el-wakil nuclear heat transport's solution manual [1 ed.]
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english Pages [203] Year 2006
Chapter 4.pdf (p.1-37)
Chapter 5.pdf (p.38-52)
Chapter 6.pdf (p.53-72)
Chapter 7.pdf (p.73-89)
Chapter 8.pdf (p.90-105)
Chapter 9.pdf (p.106-123)
Chapter 10.pdf (p.124-130)
Chapter 11.pdf (p.131-142)
Chapter 12.pdf (p.143-173)
Chapter 13.pdf (p.174-176)
Chapter 14.pdf (p.177-186)
Chapter 14-v.94.pdf (p.187-203)
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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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑, 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) 𝑇𝑒 % 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;
𝐺𝑟𝑎𝑝𝑖𝑡𝑒 − 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 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 𝑡𝑒 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 𝑛 = 10000 ; 𝑇 = 500 ℉ ; 𝐻 = 20 𝑓𝑡 ; 𝐷𝐹𝑢𝑒𝑙 = 0.6 𝑖𝑛 ; 𝐷𝐶𝑜𝑟𝑒 = 8 𝑓𝑡 ; 𝜑𝑐𝑜 = 1013 𝐻𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑 ; 𝐻𝑒𝑡𝑒𝑟𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑠 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; 𝑇𝑒𝑟𝑚𝑎𝑙 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 𝜆𝑒 = 0.186 𝑓𝑡 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 ; 𝜆𝑒 = 0.3 𝑓𝑡 (𝑖𝑛 𝑎𝑥𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛) Find; ′′′ 𝜑 @ 𝑡𝑒 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; 𝑇𝑒𝑟𝑚𝑎𝑙 𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 − 𝑓𝑢𝑒𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 ; 𝑈 𝑠𝑎𝑙𝑡 𝑖𝑛 𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 ; 𝐺𝑟𝑎𝑝𝑖𝑡𝑒 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 ; 𝑆𝑝𝑒𝑟𝑖𝑐𝑎𝑙 ; 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 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 𝑙𝑒𝑛𝑔𝑡𝑠 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; 𝑇𝑒𝑟𝑚𝑎𝑙 𝑒𝑡𝑒𝑟𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 ; 𝐶𝑢𝑏𝑒 𝑓𝑜𝑟𝑚 ; 𝑈𝑟𝑎𝑛𝑖𝑢𝑚 𝑚𝑒𝑡𝑎𝑙 ; 20% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 𝑎 = 4 𝑓𝑡 ; 𝑛 = 4900 ; 𝐷𝐹𝑢𝑒𝑙 = 0.45 𝑖𝑛 ; 𝑄𝑡 = 5 × 108 𝐵𝑡𝑢 𝑟 ; 𝑇 = 500 ℉ ; 𝐺𝑟𝑎𝑝𝑖𝑡𝑒 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑜𝑟 ; 𝑁𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑈𝑛𝑑𝑖𝑠𝑡𝑜𝑟𝑡𝑒𝑑 ; 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡𝑠 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 𝐺𝑟𝑎𝑝𝑖𝑡𝑒 , 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 𝑡𝑒 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 𝑡𝑒 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 𝑡𝑖𝑐𝑘, 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 𝑙𝑖𝑔𝑡 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 𝑙𝑖𝑔𝑡 𝑤𝑎𝑡𝑒𝑟 → 𝐿 = 2.88 𝑐𝑚 ≡ 0.095 𝑓𝑡 2 𝑓𝑡 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠 → 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠 > 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; 𝑇𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 ; 𝐻𝑒𝑎𝑣𝑦 𝑤𝑎𝑡𝑒𝑟 𝑚𝑜𝑑𝑒𝑟𝑎𝑡𝑒𝑑 ; ′′′ 𝑀𝑜𝑟𝑒 𝑡𝑎𝑛 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 𝑡𝑒 𝑓𝑖𝑟𝑠𝑡 𝑑𝑎𝑦 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑠𝑢𝑡𝑑𝑜𝑤𝑛) 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% 𝑒𝑛𝑟𝑖𝑐𝑚𝑒𝑛𝑡 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 → 𝑒𝑛𝑟𝑖𝑐𝑚𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑎𝑛𝑔𝑒 → 𝑁 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑎𝑛𝑔𝑒 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% 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 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 − 𝑠𝑒𝑙𝑙 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 % 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 ; 𝑃 = 50 𝑀𝑊 𝑡 ; 𝑞𝑎𝑣 . = 74444 𝑟𝑓𝑡 2 ;
4 % 𝑜𝑓 𝑡𝑒 𝑡𝑒𝑟𝑚𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑡𝑒 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; 𝑆𝑝𝑒𝑟𝑖𝑐𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑐𝑜𝑟𝑒 ; 20 % 𝑒𝑛𝑟𝑖𝑐𝑒𝑑 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 𝑙𝑒𝑛𝑔𝑡𝑠 ; 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 ; 𝑆𝑝𝑒𝑟𝑒 𝑓𝑢𝑒𝑙 ; 𝐷𝐹𝑢𝑒𝑙 = 0.6072 𝑓𝑡 ; 𝜌𝐹𝑢𝑒𝑙 = 12.5
𝑔𝑟
𝑐𝑚3 ;
𝑘𝐹𝑢𝑒𝑙 = 12 𝐵𝑡𝑢 𝑟 𝑓𝑡 ℉ ; 𝑇𝑆𝑒𝑙𝑙 = 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 ; 𝑆𝑝𝑒𝑟𝑖𝑐𝑎𝑙 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 𝑡𝑒 𝑐𝑜𝑟𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 =? ; Analysis;
14 Nuclear Heat Transport Problem’s Solutions M. M. EL-WAKIL
5-9 Known; 𝑆𝑝𝑒𝑟𝑖𝑐𝑎𝑙 𝑠𝑒𝑙𝑙 𝑜𝑓 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠 ∆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 𝑡𝑖𝑐𝑘 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑓𝑢𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ; 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 ; 𝐾𝑛𝑜𝑤𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 ; 𝐻𝑒𝑎𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑡𝑜 𝑓𝑙𝑜𝑤 𝑡𝑟𝑜𝑢𝑔 𝑏𝑜𝑡 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 ; Find; ′′′ 𝑟𝑚 𝑤𝑒𝑟𝑒 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 qoL
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 ( 0.2827.62) 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 ( 0.2827.62) 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 ( 0.2827.62) e ( 0.2827.62) 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 μkL 5 1.309 10 1.24in ( 0.2827.62) t m ti (e 1) (e 0.2823.15 1) 2 (8.595 ) 28 3in m
tm ti 14 .14478 F [ans]
62 Known : h 3
MeV ; 1014 ; iron ; total atenuation 90% ; 0.282cm -1 #
Analysis : total attenuatio n 90% 2 0.1o 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 0.10 ; 2 1e L o e L e L 0 e ( L L ) 0.10 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
63 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
qo 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 3.5470.5 e 3.5470.5 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 qo 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 3.5360.5 1.78 1011 12 1.50313.5470.5 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 3.5470.5 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
64 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 qoA 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 8.5950.5 1 (500 563 .3) 1 A 0.5 8.595 8.595 0.5 qo Btu 15265 .3 [ans] A hrft2
65 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
qoA 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 cm1 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 9.110 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 Lc lbm 8.9 gr 3 555 .627 3 cm ft
9.110 6 Btu
T hrft3 (1 e 0.23257.62 ) t 0.2325 cm1 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
66 Known :
67 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 cm1 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 7.5220.333 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
68 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 0.095382.54 1.44 1012 1 1 1
2 2
2 1.44 1012 2.5e 0.2792.54 0.873 1012 2 qo 0 1E 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 0.095330.48 q1t q00 1.77 10 4 L 12 (1 e ) (1 e ) A 1 0.0953 30 .48 2 0.27930.48 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
69 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 cm1 8.6 ft 1 L 3in 0.25 ft qt 10 6 A o E (1 e L ) 4.56 10 16 (1 e 8.60.25 ) photon o 7.4233 1014 [ans] sec cm2
6 10 Known : h 1
MeV ; B 3 ; 0.10 ; 0.797 cm1 #
Analysis :
0 (1 B)e L 0.10 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 x0
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 cm1 7.559 ft 1 k 63 .9W
mK
36 .92 Btu
hrftF
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 7.60.5 1 7.60.5 (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 7.60.5 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 7.60.5 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 7.60.5 7.60.5 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 0.10 #
Analysis :
1 0e L 2 1e L (0e L )e L 2 0e 2 L e 2 L 0.1 e 20.245L 0.1 2 0.10 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
q2 1h 3.162 1013 5 0.245 1.5477 10 8 Btu [ans] hrft3 q1 q1 1895932 .5 (1 e L ) (1 e 0.2454.7 ) A 0.245 0.03281 q1 Btu 32724837 [ans] A hrft2 q2 q2 Btu (1 e L ) 10346356 [ans] A hrft2
q2 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 cm1 7.523 ft 1
Table(6 2) 7.86
gm
cm3
a) 1 0e x 0.10 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 7.5230.306 ) 1.5477 10 8 A qt Btu 228652 .85 [ans] A hrft2
6 14 known: Analysis : k (1 e L ) xm Ln (t t ) q0L 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 cm1 8.6 ft 1 1 1 2 0.47 cm 14 .33 ft 1 (1 e 8.61 ) x1m Ln 0.25 ft 8.6 8.6 1 8 .6 MeV q0 0 1h1 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 8.60.25 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 14.331 14.330.186 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 ny n k 1 nx 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 kTny kTn kTnx kTn hxTf hxTn 0 4 2 2 2 2 2 x 2hx 2hx q Tny Tnx T f Tn Tn Tn 2k k k x 2 q hx , Bi t g t g Tny Tnx 2 BiT f (2 2 Bi )Tn 2k k T T 2 BiT f t g Tn ny nx [ans] 2 2 Bi 2 2 Bi
72 known: Analysis : T Tn x T T x y q 3 1 k 1 ny k y 1 nx n 2 2 2 y x T Tn y T T y x k x 1 ny k 1 nx n h 1 1(T f Tn ) 0 y 2 x 2 2 3qx 2 1 1 1 (x y ) kTny kTn kTnx kTn kTny kTn kTnx 4 2 2 2 1 kTn hxTf hxTn 0 2 3qx 2 2hx 2hx Tny 2Tnx 2Tny Tnx T f Tn 2Tn 2Tn Tn Tn 2k k k x 2 q hx , Bi t g 3t g Tny Tnx 2Tnx Tny 2 BiT f (6 2 Bi )Tn 2 k k T Tnx 2Tnx Tny 2 BiT f 3t g Tn ny [ans] 6 2 Bi 6 2 Bi
73 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 qx 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 ) hx(t f t s ) 0 2 x 2 q (t m t s ) hx 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 ) hx(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 ) hx(t f t 3 ) 0 2 2 2t1 t 4 2 Bit f 2t 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
74 known: Analysis : (x y ) qs hxTs T f q3 hxT3 T f
q4 hxT4 T f q2 hxT2 T f Method1; 2qs 2q3 2q4 q2 qtotal 2hxTs T f 2hxT3 T f 2hxT4 T f hxT2 T f qtotal hx2Ts 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
75 known: Analysis : 2
0.25 10 2 qΔx 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
76 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: . qgen. 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 qx 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
77 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 qx 2 x t n x t n t n y t n 2q 0 2k k t n x t n y qx 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 qx 2 qx t n x t n t n y t n 0 2k k t n x t n y qx 1 tn t g [ans] 2 2k 2 b) q
78 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 qx 2 12 t g 100 F 2k 22 0.24 4 2 10 qx 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 qx 1 t g 4 k 2 2t 2t5 qx 1 t 4 1 t g 4 k 2 t 2t 2 t 4 t 6 qx 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 qx 1 t t g 9 4 k 2 6
79 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 2hx hx t n x t n y t f tn 2 2 2k k k t n x t n y 2 Bix 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 2hx hx 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
qx 2 2hx 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
2hx 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 2x 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 ; 3x x 2x a f 2 2 2 2
2
2
1
y
sss ; 3x 2x 2x a f f 5 2 2
2
2
2
x
y
qx 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 qx 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 CV t f t1 ( ) Ln Ln t t hA t t f 0 f 0 CV 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 4R 200 300 0.8353 Ln 1.5635 hr 200 850 1.5635 hr[ans] t
82 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]
83 Known : Analysis : qin qout q gen
dEs dt
Ax t t n x t n t n x t n t n n kA kA qAx C V x x kt n kt n x kt n x kt n qx Cxtn Cxtn x x x x 2k k t n x tn x q Cxtn Cx t n x x C k 2k k q Fo t 1 t t t n n 2 x 2 Cx 2 Cx 2 n x n x C Cx q t n 1 2 Fo t n Fot n x t n x [ans] C
84 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 qx tg 12 x 0.125 ; t g 750 F ; tn ny nx ny nx 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
85 Known : Analysis : t ny t n t nx t n y x q x 1 k y 1 k 1 2 x 2 y t ny 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 Cx 2 nx Cx 2 n Cx 2 ny Cx 2 n k k 2h 2h t t t t t t n y n f Cx 2 Cx 2 Cx Cx n n n qx 2 q Fo 2 Fot g C k 2h 2hkx hx k 2 2 BiFo 2 k Cx 2 Cx Ckx
t n 1 4 Fo 2 BiFo t n 2 Fotnx Fot ny t ny 2 BiFot f 2 Fot g [ans]
86 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 Cx Cx Cx Cx Cx 2 n y k 2q t t n t n 2 n Cx Cx qx 2q 2qkx 2 Fo 2 C x Ck x k t n 1 4 Fo t n 2 Fotn x Fot n y t n y 2 Fo
qx 2 Fot g [ans] k
88 Known :
89 Known : Analysis : t ny t n t nx t n 3 x q x y 1 k 1 k y 1 4 2 y x t ny t n t nx 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 ny 2 n 2 n x C 3Cx 3Cx 3Cx 3Cx 2 4k 4k 2k 2k t t t tn 2 n y 2 n 2 n x 3Cx 3Cx 3Cx 3Cx 2 4hxk 4hxk t tn tn tn f 3kCx 2 3kCx 2 4 2 t n 1 4 Fo BiFo t n Fot ny t nx 2t nx 2t ny 2 Bit f 2 Fot 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,
qr0 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 qr0 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
92 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 qR 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
RC qR 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 0.0211 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 qL L t 425 t 415 t t 10 KA(t t ) qA 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:
0.289A 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
96 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 2R 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) 2R 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 qR 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 qR (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
qR t t 2 s
2
f
1 RC 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) qR 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 2R 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 7601 0.09655 761.408 R 301.408 F 2 T T F 0.89 T T 0.81T 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 )2rdr ro
ro
o
2 [(t w t fc )((
2rdr
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 2t sat t t sat g rc t t sat
4v 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 ab 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 qBurnout 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
qc (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 ; qout 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 hAT 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 1e 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 1e 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 R3 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 0.22 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 Gh 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 tableD1 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 A0g 2g c g PTP g Eqs.12 63 & 12 64
A0 f A0 f A0g A0 g
Eqs.12 61 & 12 62
A0 f m f g A0g 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 A0g 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 A0g 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 2qA 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 ? KWt 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 2.93110
4
kW Q t 33841.97kW (t )[ans ]
hr
14 2 Known :
OD of vessel 14ft ; Q t 600MWt ; 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 600MWt 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 0.06649.4 522.845 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
OD2
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 0.0488571.7 167.99 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 1.5ft 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 4ft , H ch 1.5ft
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 1.59ft H B 4 1.59 2.41ft 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 0.2941 0.12324 1 0.2941 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 45.664 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 1.5ft 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 1.5ft 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 1.5ft 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 psia144lbf
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 1.59ft H B 4 1.59 2.41ft 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 45.664 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 ? KWt 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 2.93110
4
kW Q t 33841.97kW (t )[ans ]
hr
14 2 Known :
OD of vessel 14ft ; Q t 600MWt ; 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 600MWt 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 0.06649.4 522.845 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
OD2
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 0.0488571.7 167.99 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 1.5ft 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 4ft , H ch 1.5ft
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 1.59ft H B 4 1.59 2.41ft 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 0.2941 0.12324 1 0.2941 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 45.664 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 1.5ft 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 psia144lbf
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 1.59ft H B 4 1.59 2.41ft 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 45.664 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