آموزش انتقال حرارت - بخش سوم

‫حرارت‬ ‫انتقال‬ ‫آموزش‬
faradars.org/fvmec94064
‫مدرس‬:
‫کو‬‫نقمه‬ ‫مرتضی‬
‫مکانیک‬ ‫مهندسی‬ ‫ارشد‬ ‫کارشناسی‬ ‫دانشجوی‬-‫انرژی‬ ‫تبدیل‬
‫اصفهان‬ ‫صنعتی‬ ‫دانشگاه‬
‫حرارت‬ ‫انتقال‬ ‫درس‬ ‫آموزشی‬ ‫دوره‬
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‫درس‬‫یازدهم‬
‫جابجایی‬ ‫حرارت‬ ‫انتقال‬
‫ها‬ ‫جریان‬ ‫انواع‬
‫حرارت‬ ‫انتقال‬ ‫ضریب‬ ‫آوردن‬ ‫دست‬ ‫به‬ ‫و‬
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‫مطالب‬ ‫فهرست‬
•‫حدس‬‫دما‬ ‫پروفیل‬
•‫نحوه‬‫نوسلت‬ ‫عدد‬ ‫و‬ ‫جابجایی‬ ‫حرارت‬ ‫انتقال‬ ‫ضریب‬ ‫آوردن‬ ‫دست‬ ‫به‬
•‫دمای‬‫سیال‬ ‫فیلم‬ ‫اصطالح‬ ‫به‬
•‫آنالوژی‬‫برن‬ ‫کول‬
•‫انتقال‬‫مایع‬ ‫فلزات‬ ‫در‬ ‫حرارت‬
•‫انتقال‬‫لوله‬ ‫درون‬ ‫جابجایی‬ ‫حرارت‬
•‫جریان‬‫آن‬ ‫بر‬ ‫حاکم‬ ‫روابط‬ ‫و‬ ‫کوئت‬
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‫دما‬ ‫پروفیل‬ ‫حدس‬
•‫وجود‬‫اختالف‬‫دما‬‫بین‬‫صفحه‬‫و‬‫سیال‬‫منجر‬‫به‬‫الیه‬‫مرزی‬‫حرارتی‬‫شود‬‫می‬.
•‫الیه‬‫مرزی‬‫هیدرودینامیکی‬𝛿
•‫الیه‬‫مرزی‬‫حرارتی‬𝛿𝑡
•‫حل‬‫معادالت‬‫پیوستگی‬‫و‬‫اندازه‬‫حرکت‬(‫به‬‫صورت‬‫عددی‬)«‫توزیع‬‫سرعت‬»‫را‬‫دهد‬‫می‬
•‫حل‬‫معادله‬‫انرژی‬‫همراه‬‫با‬‫این‬‫دو‬،‫معادله‬(‫به‬‫صورت‬‫عددی‬)«‫توزیع‬‫دما‬»‫را‬‫نیز‬‫دهد‬‫می‬
•‫روند‬‫حدس‬‫پروفیل‬‫دما‬‫مانند‬‫حدس‬‫پروفیل‬‫سرعت‬
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‫سرعت‬ ‫پروفیل‬ ‫حدس‬ ‫بررسی‬
𝑢 = 𝑐0 + 𝑐1 𝑦 + 𝑐2 𝑦2
+ 𝑐3 𝑦3
•‫مرزی‬ ‫شرایط‬
1) 𝑦 = 0 u = 0
2) 𝑦 = δ u = 𝑢∞
3) 𝑦 = δ
𝜕u
𝜕𝑦
= 0
𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
= 𝜗
𝜕2
𝑢
𝜕𝑦2
4) y = 0
𝜕2 𝑢
𝜕𝑦2 = 0
𝑢 =
3
2
𝑦
𝛿
−
1
2
𝑦
𝛿
3
5
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‫سرعت‬ ‫پروفیل‬ ‫حدس‬ ‫بررسی‬
𝑢 =
3
2
𝑦
𝛿
−
1
2
𝑦
𝛿
3
• Von-Karman:
𝑑
𝑑𝑥
න
0
𝛿
𝑢∞ − 𝑢 𝑢 𝑑𝑦 = 𝜐
𝑑𝑢
𝑑𝑦
| 𝑦=0
𝛿
𝑥
=
4.64
𝑅𝑒
• Exact Value:
𝛿
𝑥
=
5
𝑅𝑒
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‫حدس‬‫پروفیل‬‫دما‬
𝑇 = 𝑐0 + 𝑐1 𝑦 + 𝑐2 𝑦2
+ 𝑐3 𝑦3
•‫مرزی‬ ‫شرایط‬
1) 𝑦 = 0 T = 𝑇 𝑤
2) 𝑦 = δ 𝑡 T = 𝑇∞
3) 𝑦 = δ 𝑡
𝜕T
𝜕𝑦
= 0
𝑢
𝜕𝑇
𝜕𝑥
+ 𝑣
𝜕𝑇
𝜕𝑦
= 𝛼
𝜕2
𝑢
𝜕𝑦2
4) y = 0
𝜕2 𝑇
𝜕𝑦2 = 0
•‫متغیر‬ ‫تغییر‬𝜃 = 𝑇 − 𝑇 𝑤
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•‫جدید‬ ‫متغیر‬ ‫با‬ ‫مرزی‬ ‫شرایط‬ ‫بازنویسی‬
1) 𝑦 = 0 θ = 0
2) 𝑦 = δ 𝑡 θ = 𝑇∞ − 𝑇 𝑤 = 𝜃∞
3) 𝑦 = δ 𝑡
𝜕θ
𝜕𝑦
= 0
4) y = 0
𝜕2 𝜃
𝜕𝑦2 = 0
𝜃
𝜃∞
=
𝑇 − 𝑇 𝑤
𝑇∞ − 𝑇 𝑤
=
3
2
𝑦
𝛿𝑡
−
3
2
𝑦
𝛿𝑡
3
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•‫معادله‬ ‫از‬ ‫حرارتی‬ ‫مرزی‬ ‫الیه‬ ‫ضخامت‬ ‫آوردن‬ ‫دست‬ ‫به‬ ‫برای‬‫انتگرالی‬‫گیریم‬‫می‬ ‫بهره‬ ‫انرژی‬:
𝑑
𝑑𝑥
න
0
𝛿 𝑡
𝑇∞ − 𝑇 𝑢 𝑑𝑦 = 𝛼
𝜕𝑇
𝜕𝑦
| 𝑦=0
•‫ال‬ ‫ضخامت‬ ‫فرمول‬ ‫به‬ ‫توجه‬ ‫با‬ ‫و‬ ‫سرعت‬ ‫و‬ ‫دما‬ ‫های‬ ‫پروفیل‬ ‫حدس‬ ‫از‬ ‫آمده‬ ‫دست‬ ‫به‬ ‫معادالت‬ ‫جایگذاری‬ ‫با‬‫یه‬
‫داشت‬ ‫خواهیم‬ ‫هیدرودینامیکی‬ ‫مرزی‬:
•‫از‬ ‫صفحه‬ ‫از‬ ‫حرارت‬ ‫اینکه‬ ‫فرض‬ ‫با‬x=x0‫باشد‬ ‫شده‬ ‫آغاز‬.
𝛿𝑡
𝛿
=
1
1.026
𝑃𝑟−
1
3 1 −
𝑥0
𝑥
3
4
1
3
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‫جابجایی‬ ‫حرارت‬ ‫انتقال‬ ‫ضریب‬ ‫آوردن‬ ‫دست‬ ‫به‬ ‫نحوه‬
𝛿𝑡
𝛿
=
1
1.026
𝑃𝑟−
1
3 1 −
𝑥0
𝑥
3
4
1
3
•‫یابد‬ ‫انتقال‬ ‫صفحه‬ ‫ابتدای‬ ‫از‬ ‫حرارت‬ ‫اگر‬(x0=0)
𝛿𝑡
𝛿
=
1
1.026
𝑃𝑟−
1
3 ≅ 𝑃𝑟−
1
3
•‫عدد‬ ‫که‬ ‫کرد‬ ‫ادعا‬ ‫توان‬‫می‬‫پراندتل‬‫بین‬ ‫رابط‬‫هیدرودینامیک‬‫است‬ ‫حرارت‬ ‫و‬.
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•‫دیواره‬ ‫روی‬ ‫سیال‬ ‫به‬ ‫مربوط‬ ‫رابطه‬:
ℎ =
−𝑘∞
𝑑𝑇
𝑑𝑦
| 𝑦=0
𝑇 𝑤 − 𝑇∞
=
−𝑘∞(𝑇∞ − 𝑇 𝑤)
3
2𝛿𝑡
−
3𝑦2
2𝛿𝑡
3
𝑦=0
𝑇 𝑤 − 𝑇∞
=
3𝑘∞
2𝛿𝑡
ℎ ∝
1
𝛿𝑡
•‫عبارت‬ ‫جایگذاری‬ ‫با‬h‫فرمول‬ ‫در‬𝛿𝑡‫داشت‬ ‫خواهیم‬:
ℎ = 0.332𝑃𝑟
1
3 𝑘
𝑢∞
𝜗𝑥
1
2
1 −
𝑥0
𝑥
3
4
−
1
3
= ℎ 𝑥
• Where hx is “Local convection heat transfer coefficient”
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•‫شود‬ ‫انجام‬ ‫صفحه‬ ‫ابتدای‬ ‫از‬ ‫حرارت‬ ‫انتقال‬ ‫اگر‬:
ℎ 𝑥 = 0.332𝑃𝑟
1
3 𝑘
𝑢∞
𝜗𝑥
1
2
= 0.332
𝜇𝑐 𝑝
𝑘
1
3
𝑘
𝜌𝑢∞
𝜇𝑥
1
2
ℎ 𝑥 = 𝑓(𝜇, 𝑘, 𝑐 𝑝, 𝑢∞, 𝑥, 𝜌)
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‫نوسلت‬ ‫عدد‬
• Local Nusselt number is defined as:
𝑁𝑢 𝑥 =
ℎ𝑥
𝑘
•‫عدد‬ ‫با‬ ‫تفاوت‬‫بایو‬‫است‬ ‫آن‬ ‫طول‬ ‫و‬ ‫هدایتی‬ ‫ضریب‬ ‫در‬.
•‫داشت‬ ‫خواهیم‬ ‫قبلی‬ ‫مسئله‬ ‫برای‬ ‫پس‬:
𝑁𝑢 𝑥 = 0.332𝑃𝑟
1
3 𝑅𝑒
1
2 1 −
𝑥0
𝑥
3
4
−
1
3
•‫صفحه‬ ‫ابتدای‬ ‫از‬ ‫حرارت‬ ‫انتقال‬:
𝑁𝑢 𝑥 = 0.332𝑃𝑟
1
3 𝑅𝑒
1
2
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‫حرارت‬ ‫انتقال‬ ‫ضریب‬ ‫آوردن‬ ‫دست‬ ‫به‬ ‫نحوه‬‫نوسلت‬ ‫عدد‬ ‫و‬ ‫جابجایی‬
• So, For calculating the convective heat transfer coefficient we should
integrate hx from 0 to L and devide to the length of plane:
തℎ =
‫׬‬0
𝐿
ℎ 𝑥 𝑑𝑥
‫׬‬0
𝐿
𝑑𝑥
=
‫׬‬0
𝐿
ℎ 𝑥 𝑑𝑥
𝐿
= 2ℎ 𝑥=𝐿
• Then, For Nusselt number we have:
𝑁𝑢 = 2𝑁𝑢 𝑥=𝐿
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‫سیال‬ ‫فیلم‬ ‫دمای‬
ℎ 𝑥 = 𝑓(𝜇, 𝑘, 𝑐 𝑝, 𝑢∞, 𝑥, 𝜌)
•‫پارامترهای‬𝜇,𝑘,𝑐 𝑝,𝜌‫هستند‬ ‫دما‬ ‫به‬ ‫وابسته‬
𝑇𝑓 =
𝑇 𝑤 + 𝑇∞
2
•‫آرام‬ ‫مرزی‬ ‫الیه‬ ‫برای‬ ‫گذشته‬ ‫روابط‬(Re<5×105)‫بیش‬ ‫پراندتل‬ ‫اعداد‬ ‫و‬‫از‬0.7‫صادق‬‫است‬
•‫دارای‬ ‫که‬ ‫مایع‬ ‫فلزات‬ ‫برای‬‫پراندتل‬‫خوا‬ ‫اشاره‬ ‫آن‬ ‫به‬ ً‫ا‬‫بعد‬ ‫که‬ ‫است‬ ‫متفاوت‬ ‫ها‬ ‫فرمول‬ ‫این‬ ،‫هستند‬ ‫کمی‬‫کرد‬ ‫هیم‬
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‫مثال‬
•‫دمای‬ ‫در‬ ‫هوا‬27‫صفحه‬ ‫روی‬ ‫سلسیوس‬ ‫درجه‬‫مسطحه‬‫دهی‬ ‫حرارت‬ ‫زیر‬ ‫از‬ ‫را‬ ‫صفحه‬ ‫اگر‬ ،‫کند‬‫می‬ ‫حرکت‬ ‫افقی‬‫م‬
‫به‬ ‫صفحه‬ ‫دمای‬ ‫که‬ ‫ای‬ ‫گونه‬ ‫به‬60‫برسد؛‬ ‫سلسیوس‬ ‫درجه‬
•‫الف‬)‫در‬ ‫شده‬ ‫جابجا‬ ‫گرمای‬ ‫محاسبه‬ ‫مطلوبست‬20‫صفحه‬ ‫ابتدای‬ ‫از‬ ‫متری‬ ‫سانتی‬(‫را‬ ‫هوا‬ ‫سرعت‬2‫ثانیه‬ ‫بر‬ ‫متر‬
‫بگیرید‬ ‫نظر‬ ‫در‬)
•‫ب‬)‫در‬ ‫شده‬ ‫جابجا‬ ‫گرمای‬ ‫محاسبه‬ ‫مطلوبست‬40‫صفحه‬ ‫ابتدای‬ ‫متری‬ ‫سانتی‬
‫حل‬:
𝑇∞ = 27℃ = 273 + 27 𝐾 = 300 𝐾
𝑇 𝑤 = 60℃ = 333 𝐾
𝑢∞ = 2 Τ𝑚
𝑠
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‫مثال‬
𝑇𝑓 =
333 + 300
2
= 316.5 𝐾
𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑝𝑝𝑒𝑛𝑑𝑖𝑥 𝑡𝑎𝑏𝑙𝑒 𝑓𝑜𝑟 𝑇𝑓 = 316.5 𝑤𝑒 ℎ𝑎𝑣𝑒:
𝜗 ≅ 17.36 × 10−6 ൗ𝑚2
𝑠
𝑘 ≅ 0.02749 ൗ𝑊
𝑚℃
𝑃𝑟𝑎𝑖𝑟 ≅ 0.7
𝑐 𝑝 ≅ 1.006 ൗ
𝑘𝐽
𝑘𝑔℃
First we should check the stream that is laminar or not?
𝑅𝑒 =
𝜌𝑢∞ 𝑥
𝜇
=
𝑢∞ 𝑥
𝜗
=
(2)(0.20)
17.36 × 10−6
= 23041 < 105
So the stream is laminar.
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‫مثال‬
𝑁𝑢 𝑥 =
ℎ 𝑥 𝑥
𝑘
= 0.332𝑃𝑟 ൗ1
3 𝑅𝑒 ൗ1
2 = 0.332 0.7 ൗ1
3 23041 ൗ1
2 = 44.74
ℎ 𝑥 = 𝑁𝑢 𝑥
𝑘
𝑥
= 44.74
0.02749
0.20
= 6.15 ൗ𝑊
𝑚2℃
തℎ = 2ℎ 𝑥=𝐿 = 2 × 6.15 = 12.3 ൗ𝑊
𝑚2℃
𝑞 = തℎ𝐴 𝑇 𝑤 − 𝑇∞ = 12.3 1 0.20 333 − 300 = 81.18 𝑊
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‫مثال‬
𝑅𝑒 =
𝜌𝑢∞ 𝑥
𝜇
=
𝑢∞ 𝑥
𝜗
=
(2)(0.40)
17.36 × 10−6
= 46082 < 105
𝑁𝑢 𝑥 =
ℎ 𝑥 𝑥
𝑘
= 0.332𝑃𝑟 ൗ1
3 𝑅𝑒 ൗ1
2 = 0.332 0.7 ൗ1
3 46082 ൗ1
2 = 63.28
𝑘
𝑥
= 63.28
0.02749
0.40
= 4.349 ൗ𝑊
𝑚2℃
തℎ = 2ℎ 𝑥=𝐿 = 2 × 4.349 = 8.698 ൗ𝑊
𝑚2℃
𝑞 = തℎ𝐴 𝑇 𝑤 − 𝑇∞ = 8.698 1 0.40 333 − 300 = 114.6 𝑊
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‫آنالوژی‬‫کولبرن‬
•‫سیاالت‬ ‫مکانیک‬ ‫در‬2‫درگ‬ ‫ضریب‬ ‫که‬ ‫خواندیم‬(‫اصطکاک‬)‫آید‬‫می‬ ‫دست‬ ‫به‬ ‫شکل‬ ‫این‬ ‫به‬:
𝐶 𝑑 = 𝐶𝑓 =
𝜏 𝑤
1
2
𝜌𝑢∞
2
•‫تنش‬ ‫آن‬ ‫در‬ ‫که‬‫برشی‬‫شود‬‫می‬ ‫دیواره‬ ‫به‬ ‫مربوط‬:
𝜏 𝑤 = 𝜏 𝑦=0 = 𝜇
𝑑𝑢
𝑑𝑦
| 𝑦=0
•‫کنیم‬‫می‬ ‫محاسبه‬ ‫را‬ ‫درگ‬ ‫ضریب‬ ،‫ایم‬ ‫آورده‬ ‫دست‬ ‫به‬ ‫تابحال‬ ‫که‬ ‫روابطی‬ ‫برای‬:
𝜏 𝑤 = 𝜇
𝑑𝑢
𝑑𝑦
| 𝑦=0 = 𝜇𝑢∞
3
2𝛿
−
3𝑦2
2𝛿3
𝑦=0
=
3𝜇𝑢∞
2𝛿
=
3𝜇𝑢∞
2
𝜌𝑢∞ 𝑥
𝜇
ൗ1
2
4.64𝑥
20
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1
2
𝐶 𝑑 =
𝜏 𝑤
𝜌𝑢∞
2
=
3𝜇𝑢∞ 𝜌 ൗ1
2 𝑢∞
ൗ1
2 𝑥 ൗ1
2
2 4.64 𝑥𝜇 ൗ1
2 𝜌𝑢∞
2
=
3𝜇 ൗ1
2
2 4.64 𝑢∞
ൗ1
2 𝑥 ൗ1
2 𝜌 ൗ1
2
1
2
𝐶 𝑑 ≈ 0.323𝑅𝑒 𝑥
− ൗ1
2
I
•‫عدد‬‫استانتون‬:
𝑠𝑡 =
𝑁𝑢
𝑅𝑒 𝑃𝑟
=
ℎ𝑥
𝑘
𝑢∞ 𝑥
𝜗
𝜗
𝛼
=
ℎ
𝑘
𝜌𝑐 𝑝
𝑘𝑢∞
=
ℎ
𝜌𝑐 𝑝 𝑢∞
21
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𝑁𝑢 𝑥 =
ℎ 𝑥 𝑥
𝑘
= 0.332𝑃𝑟 ൗ1
3 𝑅𝑒 ൗ1
2
𝑠𝑡 =
𝑁𝑢 𝑥
𝑃𝑒𝑐
=
𝑁𝑢 𝑥
𝑅𝑒 𝑃𝑟
= 0.332
𝑃𝑟 ൗ1
3 𝑅𝑒 ൗ1
2
𝑅𝑒 𝑃𝑟
= 0.332𝑃𝑟− ൗ2
3 𝑅𝑒− ൗ1
2
𝑠𝑡𝑃𝑟 ൗ2
3 = 0.332𝑅𝑒− ൗ1
2 (II)
22
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1
2
𝐶 𝑑 ≈ 0.323𝑅𝑒 𝑥
− ൗ1
2
I
𝑠𝑡𝑃𝑟 ൗ2
3 = 0.332𝑅𝑒− ൗ1
2 II
1
2
𝐶 𝑑 ≅ 𝑠𝑡𝑃𝑟 ൗ2
3 = 0.332𝑅𝑒− ൗ1
2
23
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‫مثال‬
•‫در‬‫مثال‬،‫قبل‬‫اگر‬‫طول‬‫صفحه‬40‫سانتی‬‫متر‬‫بوده‬‫و‬‫عمق‬‫صفحه‬1‫واحد‬‫باشد‬‫و‬‫فشار‬‫هوا‬‫نیز‬1‫اتمسفر؛‬
‫مطلوبست‬‫محاسبه‬‫مقدار‬‫درگ؟‬
‫حل‬:
𝑃 = 𝜌𝑅𝑇 ⟹ 𝜌 =
𝑃
𝑅𝑇
=
1.0132 × 105
287 316.5
= 1.115 ൗ𝑘𝑔
𝑚3
𝑠𝑡 =
𝑁𝑢
𝑅𝑒 𝑃𝑟
=
63.28
(46082)(0.7)
= 3.88 × 10−3
1
2
𝐶 𝑑 = 𝑠𝑡𝑃𝑟 ൗ2
3 = 3.88 × 10−3
0.7 ൗ2
3 = 3.06 × 10−3
24
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‫مثال‬
𝐶 𝑑 =
𝜏 𝑤
1
2
𝜌𝑢∞
2
⟹ 𝜏 𝑤 =
1
2
𝐶 𝑑 𝜌𝑢∞
2
= 3.06 × 10−3
1.115 22
𝜏 𝑤 = 0.0136 ൗ𝑁
𝑚2
𝐷𝑟𝑎𝑔 = 𝐷 = 𝜏 𝑤 𝐴 = 0.0136 1 0.40 = 0.00544 𝑁
25
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‫تخت‬ ‫صفحه‬ ‫روی‬ ‫مغشوش‬ ‫جریان‬
• (Reminder) In laminar flow, the friction coefficient obtains from:
𝑓 =
64
𝑅𝑒
• Colburn analogy was for polished horizontal planes and a link between
heat transfer and shear stress on the wall. This analogy is used both
laminar and turbulent flows.
• In fluid mechanics mentioned that the velocity distribution for turbulent
flow is:
𝑢
𝑢∞
=
𝑦
𝛿
ൗ1
7
26
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• By replacing this formula in von-karman equation
𝑑
𝑑𝑥
න
0
𝛿
𝑢∞ − 𝑢 𝑢 𝑑𝑦 = 𝜐
𝑑𝑢
𝑑𝑦
| 𝑦=0
a) If flow from edge of plane is turbulent:
𝛿
𝑥
= 0.381𝑅𝑒 𝑥
− ൗ1
5
b) If flow is first laminar and then turbulent:
𝛿
𝑥
= 0.381𝑅𝑒 𝑥
− ൗ1
5
− 10256𝑅𝑒 𝑥
−1
27
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𝑅𝑒 𝑐𝑟 =
𝑢∞ 𝑥 𝑐
𝜗
≅ 5 × 105
⟹ 𝑥 𝑐 = 5 × 105
𝑢∞
𝜗
• Well-known equation for turbulent flow on polished horizontal plane is:
𝑁𝑢 𝐿 =
തℎ𝐿
𝑘
= 𝑃𝑟 ൗ1
3(0.037 𝑅𝑒 𝐿
0.8
− 850)
• Where ReL defines as:
𝑅𝑒 𝐿 =
𝑢∞ 𝐿
𝜗
• L is plane’s length
• Almost all problems about turbulent solve experimentaly!
28
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‫مایع‬ ‫فلزات‬ ‫در‬ ‫جابجایی‬ ‫حرارت‬ ‫انتقال‬
𝑃𝑟 ≈ 0.001 to 0.01
•‫دارند‬ ‫کاربرد‬ ‫بزرگتر‬ ‫فضای‬ ‫یک‬ ‫به‬ ‫کوچک‬ ‫فضای‬ ‫یک‬ ‫از‬ ‫حرارت‬ ‫سریع‬ ‫انتقال‬ ‫برای‬.
•‫سریع‬ ‫بسیار‬ ‫حرارت‬ ‫پخش‬
•‫مرزی‬ ‫الیه‬ ‫طول‬ ‫تمامی‬ ‫در‬ ‫سرعت‬ ‫گرفتن‬ ‫نظر‬ ‫در‬ ‫ثابت‬𝑢 = 𝑢∞
29
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•‫دما‬ ‫قبلی‬ ‫پروفیل‬ ‫از‬ ‫استفاده‬:
𝑇 − 𝑇 𝑤
𝑇∞ − 𝑇 𝑤
=
3
2
𝑦
𝛿𝑡
−
3
2
𝑦
𝛿𝑡
3
•‫معادله‬ ‫از‬‫انتگرالی‬‫داشت‬ ‫خواهیم‬ ‫انرژی‬:
𝑑
𝑑𝑥
න
0
𝛿 𝑡
𝑇∞ − 𝑇 𝑢 𝑑𝑦 = 𝛼
𝜕𝑇
𝜕𝑦
| 𝑦=0
•‫با‬ ‫است‬ ‫برابر‬ ‫حرارتی‬ ‫مرزی‬ ‫الیه‬ ‫ضخامت‬ ،‫جدید‬ ‫روابط‬ ‫براساس‬ ‫بنابراین‬:
𝛿𝑡 =
8𝛼𝑥
𝑢∞
= 𝑓(𝑢∞
ൗ1
2
, 𝑥 ൗ1
2)
30
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•‫که‬ ‫آوردیم‬ ‫دست‬ ‫به‬ ً‫ال‬‫قب‬ ‫طرفی‬ ‫از‬:
ℎ =
−𝑘∞
𝑑𝑇
𝑑𝑦
| 𝑦=0
𝑇 𝑤 − 𝑇∞
=
3𝑘∞
2𝛿𝑡
‫باال‬ ‫فرمول‬ ‫در‬ ‫رابطه‬ ‫این‬ ‫جایگذاری‬ ‫با‬𝛿𝑡 =
8𝛼𝑥
𝑢∞
‫داریم‬:
𝑁𝑢 =
ℎ𝑥
𝑘
=
3𝑘
2𝛿𝑡
𝑥
𝑘
=
3𝑥
2𝛿𝑡
=
3𝑥 𝑢∞
2 8𝛼𝑥
= 0.53 𝑅𝑒 𝑃𝑟 ൗ1
2 = 0.53𝑃𝑒𝑐 ൗ1
2
31
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•‫نیست‬ ‫مذاب‬ ‫سیال‬ ‫که‬ ‫مواقعی‬ ‫در‬ ‫نوسلت‬ ‫عدد‬ ‫کلی‬ ‫فرم‬:
𝑁𝑢 𝑥 = 𝐶𝑅𝑒 𝑚
𝑃𝑟 𝑛
•‫افقی‬ ‫مسطح‬ ‫صفحه‬ ‫روی‬ ‫سیال‬ ،‫مثال‬ ‫برای‬:
n=1/3
m=1/2
C=0.332
32
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‫مسئله‬
33
Experimental measurements of the convection heat transfer coefficient for a
square bar in cross flow yielded the following values:
തℎ1 = 50 Τ𝑊
𝑚2.𝐾 when 𝑉1 = 20 Τ𝑚
𝑠
തℎ2 = 40 Τ𝑊
𝑚2.𝐾 when 𝑉2 = 15 Τ𝑚
𝑠
Assume that the functional form of the Nusselt number is 𝑁𝑢 = 𝐶 𝑅𝑒 𝑚
𝑃𝑟 𝑛
,
where C, m, and n are constants.
(a) What will be the convection heat transfer coefficient for a similar bar with
L = 1 m when V = 15 m/s?
(b) What will be the convection heat transfer coefficient for a similar bar with
L = 1 m when V = 30 m/s?
Would your results be the same if the side of the bar, rather than its diagonal,
were used as the characteristic length?
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‫مسئله‬ ‫حل‬
34
KNOWN: Experimental measurements of the heat transfer
coefficient for a square bar in cross flow.
FIND: (a) h for the condition when L = 1m and V = 15m/s, (b) h for
the condition when L = 1m and V = 30m/s, (c) Effect of defining a
side as the characteristic length.
ASSUMPTIONS: (1) Functional form 𝑁𝑢 = C Rem Prn applies with
C, m, n being constants, (2) Constant properties.
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‫مسئله‬ ‫حل‬(‫ادامه‬)
35
ANALYSIS: (a) For the experiments and the condition L = 1m and V = 15m/s, it
follows that Pr as well as C, m, and n are constants. Hence
തℎ𝐿 ∝ 𝑉𝐿 𝑚
Using the experimental results, find m. Substituting values
തℎ1 𝐿1
തℎ2 𝐿2
=
𝑉1 𝐿1
𝑉2 𝐿2
𝑚
50 0.5
(40)(0.5)
=
20 0.5
(15)(0.5)
𝑚
giving m = 0.782. It follows then for L = 1m and V = 15m/s,
തℎ = തℎ1
𝐿1
𝐿
𝑉𝐿
𝑉1 𝐿1
𝑚
= 50
0.5
1.0
15 1.0
20 0.5
0.782
= 34.3 𝑊/𝑚2
𝐾
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36
(b) For the condition L = 1m and V = 30m/s, find
തℎ = തℎ1
𝐿1
𝐿
𝑉𝐿
𝑉1 𝐿1
𝑚
= 50
0.5
1.0
30 1.0
20 0.5
0.782
= 59.0 𝑊/𝑚2
𝐾
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37
(c) If the characteristic length were chosen as a side rather than the
diagonal, the value of C would change. However, the coefficients m
and n would not change.
COMMENTS: The foregoing Nusselt number relation is used
frequently in heat transfer analysis, providing appropriate scaling for
the effects of length, velocity, and fluid properties on the heat transfer
coefficient.
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‫مسئله‬
38
Experimental results for heat transfer over a flat plate with
an extremely rough surface were found to be correlated by
an expression of the form
𝑁𝑢 𝑥 = 0.04 𝑅𝑒 𝑥
0.9
𝑃𝑟1/3
where Nux is the local value of the Nusselt number at a
position x measured from the leading edge of the plate.
Obtain an expression for the ratio of the average heat
transfer coefficient തℎ 𝑥 to the local coefficient hx.
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39
KNOWN: Local Nusselt number correlation for flow over a roughened surface.
FIND: Ratio of average heat transfer coefficient to local coefficient.
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40
ANALYSIS: The local convection coefficient is obtained from the
prescribed correlation,
𝑘
𝑥
= 0.04
𝑘
𝑥
𝑅𝑒 𝑥
0.9
𝑃𝑟 ൗ1
3
ℎ 𝑥 = 0.04 𝑘
𝑉
𝑣
0.9
𝑃𝑟
1
3
𝑥0.9
𝑥
≡ 𝐶1 𝑥−0.1
To determine the average heat transfer coefficient for the length zero to x,
തℎ 𝑥 ≡
1
𝑥
න
0
𝑥
ℎ 𝑥 𝑑𝑥 =
1
𝑥
𝐶1 න
0
𝑥
𝑥−0.1
𝑑𝑥
തℎ 𝑥 =
𝐶1
𝑥
𝑥0.9
𝑥
= 1.11 𝐶1 𝑥−0.1
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41
Hence, the ratio of the average to local coefficient is
തℎ 𝑥
ℎ 𝑥
=
1.11 𝐶1 𝑥−0.1
𝐶1 𝑥−0.1
= 1.11
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•‫مذاب‬ ‫سیال‬ ‫برای‬ ‫نوسلت‬ ‫عدد‬ ‫کلی‬ ‫فرم‬:
𝑁𝑢 𝑥 = 𝐶𝑃𝑒𝑐 𝑚
•‫افقی‬ ‫مسطح‬ ‫صفحه‬ ‫روی‬ ‫سیال‬ ،‫مثال‬ ‫برای‬:
m=1/2
C=0.530
•‫جدول‬6-8‫کتاب‬‫هولمن‬
42
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‫مسئله‬
43
Sketch the variation of the velocity and
thermal boundary layer thicknesses with
distance from the leading edge of a flat
plate for the laminar flow of air, water,
engine oil, and mercury. For each case
assume a mean fluid temperature of 300 K.
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44
KNOWN: Air, water, engine oil or mercury at 300K in laminar,
parallel flow over a flat plate.
FIND: Sketch of velocity and thermal boundary layer thickness.
ASSUMPTIONS: (1) Laminar flow.
PROPERTIES: For the fluids at 300K:
Fluid Table Pr
Air A.4 0.71
Water A.6 5.83
Engine Oil A.5 6400
Mercury A.5 0.025
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45
ANALYSIS: For laminar, boundary layer flow over a flat plate.
𝛿
𝛿𝑡
~𝑃𝑟 𝑛
where n > 0. Hence, the boundary layers appear as shown below.
Air:
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46
COMMENTS: Although Pr
strongly influences relative
boundary layer development
in laminar flow, its influence
is weak for turbulent flow.
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‫مسئله‬
47
KNOWN: Velocity and temperature profiles and shear stress-boundary layer
thickness relation for turbulent flow over a flat plate.
FIND: (a) Expressions for hydrodynamic boundary layer thickness and average
friction coefficient, (b) Expressions for local and average Nusselt numbers.
ASSUMPTIONS: (1) Steady flow, (2) Constant properties, (3) Fully turbulent
boundary layer, (4) Incompressible flow, (5) Isothermal plate, (6) Negligible viscous
dissipation, (7) 𝛿 = 𝛿𝑡.
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48
𝜌𝑢∞
2
𝑑
𝑑𝑥
න
0
𝛿
1 −
𝑢
𝑢∞
𝑢
𝑢∞
𝑑𝑦 = 𝜏 𝑠
Substituting the expression for the wall shear stress
𝜌𝑢∞
2
𝑑
𝑑𝑥
න
0
𝛿
1 −
𝑦
𝛿
1
7 𝑦
𝛿
1
7
𝑑𝑦 = 0.228 𝜌𝑢∞
2
𝑢∞ 𝛿
𝑣
−
1
4
𝑑
𝑑𝑥
න
0
𝛿
𝑦
𝛿
1
7
−
𝑦
𝛿
2
7
𝑑𝑦 =
𝑑
𝑑𝑥
ተ
7
8
𝑦
8
7
𝛿
1
7
−
7
9
𝑦
9
7
𝛿
2
7
0
𝛿
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49
𝑑
𝑑𝑥
7
8
𝛿 −
7
9
𝛿 = 0.0228
𝑢∞ 𝛿
𝑣
−
1
4
7
72
𝑑𝛿
𝑑𝑥
= 0.0228
𝑣
𝑢∞
1
4
𝛿−
1
4
7
72
න
0
𝛿
𝛿
1
4 𝑑𝛿 = 0.0228
𝑣
𝑢∞
1
4
න
0
𝑥
𝑑𝑥
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50
7
72
×
4
5
𝛿
5
4 = 0.0228
𝑣
𝑢∞
1
4
𝑥
𝛿 = 0.376
𝑣
𝑢∞
1
5
𝑥
4
5,
𝛿
𝑥
= 0.376 𝑅𝑒 𝑥
−
1
5
Knowing𝛿, it follows
𝜏 𝑠 = 0.0228 𝜌𝑢∞
2
𝑣
𝑢∞
−
1
4
0.376 𝑥 𝑅𝑒 𝑥
−
1
5
−
1
4
𝐶𝑓,𝑥 =
𝜏 𝑠
𝜌𝑢∞
2
2
= 0.0456 0.376
𝑢∞
𝑣
𝑢∞
𝑣
−
1
5
𝑥𝑥−
1
5
−
1
4
= 0.0592 𝑅𝑒 𝑥
−
1
5
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51
The average friction coefficient is then
ҧ𝐶𝑓,𝑥 =
1
𝑥
න
0
𝑥
𝐶𝑓,𝑥 𝑑𝑥 =
1
𝑥
0.0592
𝑣
𝑢∞
−
1
5
න
0
𝑥
𝑥−
1
5 𝑑𝑥
ҧ𝐶𝑓,𝑥 =
1
𝑥
0.0592
𝑣
𝑢∞
−
1
5
𝑥
4
5
5
4
= 0.074 𝑅𝑒 𝑥
−
1
5
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52
(b) The energy integral equation for turbulent flow is
𝑑
𝑑𝑥
න
0
𝛿 𝑡
𝑢 𝑇∞ − 𝑇 𝑑𝑦 =
𝑞 𝑠
′′
𝜌𝑐 𝑝
= −
ℎ
𝜌𝑐 𝑝
𝑇𝑠 − 𝑇∞
Hence,
𝑢∞
𝑑
𝑑𝑥
න
0
𝛿 𝑡 𝑢
𝑢∞
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
𝑑𝑦 = 𝑢∞
𝑑
𝑑𝑥
න
0
𝛿 𝑡 𝑦
𝛿
1
7
1 −
𝑦
𝛿𝑡
1
7
𝑑𝑦 =
ℎ
𝜌𝑐 𝑝
𝑢∞
𝑑
𝑑𝑥
7
8
𝛿𝑡
8
7
𝛿
1
7
−
7
9
𝛿𝑡
8
7
𝛿
1
7
=
ℎ
𝜌𝑐 𝑝
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53
or, with 𝜉 ≡ 𝛿𝑡/𝛿,
𝑢∞
𝑑
𝑑𝑥
7
8
𝛿𝜉
8
7 −
7
9
𝛿𝜉
8
7 =
ℎ
𝜌𝑐 𝑝
𝑢∞
𝑑
𝑑𝑥
7
72
𝛿𝜉
8
7 =
ℎ
𝜌𝑐 𝑝
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54
or, with 𝜉 ≡ 𝛿𝑡/𝛿,
𝑢∞
𝑑
𝑑𝑥
7
8
𝛿𝜉
8
7 −
7
9
𝛿𝜉
8
7 =
ℎ
𝜌𝑐 𝑝
𝑢∞
𝑑
𝑑𝑥
7
72
𝛿𝜉
8
7 =
ℎ
𝜌𝑐 𝑝
Hence, with 𝜉 ≈ 1 and /𝑥 = 0.376 𝑅𝑒 𝑥
−1/5
,
7
72
𝑢∞ 0.376
𝑢∞
𝑣
−
1
5 𝑑 𝑥
4
5
𝑑𝑥
=
ℎ
𝜌𝑐 𝑝
ℎ = 0.0292 𝜌 𝑐 𝑝 𝑢∞ 𝑅𝑒 𝑥
−
1
5
= 0.0292
𝑘
𝑥
𝑣
𝛼
𝑢∞ 𝑥
𝑣
𝑅𝑒 𝑥
−
1
5
𝑁𝑢 𝑥 =
ℎ𝑥
𝑘
= 0.0292 𝑅𝑒 𝑥
4
5
𝑃𝑟
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55
Hence,
തℎ 𝑥 =
1
𝑥
න
0
𝑥
ℎ 𝑑𝑥 =
0.0292 𝑃𝑟
𝑥
𝑘
𝑢∞
𝑣
4
5
න
0
𝑥
𝑥−
1
5 𝑑𝑥 = 0.0292
𝑘
𝑥
𝑃𝑟
𝑢∞ 𝑥
𝑣
4
5 5
4
𝑁𝑢 𝑥 =
തℎ 𝑥 𝑥
𝑘
= 0.037 𝑅𝑒 𝑥
4
5
𝑃𝑟
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‫جریان‬‫خزشی‬‫کره‬ ‫روی‬
• Creeping flow (Re<1) on sphere:
𝑁𝑢 =
ℎ𝑑
𝑘
= 2
ℎ =
𝑁𝑢 𝑘
𝑑
=
2𝑘
𝑑
⟹ 𝑞 = ℎ𝐴 𝑠𝑝ℎ𝑒𝑟𝑒 𝑇 𝑤 − 𝑇∞ =
2𝑘
𝑑
4𝜋𝑅2
𝑇 𝑤 − 𝑇∞
𝑞 = 2𝜋𝑑𝑘 𝑇 𝑤 − 𝑇∞
56
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‫لوله‬ ‫درون‬ ‫حرارت‬ ‫انتقال‬
• (Re<2200) Laminar
• (2200<Re<2800) Transition
• (Re>2800) Turbulant
• Laminar stream in steady state condition and fully developed and
incompressible flow:
𝑢 = −
𝑅2
4𝜇
𝜕𝑃
𝜕𝑥
1 −
𝑟
𝑅
2
𝑁𝑢 𝑑 =
ℎ𝑑
𝑘
= 4.364
57
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• Bulk temperature(Tb) is the mean temp. of flow in each section.
• For calculating Tb, solve the Energy equation in cylindrical coordinates
and obtain temperature distribution and get mean of each section’s
temperature.
• We will explain how to calculate the Tb.
• So, we have:
𝑞 = ℎ𝐴 𝑇 𝑤 − 𝑇𝑏 = 4.364
𝑘
𝑑
𝜋𝑑𝐿 𝑇 𝑤 − 𝑇𝑏 = 4.364 𝑘𝜋𝐿 𝑇 𝑤 − 𝑇𝑏
• Convected heat transferred per length unit:
𝑞
𝐿
= 4.364 𝑘𝜋 𝑇 𝑤 − 𝑇𝑏
58
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• In other way, we know that
the Enthalpy changes from
one section to other. Then:
• ሶ𝑞 = ሶ𝑚𝑐 𝑝(𝑇𝑏1
− 𝑇𝑏2
)
59
𝑇𝑏1
𝑇𝑏2
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‫دمای‬ ‫محاسبه‬‫بالک‬
𝑇𝑏 = ത𝑇 =
‫׬‬0
𝑅
ሶ𝑚(𝑐 𝑝 𝑇)
‫׬‬0
𝑅
ሶ𝑚(𝑐 𝑝)
• ሶ𝑚 = 𝜌𝐴𝑢 = 𝜌 2𝜋𝑟𝑑𝑟 𝑢
𝑇𝑏 = ത𝑇 =
‫׬‬0
𝑅
𝜌 2𝜋𝑟𝑑𝑟 𝑢(𝑐 𝑝 𝑇)
‫׬‬0
𝑅
𝜌 2𝜋𝑟𝑑𝑟 𝑢(𝑐 𝑝)
𝑇𝑏 =
‫׬‬0
𝑅
𝑢𝑇𝑟 𝑑𝑟
‫׬‬0
𝑅
𝑢𝑟 𝑑𝑟
60
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‫دما‬ ‫توزیع‬ ‫محاسبه‬ ‫برای‬ ‫لوله‬ ‫در‬ ‫سرعت‬ ‫توزیع‬ ‫معادالت‬
• For laminar flow:
𝑢 = −
𝑅2
4𝜇
𝜕𝑃
𝜕𝑥
1 −
𝑟
𝑅
2
𝑢 𝑚𝑎𝑥 = −
𝑅2
4𝜇
𝜕𝑃
𝜕𝑥
𝑢
𝑢 𝑚𝑎𝑥
= 1 −
𝑟
𝑅
2
⟹ 𝑢∗
= 1 − 𝑥∗2
61
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‫دما‬ ‫توزیع‬ ‫محاسبه‬ ‫برای‬ ‫لوله‬ ‫در‬ ‫سرعت‬ ‫توزیع‬ ‫معادالت‬
• For turbulent flow:
𝑢
𝑢 𝑐𝑒𝑛𝑡𝑒𝑟
= (1 −
𝑟
𝑅
) ൗ1
7
62
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‫دمای‬ ‫محاسبه‬‫بالک‬‫نوسلت‬ ‫و‬
• Temperature distribution in a pipe with radius R:
𝑇 − 𝑇𝑐𝑒𝑛𝑡𝑒𝑟 =
1
𝛼
𝜕𝑇
𝜕𝑥
𝑢 𝑚𝑎𝑥 𝑅2
4
𝑟
𝑅
2
−
1
4
𝑟
𝑅
4
• So, the bulk temperature is:
𝑇𝑏 = 𝑇𝑐 +
7
96
𝑢 𝑚𝑎𝑥
𝛼
𝑅2
𝜕𝑇
𝜕𝑥
• For turbulent flow and polished pipe, best formula for Nusselt is:
𝑁𝑢 𝑑 =
ℎ𝑑
𝑘
= 0.023 𝑅𝑒 𝑑
0.8
𝑃𝑟 𝑛
𝑛 = ቊ
0.4 𝑖𝑓 𝑓𝑙𝑢𝑖𝑑 𝑖𝑠 ℎ𝑒𝑎𝑡𝑒𝑑
0.3 𝑖𝑓 𝑓𝑙𝑢𝑖𝑑 𝑖𝑠 𝑐𝑜𝑜𝑙𝑒𝑑
63
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‫لوله‬ ‫داخل‬ ‫مغشوش‬ ‫جریان‬
• Eddy prandtl number is defined for turbulent flow as:
𝑃𝑟𝑡 =
𝜀 𝑚
𝜀 𝐻
=
𝑒𝑑𝑑𝑦 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑒𝑑𝑑𝑦 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡
• For rough tube and flow with pandtl number about one:
𝑠𝑡 =
ℎ
𝜌𝑐 𝑝 ത𝑢
=
𝑁𝑢
𝑃𝑒𝑐
=
𝑓
8
= 𝐴𝑛𝑎𝑙𝑜𝑔𝑦 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠
• Where f is friction coefficient obtains from moody diagram.
• For using the moody diagram, we need to know the Reynolds number and
the rate of “rough coefficient” to “diameter”.
• Then the Reynolds analogy shows the relation between flow friction with
surface and heat transfer in turbulent flows.
64
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‫مسئله‬
65
Fully developed conditions are known to
exist for water flowing through a 25-
mm-diameter tube at 0.01 kg/s and 27
ºC. What is the maximum velocity of the
water in the tube? What is the pressure
gradient associated with the flow?
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66
KNOWN: Flowrate and temperature of water in fully developed flow through
a tube of prescribed diameter.
FIND: Maximum velocity and pressure gradient.
ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal flow.
PROPERTIES: Table A-6, Water (300K): ρ = 998 kg/m3, µ = 855×10-6
N×s/m2.
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67
ANALYSIS: From Eq. 8.6,
𝑅𝑒 𝐷 =
4 ሶ𝑚
𝜋𝐷𝜇
=
4 0.01
𝜋 0.025 855 × 10−6
= 596
Hence the flow is laminar and the velocity profile is given by Eq. 8.15,
𝑢 𝑟
𝑢 𝑚
= 2 1 −
𝑟
𝑟𝑜
2
The maximum velocity is therefore at r = 0, the centerline, where
𝑢 0 = 2𝑢 𝑚
From Eq. 8.5
𝑢 𝑚 =
ሶ𝑚
𝜌𝜋 𝐷2/4
=
4 0.001
998 𝜋 0.025
= 0.020 𝑚/𝑠
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68
hence
𝑢 0 = 0.041 𝑚/𝑠
Combining Eqs. 8.16 and 8.19, the pressure gradient is
𝑑𝑝
𝑑𝑥
= −
64
𝑅𝑒 𝐷
𝜌𝑢 𝑚
2
2𝐷
𝑑𝑝
𝑑𝑥
= −
64
596
×
998 0.020
2 0.025
= −0.86 𝑘𝑔/𝑚2
𝑠2
𝑑𝑝
𝑑𝑥
= −0.86 𝑁/𝑚2
𝑚 = −0.86 × 10−5
𝑏𝑎𝑟/𝑚
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‫مسئله‬
69
For fully developed laminar flow through a parallel-plate channel, the x-momentum
equation has the form
𝜇
𝑑2
𝑢
𝑑𝑦2
=
𝑑𝑝
𝑑𝑥
= constant
The purpose of this problem is to develop expressions for the velocity distribution and
pressure gradient analogous to those for the circular tube in Section 8.1.
(a) Show that the velocity profile, u(y), is parabolic and of the form
𝑢 𝑦 =
2
3
𝑢 𝑚 1 −
𝑦2
ൗ𝑎
2
2
Where um is the mean velocity
𝑢 𝑚 = −
𝑎2
12𝜇
𝑑𝑝
𝑑𝑥
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‫مسئله‬(‫ادامه‬)
70
And − ൗ𝑑𝑝
𝑑𝑥 = ൗ∆𝑝
𝐿, where Δp is the
pressure drop across the channel of
length L.
(a) Write an expression defining the friction
factor, f, using the hydraulic diameter Dh
as the characteristic length. What is the
hydraulic diameter for the parallel-plate
channel?
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‫مسئله‬(‫ادامه‬)
71
(c) The friction factor is estimated from the expression 𝑓
= 𝐶/𝑅𝑒 𝐷ℎ
, where C depends upon the flow cross section, as
shown in Table 8.1. What is the coefficient C for the parallel-
plate channel?
(d) Airflow in a parallel-plate channel with a separation of 5
mm and a length of 200 mm experiences a pressure drop of
Δp = 3.75 N/m2. Calculate the mean velocity and the
Reynolds number for air at atmospheric pressure and 300 K.
Is the assumption of fully developed flow reasonable for this
application? If not, what is the effect on the estimate for um?
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72
KNOWN: The x-momentum equation for fully developed laminar flow in a parallel-
plate channel
𝑑𝑝
𝑑𝑥
= constant = 𝜇
𝑑2
𝑢
𝑑𝑦2
FIND: Following the same approach as for the circular tube in Section 8.1: (a) Show
that the velocity profile, u(y), is parabolic of the form
𝑢 𝑦 =
2
3
𝑢 𝑚 1 −
𝑦2
ൗ𝑎
2
2
Where um is the mean velocity expressed as
𝑢 𝑚 = −
𝑎2
12𝜇
𝑑𝑝
𝑑𝑥
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73
and -dp/dx = Δp/L where Δp is the pressure drop across the channel of length L;
(b) Write the expression defining the friction factor, f, using the hydraulic
diameter as the characteristic length, Dh; What is the hydraulic diameter for the
parallel-plate channel? (c) The friction factor is estimated from the expression f
= C ReDh where C depends upon the flow cross-section as shown in Table 8.1;
What is the coefficient C for the parallel-plate channel (b/a→ ∞)? (d) Calculate
the mean air velocity and the Reynolds number for air at atmospheric pressure
and 300 K in a parallel-plate channel with separation of 5 mm and length of 100
mm subjected to a pressure drop of ΔP = 3.75 N/m2; Is the assumption of fully
developed flow reasonable for this application? If not, what effect does this have
on the estimate for um?
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74
ANALYSIS: (a) The x-momentum equation for fully developed
laminar flow is
𝜇
𝑑2
𝑢
𝑑𝑦2
=
𝑑𝑝
𝑑𝑥
= constant
Since the longitudinal pressure gradient is constant, separate
variables and integrate twice,
𝑑
𝑑𝑦
𝑑𝑢
𝑑𝑦
=
1
𝜇
𝑑𝑝
𝑑𝑥
𝑑𝑢
𝑑𝑦
=
1
𝜇
𝑑𝑝
𝑑𝑥
𝑦 + 𝐶1
𝑢 =
1
2𝜇
𝑑𝑝
𝑑𝑥
𝑦2
+ 𝐶1 𝑦 + 𝐶2
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75
ASSUMPTIONS: (1) Fully developed laminar flow, (2) Parallel-plate
channel, a << b.
PROPERTIES: Table A-4, Air (300 K, 1 atm): μ = 184.6 × 10-7 N⋅s/m2, ν =
15.89 × 10-6 m2/s.
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76
The integration constants are determined from the boundary conditions,
ቤ
𝑑𝑢
𝑑𝑦 𝑦=0
= 0 𝑢(𝑎/2) = 0
to find
𝐶1 = 0 𝐶2 = −
1
2𝜇
𝑑𝑝
𝑑𝑥
𝑎/2 2
giving
𝑢 𝑦 = −
𝑎/2 2
2𝜇
𝑑𝑝
𝑑𝑥
1 −
𝑦2
𝑎/2 2
The mean velocity is
𝑢 𝑚 =
2
𝑎
න
0
𝑎/2
𝑢 𝑦 𝑑𝑦 = −
2
𝑎
𝑎/2 2
2𝜇
𝑑𝑝
𝑑𝑥
𝑦 −
𝑦3
𝑎/2 2
0
𝑎/2
=
𝑎2
12𝜇
−
𝑑𝑝
𝑑𝑥
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77
(c) Substituting for the pressure gradient, Eq. (3), and rearranging, find using Eq.
(6)
𝑓 =
𝑢 𝑚
𝑎2/12𝜇
𝐷ℎ
𝜌𝑢 𝑚
2
/2
=
96
𝑢 𝑚 𝐷ℎ/𝑣
=
96
𝑅𝑒 𝐷ℎ
where the Reynolds number is
𝑅𝑒 𝐷ℎ
= 𝑢 𝑚 𝐷ℎ/𝑣
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78
Substituting Eq. (3) for dp/dx into Eq. (2) find the velocity distribution in terms
of the mean velocity
𝑢 𝑦 =
3
2
𝑢 𝑚 1 −
𝑦2
𝑎/2 2
(b) The friction factor follows from its definition, Eq. 8.16,
𝑓 = −
𝑑𝑝/𝑑𝑥 𝐷ℎ
𝜌𝑢 𝑚
2
/2
where the hydraulic diameter for the channel using Eq. 8.67 is
𝐷ℎ =
4𝐴 𝑐
𝑃
=
4 𝑎 × 𝑏
2 𝑎 + 𝑏
= 2𝑎
Since a<<b.
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79
This result is in agreement with Table 8.1 for the cross-section with b/a → ∞ where
C = 96
(d) For the conditions shown in the schematic, with air properties evaluated at 300 K,
using Eqs. (3) and (8), find
𝑢 𝑚 =
0.005
12 184.6 × 10−7
3.75
0.100
= 1.06 𝑚/𝑠
𝑅𝑒 𝐷 =
1.06 2 0.005
15.89 × 10−6
= 667
The flow is laminar as ReDh < 2300, and from Eq. 8.3, the entry length is
𝑥 𝑓𝑑,ℎ
𝐷ℎ 𝑙𝑎𝑚
= 0.05 𝑅𝑒 𝐷ℎ
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80
𝑥 𝑓𝑑,ℎ = 2 0.005 0.05 667 = 0.334 𝑚 = 334 𝑚𝑚
We conclude that the flow is not fully developed, and the friction factor in the entry
region will be higher than for fully developed conditions. Hence, for the same pressure
drop, the mean velocity will be less than our estimate.
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‫خارجی‬ ‫های‬ ‫جریان‬
• From the perspective of heat transfer, flows are two parts:
• Internal flows (like tubes,…)
• External flows:
• General forms for Nusselt number is:
𝑁𝑢 = 𝐶𝑅𝑒 𝑚
𝑃𝑟 𝑛
• Where m,n,C are constant and obtain from tables.
• Fand equation is a well-known equation for external liquid flows on
cylinder that is defined for Re between 10-1 and 105 as:
𝑁𝑢 𝑓 = 0.35 + 0.56𝑅𝑒𝑓
0.52
𝑃𝑟𝑓
0.3
• The f index is the properties in film temperature (Tf)
81
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‫خارجی‬ ‫های‬ ‫جریان‬(‫کره‬ ‫حول‬)
• For very low speed velocities:
𝑁𝑢 𝑑 =
ℎ𝑑
𝑘
= 2
• For wide range of Re(17 to 70000), MC Adams equation is used:
𝑁𝑢 =
ℎ𝑑
𝑘
= 0.37
𝑢∞ 𝑑
𝜗𝑓
0.6
• Note that above equation is only for gases.
• For both liquids and gases, whitaker’s equation is defined as:
𝑁𝑢 𝑑 =
ℎ𝑑
𝑘
= 2 + (0.4 𝑅𝑒 𝑑
ൗ1
2
+ 0.06 𝑅𝑒 𝑑
ൗ2
3
)𝑃𝑟0.4
𝜇∞
𝜇 𝑤
ൗ1
4
• Index ω means µ in wall temperature.
82
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‫کوئت‬ ‫های‬ ‫جریان‬
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
= 0 ⟹
𝜕𝑢
𝜕𝑥
= 0 ⟹ 𝑢 = 𝑓 𝑦
𝜌 𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
= −
𝜕𝑃
𝜕𝑥
+ 𝜇
𝜕2
𝑢
𝜕𝑥2
+
𝜕2
𝑢
𝜕𝑦2
+ 𝜌𝑔 𝑥
𝜕2
𝑢
𝜕𝑦2
=
1
𝜇
𝜕𝑃
𝜕𝑥
⟹ 𝑓 𝑦 = 𝑔 𝑥 ⟹
𝜕2
𝑢
𝜕𝑦2
=
1
𝜇
𝜕𝑃
𝜕𝑥
= 𝑐𝑡𝑒
• For coquette flow
𝜕𝑃
𝜕𝑥
= 0 ⟹ 𝑢 = 𝑐1 𝑦 + 𝑐2
83
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𝑢 = 𝑐1 𝑦 + 𝑐2
• Boundary conditions:
• 𝑦 = 0 ∶ 𝑢 = 0
• 𝑦 = 𝐻 ∶ 𝑢 = 𝑢∞
• Then:
𝑢 =
𝑢∞
𝐻
𝑦 ⟹ 𝑢∗
= 𝑦∗
84
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• After calculating velocity distribution, we should examine energy
equation:
𝜌𝑐 𝑝 𝑢
𝜕𝑇
𝜕𝑥
+ 𝑣
𝜕𝑇
𝜕𝑦
= 𝑘
𝜕2
𝑇
𝜕𝑥2
+
𝜕2
𝑇
𝜕𝑦2
+ 𝜇
𝜕𝑢
𝜕𝑦
2
0 = 𝑘
𝜕2
𝑇
𝜕𝑦2
+ 𝜇
𝑢∞
𝐻
2
𝜕2
𝑇
𝜕𝑦2
= −
𝑢∞
2
𝜇
𝑘𝐻2
• Make dimensionless:
𝜕2
𝑇∗
𝜕𝑦∗2 = −𝐸𝐶 𝑃𝑟
𝜕𝑢∗
𝜕𝑦∗
2
85
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‫مسئله‬
86
KNOWN: Conditions associated with the Couette flow of air or water.
FIND: (a) Force and power requirements per unit surface area, (b) Viscous
dissipation, (c) Maximum fluid temperature.
ASSUMPTIONS: (1) fully-developed Couette flow, (2) Incompressible fluid with
constant properties.
PROPERTIES: Table A-4, Air (300K): μ =184.6×10-7N⋅s/m2, k=26.3×10-3W/m⋅K;
Table A-6,Water(300K): μ = 855×10-6 N⋅s/m2, k = 0.613 W/m⋅K.
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87
ANALYSIS: (a) The force per unit area is associated with the shear stress. Hence, with the linear
velocity profile for Couette flow, 𝜏 = 𝜇(𝑑𝑢/𝑑𝑦) = 𝜇(𝑈/𝐿).
Air:
𝜏 𝑎𝑖𝑟 = 184.6 × 10−7
200
0.005
= 0.738 𝑁/𝑚2
Water:
𝜏 𝑤𝑎𝑡𝑒𝑟 = 855 × 10−6
200
0.005
= 34.2 𝑁/𝑚2
With the required power given by P/A= τ.U,
Air:
( ൗ𝑃
𝐴) 𝑎𝑖𝑟= 0.738 200 = 147.6 𝑊/𝑚2
Water:
( ൗ𝑃
𝐴) 𝑤𝑎𝑡𝑒𝑟= 34.2 200 = 6840 𝑊/𝑚2
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88
(b) The viscous dissipation is 𝜇Φ = 𝜇 Τ𝑑𝑢 𝑑𝑦 2
= 𝜇 Τ𝑈 𝐿 2
. Hence,
Air:
(𝜇Φ) 𝑎𝑖𝑟= 184.6 × 10−7
200
0.005
2
= 2.95 × 104 Τ𝑊 𝑚3
Water:
(𝜇Φ) 𝑤𝑎𝑡𝑒𝑟= 855 × 10−6
200
0.005
2
= 1.37 × 106 Τ𝑊 𝑚3
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89
(c) From the solution to Part 4 of the text Example, the location of the maximum
temperature corresponds to ymax=L/2. Hence, 𝑇 𝑚𝑎𝑥 = 𝑇𝑜 + 𝜇𝑈2
/8𝑘 and
Air:
(𝑇 𝑚𝑎𝑥) 𝑎𝑖𝑟= 27 +
184.6 × 10−7
200
8 0.0263
= 30.5℃
Water:
(𝑇 𝑚𝑎𝑥) 𝑤𝑎𝑡𝑒𝑟= 27 +
855 × 10−6
200
8 0.613
= 34℃
COMMENTS: (1) The viscous dissipation associated with the entire fluid layer,
μΦ(LA), must equal the power, P. (2) Although (μΦ)water >> (μΦ)air , kwater >> kair .
Hence, Tmax,water ≈ Tmax,air.
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‫مسئله‬
90
KNOWN: Velocity and temperature difference of plates maintaining Couette flow.
Mean temperature of air, water or oil between the plates.
FIND: (a) Pr⋅Ec product for each fluid, (b) Pr⋅Ec product for air with plate at sonic
velocity.
ASSUMPTIONS: (1) Steady-state conditions, (2) Couette flow, (3) Air is at 1 atm.
PROPERTIES: Table A-4, Air (300K, 1atm), cp = 1007 J/kg⋅K, Pr = 0.707, γ = 1.4,
R= 287.02 J/kg⋅K; Table A-6, Water (300K): cp = 4179 J/kg⋅K, Pr = 5.83; Table A-5,
Engine oil (300K), cp = 1909 J/kg⋅K, Pr = 6400.
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91
PROPERTIES: Table A-4, Air (300K, 1atm), cp = 1007 J/kg⋅K, Pr = 0.707, γ = 1.4,
R= 287.02 J/kg⋅K; Table A-6, Water (300K): cp = 4179 J/kg⋅K, Pr = 5.83; Table A-5,
Engine oil (300K), cp = 1909 J/kg⋅K, Pr = 6400.
ANALYSIS: The product of the Prandtl and Eckert numbers is dimensionless,
𝑃𝑟. 𝐸𝑐 = 𝑃𝑟
𝑈2
𝑐 𝑝∆𝑇
→
Τ𝑚2
𝑠2
( Τ𝐽 𝑘𝑔. 𝐾)𝐾
=
Τ𝑚2
𝑠2
(𝑘𝑔.
𝑚2
𝑠2 )/𝑘𝑔
Substituting numerical values, find
Air Water Oil
Pr.Ec 0.0028 0.0056 13.41
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92
(b) For an ideal gas, the speed of sound is
𝑐 = 𝛾𝑅𝑇 1/2
where R, the gas constant for air, is Ru/Μ = 8.315 kJ/kmol⋅K/(28.97 kg/kmol) =
287.02 J/kg⋅K. Hence, at 300K for air,
𝑈 = 𝑐 = 1.4 × 287.02 × 300 1/2
= 347.2 𝑚/𝑠
For sonic velocities, it follows that
𝑃𝑟. 𝐸𝑐 = 0.707
347.2
1007 25
= 3.38
COMMENTS: From the above results it follows that viscous dissipation effects
must be considered in the high speed flow of gases and in oil flows at moderate
speeds. For Pr⋅Ec to be less than 0.1 in air with ΔT = 25°C, U should be < ~ 60
m/s.
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‫مسئله‬
93
KNOWN: Couette flow with moving plate isothermal and stationary
plate insulated.
FIND: Temperature of stationary plate and heat flux at the moving
plate.
ASSUMPTIONS: (1) Steady-state conditions, (2) incompressible fluid
with constant properties, (3) Couette flow.
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94
ANALYSIS: The energy equation is given by
0 = 𝑘
𝜕2
𝑇
𝜕𝑦2
+ 𝜇
𝜕𝑢
𝜕𝑦
2
Integrating twice find the general form of the temperature distribution,
𝜕2
𝑇
𝜕𝑦2
= −
𝜇
𝑘
𝑈
𝐿
2
𝜕𝑇
𝜕𝑦
= −
𝜇
𝑘
𝑈
𝐿
2
𝑦 + 𝐶1
𝑇 𝑦 = −
𝜇
2𝑘
𝑈
𝐿
2
𝑦2
+ 𝐶1 𝑦 + 𝐶2
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95
Consider the boundary conditions to evaluate the constants,
ቤ
𝜕𝑇
𝜕𝑦 𝑦=0
= 0 → 𝐶1 = 0
And
𝑇 𝐿 = 𝑇𝐿 → 𝐶2 = 𝑇𝐿 +
𝜇
2𝑘
𝑈2
Hence, the temperature distribution is
𝑇 𝑦 = 𝑇𝐿 +
𝜇𝑈2
2𝑘
1 −
𝑦
𝐿
2
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96
The temperature of the lower plate (y = 0) is
𝑇 0 = 𝑇𝐿 +
𝜇𝑈2
2𝑘
The heat flux to the upper plate (y = L) is
𝑞′′
𝐿 = −𝑘 ቤ
𝜕𝑇
𝜕𝑦 𝑦=𝐿
=
𝜇𝑈2
𝐿
COMMENTS: The heat flux at the top surface may also be obtained by
integrating the viscous dissipation over the fluid layer height. For a control
volume about a unit area of the fluid layer,
ሶ𝐸𝑔
′′
= ሶ𝐸 𝑜𝑢𝑡
′′
න
0
𝐿
𝜇
𝜕𝑢
𝜕𝑦
2
𝑑𝑦 = 𝑞′′
𝐿 𝑞′′
𝐿 =
𝜇𝑈2
𝐿
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