Question 13.5: In an air-conditioning system, 1 kg/s air at 350 K and 10 pe......

From Figure 13.4:

at \theta _{1} = 350 K and humidity = 10 per cent; \mathscr{H} _{1} = 0.043 kg/kg
at \theta _{2} = 300 K and humidity = 30 per cent; \mathscr{H} _{2} = 0.0065 kg/kg

Thus, in equation 13.23:

m_{1}\mathscr{H} _{1}+m_{2}\mathscr{H} _{2}=m\mathscr{H} (13.23)

(1 x 0.043) + (5 x 0.0065) = (1 + 5)\mathscr{H}
and: \underline{\underline{\mathscr{H} =0.0125\ kg/kg}}

From Figure 13.5:

at \theta _{1} = 350 K and \mathscr{H} _{1} = 0.043 kg/kg; H_{1} = 192 kJ/kg
at \theta _{2} = 300 K and \mathscr{H} _{2} = 0.0065 kg/kg; H_{1} = 42 kJ/kg
Thus, in equation 13.25:

m_{1}(\mathscr{H} -\mathscr{H} _{1})=m_{2}(\mathscr{H} _{2}-\mathscr{H} ) (13.25)

1(H – 192) = 5(42 – H)
and: \underline{\underline{H=67\ kJ/kg}}

From Figure 13.5:
at H = 67 kJ/kg and \mathscr{H} = 0.0125 kg/kg

The data used in this example are shown in Figure 13.8.