Part A: A room with a pine ceiling measured 12 m x 6 m x 2 cm thick. On a cold d
ID: 1270856 • Letter: P
Question
Part A: A room with a pine ceiling measured 12 m x 6 m x 2 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 28 oC and in the attic T2 = 6 oC. kpine = 0.120 J/msC kinsul = 0.042 J/msC.
PArt B: A room with a pine ceiling measured 12 m x 6 m x 2 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 28 oC and in the attic T2 = 6 oC. If a 4 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC. If a 4 cm piece of insulation is added to the pine ceiling , find the interface temperature for the pine and insulation.
Part C: A room with a pine ceiling measured 12 m x 6 m x 2 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 28 oC and in the attic T2 = 6 oC. If a 4 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC. If 4 cm of insulation is added to the pine ceiling how much heat does the layer of pine / insulation transmit in one hour.
Part D: A room with a pine ceiling measured 12 m x 6 m x 2 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 28 oC and in the attic T2 = 6 oC. If a 4 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC. If 4 cm piece of insulation is added to the pine ceiling, find the % decrease in heat loss , once the insulation has been installed.
Please write the steps and Answer. Thank you!
Explanation / Answer
A)
When only pine was there,
Q = K * A * ( T1- T2) /l = 0.12 * 12 * 6 * ( 28-6) /0.02 = 9504 W
Heat loss per hour = 9504 * 3600 = 34214400 J
B)
When insulation and pine were there, T be the temp at interface of pine and insulation
we have Q = Kpine * A * ( T 1 - T) /lpine = Kins * A* ( T - T2) /lins
so, 0.12 * 12 * 6 * ( 28 - T) /0.02 = 0.042 * 12 * 6 * ( T - 6) /0.04
so, T =24.7234 C is the interface temperature for the pine and insulation.
C)
we have Q = Kpine * A * ( T 1 - T) /lpine = 0.12 * 12 * 6 * ( 28-24.7234) /0.02 = 1415.4912 W
For one hour, heat transmitted = 1415.4912 * 3600 = 5095768.32 J
D)
% decrease in heat loss = 100 * (34214400 - 5095768.32 ) / 34214400 = 85.10636 % is the answer