Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle heater made from copper whose dimensions are 450 cm by 384 cm. Outside and inside temperature are 30° C and 50° C respectively

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal  

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside  temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal  = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P =

Rate of heat transfer,

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

L = 6.23 cm

Therefore, the thickness of the material is 6.23 cm

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Explanation:

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