Thermal Conductivty

1.0 Objectives: To calculate the coefficient of thermal conductivity for a good conductor, plus, to find the rate of heat transfer.

2.0 Theory: The science that deals with the determination of the rates of energy transfers is the heat transfer. Heat can be transferred in three different ways: conduction, convection, and radiation. All modes of heat transfer require the existence of a temperature difference. In the lab we studied the conduction method, which is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones, as a result of interaction between particles. The rate at which heat is transferred (Q) can be calculated using the following expression: Q = kA ∆T/∆x, in watts (W). Where; • Q: is the rate of heat transfer by conduction. • k: is the thermal conductivity coefficient. • A: is the area of the surface through which heat is transferred, and is always perpendicular to the direction of transfer. • ∆T/∆x: is the temperature gradient.

3.0 Calculations & Results: 3.1) to calculate the thermal conductivity coefficient for the two specimens, stainless steel & aluminum, we use: k = {J*m*L(Tout – Tin)} / {A*t(T4 – T3)}, in W/m.K. 1

Where; • k: is the thermal conductivity. • J: is the specific heat for water at constant pressure (CP) = 4.186 kJol. • • • • • • •

m/t: is the mass flow rate, which equals, here, 5*10-7 kg/s. Tin: is the inlet water temperature (OC). Tout: is the outlet water temperature (OC). A: is the area of the specimen’s surface (m2). T3: is the thermocouple temperature, the cold end (OC). T4: is the thermocouple temperature, the hot end (OC). L: is the distance between the thermocouples, and is 0.05 (m).

ksteel = 0.147 W/mK. kCu = 0.030 W/mK. 3.2) to calculate the rate of heat transfer through out the two specimens: Q = (x1 – x2) / (L1/Ak1) + (L2/Ak2). Where; • x1: temperature at the element’s end, (oC). • x2: temperature at the water’s end, (oC). • L1: is the length of the short element, (m). • L2: is the length of the long element, (m). • A1 & A2: is the area of the short element, which equals the area of the other element (A2), (m2). • k1: is the thermal conductivity for the short element, (W/mK). • k2: is the thermal conductivity for the long element, (W/mK). Q = 0.026 W/mK.

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4.0 Conclusion: • As expected (k) for the steel is higher than that for the copper. • For good results a steady state should first be reached. • Isolation is very important in the experiment, as it is the only way to force the heat to move in one direction

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