qt 2   H

   

qs 1    H0  1  cos qt 2   6  H 0  2.56 ft [ans ] 

lbm

    0.386 

 H B  H  H 0  6  2.56  3.44 ft [ans ]

12  7 Known : BWR ; H  6 ft ; P  700 psia ; Tsub  23 .1 F ; xe  0.06 ; inlet

  Ce

z

H

sin

z H

Analysis : d a) 0 dz z z    z   z   Ce H sin    Ce H cos   0 H H H  H   z   tan   1 H  H  z m  tan 1 1



6      4  z m  1.5 ft[ans]  zm 

b)  max   ( z m )   max  Ce  tan C 

1

1 

sin tan 1 1

    4

 max e  

sin   4  C  0.65 max [ans]

12  8 Known : BWR ; H  5 ft ; P  600psia ; A  4.25in  0.45in ; Sinusoidal ly ; h subinlet  5 Btu ; Vi  2 fps ;  e  32.9% lbm Analysis : Tsat  486.21 F  3 v  0.7698 ft g lbm  3  P  600 psia  v f  0.0201 ft lbm  h f  471.6 Btu lbm  h fg  731.6 Btu lbm  hi  h f  hsub  hi  471.6  5  466.6 Btu

lbm

1 1 lbm  ft 3 vi 0.0196

 i 

1 4.25 0.45  2  3600   0.0196 12 12  m  4878.83 lbm hr m   iVi Ai  m 

a ) q  m h h  h f  xe h fg  hi 1  0.7698  Eq.(12  12)     1    xe  0.0201 

0.67

 e  0.329

 0.7698     0.0201 1   1      0.329  

 xe  0.0215  h  471.6  0.0215  731.6  466.6  20.66 Btu  q  4878.83  20.66  100778.2834 Btu

lbm

hr 4.25 0.45 V    5  0.0664 ft 3 12 12 q q   1.52  10 6 Btu  15.71 kW [ans ] 3  lit hrft V V

0.1

   

b) q H B  m xh fg  q H B  76741.07 Btu

hr

4.25 0.45   HB 12 12 1 h  H   1  cos 0    2  H   h  xh fg

V 

 

1  H 1  cos 0 2  H

5   0.241      5  0.0215  731.6

H 0

 58.83  H 0  1.634 ft 5  H B  H  H 0  H B  3.366 ft

 V  0.0447 ft 3 

q q  1.72  10 6 Btu  17.77 kW [ans ] 3  lit hrft V V

12  9 Known : P  1000 psia ; Tsub  19.6  F ; xe  0.10 ; Pa  0.1 psi ; S  2 ; A s  3in 2 Analysis : Ts   v f P  1000 psia   v g  h  f Ti  Ts  Tsub

 544.6  F  0.02165 ft  0.4456 ft

3

lbm

h fg  649.4 Btu

lbm

h g  1191.8 Btu

lbm

3

 542.4 Btu

lbm lbm

hi  518.075 Btu lbm   525.01   3 v i  0.021 ft lbm 

h  h f  xe h fg  hi  h  542.4  0.10  649.4  518.075  h  89.25 Btu

lbm

Eq.(12  40)  Pa  r

G2 gc

(1  xe ) 2 x Eq.(12  41)  r  v f  e v g  vi 1e e 2

Eq.(12  9)   

1 1   1  0.1  0.0216 1 x  v f 2 1    S 1   0.1  0.4456  x  vg

 a  0.534 (1  0.1) 2 0.12  0.0216   0.4456  0.021 1  0.534 0.534 3  r  0.025 ft lbm r

Pa  r

G2 G  gc

 G  40931.6 lbm

Pa  g c  r hrft 2

0.1  4.17  10 8 0.025

m  GA  m  40931.6 

3  852.47 lbm hr 144

q  m h  q  852.74  89.25  76107.2 Btu

hr

[ans ]

12  10 Known : P  1200 psia ; Vi  2 fps ; Saturated ; f  0.03 ; S  2 ; A  4.5in  0.5in ; L  4 ft ; q   3  10 5 Btu

hrft 2

Analysis : Ts  567.22  F  3  P  1200 psia  v f  0.0223 ft lbm  3 v  0.3619 ft lbm  g  H B  f V f20   Pf H  R  f B  D 2 g e c   1  1 Uniform  R  1   3  1 e  h  h f  xe h fg  hi

  1     1 e

h h

f Saturated  i  h  xe h fg  xe 

q  m h  h 

  

2

h fg  611.7 Btu

lbm

h f  571.7 Btu

lbm

  

h h fg

q m

 4.5  q  q   A  3  10 5  2    4   9  10 5 Btu hr  12  1 4.5 0.5 m   i AiVi   2  3600    5044.84 lbm hr 0.0223 12 12 5 9  10  h   178.4 Btu lbm 5044.84 178.4  xe   0.29 611.7 1 1 Eq.(12  9)   e    1  xe  v f  1  0.29  0.0223 2  S 1    1    0.29  0.3619  xe  v g   e  0.768 1  1 1     R  1     3   1  0.768   1  0.768   R  7.963

2

  

De  4

Ac P

4.5 0.5  12 12  1 ft  De  4 4.5 12 2 12 1  2  2  3600  4 0 . 0223 Pf H  7.963   0.03  1   2  4.17  10 8  12  G2 Eq.(12  40)  Pa  r gc

    31.96 lbf 2  0.222 psi[ans ] ft   

(1  xe ) 2 x Eq.(12  41)  r  v f  e v g  vi 1e e 2

(1  0.29) 2 0.29 2  0.0223   0.3619  0.0223 1  0.768 0.768 3  r  0.0658 ft lbm G   iVi r

1  2  3600  322869.9552 lbm hrft 2 0.0223 322869.9552 2  Pa  0.0658  4.17  10 8  Pa  16.45 lbf 2  0.114 psi[ans ] ft G

12  11 Known : P  900 psia ; Vi  10 fps ; m  2  10 5 lbm

hr

; ΔTsub  22  F ; S  1.8 ;

De  0.5in ; L  5 ft ; q  5MWt Analysis : Ts  531.98  F    0.2395 lbm hrft  3  P  900 psia  v f  0.0212 ft h f  526.6 Btu lbm lbm  3 v  0.5006 ft h fg  668.8 Btu  g lbm lbm Ti  Tsat  Tsub hi  499.85 Btu lbm  3   Ti  531.98  22  509.98  F  vi  0.02065 ft lbm  lbm  i  0.251 hrft  2  H B  f V f20  H 0  0V 0  Pf  f 0  R fB   De 2 g c D 2 g e c   q q  m h  h  m q  5MWt  17.06  10 6 Btu hr 6 17.06  10  h   85.3 Btu lbm 2  10 5 h  h f  xe h fg  hi 1 85.3  526.6  499.85 668.8  xe  0.0875  xe 

Eq.(12  9)   e 

  e  0.557

1  1  xe 1    xe

 vf  S  vg



1  1  0.0875  0.0212 1   1.8   0.0875  0.5006

h f  hi qs 1   H    1  cos 0    qt 2   H   h f  xh fg  hi 1 526 .6  499 .85    H0   1  cos  0.314   2  5   526 .6  0.0875  668 .8  499 .85  H 0  19 ft  H B  H  H 0  H B  3.1 ft De 

0.5 ft 12

g c  4.17  10 8 1 1

lbmft lbfhr2

1

1

1

1



 0          47 .798 lbm ft 3 2  vi v f  2  0.02065 0.0212   10  3600  0.0212  ft   1    36479 .42 hr 2  0.02065   1 0.5  10  3600  V D 12  296184 .4044  2.9  10 5 Re 0  i i e  0.02065 1 0 0.2395  0.251 2 0.000005 drawn tubi ng.   0.000005     1.2  10 4  0.0001 D 0.5 12 Fig F_1  f 0  0.0155 Eq.12  19b   V0 

f 

Vi 2

 vf 1  vi 

1 1   47 .17 lbm 3 ft v f 0.0212

m  cte   iVi   f V f  V f  0

0

v i 0.0212 Vi  f Vi   10  3600  36958 .84 ft hr f vi 0.02065

0.5 1   10  3600 De  iVi 12 0.02065 Re B    303295 .3  3  10 5 f 0.2395



 1.2  10 4  0.0001

D F_1 Fig   f B  0.0154

 x  8.75 % Fig 12 _ 13   e R3 P  900 psia  2     1 1 1    Also; Eq.12  33b   R  1     2 . 98     3 1  0 . 577 1  0 . 577         2 2   1.9 47 .798  36479 .42  3.1 47 .17  36958 .84     Pf  0.0155    3  0 . 0154   8 0.5 0 . 5 2  4.17  10 8 2  4 . 17  10       12 12  Pf  319 .46 lbf 3  2.22 lbf 2 [ans] ft in G2 Pa  r gc 3  1  0.0875 2  0.0875 2 r  0.0212   0.5006  0.02065   0.02608 ft lbm 0.557  1  0.557  1 G   i Vi   10  3600  1743341 .404 lbm 2 ft .hr 0.02065 2  1743341 .404   Pa  0.02608   Pa  190 .08 lbf 3  1.32 lbf 2 [ans] 18 ft in 4.17  10

12  12 Known : P  800 psia ; ΔTsub  18.23  F ; De  0.14 ft ; f  0.015 ; S  2 ; H 0  4 ft ; L  12 ft ; xe  0.2 ; ΔPa  10 lbf

ft 2

Analysis :

 f V f2    H 0  R H B  Eq.(12  24 )  Pf   f B 2 g D  c e   Vf 

2 g c De Pf

f B  f H 0  R H B 

Ts  518 .23  F  3  P  1200 psia  v f  0.0209 ft lbm  3 v g  0.5687 ft lbm 

h fg  688 .9 Btu h f  509 .7 Btu

lbm

lbm

Ti  Tsat  Tsub hi  487 .8 Btu lbm   Ti  518 .23  18 .23  500 F   3 vi  0.0204 ft lbm  1 1 Eq.(12  9)   e    1  xe  v f  1  0.2  0.0209 2  S 1    1   0 . 2 0 . 5687   x v  e  g 

  e  0.773 2 1  1 1     Eq.(12  33b)  R  1      3   1  0.773   1  0.773    R  8.258

2  4.17  10 8  0.14  10  Vf  1 0.015   4  8.258  8 0.0209  V f  4818 .67 ft  1.34 fps hr 2 G Pa  r gc  1  0.22  0.2 2 r  0.0209   0.5687  0.0204   0.068 lbm 2 hrft 1  0 . 773 0 . 773   G   iVi

 1  4818 .67  230558 .4 lbf 2 G  ft  iVi   f V f  0.0209 2  230558 .4   Pa  0.068   8.67 lbf 2  0.06 lbf 2 [ans] 8 ft in 4.17  10 P  Gh  G h f  xhfg  hi   230558 .4  509 .7  0.2  688 .9  487 .8  P  36815565 .31 Btu

hrft

2

 2.931  10 4

kW MW t   10 .79 [ans] Btu ft 2 hr

12  13 Known : BWR ; Ac  0.025 ft 2 ; P  1000 psia ; xe  0.1 ; Satueated ; S  3 ; Pa  0.1 psi Analysis : Tsat  544 .61 F    3 ft v f  0.0216 v lbm i   P  1000 psia   3  ft v g  0.4456  lbm   Btu h fg  649 .9  lbm   2 G pa  r gc Eq.12 - 9    

1  0.433  1 - 0.1  0.0216  1   3   0.1  0.4456 

(0.1) 2 ft 3  (1  0.1) 2  r  0.0216   0.4456  0.0216   0.01955 1  0 . 433 0 . 433 lbm   g c pa 4.17  10 8  0.1 G  G  46186 .53 lbm hrft2 r 0.01955 2

.

xh fg   0.1 46186 .53  649 .4  0.025  74983 .83 Btu [ans] hr

 h GA qt  m

12  14 Known : P  1000 psia ; qt  2  10 5 ΔΔsub  24 .61 F ; m  2000

Btu ; De  0.15 ft ; Vi  6 fps ; hr

lbm ; S  3 ; f  0.015 ; L  10 ft hr

Analysis: Tsat  544 .61  Ti  Tsat  Tsub  520  F (p f ) H  ( f B

p f V f2 2 g c De

)( H o  R H B )

1  1   1    R  1   3   1   e   1   e 

2

  

qt  m i [h f  xe h fg  hi ]  2  10 5  2000 [542 .4  xe  649 .4  511 .9]  xe  0.107 1  0.452  1  0.107  0.0216  1   3   0.107  0.4456    fVf 2   H 0  RH B Eq.12  24   Pf   f B  2 g D c e   v 0.0216  f V f   iVi  V f  f Vi   6  3600  22323 .45 ft  6.2 fps hr vi 0.0209

e 





H 0 h f  hi H 542 .4  511 .9   0   H 0  3.05 ft  H B  6.95 ft H h 10 100 2 1 1 1    uniform  R  1      2.052 3  1  0.452  1  0.452   1 2  22323 .45   Pf  0.015  0.0216  3.05  2.052  6.95   47 .89 lbf 2 [ans] 8 ft 2  4.17  10  0.15 uniform 

12  15 Known : P  900 psia ; Vi  3 fps ; ΔTsub  22  F ; S  2 ; D1  3in ; D2  1.7in ; q  10 6 Btu

hr

Analysis : Ts  531 .98  F  3  1a P  900 psia TableD   v f  0.0212 ft lbm  3 v g  0.5006 ft lbm 

h fg  668 .8 Btu h f  526 .6 Btu

lbm

lbm

v  0.0206 ft 3  i lbm Tsub.  22  F  Ti  Tsat  Tsub.  509 .98  F tableD    1b hi  500 Btu lbm  2 2 0.6  1 1  m t 1  x  Eq.(12  57 b)  Pcontraction.   2  2 g c  A1 A2   f 1    m t   iVi Ai 2

2   1 12 Ai  A1      0.0218 ft 2  m t   3  3600  0.0218  11437 .84 lbm hr 4 0.0206 2  1.7    12 A2      0.0158 ft 2 and A3  A1  0.0218 ft 2 4 10 6 1 87.43  526 .6  500   0.091 q  m h  h   87 .43 Btu x lbm 11437 .84 668 .8 1   0.542  1  0.091  0.0212  1   2   0.091  0.5006   Pcontraction.  13 .69 lbm 2  0.0936 psi[ans] ft

Eq.(12  51)  Pexpansion. 

1 1 1  2       11437 . 84   4.17  10 8  0.0158  0.0218 0.0218 2 

  2 2   1  0.091 0.091    1 1   1  0.542   0.542  0.5006  0.0212   Pexpansion.  11.51 lbf 2  0.08 psi[ans] ft P  Pcontraction.  Pexpansion.  0.014 psi[ans]

12  16 Known : P  600 psia ; m  124 .3 lbm

hr

; PTP  0.1lbf

ft 2

;

C D  0.6 ; D  2in ; D  1in Analysis :  2 At  0.0218 ft 2  A0  0.00545 ft  3 A0 tableD1 a P  600 psia  v f  0.0201 ft  0.25 lbm A t  3  ft v g  0.7698 lbm A0 f   Eq 12  63   A0 f  0.6  A0 f  0.6197 A0 f 2 1  0.25   Eq 12  64   A0 g  0.6

  A0 g  0.6197 A0 g

A0 g 1  0.25 

2



A0 f  A0 g  A0  0.00545  A0 f  A0 g



00545    0.   0.6197 A0 f  A0 g     

   A0 f  A0 g  0.0034 [1] PTp  PTp ) f  PTp ) g  0.1lbf m t  m f  m g  A0 f



ft 2

2  4.17  10 8 

1 1   0.1  A0 g 2  4.17  10 8   0.1 0.0218 0.7698

   124 .3  61852 .15 A0 f  10408 A0 g      497 .6 A0 f  83 .74 A0 g  1  5.94 A0 f  A0 g  0.012 [2]   [1] and [2]  4.94 A0 f  0.0086  A0 f  0.00174 ft 2   A0 g  0.012  5.94  0.00174  A0 g  0.00166 ft 2

Eq 12  61  m g  0.00166  2  4.17 10 8  m g  xm t  x 

17.3  0.14  x  14%[ans] 124 .3

1  0.1  17.3 lbm hr 0.7698

12  17 Known : P  1000 psia ; m  100 lbm D2  0.2in ; qOrifice  3247 Btu

hr

; Saturated ; C D  0.6 ; D1  1in ;

hr

Analysis : Ts  544 .61 F  3  1 a P  1000 psia tableD  v f  0.0216 ft lbm  3 v g  0.4456 ft lbm  qor.  m t h  m xh fg  x 

h fg  542 .4 Btu

lbm

h f  649 .4 Btu

lbm

3247  x  0.05 100  649 .4

m g  xm  5 lbm  hr  m f  1  x m  95 lbm hr 2 A0 D0 2  2  0.2  Ac Dc Eq.(12  59 )  Psp ) g 

m g

2

2 A0  2 g c  g 2

 0.2     12  A0 4 A0  C D  0.6   1.34  10 4 ft 2 2 2 1  0.2  A  1   0   Ac  25  Psp ) g   0.748 lbf 2 ft 1.34  10 4 2  2  4.17  10 8  1 0.4456

95 

2

Eq.(12  58)  Psp ) f 

 13.02 lbf

1.34 10   2  4.17 10  1 0.0216 2 Eq.(12  67 )  PTp  1.26 13.02  0.748  29 .28 lbf 2 [ans] ft



4 2

8



ft 2

12  18 Known : P0  2000 psia ; x  0.05 ; Analysis : Eq.(12  87 )  S  

vg vf

v  0.0223 ft 3  f lbm P  2000 psia   3 v g  0.3619 ft lbm   S   2.7 Eq.(12  9)   

1 1  x  v f 1   S  x  vg

1  1  0.05  0.0257 1   2.7   0.05  0.1887    0.125  12 .5%[ans]  

12  19 Known : Pb  799 psia ; C D  0.6 ; P0  80 0 psia ; x  0.1 ; Ahole  0.02 ft 2 Analysis :  v  0.0.029 ft 3  f lbm P  800 psia   3 v g  0.5687 ft lbm  PTP  P0  Pb Eqs.12  61 & 12  62   m t  m f  m g  A0 f 2g c  f PTP  f  A0g 2g c  g PTP g Eqs.12  63  & 12  64  

A0 f A0 f  A0g A0 g

Eqs.12  61 & 12  62  

A0 f m f g   A0g m g f

and A0 f  A0 g  Ahole [1]

m f 1  x m 1  0.1 A A 0.029    9  0f  9  2.03  0 f  2.03 [2] x m m g 0.1 A0g 0.5687 A0 g 2.03 A0 g  A0 g  0.02  A0 g  0.0066 ft 2 [1]and [2]   2  A0 f  2.03  A0 g  A0 f  0.0134 ft Ac  A0  A0g  C D A0 g  0.00396 ft 2 and A0 f  0.00804 ft 2 m t  0.00804  2  4.17  10 8   m t  1515 .1lbm

hr

[ans]

1 1  1  0.00396  2  4.17  10 8  1 0.029 0.5687

12  20 Known:P0  2000 psia ; T  560  F ; L  2 ft ; D  2 ft Analysis : 3 T  560  F  v f  0.0221 ft

lbm

L 2   2  Short channel (region1) D 12 12 G  0.61 2  4.17  10 8  Eq.(12  93)   G  4527861 .681 lbm

1  (2000  540 ) 0.0221

hrft2 P L Fig12  21   2  C  0.27  PC  540 psia D P0

m  GA  3556174 .248 lbm 

A

2

1 4

 m  987 .83 lbm

hr

[ans]

hr

Chapter 13 Core Thermal Design

13  1 Known :

13  2 Known : m

water

h  3080 Btu

 3080 lbm hrft

2

; t  551.8 F ; t  650.7 F ; 

hr



f1

f2

;   222b f

Analysis:

m C t  t  2qA H    Eq.(13  7); t  t  q   C m 2A H   q  C

f2

f1

f2

C

P

C

P

C

C

f1

Chapter 14 The Boiling Core

14  1 Known : boiling  sodium ; Tsub i  100  F ; Vi  3 ft  ; Pav.  38 psia   s A fiow  100in 2 ; Tdown comer  1300  F ; Q  ? KWt  Analysis:



Eq.(14  1) ; Q t  m i h f  xe h fg  _ hi







h f  712 Btu

lbm m i   iVi Ai  Pav.  38 psia   hg  2330.7 Btu lbm



 



 

R  F  460 Tsat  2300  R   Tsat  1840  F

1 1  lbm   3 2    ft vi 2.1972 10   Ti  Tsat  Tsub  1840  100  1740 F  2200 R  hi  680.7 Btu lbm 1 100 m i   3  3600   341343.5 lbm 2 hr 12  12 2.1972 10 h f  hi Eq.(14  1) ; xe   Td  1300  F  1760  R  hd  546.56 Btu lbm h h

i 











f

 xe 



d

712  680.7  0.1892 712  546.56



 Q t  341343.5  712  0.1892  2330.7  712 _ 680.7  115.46  10 6 Btu

1Btu hr  2.93110

4



kW  Q t  33841.97kW (t )[ans ]

hr



14  2 Known :

OD of vessel 14ft  ; Q t  600MWt  ; xe  6%  0.06 ; P  1000 psia  A fiow  100in 2 ; Td  250  F ; ID of vesse  ? Analysis: Eq.(14  1) ; Q t  m i h f  xe h fg  _ hi







Q t  600MWt   600 10 6  3.412  2.0472 10 9 Btu





h f  542.4 Btu

lbm h  1191.8 Btu lbm P  1000 psia   g  Tsat  544.61 F



h fg  649.4



hd  216.48 Btu Td  250 F  

lbm

hr



v f  0.0216



v g  0.4456



 3  v f  0.01700 ft  Ibm   h f  hi 542.4  hi Eq.(14  1) ; xe    0.06  hi  522.845 Btu lbm h f  hd 542.4  216.48





 Q t  m i h f  xe h fg  _ hi





2.0472 10 9  m i 542.4  0.06649.4   522.845  m i  3.498 10 7 Ibm   1.5  3600  5400 ft  Vdc  1.5 ft sec    hr  m i   dc AdcVdc  1  dc  vi hi  522.845  t i  526.86  vi  0.6211

 dc 

1 1     47.39 Ibm 3  ft vi 0.0211  

m i   dc AdcVdc  Adc  Adc 

ID 2

 OD2



 ID 2

3.498  10 7  136.68 ft 2 5400  43.39

 



14 4



2

 ID 

2

4 4 2  21.97 ft  ID  4.68 ft  ans

 





hr



14  3   Known : q  10 5  Btu 2  ; Tsub  ? ; Vi  2 fps ; Pav.  1200 psia  hrft   m f xe  ? ; Td  200  F ; ? m g Analysis:

Eq.(14  11) ; Q t  m g hg  hd  Q t  q  2A w  10 5  2 



4.5  4  300000 Btu hr 12



h f  571.7 Btu



 lbm

Pav.  1200 psia   hg  1183.4 Btu

lbm





 3  v g  0.3619 ft   hr  1 1  lbm  i   f     ft 3  v f 0.0223 

Tsat  567.22  F



Td  200 F  hd  167.99 Btu



lbm 1 0.6 4.5 m i   i vi Ai   2  3600    6053.8 lbm hr 0.0223 12 12 m g m g m g 295.45 m g  m d Eq.(14  3) ; xe        xe  0.0488  4.88% a ans m g  m f m d  m f m i 6053.8



Eq.(14  4) ; recirculat ion raito ;

m f 1 - xe   19.49 m g xe



b

ans



Eq.(14  8b) ; hsub  xe h f  hd   0.0488571.7  167.99   19.7 Btu



Eq.(14  8a) ; h i  h f  hsub  571.7  19.7  552 Btu  Tsub  Tsat  Ti  567.22  552.1  15.12  F

c

ans

lbm

  T  552.1 F 

lbm



i

14  4

Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

g gc

Analysis: problem 14 - 4 ; Ti  552.1   i  H  4ft  , H ch  1.5ft 

1 1       45.66 lbm 3    dc  45.66 lbm 3  f ft ft vi 0.0219    

1  1 1  1  1 1   lb       45.25 m 3   ft  2  vi v f  2  0.0219 0.0223   h f  hi H 571.7  552 Eq.(14  13) ; 0    0.398 H h f  xe h fg   hi 571.7  0.0488  1183.4  571.71  552 Eq.(14  18) ;  0 

 H 0  0.398  4  1.59ft   H B  4  1.59  2.41ft  Eq.(14  38) ;  B   f -  f   g  Eq.(12  11) ;  Eq.(12  9) ;  e 

 B 

vf vg

S

1 1 -

    1 1  1 ln 1 -     e 1     1 -  e 1    

0.0223  2  0.12324 0.3619

1  1  xe   1    x  e 



1  0.294 1  19.49  0.12324

    1 1  1 1 1  1    1 ln  1    0.0223  0.0223 0.3619  1  0.12324   0.2941  0.12324  1  0.2941  0.12324 

    B  38.04448 lbm 3  ft  

 e  1    f   g  1  0.294

1 1  lb   0.294   32.47 m 3  ft  0.0223 0.3619 

 lb   Pd  45.664  1.5  45.25  1.59  38.04448  2.41  32.47  1.5  1  38.79 f 2  ft    lb   lb  1 f 2   0.006944 psi  2   Pd  0.269 psi  ans ft in    

14  5 Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

Analysis:

g gc

14  6 Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

Analysis:

g gc

14  7 Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

Analysis:

g gc

14  8 Known : BWR ; Vi  3 ft

sec

; S  1.9 ; Pav.  1000 psia ; 4.85in  4.85in  6 ft ; n  36 ; d  0.55in ;

Plosses  0.301 psia (channel , chimney ) ; Plosses  0.182 psia (downcomer ) ; Pf  0.08 psi

ft

(in downcom

Analysis: Totalloss  0.301  0.182  0.008(6  H ch )  0.531  0.008 H ch ( psia ) 1 psia144lbf

ft      76.464  1.152 H ch [1] 2

Eq .(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch  Eq .(14  8b) ; h sub  xe h f  hd   Eq .(12  10) ; xe 

e 1    e

 

g gc

1  1-  1

1  0.5 

 0.916

1  0.0921  0.5 

Eq .(12  10) ; xe 

1 1-e 1    e

1  



1  0.916  0.5  1  0.0921  0.5 

Tsat  544 .61 F

   h f   hi H0 542 .4  389 .09 Pav  1000  psi   v f  0.0216  lbm 3  , v g  0.4456  lbm   ft    h f  xe h fg   hi 542.4  0.416  649.4  389.09 H h f  542 .4 Btu , hg  1191 .8 Btu lbm 0.0216    1.9  0.0921 0.4456



Td  400





1  1 1  1  1 1   lb       45.25 m 3   ft  2  vi v f  2  0.0219 0.0223   h f   hi H 571.7  552  Eq .(14 13) ; 0  h f  xe h fg   hi 571.7  0.0488  1183.4  571.71  552   0.398 H

Eq .(14 18) ;  0 

 H 0  0.398  4  1.59ft   H B  4  1.59  2.41ft  1 Eq.(14  38) ;  B   f -  f   g  1- Eq.(12 11) ; 

vf

Eq.(12  9) ; e 

1  1  xe 1    xe

 B 

vg

S

    1 1  1 ln 1-     e 1    1-  e 1   

0.0223  2  0.12324 0.3619   



1  0.294 1  19.49  0.12324

    1 1  1 1 1  1    1 ln   1   0.0223  0.0223 0.3619  1  0.12324   0.294 1  0.12324   1  0.294 1  0.12324  

    B  38.04448  lbm 3  ft  

 e  1    f   g  1  0.294 

1 1  lb   0.294   32.47 m 3  ft 0.0223 0.3619  

  lb  Pd  45.664  1.5  45.25  1.59  38.04448  2.41  32.47  1.5 1  38.79 f 2  ft    lb   lb  1 f 2   0.006944 psi  2   Pd  0.269  psi  ans ft in    

Chapter 14 The Boiling Core

14  1 Known : boiling  sodium ; Tsub i  100  F ; Vi  3 ft  ; Pav.  38 psia   s A fiow  100in 2 ; Tdown comer  1300  F ; Q  ? KWt  Analysis:



Eq.(14  1) ; Q t  m i h f  xe h fg  _ hi







h f  712 Btu

lbm m i   iVi Ai  Pav.  38 psia   hg  2330.7 Btu lbm



 



R  F  460 Tsat  2300  R   Tsat  1840  F 



1 1  lbm   3 2    ft vi 2.1972 10   Ti  Tsat  Tsub  1840  100  1740 F  2200 R  hi  680.7 Btu lbm 1 100  3  3600   341343.5 lbm m i  2 hr 12  12 2.1972 10 h f  hi Eq.(14  1) ; xe   Td  1300  F  1760  R  hd  546.56 Btu lbm h h

i 











f

 xe 



d

712  680.7  0.1892 712  546.56



 Q t  341343.5  712  0.1892  2330.7  712 _ 680.7  115.46  10 6 Btu

1Btu hr  2.93110

4



kW  Q t  33841.97kW (t )[ans ]

hr



14  2 Known :

OD of vessel 14ft  ; Q t  600MWt  ; xe  6%  0.06 ; P  1000 psia  A fiow  100in 2 ; Td  250  F ; ID of vesse  ? Analysis: Eq.(14  1) ; Q t  m i h f  xe h fg  _ hi







Q t  600MWt   600 10 6  3.412  2.0472 10 9 Btu





h f  542.4 Btu

lbm h  1191.8 Btu lbm P  1000 psia   g  Tsat  544.61 F



h fg  649.4



Td  250 F 

hd  216.48 Btu



lbm

hr



v f  0.0216



v g  0.4456



 3  v f  0.01700 ft  Ibm   h f  hi 542.4  hi Eq.(14  1) ; xe    0.06  hi  522.845 Btu lbm h f  hd 542.4  216.48





 Q t  m i h f  xe h fg  _ hi





2.0472 10 9  m i 542.4  0.06649.4   522.845  m i  3.498 10 7 Ibm   1.5  3600  5400 ft  Vdc  1.5 ft sec    hr  m i   dc AdcVdc  1  dc  vi hi  522.845  t i  526.86  vi  0.6211

 dc 

1 1     47.39 Ibm 3  ft vi 0.0211  

m i   dc AdcVdc  Adc  Adc 

ID 2

 OD2



 ID 2

3.498  10 7  136.68 ft 2 5400  43.39

 



14 4



2

 ID 

2

4 4 2  21.97 ft  ID  4.68 ft  ans

 





hr



14  3   Known : q  10 5  Btu 2  ; Tsub  ? ; Vi  2 fps ; Pav.  1200 psia  hrft   m f xe  ? ; Td  200  F ; ? m g Analysis:

Eq.(14  11) ; Q t  m g hg  hd  Q t  q  2A w  10 5  2 



4.5  4  300000 Btu hr 12



h f  571.7 Btu



 lbm

Pav.  1200 psia   hg  1183.4 Btu

lbm





 3  v g  0.3619 ft   hr  1 1  lbm  i   f     ft 3  v f 0.0223 

Tsat  567.22  F



Td  200 F  hd  167.99 Btu



lbm 1 0.6 4.5 m i   i vi Ai   2  3600    6053.8 lbm hr 0.0223 12 12 m g m g m g 295.45 m g  m d Eq.(14  3) ; xe        xe  0.0488  4.88% a ans m g  m f m d  m f m i 6053.8



Eq.(14  4) ; recirculat ion raito ;

m f 1 - xe   19.49 m g xe



b

ans



Eq.(14  8b) ; hsub  xe h f  hd   0.0488571.7  167.99   19.7 Btu



Eq.(14  8a) ; h i  h f  hsub  571.7  19.7  552 Btu  Tsub  Tsat  Ti  567.22  552.1  15.12  F

c

ans

lbm

lbm



  T  552.1 F i



14  4

Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

g gc

Analysis: problem 14 - 4 ; Ti  552.1   i  H  4ft  , H ch  1.5ft 

1 1       45.66 lbm 3    dc  45.66 lbm 3  f ft ft vi 0.0219    

1  1 1  1  1 1   lb       45.25 m 3   ft  2  vi v f  2  0.0219 0.0223   h f  hi H 571.7  552 Eq.(14  13) ; 0    0.398 H h f  xe h fg   hi 571.7  0.0488  1183.4  571.71  552 Eq.(14  18) ;  0 

 H 0  0.398  4  1.59ft   H B  4  1.59  2.41ft  Eq.(14  38) ;  B   f -  f   g  Eq.(12  11) ;  Eq.(12  9) ;  e 

 B 

vf vg

S

1 1 -

    1 1  1 ln  1 -    e 1     1 -  e 1    

0.0223  2  0.12324 0.3619

1  1  xe   1    x  e 



1  0.294 1  19.49  0.12324

    1 1  1 1 1  1    1 ln  1    0.0223  0.0223 0.3619  1  0.12324   0.2941  0.12324  1  0.2941  0.12324 

    B  38.04448 lbm 3  ft  

 e  1    f   g  1  0.294

1 1  lb   0.294   32.47 m 3  ft  0.0223 0.3619 

 lb   Pd  45.664  1.5  45.25  1.59  38.04448  2.41  32.47  1.5  1  38.79 f 2  ft    lb   lb  1 f 2   0.006944 psi  2   Pd  0.269 psi  ans ft in    

14  5 Known : H ch  1.5ft same corss section  ; S  2 ; driving pressure  ? Eq.(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch 

Analysis:

g gc

14-6

14-7

14  8 Known : BWR ; Vi  3 ft

sec

; S  1.9 ; Pav.  1000 psia ; 4.85in  4.85in  6 ft ; n  36 ; d  0.55in ;

Plosses  0.301 psia (channel , chimney ) ; Plosses  0.182 psia (downcomer ) ; Pf  0.08 psi

ft

(in downcom

Analysis: Totalloss  0.301  0.182  0.008(6  H ch )  0.531  0.008 H ch ( psia ) 1 psia144lbf

2

ft      76.464  1.152 H ch [1]

Eq .(14  20) ; Pd   dc H  H ch    0 H 0   B H B   e H ch  Eq .(14  8b) ; h sub  xe h f  hd   Eq .(12  10) ; xe 

g gc

1 1   0.916 1- e  1  0.5  1   0.0921 1   0.5      e 

h f   hi H0 542 .4  389 .09   h f  xe h fg   hi 542.4  0.416  649.4  389.09 H Eq .(12  10) ; xe 

1 1-e 1    e

1  



1  0.916  0.5  1  0.0921  0.5 

Tsat  544 .61 F     Pav  1000  psi   v f  0.0216  lbm 3  , v g  0.4456  lbm 3  ft  ft    h f  542 .4 Btu , hg  1191 .8 Btu , h fg  649.4 Btu lbm lbm lbm 0.0216    1.9  0.0921 0.4456



Td  400











1  1 1  1  1 1   lb       45.25 m 3   ft  2  vi v f  2  0.0219 0.0223   h f   hi H 571.7  552  Eq .(14 13) ; 0  h f  xe h fg   hi 571.7  0.0488  1183.4  571.71  552   0.398 H

Eq .(14 18) ;  0 

 H 0  0.398  4  1.59ft   H B  4  1.59  2.41ft  1 Eq.(14  38) ;  B   f -  f   g  1- Eq.(12 11) ;  Eq.(12  9) ; e 

 B 

vf vg

S

    1 1  1 ln 1-     e 1    1-  e 1   

0.0223  2  0.12324 0.3619

1  1  xe 1    xe

  



1  0.294 1  19.49  0.12324

    1 1  1 1 1  1    1 ln   1   0.0223  0.0223 0.3619  1  0.12324   0.294 1  0.12324   1  0.294 1  0.12324  

    B  38.04448  lbm 3  ft  

 e  1    f   g  1  0.294 

1 1  lb   0.294   32.47 m 3  ft 0.0223 0.3619  

  lb  Pd  45.664  1.5  45.25  1.59  38.04448  2.41  32.47  1.5 1  38.79 f 2  ft    lb   lb  1 f 2   0.006944 psi  2   Pd  0.269  psi  ans ft in    

14-9

14-11

14-12

14-13

14-14