Question

A hot plane surface at 100°C is exposed to air at 25°C with a combined heat transfer coefficient of 20W/m2*K The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of 0.1 W/m*K. Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is

A. 0.1 cm
B. 0.5 cm
C. 1.0 cm
D. 2.0 cm
E. 5.0 cm

Answer 1

To reduce the heat loss from a hot plane surface by half, the required thickness of insulation can be calculated using the given heat transfer coefficient and thermal conductivity.

Step-by-step explanation:

To reduce the heat loss from a hot plane surface by half, we need to cover it with sufficient insulation. The formula to calculate the heat loss is given by:

Q = (h * A * ΔT) / k

Where:
Q is the heat loss
h is the heat transfer coefficient
A is the surface area
ΔT is the temperature difference
k is the thermal conductivity

First, we calculate the initial heat loss using the given values:
Qinitial = (20 * A * (100 - 25))

Then, we calculate the required heat loss by dividing the initial heat loss by 2:
Qrequired = Qinitial / 2

Next, we rearrange the formula to solve for the thickness of insulation:
Qrequired = (h * A * ΔT) / k

Plugging in the values, we get:
(20 * A * ΔT) / k = Qrequired

Simplifying and solving for ΔT, we get:
ΔT = (Qrequired * k) / (20 * A)

Finally, we can calculate the required thickness of insulation:
Thickness = ΔT / (k)