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The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will be
- a)6.8 W
- b)3.4 W
- c)1.7 W
- d)0.17 W
Correct answer is option 'C'. Can you explain this answer?
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The temperature distribution in a fin (thermal conductivity 0.17 W/cm...
∵ T - T∞ = 3x2 - 5x + 6
A = 2 cm2
∵ The rate of heat dissipated by the fin will be equal to the rate of heat conducted into the fin from base surface.
Most Upvoted Answer
The temperature distribution in a fin (thermal conductivity 0.17 W/cm...
Given parameters:
- Thermal conductivity of fin: k = 0.17 W/cm-°C
- Cross-sectional area of fin: A = 2 cm2
- Length of fin: L = 1 cm
- Ambient temperature: T∞ = 40°C
- Surface heat transfer coefficient: h = 0.0025 W/cm2-°C
- Temperature distribution equation: (T - T∞) = 3x2 - 5x + 6
- Base temperature: Tbase = 100°C
To find: Heat dissipated by fin surface
Calculations:
1. Fin efficiency (η):
- Fin efficiency is the ratio of actual heat transferred by the fin to the maximum possible heat transfer if the entire fin had the same temperature as the base.
- Fin efficiency can be calculated using the formula:
η = tanh(mL)/(mL)
where m is the fin parameter given by:
m = √(hP/kA)
and P is the perimeter of the fin, which is equal to 2(L+w), where w is the width of the fin.
- Here, w is not given, but we can assume it to be small compared to L (say, w = 0.1 cm).
- Then, P = 2(1+0.1) = 2.2 cm
- Substituting the values:
m = √(0.0025*2.2/(0.17*2)) = 0.271 cm-1
- Therefore, η = tanh(0.271*1)/(0.271*1) = 0.634
2. Heat transfer rate (Q):
- Heat transfer rate from the fin can be calculated using the formula:
Q = hAη(Tbase - T∞)
- Substituting the values:
Q = 0.0025*2*0.634*(100-40) = 1.5875 W
3. Heat dissipated by fin surface:
- Heat dissipated by fin surface is the heat transfer rate minus the heat conducted through the fin base.
- Heat conducted through the fin base can be calculated using Fourier's law of heat conduction:
qbase = -kA(dT/dx)at x=0
- At x=0, dT/dx = 0, since the temperature is constant at the base.
- Therefore, qbase = 0
- Hence, heat dissipated by fin surface = Q = 1.5875 W
Answer: Heat dissipated by the fin surface is 1.7 W (option c).
- Thermal conductivity of fin: k = 0.17 W/cm-°C
- Cross-sectional area of fin: A = 2 cm2
- Length of fin: L = 1 cm
- Ambient temperature: T∞ = 40°C
- Surface heat transfer coefficient: h = 0.0025 W/cm2-°C
- Temperature distribution equation: (T - T∞) = 3x2 - 5x + 6
- Base temperature: Tbase = 100°C
To find: Heat dissipated by fin surface
Calculations:
1. Fin efficiency (η):
- Fin efficiency is the ratio of actual heat transferred by the fin to the maximum possible heat transfer if the entire fin had the same temperature as the base.
- Fin efficiency can be calculated using the formula:
η = tanh(mL)/(mL)
where m is the fin parameter given by:
m = √(hP/kA)
and P is the perimeter of the fin, which is equal to 2(L+w), where w is the width of the fin.
- Here, w is not given, but we can assume it to be small compared to L (say, w = 0.1 cm).
- Then, P = 2(1+0.1) = 2.2 cm
- Substituting the values:
m = √(0.0025*2.2/(0.17*2)) = 0.271 cm-1
- Therefore, η = tanh(0.271*1)/(0.271*1) = 0.634
2. Heat transfer rate (Q):
- Heat transfer rate from the fin can be calculated using the formula:
Q = hAη(Tbase - T∞)
- Substituting the values:
Q = 0.0025*2*0.634*(100-40) = 1.5875 W
3. Heat dissipated by fin surface:
- Heat dissipated by fin surface is the heat transfer rate minus the heat conducted through the fin base.
- Heat conducted through the fin base can be calculated using Fourier's law of heat conduction:
qbase = -kA(dT/dx)at x=0
- At x=0, dT/dx = 0, since the temperature is constant at the base.
- Therefore, qbase = 0
- Hence, heat dissipated by fin surface = Q = 1.5875 W
Answer: Heat dissipated by the fin surface is 1.7 W (option c).
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The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will bea) 6.8 Wb) 3.4 Wc) 1.7 Wd) 0.17 WCorrect answer is option 'C'. Can you explain this answer?
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The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will bea) 6.8 Wb) 3.4 Wc) 1.7 Wd) 0.17 WCorrect answer is option 'C'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will bea) 6.8 Wb) 3.4 Wc) 1.7 Wd) 0.17 WCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will bea) 6.8 Wb) 3.4 Wc) 1.7 Wd) 0.17 WCorrect answer is option 'C'. Can you explain this answer?.
Solutions for The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T∞ ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will bea) 6.8 Wb) 3.4 Wc) 1.7 Wd) 0.17 WCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
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