ME354
Thermodynamics 2
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Quiz #5 - T02:
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Problem: An air-conditioning system operates at a total pressure of 1 atm and consists of a heating section and a humidifier
that supplies wet steam (saturated water
vapor) at 100 ◦ C. Air enters the heating
section at 10 ◦ C and 70% relative humidity at a rate of 35 m3 /min, and it
leaves the humidifying section at 20 ◦ C
and 60% relative humidity. Determine:
(a)
(b)
(c)
the temperature and relative humidity of air when it leaves the heating section,
the rate of heat transfer in the heating section, and
the rate at which water is added to the air in the humidifying section.
Assumptions
1.
2.
3.
4.
steady flow
mass flow rate of dry air remains constant throughout
dry air and water vapor (at this temperature) are ideal gases
KE = P E = 0
Part a)
There is sufficient information at the inlet and the exit to determine the state points on a psychrometric
chart.
h∗1 = 23.5 kJ/kgdry air
ω1 = 0.0053 kg H2 O/kgdry air
v1 = 0.808 m3 /kgdry air
h∗3 = 42.3 kJ/kgdry air
ω3 = 0.0087 kg H2 O/kgdry air
We note that
ṁa,1 = ṁa,2 = ṁa,3 = ṁa
and
ṁa =
V̇1
v1
=
35 m3 /min
0.809 m3 /kg
= 43.3 kg/min
Performing an energy balance between state point 2 and 3
(ṁh)a,2 + (ṁh)w,2 + ṁw hw = (ṁh)a,3 + (ṁh)w,3
ṁa h∗2 + ṁw hw = ṁa h∗3
But
mw
ma
= ω1 − ω2
Therefore
ṁa h∗2 = ṁa h∗3 + (ω2 − ω3 )hw
h∗2 = h∗3 + (ω2 − ω3 )hg@100 ◦ C
= 42.3 kJ/kg + (0.0053 − 0087) kg H2 O/kgdry air × 2675.6 kJ/kg
= 33.2 kJ/kgdry air
Therefore at state 2 we now have the enthalpy and ω2 = ω1 = 0.0053 kg H2 O/kgdry air . From
the psychrometric chart
T2 = 19.5 ◦ C ⇐ part a)
φ2 = 0.378 = 37.8% ⇐ part a)
Part b)
In the heating section
Q̇in = ṁa (h∗2 − h∗1 ) = (43.3 kg/min)(33.2 kJ/kg − 23.5 kJ/kg) = 420 kJ/min ⇐ part b)
Part c)
Applying conservation of mass over the humidification section
ṁw = ṁa (ω3 − ω2 ) = (43.3 kg/min)(0.0087 − 0.0053) kg H2 O/kgdry air
= 0.15 kg/min ⇐ part c)