QUESTION 2 The wall of a refrigerated truck consists of 1.5mm sheet steel outer surface, 10mm plywood at the inner surface and 20mm of glass wool in between. The inside temperature is -15°C and outside temperature is 24°C. Take the thermal conductivities of the materials as follows: - k for plywood k for steel k for glass-wool = 0,052W/mK = 23,2W/mK = 0,14W/mK %3D %3D Calculate: 2.1. the rate of heat flow per unit area; 2.2. the interface temperature. If the glass-wool is replaced by a 5mm cork board with a thermal conductivity of 0.043W/mK; 2.3. What percentage change in heat flow is obtained? 2.4. What must be the thickness of the cork board be, to achieve the same heat flow as in (2.1.).

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter1: Basic Modes Of Heat Transfer
Section: Chapter Questions
Problem 1.1P: 1.1 On a cold winter day, the outer surface of a 0.2-m-thick concrete wall of a warehouse is exposed...
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QUESTION 2
The wall of a refrigerated truck consists of 1.5mm sheet steel
outer surface, 10mm plywood at the inner surface and 20mm
of glass wool in between. The inside temperature is -15°C
and outside temperature is 24°C. Take the thermal
conductivities of the materials as follows: -
=0,052W/mK
= 23,2W/mK
k for glass-wool = 0,14W/mK
k for plywood
k for steel
%3D
Calculate:
2.1. the rate of heat flow per unit area;
2.2. the interface temperature.
If the glass-wool is replaced by a 5mm cork board with a
thermal conductivity of 0.043W/mK;
2.3. What percentage change in heat flow is obtained?
2.4. What must be the thickness of the cork board be, to
achieve the same heat flow as in (2.1.).
Transcribed Image Text:QUESTION 2 The wall of a refrigerated truck consists of 1.5mm sheet steel outer surface, 10mm plywood at the inner surface and 20mm of glass wool in between. The inside temperature is -15°C and outside temperature is 24°C. Take the thermal conductivities of the materials as follows: - =0,052W/mK = 23,2W/mK k for glass-wool = 0,14W/mK k for plywood k for steel %3D Calculate: 2.1. the rate of heat flow per unit area; 2.2. the interface temperature. If the glass-wool is replaced by a 5mm cork board with a thermal conductivity of 0.043W/mK; 2.3. What percentage change in heat flow is obtained? 2.4. What must be the thickness of the cork board be, to achieve the same heat flow as in (2.1.).
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