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3 years ago

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A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur

face of the tank wall is held constant at 25 oC and the outside surface heat transfer coefficient is 6 W m2 K. Calculate the rate of heat loss from the tank when the outside air temperature is 15°C.

1 answer:

SOVA2 [1]3 years ago

8 0

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W $m^{-1}K^{-1}$

diameter (d1) = 10 cm

so radius (r1) = 10 /2 = 5 cm = 5 × $10^{-2}$

radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × $10^{-2}$

temperature = 25°C

surface heat transfer coefficient = 6 6 W $m^{-2}K^{-1}$

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to 15°C

first pass through through resistance R1 i.e.

R1 = ( r2 - r1 ) / 4 $\pi$ × r1 × r2 × K

R1 = ( 5.2 - 5 ) $10^{-2}$ / 4 $\pi$ × 5 × 5.2 × 16 × $10^{-4}$

R1 = 3.825 × $10^{-3}$

same like we calculate for resistance R2 we know i.e.

R2 = 1 / ( h × area )

here area = 4 $\pi$ r2²

area = 4 $\pi$ (5.2 × $10^{-2}$ )² = 0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 × $10^{-3}$ + 4.90499

heat loss = 2.037152 W

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