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A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.
- a)88
- b)89
Correct answer is between '88,89'. Can you explain this answer?
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A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two ...
Because of symmetry, we would consider half-length of the bar (figure).
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P = 2(a+b) = 2(50 + 50) = 200 mm = 0.2 m
A = 50 mm X 50 mm = 2500 mm2 X 10-6
= 25 X 10-4m2
= 5.174m-1
ml = 5.174 X 0.2 = 1.0348
Tanh ml = 0.776
= 5.174 X 52 X 0.0025 X (120 - 35 ) X 0.776
= 44.37 W
Rate of heat loss from the bar is
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A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two ...
Given Data:
Length of the bar (L) = 0.4 m
Cross-sectional area of the bar (A) = (50 mm)² = 0.0025 m²
Temperature of the heated reservoirs (T1) = 120°C
Ambient temperature (T∞) = 35°C
Convective heat transfer coefficient (h) = 17.4 W/m²-K
Thermal conductivity of iron (k) = 52 W/m-K
Formula:
Rate of heat transfer through convection (qconv) = h × A × (T1 - T∞)
Rate of heat transfer through conduction (qcond) = k × A × (T1 - T∞)/L
Rate of heat loss from the bar (q) = qconv + qcond
Calculation:
qconv = 17.4 × 0.0025 × (120 - 35) = 0.72525 W
qcond = 52 × 0.0025 × (120 - 35)/0.4 = 5.56375 W
q = 0.72525 + 5.56375 = 6.289 W
Therefore, the rate of heat loss from the bar is 6.289 W, which is between 88 and 89 in rounding off the answer to two decimal places.
Length of the bar (L) = 0.4 m
Cross-sectional area of the bar (A) = (50 mm)² = 0.0025 m²
Temperature of the heated reservoirs (T1) = 120°C
Ambient temperature (T∞) = 35°C
Convective heat transfer coefficient (h) = 17.4 W/m²-K
Thermal conductivity of iron (k) = 52 W/m-K
Formula:
Rate of heat transfer through convection (qconv) = h × A × (T1 - T∞)
Rate of heat transfer through conduction (qcond) = k × A × (T1 - T∞)/L
Rate of heat loss from the bar (q) = qconv + qcond
Calculation:
qconv = 17.4 × 0.0025 × (120 - 35) = 0.72525 W
qcond = 52 × 0.0025 × (120 - 35)/0.4 = 5.56375 W
q = 0.72525 + 5.56375 = 6.289 W
Therefore, the rate of heat loss from the bar is 6.289 W, which is between 88 and 89 in rounding off the answer to two decimal places.
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A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. Can you explain this answer?
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A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. 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 A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. 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 A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. Can you explain this answer?.
A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. 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 A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. 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 A 50 mm X 50 mm iron bar 0.4 m long is connected to the walls of two heated reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective heat transfer coefficient is 17.4 W/m2-K. Calculate the rate of heat loss from the bar. The thermal conductivity of iron is 52W/m-K.a)88b)89Correct answer is between '88,89'. Can you explain this answer?.
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