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A hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins is
- a)325 W
- b)707 W
- c)566 W
- d)424 W
Correct answer is option 'D'. Can you explain this answer?
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Given:
- Temperature of the hot surface (Th) = 80°C
- Temperature of the air (T∞) = 20°C
- Length of the fin (L) = 10 cm = 0.1 m
- Diameter of the fin (D) = 1 cm = 0.01 m
- Combined heat transfer coefficient (h) = 30 W/m²·K
- Fin efficiency (η) = 0.75
To find:
Rate of heat loss from 100 fins
Solution:
Step 1: Calculate the surface area of the fin:
The surface area of a cylinder can be calculated using the formula:
A = 2πrh + πr²
Where:
r = radius of the fin = D/2
h = height of the fin = L
Given that D = 0.01 m and L = 0.1 m, we can calculate the radius and height of the fin:
r = 0.01/2 = 0.005 m
h = 0.1 m
Now, substitute the values in the formula to calculate the surface area of the fin:
A = 2π(0.005)(0.1) + π(0.005)²
A ≈ 0.0314 m²
Step 2: Calculate the heat transfer rate from a single fin:
The heat transfer rate from a fin can be calculated using the formula:
Q = hA(Th - T∞)
Where:
Q = heat transfer rate from a single fin
h = combined heat transfer coefficient
A = surface area of the fin
Th = temperature of the hot surface
T∞ = temperature of the air
Substitute the given values in the formula to calculate the heat transfer rate from a single fin:
Q = 30(0.0314)(80 - 20)
Q ≈ 75.36 W
Step 3: Calculate the total heat transfer rate from 100 fins:
Since there are 100 fins, the total heat transfer rate from all the fins can be calculated by multiplying the heat transfer rate from a single fin by the number of fins:
Total heat transfer rate = Q × Number of fins
Total heat transfer rate = 75.36 × 100
Total heat transfer rate ≈ 7536 W
Step 4: Calculate the rate of heat loss from 100 fins:
The rate of heat loss is the same as the total heat transfer rate from 100 fins, so the rate of heat loss from 100 fins is approximately 7536 W.
Therefore, the correct option is (D) 424 W.
- Temperature of the hot surface (Th) = 80°C
- Temperature of the air (T∞) = 20°C
- Length of the fin (L) = 10 cm = 0.1 m
- Diameter of the fin (D) = 1 cm = 0.01 m
- Combined heat transfer coefficient (h) = 30 W/m²·K
- Fin efficiency (η) = 0.75
To find:
Rate of heat loss from 100 fins
Solution:
Step 1: Calculate the surface area of the fin:
The surface area of a cylinder can be calculated using the formula:
A = 2πrh + πr²
Where:
r = radius of the fin = D/2
h = height of the fin = L
Given that D = 0.01 m and L = 0.1 m, we can calculate the radius and height of the fin:
r = 0.01/2 = 0.005 m
h = 0.1 m
Now, substitute the values in the formula to calculate the surface area of the fin:
A = 2π(0.005)(0.1) + π(0.005)²
A ≈ 0.0314 m²
Step 2: Calculate the heat transfer rate from a single fin:
The heat transfer rate from a fin can be calculated using the formula:
Q = hA(Th - T∞)
Where:
Q = heat transfer rate from a single fin
h = combined heat transfer coefficient
A = surface area of the fin
Th = temperature of the hot surface
T∞ = temperature of the air
Substitute the given values in the formula to calculate the heat transfer rate from a single fin:
Q = 30(0.0314)(80 - 20)
Q ≈ 75.36 W
Step 3: Calculate the total heat transfer rate from 100 fins:
Since there are 100 fins, the total heat transfer rate from all the fins can be calculated by multiplying the heat transfer rate from a single fin by the number of fins:
Total heat transfer rate = Q × Number of fins
Total heat transfer rate = 75.36 × 100
Total heat transfer rate ≈ 7536 W
Step 4: Calculate the rate of heat loss from 100 fins:
The rate of heat loss is the same as the total heat transfer rate from 100 fins, so the rate of heat loss from 100 fins is approximately 7536 W.
Therefore, the correct option is (D) 424 W.
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A hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. Can you explain this answer?
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A hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. 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 hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. 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 hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. Can you explain this answer?.
A hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. 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 hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. 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 hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A hot surface at 80°C exposed to air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylinder fins. The combined heat transfer coefficient is 30 W/m2 . K, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins isa) 325 Wb) 707 Wc) 566 Wd) 424 WCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
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