Let us represent the body's surface tissue as layer of thickness X1=5 cm which has a thermal conductivity K,= o.59 J/m.s. ºC . This is surrounded by a layer of tissues fat of thickness X,=2 cm with K,=0.21 J/m.s. °C .The body's interior is at temperature Tj=37ºC and the skin temperature is T2=33.5°C .What is the heat loss per unit area?

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Chapter1: Units, Trigonometry. And Vectors
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Let us represent the body's surface tissue as layer of thickness X1=5 cm
which has a thermal conductivity K= 0.59 J/m.s. ºC . This is surrounded
by a layer of tissues fat of thickness X,=2 cm with K,=0.21 J/m.s. °C
.The body's interior is at temperature Tj=37ºC and the skin temperature is
T,=33.5°C .What is the heat loss per unit area?
Sol.
1/A*AQ/ At= (T2-T¡) /(X,/K,) +(X,/K,)
= (33.5 – 37.5) /(0.05/0.59) + (0.02/0.21)
= - 19.4 watt/m
Negative sign indicates heat loss.
Transcribed Image Text:Example:- Let us represent the body's surface tissue as layer of thickness X1=5 cm which has a thermal conductivity K= 0.59 J/m.s. ºC . This is surrounded by a layer of tissues fat of thickness X,=2 cm with K,=0.21 J/m.s. °C .The body's interior is at temperature Tj=37ºC and the skin temperature is T,=33.5°C .What is the heat loss per unit area? Sol. 1/A*AQ/ At= (T2-T¡) /(X,/K,) +(X,/K,) = (33.5 – 37.5) /(0.05/0.59) + (0.02/0.21) = - 19.4 watt/m Negative sign indicates heat loss.
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