Solution Manual Heat And Mass Transfer Cengel 5th Edition Chapter 3 New -

Since the wall is large, we can assume one-dimensional heat conduction. The temperature distribution through the wall is linear, and the temperature gradient is:

A large plane wall of thickness 40 cm has a thermal conductivity of 1.2 W/m°C. One side of the wall is maintained at a temperature of 80°C, while the other side is maintained at 40°C. Determine the heat flux through the wall.

dT/dx = (80 - 40) / 0.4 = 100°C/m

q = (20 - 0) / 0.5625 = 35.56 W/m²

A composite wall consists of three layers: a 2-cm thick layer of insulation, a 5-cm thick layer of concrete, and a 1-cm thick layer of plywood. The thermal conductivities of the materials are 0.05 W/m°C, 0.8 W/m°C, and 0.1 W/m°C, respectively. The inner surface of the wall is maintained at 20°C, while the outer surface is maintained at 0°C. Determine the heat transfer through the wall. Since the wall is large, we can assume

q = -k * A * (dT/dx)

R = L / k * A

where q is the heat flux, k is the thermal conductivity, A is the area, and dT/dx is the temperature gradient.