Exam memo for MTX311 2015 with worked out solutions. The memorandum has solutions are 100% correct including alternative working methods. The solutions are done in a step by step matter
10pt. 1)* To create comfortable living conditions, we want to produce 400 g/s of air of 25°C, 40%
relative humidity at 1 bar. The process starts with condensing water from hot, moist air (which has a
temperature of 45°C and a relative humidity of 45%). When the humidity of the air is sufficiently
low, the air is heated to 25°C. To what temperature do we have to cool the air? How much heat has
to be extracted in the cooling process (per kg of air), and how much condensate is formed?
Use the psychrometic chart for this question.
The following process is followed:
1) cool down at constant humidity ratio until air is saturated (φ=100%)
2) cool down and condensate at constant rel. humidity until end humidity ratio is reached
3) heat up at constant humidity ratio until T=25°C
First find the begin and end conditions (using the graph I used in class)
Begin =45°C, 45% rel. humidity. Read off humidity ratio and enthalpy: ω=0.0274, h=117 kJ/kg
End: 25°C, 40% rel. humidity. ω=0.010, h=54.5kJ/kg
End of cooling: ω=ωend=0.010, φ=100% h=34.5 kJ/kg
Read off Tend off cooling: 12.2°C
Q=mairΔh=0.4*(117-34.5)=33 kW
mc= mairΔω=0.4*(0.0274-0.0100)=0.00696kg/s=6.96g/s
Begin =45°C, 45% rel. humidity. Read off humidity ratio and enthalpy: ω=0.0272 (0.0265-0.0275),
h=134 (132-136)kJ/kg
End: 25°C, 40% rel. humidity. ω=0.0075 (0.007-0.008), h=65.2kJ/kg
End of cooling: ω=ωend=0.0075, φ=100% h=50 kJ/kg
Read off Tend off cooling: 10.0°C
Q=mairΔh=0.4*(134-50)=0.4*84=33.6 kW
mc= mairΔω=0.4*(0.0272-0.0075)=0.4*0.0197=0.00788kg/s=7.88 g/s
,
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