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A metal ball of diameter 60 mm is initially at 220°C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2K. The specific heat, thermal conductivity and density of the metal ball are 400 J/kgK, 400 W/mK and 9000 kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately
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- 141
- 163
- 189
- 210
- 141
Correct Option: A
CP = 400 J /kg-K
K = 400 W/mK
ρ = 9000 kg/m3
Time (τ) = 90 sec
τ | ||||
ρVCP |
= e | ||
T - T∞ |
Put | = | = 100 | ||
v | R |
τ = | 200 × | × | × | × 90 = 0.5 | ||||||||
ρVCP | 0.03 | 9000 | 400 |
= e0.5 | ||
T - 20 |
By solving, we get T = 141.3°C