The value of thermal conductivity (k) is theoretically a constant. In practice,
ID: 978868 • Letter: T
Question
The value of thermal conductivity (k) is theoretically a constant. In practice, however, it may be a (relatively weak) function of temperature. Assume that k increases linearly with distance into the wall, i.e.:
k = k0 [1 + 1.8 z]
where k0 = value of k at T0 = 20 oC. The temperature on one side of a 5 cm thick plane wall is held constant at 20 oC and on the other side at 100 oC.
a) Write an energy balance for a differential thickness of the wall. Integrate the equation to yield an expression for the temperature profile in the wall.
b) If k0 = 5.2 W/m-K, calculate a value of the heat flux (q”) through the wall.
c) If k were assumed constant (and equal to k0), what would be the value of the heat flux?
Explanation / Answer
a) we can assume steady state so Qin-Qout=0
q=k*A*dT/dz---> k=ko*[1+1.8z]
dT=(ko*A/q)*dz/(1+1.8z)
Integrating equation above
T2-T1=(ko*A/q) *0.556log(9z+5)
b)if ko=5.2W/m-K
q=(T2-T1)/(ko*A)/0.556log(9z+5) assuming A=1m2
q=(80)/(5.2*1)/0.556log(9.0.05+5)=37.595W
c) if k=ko
q=kA(T2-T1)/z=5.2*(80)/0.05=8320W
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.