Question 13.6: 0.15 kg/s steam at atmospheric pressure and superheated to 4......
0.15 kg/s steam at atmospheric pressure and superheated to 400 K is bled into an air stream at 320 K and 20 per cent relative humidity. What is the temperature, enthalpy, and relative humidity of the mixed stream if the air is flowing at 5 kg/s? How much steam would be required to provide an exit temperature of 330 K and what would be the humidity of this mixture?
Steam at atmospheric pressure is saturated at 373 K at which the latent heat
= 2258 kJ/kg
Taking the specific heat of superheated steam as 2.0 kJ/kg K;
enthalpy of the steam: H_{3} = 4.18(373 – 273) + 2258 + 2.0(400 – 373)
= 2730 kJ/kg
From Figure 13.5:
at θ_{1} = 320 K and 20 per cent relative humidity; \mathscr{H} _{1} = 0.013 kg/kg and H_{1} = 83 kJ/kg
The line joining the axis and slope H_{3} = 2730 kJ/kg at the edge of the chart is now drawn in and a parallel line is drawn through (H_{1}, \mathscr{H} _{1}).
Thus: (\mathscr{H} -\mathscr{H} _{1})=\frac{m_{3}}{m_{1}}=\left(\frac{0.15}{5}\right)=0.03~\mathrm{kg/kg}
and: \mathscr{H} = (0.03 + 0.013) = 0.043 kg/kg
At the intersection of \mathscr{H} = 0.043 kg/kg and the line through (\mathscr{H} _{1}, H_{1})
\underline{\underline{H=165\ kJ/kg\ and\ \theta =324\ K}}When θ = 330 K the intersection of this isotherm and the line through (\mathscr{H} _{1}, H_{1}) gives an outlet stream in which \mathscr{H} = 0.094 kg/kg (83 per cent relative humidity) and H = 300 kJ/kg.
Thus, in equation 13.28:
m_{1}(\mathscr{H} -\mathscr{H} _{1})=m_{3} (13.28)
m_{3} = 5(0.094 – 0.013) = \underline{\underline{0.41\ kg/s}}
The data used in this example are shown in Figure 13.9.
