A composite wall consisting of 3 blocks of materials A B C

A composite wall consisting of 3 blocks of materials A, B & C (k = 1.5 & 10W/m degree C, respective) and Air (k=0.01 W/m degree C and h=0.5 W/m degree C) divides a hot spot at 25 degree C and a cold spot - 20 degree C. The price of materials A, B & C are $50. $20 and $10/meter (d: thickness) respectively. When the d2 and d4 are wider than 30 cm, the heat transfer mechanism would be governed by beat convection otherwise beat conduction is applicable. Design the most economical natural-refrigerator (T_4: 5 degree C and T_9 = -5 degree C and d4 should be wider than 50cm), in terns of a) material price and b) energy consumption. (Assumed at steady state and you could state any assumptions if necessary)

Solution

a) The heat flux from Left to right is governed by the composite overall coefficient

d1/k1 + d2/k2 + d3/k3 + d5+k5 + 1/h ( where the gap at d4 is greater= than 50 hence convective)

b) the overall length is to be lower than 200 cm

so d1+d2+d3+d5 <= 150

c) The price should be minimised, so

50d1+20d3+10d5 should be minimum

solution: wall C is the least costly and has the largest conductivity

A is most costly and has lowest conductivity

Air is free and has lowest condutivity

Try and maximize the air gap as much as possible within the space constraint. However, this will raise the heat transfer since h is much large than k

Taking d2 =120 (maximum) we get d1+d3+d5 =30 cm

at this extreme, d1=d3=d5 =10 cm

Price for this option 50 *.1 + 20*.1 + 10*.1 =8

b) Evaluation of the steady state flux at RHS wall

kc(15)/d5 = koverall (25--20)/200

koverall = 666.7/d5 varies from 66.7 to 5.55

Hence d5 to be maximised to lower heat loss.

Taking d5 =30 is the other alternative