Water is to be cooled by passing through the tube in an insulated cross-flow heat exchanger, while air flows over the outside surface of the tube. The water enters the 4.0 cm-diameter pipe with a velocity of 15 m/s, a temperature of 150oC, and a quality of 0.10. The water exits as a liquid at 100oC. (This can be approximated as a saturated liquid at 100oC.) The air enters the heat exchanger at 15oC. (a) For a mass flow rate of the air of 5.0 kg/s, determine the air exit temperature. (b) Using your heat exchanger model to acquire the data, plot the exit temperature of the air as the air mass flow rate varies from 4.0 kg/s to 25 kg/s.
Consider State 1 to be the water inlet and State 2 to be the water outlet. Consider State 3 to be the air inlet and State 4 to be the air outlet.
Given: T1 = 150oC; x1 = 0.10; D1 = 4.0 cm = 0.04 m; V1 = 15 m/s; T2 = 100oC (take x2 = 0)
T3 = 15oC, m?air = 5.0 kg/s
Assume: Q? = 0 (insulated). Also, given no other information regarding the heat exchanger, make the following common heat exchanger assumptions: W? = ?KE =?PE = 0
Also, assume the heat exchanger is a multiple-inlet, multiple-outlet, steady-state, steady-flow device.
Treat the air as an ideal gas with constant specific heats: cp = 1.005 kJ/kg-K; ?hair = cp,air ?Tair
What will be an ideal response?
(a) The properties for the water: v1 = 0.04026 m3/s
h1 = 843.63 kJ/kg
h2 = 419.04 kJ/kg
A1 = ?D12/4 = 0.001257 m2
m?w = V1A1/v1 = 0.4682 kg/s
Then, from the First Law:
T4 = mw (h1-h2)/maircpair + T3 = 54.6 °C
(b) For the changing air mass flow rate:
Larger air mass flow rates result in a lower air exit temperature (less of a temperature increase for the air).
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