Heat transfer ceiling Building insulation refers broadly to any object in a buil
ID: 1621777 • Letter: H
Question
Heat transfer ceiling
Building insulation refers broadly to any object in a building used as insulation for any purpose. While the majority of insulation in buildings is for thermal purposes, the term also applies to acoustic insulation, fire insulation, and impact insulation (e.g. for vibrations caused by industrial applications). Often an insulation material will be chosen for its ability to perform several of these functions at once.
Part A
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. kpine = 0.120 J/msC kinsul = 0.042 J/msC
Find the rate the heat is moving through the pine ceiling.
Q/t =___ joules/sec
Part B
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If a 6 cm piece of insulation is added to the pine ceiling , find the interface temperature for the pine and insulation.
Ti= ___ Cdegrees
Part C
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If 6 cm of insulation is added to the pine ceiling how much heat does the layer of pine / insulation transmit in one hour.
Heat lost in one hour =__Joules
Part D
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If 6 cm piece of insulation is added to the pine ceiling, find the % decrease in heat loss , once the insulation has been installed.
Efficiency = __%
Q/t =___ joules/sec
Part B
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If a 6 cm piece of insulation is added to the pine ceiling , find the interface temperature for the pine and insulation.
Ti= ___ Cdegrees
Part C
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If 6 cm of insulation is added to the pine ceiling how much heat does the layer of pine / insulation transmit in one hour.
Heat lost in one hour =__Joules
Part D
A room with a pine ceiling measured 16 m x 6 m x 4 cm thick. On a cold day the temperature on inside of the ceiling is T1 = 22 oC and in the attic T2 = 5 oC. If a 6 cm piece of insulation is added to the pine ceiling kpine = 0.120 J/msC kinsul = 0.042 J/msC
If 6 cm piece of insulation is added to the pine ceiling, find the % decrease in heat loss , once the insulation has been installed.
Efficiency = __%
Cm (a) Heat flow (b)Explanation / Answer
A) A = 16*6 = 96 m2
L = 4 cm = 0.04m
(dQpine/dt) = (kpineA/L)(Th - Tc)
=> (dQpine/dt) = (0.12*96/0.04)(22 - 5)
=> (dQpine/dt) = 4896 J/s
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B) (dQ/dt)pine = (dQ/dt)insul
(dQ/dt)pine = (kA/L)pine(Th-Ti)
=> (dQ/dt)pine = (0.12*96/0.04)(22 - Ti)
(dQ/dt)ins = (kA/L)ins(Ti-Tc)
=> (dQ/dt)ins = (0.042*96/0.06)(Ti-5)
(0.12*96/0.04)(22 - Ti) = (0.042*96/0.06)(Ti - 5)
=> Ti = 18.780C
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C) R = Rpine + Rins
=> R = (L/kA)pine + (L/kA)ins
=> R = (0.04/0.12*96) + (0.06/0.042*96)
=> R = 0.018
dQ/dt = (Th - Tc)/R
=> dQ/dt = (22 - 5)/0.018
=> dQ/dt = 925.52 J/s
Heat Lost in one hour = (dQ/dt)*3600 s = 3331.87 kJ
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D)
(dQ/dt)only pine = 4896 J/s
Heat Lost in one hour = 4896*3600 s = 17625.6 kJ
(dQ/dt)pine+insulation = 925.52 J/s
Heat Lost in one hour = (dQ/dt)*3600 s = 3331.87 kJ
% decrease in heat loss = [(17625.6 - 3331.87)/17625.6]*100 = 81.1%
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