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.

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Chapter6: Forced Convection Over Exterior Surfaces
Section: Chapter Questions
Problem 6.41P
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i dont know how to equate the 3 equations to each other to solve for heat loss

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(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
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

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