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Calculate the rate of heat flow through the wall of a refrigerated van of 1.5 mm of steel sheet at outer surface, 100 mm plywood at the inner surface and 2 cm of glass-wool in between, if the temperatures of the inside and outside surfaces are 15°C and 24°C respectively. Take thermal conductivities of steel, glass-wool and plywood as 23.2 W/m°C, 0.014 W/m°C and 0.052 W/m°C respectively.
Asked by ogatheo2000 | 08 Mar, 2023, 20:59: PM
answered-by-expert Expert Answer
Figure shows the insulation arrangement made by plywood , Glass-wool and steel sheet in a referegirated van.
 
Ambient temperature is 24o C and maintained temperature inside van is 15o C inside van.
 
Let T1 be the temperature at plywood-Glasswool interface and T2 be the temperature at Glasswool-steel interface .
 
At steady state , heta flux per unit area is written as
 
begin mathsize 14px style k subscript s cross times fraction numerator left parenthesis 24 minus T subscript 2 right parenthesis over denominator 1.5 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space k subscript g fraction numerator T subscript 2 minus T subscript 1 over denominator 0.2 end fraction equals k subscript p fraction numerator T subscript 1 minus 15 over denominator 0.1 end fraction end style
where ks , kg and kp are thermal conductivities of steel, glasswool and plywood respectively
 
If we consider heat flow through plywood and glasswool , we get
 
begin mathsize 14px style T subscript 2 space end subscript minus T subscript 1 space equals space 2 space cross times k subscript p over k subscript g open parentheses T subscript 1 space minus space 15 close parentheses end style
 
begin mathsize 14px style T subscript 2 space end subscript minus T subscript 1 space equals space 2 space cross times fraction numerator 0.052 over denominator 0.014 end fraction open parentheses T subscript 1 space minus space 15 close parentheses space equals space 7.428 space left parenthesis T subscript 1 minus 15 right parenthesis end style
 
8.428 T1 - T2 = 7.428 × 15 =  111.4 ..................................(1)
 
If we consider heat flow through plywood and glasswool , we get
 
begin mathsize 14px style T subscript 2 minus T subscript 1 space equals open parentheses fraction numerator 0.2 over denominator.0015 end fraction close parentheses cross times k subscript s over k subscript g cross times left parenthesis 24 minus T subscript 2 right parenthesis end style
begin mathsize 14px style T subscript 2 minus T subscript 1 space equals 133.33 cross times open parentheses fraction numerator 23.2 over denominator 0.014 end fraction close parentheses cross times left parenthesis 24 minus T subscript 2 right parenthesis end style
begin mathsize 14px style T subscript 2 minus T subscript 1 space equals 2.21 space cross times 10 to the power of 5 space cross times space left parenthesis 24 minus T subscript 2 right parenthesis end style
  From above expression , we get T2 ≈ 24oC
 
If we substitute T2 = 24o C in eqn.(1) , we get T1 = 16oC
 
 
if Rate of heat flow per unit area  Q/A , then we get

 
begin mathsize 14px style Q over A space equals space k subscript g fraction numerator T subscript 2 minus T subscript 1 over denominator 0.2 end fraction space equals space fraction numerator 0.014 over denominator 0.2 end fraction cross times left parenthesis 24 minus 16 right parenthesis space equals space 0.56 space W space divided by space m squared end style
Answered by Thiyagarajan K | 09 Mar, 2023, 09:21: AM
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