1. Hot air flowing through a counter-flow double-pipe heat exchanger at a rate o
ID: 2078726 • Letter: 1
Question
1. Hot air flowing through a counter-flow double-pipe heat exchanger at a rate of 0.5 kg/s must be cooled from 250ºC to about 100ºC by water in the annulus initially at 15ºC. The thermal resistance of the heat exchanger is 1.845x10-3 K/W. The specific heat capacities of the air and water may be taken as 1050 and 4190 J/kgK respectively. Will this exchanger do the job when cooling water flow is 1.2 kg/s. Use the NTU-Effectiveness Method.
Step by step breifly it is sometimes difficult to understand
Explanation / Answer
Heat lost by air + heat lost by resistance of heat exchanger <= heat gained by water
Maximum heat lost or gained = specific heat capacity x mass x (difference in temperature)
q = cm delta(T)
Heat lost by air 1sec = 1050 x 0.5 x (273+(250-100))
= 222075 J
Heat lost by resistance of heat exchanger in 1sec = delta(T) / resistance
= ((250-100)+273) / 1.845 x 10-3 = 229.3 x 10-3 J
Heat gained by water in 1sec = 4190 x 1.2 x ((15-100)+273)
= 1800024 J
As we can see (222075+229.3) J < 1800024 J
So the exchanger will do the job in cooling the water.
Here I used the flow rate as the mass because then the heat we calculate would be that in one second. Since we did the same in the calculation of heat everywhere in the equation, the calculation is correct.
Note, we should use temperature in Kelvin everywhere, hence the addition of a factor of 273.
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