3-13 2 80 T 60 T1 SAMPLE 20 INSULATION Time s The above sketch shows an apparatu

Question

3-13 2 80 T 60 T1 SAMPLE 20 INSULATION Time s The above sketch shows an apparatus being used to measure a value of the thermal conductivity of a solid material and the accompanying plot shows the measured values of T,. T and ambient temperature, T, (in degrees C) as a function of time. The sample is configured as a large, thin slab with thickness, Lx, and it is placed firmly on top of a heater strip and there is a thermocouple at the very bottom of the strip (T) as well as at the top surface (T If the heater power is W watts (a) Derive an equation for the steady-state temperature distribution in the sample slab and then derive an equation which can be used to calculate the thermal conductivity of the sample from the measured temperatures (b) Using the thermocouple data (it should be obvious which temperatures are plotted), calculate the thermal conductivity of the sample if its thickness is 0.3 cm and the heater strip power is 30 watts

Explanation / Answer

a) At the steady state , the heat generated by the heater is emitted by the slab.

at steady state;

Qh= T1-T2/L/KA ; A=1m2

Qh=K(T1-T2)/L.................... (1)

The temperature profile in the slab will be linear.

as T= T1-x(T1-T2)/L

Equation(1) can be used to calculate the thermal conductivity of the slab.since Q,T1,T2 are known/measured, a plot against Land K be made

K= (Q/T1-T2)*L

By changing the value of L, K changes, hence the thermal conductivity can be calculated and the average value can be considered for further needs.

b) given Q=30, L=0.3cm=0.003m

given at steady state

T1=55 degree (from graph)

t2=65 degree

K= (30/65-55)*0.003=0,9W/M degree celsius

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