A highly corrosive liquid is cooled by passing it through a circular channel with a diameter of 5 cm drilled in a rectangular block of graphite (k = 800 W/mK). Ambient air is blown over the exterior of the block as shown in the sketch below. The heat-transfer coefficient for the air is 100 W/m² K, based on the total exterior surface area of the block. Calculate the rate of heat loss from the liquid per meter of channel length.
A highly corrosive liquid is cooled by passing it through a circular channel with a diameter of 5 cm drilled in a rectangular block of graphite (k = 800 W/mK). Ambient air is blown over the exterior of the block as shown in the sketch below. The heat-transfer coefficient for the air is 100 W/m² K, based on the total exterior surface area of the block. Calculate the rate of heat loss from the liquid per meter of channel length.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
i dont know how to equate the 3 equations to each other to solve for heat loss
answer is given.
![(2.12) A highly corrosive liquid is cooled by passing it through a circular channel with a diameter of 5 cm drilled in a rectangular
block of graphite (k = 800 W/mK). Ambient air is blown over the exterior of the block as shown in the sketch below. The
heat-transfer coefficient for the air is 100 W/m² K, based on the total exterior surface area of the block. Calculate the rate of
heat loss from the liquid per meter of channel length.
Ans. 10,500 W/m of length.
Ssquare=
2πL
In (1.08w)
q=kSAT
Air
T = 25°C
h = 100 W/m².K
Conduction
S = 3.726m & q = 2980.8(Ts T₂) [¹]
Convection from Liq→ Surface
q = 125.66(573K - Ts) [2]
Convection Blocksurface → air
q = h∞AAT
q = 100(T₂-298) [3]
Steady state all should (=)
Graphite
block
25 cm
25 cm
Corrosive liquid
T = 300°C
h = 400 W/m².K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2F0710b4a1-e40c-415d-9368-e38bd836c7b8%2F5ppir5y_processed.png&w=3840&q=75)
Transcribed Image Text:(2.12) A highly corrosive liquid is cooled by passing it through a circular channel with a diameter of 5 cm drilled in a rectangular
block of graphite (k = 800 W/mK). Ambient air is blown over the exterior of the block as shown in the sketch below. The
heat-transfer coefficient for the air is 100 W/m² K, based on the total exterior surface area of the block. Calculate the rate of
heat loss from the liquid per meter of channel length.
Ans. 10,500 W/m of length.
Ssquare=
2πL
In (1.08w)
q=kSAT
Air
T = 25°C
h = 100 W/m².K
Conduction
S = 3.726m & q = 2980.8(Ts T₂) [¹]
Convection from Liq→ Surface
q = 125.66(573K - Ts) [2]
Convection Blocksurface → air
q = h∞AAT
q = 100(T₂-298) [3]
Steady state all should (=)
Graphite
block
25 cm
25 cm
Corrosive liquid
T = 300°C
h = 400 W/m².K
Expert Solution
Step 1
To Find :
The rate of heat loss from the liquid.
Given :
The diameter of hole is
The thermal conductivity of graphite block is
The heat transfer coefficient for the air is
Step by step
Solved in 3 steps with 1 images
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