-
Consider steady-state heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides.
Given: hi = 20 W/m²K; h0 = 50 W/m²K; T∞i = 20°C; T*infin;,0 = –2°C; k1 = 20 W/mK; k2 = 50 W/mK; L1 = 0.30 m and L2 = 0.15 m.
Assuming negligible contact resistance between the wall surfaces, the interface termperature, T(in °C), of the two walls will be
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- – 0.50
- 2.75
- 3.75
- 4.50
Correct Option: C
| Req = | + | + | + | ||||
| h2A | h1A | h2A | h2A |
| = |  | + | + | + |  | = | ||||||
| A | 20 | 20 | 50 | 50 | A |
| Now Q2 = | = | |||||
| Req | 1 | + | L1 | |||
| h | h | |||||
| ∴ | = | ||||||
| 0.8 | 0.065 | ||||||
| A | A | ||||||
| ⇒ | = 293 - t | |
| 0.088 |
⇒ t = 3.75°C
Alternately
Under steady state conditions
| = | ||||||||
| 1 | + | l1 | l2 | + | 1 | |||
| hiA | k1A | k2A | h0A | |||||
| ∴ | = | |||||||
| 1 | + | 0.3 | 0.15 | + | 1 | |||
| 20 | 20 | 50 | 50 | |||||
| or | = | ||||||
| 1.3 | 1.15 | ||||||
| 20 | 50 | ||||||
| or 20 - ti = | × | {ti + 2} | ||
| 1.05 | 20 |
or 20 – ti = 2.826 (ti + 2)
or 3.826 ti = 20 – 2 × 2.826
or ti = 3.75°C
