Air enters a 30-cm-diameter cooling section at 1 atm, 35°C, and 45 percent relative humidity at 18 m/s. Heat is removed from the air at a rate of 750 kJ/min. Determine (a) the exit temperature, (b) the exit relative humidity of the air, and (c) the exit velocity.

T2=26C

Φ2=74%

V2=17.4 m/s

Explanation:

Determine first mass flow rate

mass flow rate=V1*r^2*pi/v1

V1 is the entrance velocity and v1 is the specific volume

mass flow rate=1.41 kg/s

using mass flow rate calculate h2

h2=h1-q/mass flow rate

h2=67.23 kJ/kg

Now using psychrometric chart calculate exist temperature

T2=26 C

V2=v2*V1/v1

V2=0.87*0.18/0.9

V2=17.4 m/s

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