Basic Heat Transfer Modes
Fourier’s Law:
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Where q = heat flux (W/m )
K = thermal conductivity (W/m K) (material property, Table 5-1 of the text) dT/dx = temperature gradient (K/m)
• Heat transfer in a single-layer wall
In most practical uses, k is approximated as constant. For steady-state heat transfer, q is constant. Then the equation can be integrated
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Heat transfer in a multi-layer wall T. |
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X, |
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X4 |
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X2 X, |
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X |
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K(T2-T1)_ (T2-T,) (x2 ~xi) |
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Q = — |
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R |
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„ X, —X9 R? = -*—— 1 K2 X4 ~X3 |
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R, |
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Q-T,) R, Q-T7.) R2 (T4-T,) R, |
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Qi =“ |
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Q3 =- |
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Qi = q2 = q3 = q Re-arrange the equations |
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Since |
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QiRi = — (T2 — Ti) q2R2 = — (T3 — T2) q3R3 = — (T4 — T3) |
So that q(Ri + R2 + R3) = — (T4 — Ti)
Or
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(T4-T,) Rj + R-, + R3 |
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Q = |
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Here
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R — Ri + R2 + R3 |
Is the total thermal resistance due to conduction.
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C= 1/R Example 5.1 |
Thermal conductance (W/m K).
An exterior wall of an PSU classroom consists of 0.24 m thick face brick, 0.09 m thick mineral fiber, and 0.013 m thick plasterboard. If the exterior surface temperature of the wall is 0 °C and the interior surface 20 °C. Determine the heat flux and temperature distribution in the wall.
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Common Mineral Plaster |
Solution:
From Table 5-1, k^ck = 1.30 W/m K, Rflber = 1.94 m2 K/W, Rb0ard =0.078 m2 K/W. O
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Q = |
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R |
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Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning |