A composite wall consists of 20 mm thick steel plate backed by insulation brick (k = 0.39 W/mK) of 50 cm thickness and overlaid
by mineral wool of 20 cm thickness (k = 0.05 W/mK) and 70 cm layer of brick of (k = 0.39 W/mK). The inside is exposed to convection at 650°C with h = 65 W/ m2K. The outside is exposed to air at 35°C with a convection coefficient of 15 W/m2K. Determine the heat loss per unit area, interface temperatures and temperature gradients in each materials.
1 answer:
Answer:
Heat loss=85.9W/m^2
ΔT1(Steel)=0.04C
ΔT2(Brick1)=110.13C
ΔT3(Mwood)=343.6C
ΔT1(Brick2)=154.18C
Explanation:
raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.
To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.
I attached the procedure
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Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
