Ambient air at 1 atm, 32degree C and 60 % relative humidity enters a 400 mm internal diameter air-cooling duct with a velocity of 120 m/min. Inside the duct there is a cooling coil. The air leaves the duct saturated at 20 degree C and 1 atm. You also know that the cooling coil is receiving cold water from the local chilled water plant and the temperature of the chilled water when it leaves the cooling coil increases by 6 degree C with respect to its inlet temperature. Without using a psychrometric chart determine the amount of water that needs to be supplied to the cooling coil to produce the given exit conditions for the air. Your solution must include hand-calculations and show all the steps required to get the answer. If you need a value for the specific heat of water for the water going inside the cooling coil use 4.16 kJ/kg-K. You are interested in analyzing how changes in the inlet relative humidity of air affect both the required mass flow rate of chilled water needed for the cooling coil and the heat removed from the air. All of the other operating conditions remain unchanged. For your analysis consider the relative humidity of air at the inlet varying from 50% to 90%. Use EES or any other suitable software to study the effect of varying the relative humidity of ambient air that enters the duct. The result of your analysis should be a table that shows the heat removed in kJ/s and chilled water mass flow rate in kg/s ranging from relative humidity 50% to 90% in 5% increments and a plots for heat removed and mass flow rate as a function of relative humidity.
solution:
1)volume flow rate s given by=Q=A*V=.2513 m3/s
2)density of air at inlet is
dnsity=P1/RT1=1.1575*10^-5 kg/m3
mass flow rate=density*Q=2.9088*10^-6 kg/s
1)relative humidity of air at inlet is a=.6
hence a=Pv/Pvs
Pvs=.05 bar
Pv=.03 bar
where specific humidity is
w1=.622*Pv/Pa=.01841 kg/kg of dry air
entthalphy at inlet is
h1=m1\'(Cpa*T1+w(hg1+Cpv(T1-T1dp)))
at pv=.03 bar,T1dp=24.8
hence h1=79.2739m1\' kj
4)for second case
a=1
Pv=Pvs=.02337 bar
w2=.622*Pv/Pa=.01434 kg/kg of air
h2=m1\'(Cpa*t2+w2hg2)
h2=56.4963m1\' kj
4)heat rejected is
h1-h2=Mw*Cpw*6
we get on putting all value as
mw=2.6544*10^-6 kg/s
5)for varying inlet relative humidity like a=.9
we have
Pv=.9*.03=.045 bar
w3=.02762
and h3=103.m1\'
hence heat rejected is
h3-h2=mwcpw*6
mw=5.41*10^-6 kg/s
7)hence it is observe that with increasing relative humudity at inlet total amount of water neededis increases
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