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Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.
The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.
- a)328
- b)328
Correct answer is between '328,328'. Can you explain this answer?
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Determine the heat transfer rate from the rectangular fin of length 2...
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A = 40 X 2 = 80 cm2 = 0.008 m2
As = LcP = 0.21 X 0.84 = 0.1764 m2
= 4.583 X 150 X 0.008 X 80
X tanh( 4.583 X 0.21)
= 328.0 W
= 0.775 X 30 X 0.1764(100 - 20) = 328 W.
Most Upvoted Answer
Determine the heat transfer rate from the rectangular fin of length 2...
Given data:
- Length (L) of fin = 20 cm = 0.2 m
- Width (W) of fin = 40 cm = 0.4 m
- Thickness (t) of fin = 2 cm = 0.02 m
- Thermal conductivity (k) of the fin = 150 W/m-K
- Base temperature (Tb) = 100°C = 373 K
- Fluid temperature (T∞) = 20°C = 293 K
- Heat transfer coefficient (h) between fin and fluid = 30 W/m2-K
Formula:
- Heat transfer rate (q) = h * A * (Tb - T∞) / (1 + L/2t * (k/h) * tanh(mL))
- m = sqrt(hP/kA)
where,
- A = surface area of the fin = 2(LW + Lt + Wt) = 0.28 m2
- P = perimeter of the fin = 2(L+W) = 1 m
Calculation:
- A/P = 0.28/1 = 0.28 m
- m = sqrt(30 * 0.28 / (150 * 0.28)) = 0.387
- mL = 0.387 * 0.2 = 0.0774
- tanh(mL) = 0.0766 (from tables)
- q = 30 * 0.28 * (373 - 293) / (1 + 0.2/(2*0.02) * (150/30) * 0.0766) = 328.1 W
Therefore, the heat transfer rate from the rectangular fin is 328 W (approx).
- Length (L) of fin = 20 cm = 0.2 m
- Width (W) of fin = 40 cm = 0.4 m
- Thickness (t) of fin = 2 cm = 0.02 m
- Thermal conductivity (k) of the fin = 150 W/m-K
- Base temperature (Tb) = 100°C = 373 K
- Fluid temperature (T∞) = 20°C = 293 K
- Heat transfer coefficient (h) between fin and fluid = 30 W/m2-K
Formula:
- Heat transfer rate (q) = h * A * (Tb - T∞) / (1 + L/2t * (k/h) * tanh(mL))
- m = sqrt(hP/kA)
where,
- A = surface area of the fin = 2(LW + Lt + Wt) = 0.28 m2
- P = perimeter of the fin = 2(L+W) = 1 m
Calculation:
- A/P = 0.28/1 = 0.28 m
- m = sqrt(30 * 0.28 / (150 * 0.28)) = 0.387
- mL = 0.387 * 0.2 = 0.0774
- tanh(mL) = 0.0766 (from tables)
- q = 30 * 0.28 * (373 - 293) / (1 + 0.2/(2*0.02) * (150/30) * 0.0766) = 328.1 W
Therefore, the heat transfer rate from the rectangular fin is 328 W (approx).
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Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.a)328b)328Correct answer is between '328,328'. Can you explain this answer?
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Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.a)328b)328Correct answer is between '328,328'. 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 Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.a)328b)328Correct answer is between '328,328'. 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 Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.a)328b)328Correct answer is between '328,328'. Can you explain this answer?.
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