# Basic Heat Transfer Modes

Fourier’s Law: O

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 Heat transfer in a multi-layer wall T.
 1 j ► k, K2 K3 41
 X,
 X4
 X2 X,
 X
 K(T2-T1)_ (T2-T,) (x2 ~xi)
 Q = —
 R
 To yield

 Q (x2 — xi) = — k (T2 — Ti)

 Or

 T, O—

 A/W R Where

 R = ——— L =thermal resistance (K m2/W) or (hr ft °F/Btu). K U U O

L|/ki L2/k2 L3/k3

 „ X, —X9 R? = -*—— 1 K2 X4 ~X3
 R,
 Q-T,) R, Q-T7.) R2 (T4-T,) R,
 Qi =“
 Q3 =- X2 X1

 R: Qi = q2 = q3 = q Re-arrange the equations
 Since QiRi = — (T2 — Ti) q2R2 = — (T3 — T2) q3R3 = — (T4 — T3) So that q(Ri + R2 + R3) = — (T4 — Ti)

Or

 (T4-T,) Rj + R-, + R3
 Q = T — T 4 xi R

 N N

R

Here

 R — Ri + R2 + R3 Is the total thermal resistance due to conduction.

 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.

 Common Mineral Plaster Solution:

Q =

From Table 5-1, k^ck = 1.30 W/m K, Rflber = 1.94 m2 K/W, Rb0ard =0.078 m2 K/W. O

 Q =
 R