Q1: The inner wall of a furnace is at a temperature of 700°C. The composite wall is made of two substances, 10 and 20 cm thick with thermal conductivities of 0.05 and 0.1 W.m^-1.°C^-1 respectively. The ambient air is at 30°C and the heat transfer co-efficient between the outer surface of wall and air is 20 W.m^-2.°C^-1. The rate of heat loss from the outer surface in W.m^-2is

ANS:A - 165.4

To find the rate of heat loss from the outer surface of the wall, we can use the concept of thermal resistance and Fourier's law of heat conduction. The total thermal resistance (𝑅𝑡𝑜𝑡𝑎𝑙Rtotal​) of the composite wall is the sum of the thermal resistances of its individual layers. The thermal resistance (𝑅R) of a layer is given by: 𝑅=𝐿𝑘×𝐴R=k×AL​ Where:

• 𝐿L is the thickness of the layer,
• 𝑘k is the thermal conductivity of the material, and
• 𝐴A is the cross-sectional area perpendicular to the direction of heat flow.

Given that the composite wall is made of two layers, we can calculate the total thermal resistance as follows: 𝑅𝑡𝑜𝑡𝑎𝑙=𝑅1+𝑅2Rtotal​=R1​+R2​ 𝑅𝑡𝑜𝑡𝑎𝑙=𝐿1𝑘1×𝐴+𝐿2𝑘2×𝐴Rtotal​=k1​×AL1​​+k2​×AL2​​ Since the area and the thermal conductivity remain the same for both layers, we can simplify the equation: 𝑅𝑡𝑜𝑡𝑎𝑙=𝐿1𝑘1+𝐿2𝑘2Rtotal​=k1​L1​​+k2​L2​​ Now, we use the formula for heat transfer rate through a composite wall: 𝑄=𝑇1−𝑇2𝑅𝑡𝑜𝑡𝑎𝑙Q=Rtotal​T1​−T2​​ Where:

• 𝑄Q is the heat transfer rate (W/m²),
• 𝑇1T1​ is the temperature on one side of the wall (in Kelvin),
• 𝑇2T2​ is the temperature on the other side of the wall (in Kelvin), and
• 𝑅𝑡𝑜𝑡𝑎𝑙Rtotal​ is the total thermal resistance of the wall.

First, we need to convert all temperatures to Kelvin:

• Inner wall temperature (𝑇1T1​): 700°𝐶=700+273.15=973.15𝐾700°C=700+273.15=973.15K
• Ambient air temperature (𝑇2T2​): 30°𝐶=30+273.15=303.15𝐾30°C=30+273.15=303.15K

Now, let's calculate the total thermal resistance: 𝑅𝑡𝑜𝑡𝑎𝑙=0.100.05+0.200.1=2+2=4Rtotal​=0.050.10​+0.10.20​=2+2=4 Now, plug the values into the formula for heat transfer rate: 𝑄=973.15−303.154=6704=167.5 W/m²Q=4973.15−303.15​=4670​=167.5W/m² So, the rate of heat loss from the outer surface of the wall is approximately 167.5 W/m²167.5W/m². Therefore, the closest option is 167.5.