Basic Heat Transfer Modes

Fourier’s Law:

Basic Heat Transfer Modes

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

Basic Heat Transfer Modes

Heat transfer in a multi-layer wall

T.

1 j

► k,

K2

K3

41

<ii

Q2

Q3

4

• T3

4

. T4

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

 

Basic Heat Transfer Modes

Where

 

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

K

 

Basic Heat Transfer Modes

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 =-

Basic Heat Transfer Modes

X2 X1

 

R:

 

Basic Heat Transfer Modes

Qi = q2 = q3 = q

Re-arrange the equations

Since

Basic Heat Transfer Modes

QiRi = — (T2 — Ti) q2R2 = — (T3 — T2) q3R3 = — (T4 — T3)

подпись: qiri = - (t2 - ti) q2r2 = - (t3 - t2) q3r3 = - (t4 - t3)So that q(Ri + R2 + R3) = — (T4 — Ti)

Or

(T4-T,) Rj + R-, + R3

Q =

Basic Heat Transfer Modes

T — T

4 xi

R

 

N

N

R

 

Here

R — Ri + R2 + R3

подпись: r — ri + r2 + r3Is the total thermal resistance due to conduction.

C= 1/R Example 5.1

подпись: c= 1/r example 5.1Thermal 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

подпись: 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

Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning


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