© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-1 A mixture of helium and water vapor is flowing through a pipe at T= 90°C and P = 150
kPa. The mole fraction of helium is yHe = 0.80.
a.) What is the relative humidity of the mixture?
b.) What is the humidity ratio of the mixture?
c.) What is the dew point of the mixture?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-2 Figure 12.A-2 shows a humidifier that takes in an air-water vapor mixture with 10%
relative humidity, 1 = 0.1, at temperature T1 = 18°C and pressure P1 = 1 atm and mixes it
with a source of steam at P2 = 50 psia in order to generate an air-water mixture that has
70% relative humidity, 3 = 0.7, with temperature T3 = 20°C and pressure P3 = 1 atm.
The mass flow rate of the inlet air-water mixture is m 1 = 0.25 kg/s.
steam at P2  50 psia
m 1  0.25kg/s
T1  18C
P1  1 atm
1  0.10
humidifier
T3 = 20°C
P3 = 1 atm
3 = 0.7
Figure 12.A-2: Humidifier.
The humidifier is at steady-state and is externally adiabatic.
a.) Determine the humidity ratio of the air-water vapor mixture entering the humidifier at
state 1 (1) and the humidity ratio at state 3 (3).
b.) Determine the mass flow rates of air and the mass flow rate of water vapor entering
the humidifier at state 1 and leaving at state 3.
c.) Determine the mass flow rate of steam entering the humidifier at state 2.
d.) What is the temperature of the steam entering the humidifier at state 2?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-3 Figure 12.A-3 shows an air compressor system that is used in a factory. At state 1, the
compressor draws in ambient air (which is actually an air/water vapor mixture) with To =
20ºC and Po = 1 atm. The relative humidity of the ambient air/water mixture is 1 = 0.5.
The mass flow rate of the incoming air/water mixture is m 1 = 0.1 kg/s. The rate of heat
transfer from the compressor to the atmosphere is Q = 50 kW. The air/water mixture
c
leaves the compressor at T2 = 175ºC and P2 = 90 psig. The hot air/water mixture is
cooled in an aftercooler that rejects heat to ambient. The air/water mixture leaving the
aftercooler is at T3 = To and P3 = P2. The aftercooler also serves as a filter and drier.
Condensate (i.e., liquid water condensed out of the air) leaves the aftercooler at state 4, T4
= To and P4 = Po.
aftercooler
T3 = To
P3 = P2
T2 = 175°C
P2 = 90 psig
2
Wc
compressor
Q ac
1
4 condensate
T4 = To
P4 = Po
Q c  50 kW
T1  To  20C
P1  Po  1 atm
1  0.5
m 1  0.1 kg/s
Figure 12.A-3: Air compressor with aftercooler.
a.) Determine the humidity ratio of the air at state 1. What is the mass flow rate rate of
dry air and water vapor entering the compressor ( m a ,1 and m v ,1 )?
b.) Assume that no condensation of water occurs in the compressor. Determine the
relative humidity and humidity ratio at state 2. Is the assumption correct? How do
you know?
c.) Determine the power consumed by the compressor ( Wc ) and the rate of entropy
generation within the compressor ( S ).
gen , c
d.) Determine the humidity ratio of the air at state 3.
e.) What is the mass flow rate rate of dry air and water vapor leaving the aftercooler
( m a ,3 and m v ,3 )? What is the mass flow rate of condensate leaving the aftercooler
( m 4 )? What is volumetric flow rate of condensate (in gallons/day)?
f.) What is the rate of heat transfer from the aftercooler to the environment ( Q ac )?
g.) Use EES to draw a T-s diagram for the states of the water in the problem (i.e., the
water vapor at states 1, 2, and 3 and the water leaving at state 4). Label the states.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-4 During the winter, indoor air with relative humidity 1 = 0.7, temperature T1 = 22°C and
volumetric flow rate V1 = 0.04 m3/s is used to pre-heat air drawn from the outdoors at 3
= 0.2, temperature T3 = 10°C, and volumetric flow rate V = 0.06 m3/s in a heat
3
exchanger as shown in Figure 12.A-4. The outdoor air leaving the heat exchanger is
heated to T4 = 20°C. The process occurs at atmospheric pressure and the heat exchanger
is externally adiabatic. Use the Psychrometric Chart to solve this problem.
indoor air
3
V1  0.04 m /s
T1  22C (1)
1  0.7
heat exchanger
(2)
Q
T4 = 20°C
(3) outdoor air 3
V3  0.06 m /s
T3  10C
3  0.2
Figure 12.A-4: Heat exchanger.
(4)
a.) What is the relative humidity of the air leaving the heat exchanger at state (4)?
b.) What is the rate of heat transfer between the air streams in the heat exchanger, Q ?
c.) Will there be any condensation as the indoor air flows through the heat exchanger?
Justify your answer.
d.) Assume that your answer to (c) is yes. At what temperature will the indoor air begin
to condense?
e.) What is the exit temperature of the indoor air (T2)? What is the mass flow rate of the
condensate from the indoor air? You may neglect the enthalpy carried by the
condensate to answer this question.
f.) Locate and label all four states on the Psychrometric chart.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-5 Two flows of air/water vapor mixtures enter a steady flow device, as shown in Figure
12.A-5.
Flow 2
3
V2  0 .03 m /s
T2  10C
2  0.3
Flow 1
3
V1  0 .01 m /s
T1  25C
1  0.8
steady flow
device
Flow 3
Q  100 W
Figure 12.A-5: Steady flow device.
Flow 1 enters the device with temperature T1 = 25°C and relative humidity 1 = 0.8. The
volumetric flow rate at state 1 is V1 = 0.01 m3/s. Flow 2 enters the device with
temperature T2 = 10°C and relative humidity 2 = 0.3. The volumetric flow rate at state 2
is V2 = 0.03 m3/s. Heat is added to the device at a rate of Q = 100 W. A single flow of
air/water vapor mixture leaves at state 3 and there is no condensation in the device. The
pressure of each stream (entering and leaving) is P = 1 atm. Use the psychrometric chart
to determine the temperature and relative humidity of the air leaving the device. Indicate
each of the states on the chart.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-6 A flow of 400 cfm of air at 80°F, 14.7 psia is cooled to a saturated state at 50°F.
Calculate and plot the following quantities as a function of the relative humidity of air for
relative humidities between 0.35 and 0.75.
a.) humidity ratio of the stream
b.) the rate of moisture removal in lbm/hr
c.) the minimum energy required to cool the air in Btu/hr
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-7 An apparatus that can supply air at a controlled dry-bulb temperature and relative
humidity has been designed as indicated in the schematic in Figure 12.A-7. Dry air is
supplied at locations 1 and 2. One stream is isothermally humidified to saturation (state
3) and then mixed with the other stream to produce the supply air at location 4. The
entire operation occurs isothermally with a temperature-regulated enclosure. The dry air
flow rates at locations 1 and 2 are adjusted in proportion to the desired relative humidity
at location 4. In the case at hand, 50% relative humidity is desired so the dry air mass
flow rates at location 1 and 2 are equal.
Table 12.A-7: Measurements
Constant Temperature Enclosure
Dry air
Humidifer
1
3
4
Dry air
2
T
°C
10
20
30
40
50
60
70
80
Relative
Humidity
49
49
50
50
51
52
57
61
Figure 12.A-7: Apparatus for supplying controlled humidity air
Table 12.A-7 shows the measured conditions at state 4 for several temperatures. For
temperatures below 60°C, the apparatus functions well in achieving the desired 50%
relative humidity. However, significant errors occur for temperatures above 60°C. The
technicians have checked the equipment and calibrated all measuring devices, but cannot
find a problem. Can you help?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A.8 In an HVAC (heating, ventilating, and air conditioning) system, atmospheric air at a
volumetric flow rate of 0.35 m3/s at 25C and 60% relative humidity is dehumidified and
cooled with a water spray in a counter-current arrangement. The water is supplied to the
spray head at 0.5 kg/s and 5C. The dry-bulb temperature, wet-bulb temperature, and
relative humidity of the air exiting the spray are to be determined. Heat and mass transfer
theory indicate the process may be described in terms of an effectiveness factor defined
by
 hair ,in  hair ,out    air ,in  air ,out 

air ,in  air ,out , sat ' d at Tw ,in
hair ,in  hair ,out , sat ' d at Tw ,in

 

where hair ,in and hair ,out are the specific enthalpies of the entering end exiting air-water
vapor mixtures per unit masss of dry air, air ,in and air ,out are the humidity ratios of
entering and exiting air, and hair ,out , sat ' d at Tw ,in and air ,out , sat ' d at Tw ,in are the specific enthalpy
and humidity of saturated air at the water inlet temperature. Prepare a plot of the drybulb and wet-bulb temperatures and the relative humidity as a function of the effectivenss
factor for values between 0 and 1.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-9 A building space having a volume of 140 m3 is to be maintained at 25°C, 50% relative
humidity on a winter day. Ventilation air is provided at 0.02 m3/s, 5°C, and 30% relative
humidity. Heat losses from the building space to outdoors occur at a rate of 1.5 kW. A
humidifier provides the necessary moisture to the building space in the form of saturated
vapor at 100°C. Space heaters within the building space maintain the required room
temperature. Assuming the air in the room to be fully mixed, determine:
a.) the rate at which moisture must be added
b.) the rate at which heating must be provided in the building space
c.) the mass of water vapor in the room at any time
d.) the temperature of the walls in the room that would produce condensate
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-10 Ambient air at 85°F and 75°F wetbulb enters a cooling tower at a volumetric flowrate of
54,440 cfm. The air exits the cooling tower at 91°F. The relative humidity of the air
exiting the cooling tower is difficult to measure, but it is believed to be somewhere
between 90% and saturated. Condenser water from a chiller plant enters the cooling
tower at 103°F at a flowrate of 2,848 lbm/min.
a.) What is the relative humidity of the ambient air?
b.) What is the inlet air flow rate?
c.) Calculate and plot the water outlet temperature and make-up water required as a
function of the relative humidity exiting the cooling tower for relative humidities
between 0.9 and 1.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-11 Two tanks, A and B, are connected by piping and a closed valve. Tank A contains 4 kg
of butane (C8H18) at 3.5 bar and 175°C. Tank B contains 1.4 kg of water vapor at 0.5 bar
and 95°C. The valve is opened and after some time, the contents of the tanks completely
mix. A temperature measurement indicates that the mixture is at 126°C.
a.) What is the final pressure of the gas mixture?
b.) What is the humidity ratio of the gas mixture?
c.) Determine the value of the relative humidity of the final gas mixture.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-12 A manufacturing plant requires a supply of 100 cfm of air at 70°F, 20% relative humidity
and 1 atm for purging equipment. One way to obtain this relatively dry air is to
compressed atmospheric air which is available at 70°F dry-bulb, 66°F wet-bulb at 1 atm
to a suitable pressure. The compressed air is cooled to 70°F in a heat exchanger with
cooling water and then throttled to 1 atm as shown in Figure 12.A-12. The compressor
operates adiabatically with an isentropic efficiency of 0.72.
Figure 12.A-12: System for drying air
a.) Determine the necessary pressure at the compressor outlet (state 2).
b.) Determine the power required to operate the compressor at steady conditions.
c.) Determine the rate at which condensate must be removed.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-13 A rigid tank with volume V = 1 m3 initially contains pure refrigerant R134a vapor at P1=
100 kPa and T1 = Tamb = 20ºC. The tank is connected to a source of high pressure
nitrogen at Ps = 500 kPa and Ts = 200ºC, as shown in Figure 12.A-13.
pure nitrogen
Ps = 500 kPa
Ts = 200°C
tank
initially filled with pure R134a
P1 = 100 kPa
T1 = 20°C
V = 1 m3
Figure 12.A-13: Tank containing refrigerant vapor connected to a source of nitrogen.
The valve is opened, allowing pure nitrogen to flow into the tank. At the conclusion of
the flow process, the pressure in the tank is P2 = Ps and the temperature in the tank is T2 =
100ºC. Model both the nitrogen and the refrigerant as an ideal gas and assume that they
form an ideal gas mixture. The properties of the refrigerant should be evaluated using Rr
= 81.5 J/kg-K, cP,r = 840 J/kg-K, and cv,r = 758.5 J/kg-K. The properties of nitrogen
should be evaluated using RN2 = 297 J/kg-K, cP , N2 = 1040 J/kg-K, and cv , N2 = 743 J/kgK.
a.) Determine the amount of heat transferred from the tank to the surroundings at Tamb
during the filling process.
b.) Determine the entropy generated by the filling process.
At the conclusion of the filling process, the valve is closed and the temperature of the
tank is gradually reduced.
c.) At what temperature will R134a begin to condense? Note that your text includes
tables for the properties of saturated R134a.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.A-14 A closed rigid tank having a volume of 3 m3 contains air at 100°C, 4.4 bar, and 40%
relative humidity. The tank contents cool to 20° C as a result of heat transfer with the
surroundings. The gas mixture may be assumed to behave as an ideal gas.
a.) The mass of water in the tank is ________ kg.
b.) The temperature at which condensation begins is _________ °C
c.). The mass of liquid water in the tank at 20° C is __________ kg.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-1 In an air conditioning system that operates at steady-state, return air at V1 = 35 m3/min, T1
= 22°C,  54% relative humidity is mixed with outdoor ventilation air at V = 12
2
m /min, T2 = 30°C, 2 = 55% relative humidity, as shown in Figure 12.B-1. The mixed
air at state (3) first passes over a cooling coil. The air-water mixture exits in a saturated
condition at state (4). Some water vapor is condensed in the cooling coil and leaves at
state (6). The saturated moist air and condensate streams exit the dehumidifier at the
same temperature (T4 = T6). The moist air at state (4) then passes through a heating coil.
The air is exhausted to the building at T5 = 24°C and 5 = 40% relative humidity. The
entire process occurs at atmospheric pressure.
3
building return air
3
V1  35 m /min
T1  22C
1  0.54
Q h
(1)
(2)
(3)
cooling
coil
mixer
outdoor air
3
V2  12 m /min
T2  30C
2  0.55
(4)
heating
coil
(6)
condensate
Q c
(5)
air to building
T5  24C
5  0.40
Figure 12.B-1: Air conditioning system.
a.) Determine the temperature and relative humidity of the air at state (3).
b.) Determine the temperature of the moist air exiting the cooling coil at state (4).
c.) Determine the rate at which cooling must be provided in the cooling coil in kW and in
tons. Also determine the mass flow rate of condensate leaving the cooling coil.
d.) Plot the cooling required as a function of the outdoor air relative humidity.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-2 One type of residential humidifier is shown in Figure 12.B-2. The humidifier is directly
connected to the furnace. In a particular case, air from the building at T1 = 20°C and 
30% relative humidity (state 1) is heated in the furnace to T2 = 95°C. A fraction of
furnace outlet air (f) passes through a humidifier; it exits as saturated air at T3 = 50°C.
The humidified air mixes with the air from the furnace and the mixture reenters the
building at state (4). The entire process occurs at atmospheric pressure.
water
T2 = 95°C
T3 = 50°C,  = 1
f
humidifier
(2)
(3)
mixer
furnace
(1-f)
air from building
T1 = 20°C, 1 = 0.3
(2)
(4)
Figure 12.B-2: Humidifier.
a.)
b.)
c.)
d.)
What is the humidity ratio of the air in the building at state 1?
What is the relative humidity of the air at state (2)?
What is the humidity ratio of the air at state 3?
What is the ratio of the mass flow rate of liquid water that must be added to the
humidifier to the mass flow rate of the air entering the humidifier?
e.) Determine the temperature and relative humidity of the air leaving the mixer that is
supplied to the building if f = 0.5 (i.e., 50% of the flow passes through the
humidifier).
f.) Plot the temperature and relative humidity of the air returning to the building as a
function of the fraction of the flow diverted through the humidifier, f.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-3 Your cabin has no air conditioning system. Therefore you have installed a swamp cooler,
as shown in Figure 12.B-3.
makeup water
3
V1  0.035 m /s
T1  30C
1  0.2
2 = 0.9
Q  250 W
(3)
swamp cooler
cabin
(3)
Figure 12.B-3: Swamp cooler.
The cooler draws in air at a rate of V1 = 0.035 m3/s from outdoors. Your cabin is located
in a hot but dry climate; the outdoor air is at T1 = 30ºC and 1 = 0.2. The air is forced to
flow through a chamber that is filled with a porous structure that is saturated with liquid
water. The chamber is adiabatic and the air exits from the chamber at 2 = 0.90. The
air/water mixture passing through the chamber is being adiabatically saturated; therefore,
it follows a line of constant adiabatic saturation (or wet bulb) temperature. The air
leaving the chamber is directed to your cabin. The cabin has no significant moisture gain
but does experience heating at a rate of Q = 250 W. Air leaves the cabin at temperature
T3, the indoor air temperature of the cabin.
a.) Determine the humidity ratio of the outdoor air.
b.) Determine the mass flow rate of dry air passing through the system.
c.) Determine the humidity ratio and temperature of the air leaving the chamber.
d.) Determine the volume of makeup water (in gallon) that must be provided to the
chamber each day.
e.) Determine the temperature and relative humidity of the air leaving the cabin.
f.) Sketch the three states on the psychrometric chart - make sure that you label each of
the states.
g.) If you wanted to keep your cabin at the temperature that you calculated in (e) and the
outdoor air temperature remained at 30ºC, then estimate the outdoor air relative
humidity at which the swamp cooler would no longer provide any useful cooling to
the cabin (assume the air leaving the swamp cooler will always be at 2 = 0.9).
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-4 Figure 12.B-4 illustrates a system used to humidify building air.
saturated steam
Ts  110 C
m s  0.0015kg/s
3
V1  0 .10 m /s
Building air
Tb = 20°C
Twb,b = 10°C
1
2
steam-spray
humidifier
3
V3  0 .15m /s
3
mixer
4
Figure 12.B-4: Steady flow device.
Two streams of air are removed from the building space. The building air has
temperature (i.e., dry bulb temperature) Tb = 20°C and adiabatic saturation temperature
(i.e., wet bulb temperature) Twb,b = 10°C. The stream of air at state 1 has volumetric flow
rate V1 = 0.1 m3/s and enters a steam-spray humidifier where it is humidified through the
injection of a saturated steam flow. The saturated steam is injected at Ts = 110°C with
mass flow rate m s = 0.0015 kg/s. A separate flow of air is pulled from the building with
volumetric flow rate V = 0.15 m3/s and mixes with the air leaving the humidifier. The
3
mixed air flow at state 4 re-enters the building. The pressure of each air stream is at
ambient pressure. Use the Psychrometric chart to do this problem.
a.) Determine the temperature and relative humidity of the air leaving the steam spray
humidifier at state 2.
b.) Determine the temperature and relative humidity of the air that is provided to the
building at state 4.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-5 In an air conditioning system that operates at steady-state, return air at 35 m3/min, 22°C,
54% relative humidity (state 1) is mixed with outdoor ventilation air at 12 m3/min, 30°C,
55% rh (state 2), as shown in Figure 12.B-5. The mixture (state 3) air first passes over a
cooling coil such that the air exits in a saturated condition (state 4) and some water vapor
is condensed (state 6). The saturated moist air and condensate streams exit the
dehumidifier at the same temperature. The moist air then passes through a heating unit,
exiting at 24°C, 40% rh (state 5).
Building return air
35 m3/min
22°C, 54% rh
1
2
Outdoor air
12 m3/min
30°C, 55% rh
heat
mixing
chamber
3
cooling coil
heat
4
heating coil
5
24°C, 40% rh
6
condenstate
Figure 12.B-5: Air conditioning system
Determine:
a) the temperature and relative humidity of the air state 3
b) the temperature of the moist air exiting the cooling coil (state 4)
c) the rate at which cooling must be provided in the cooling coil in kW and in tons
d) the rate of condensate in kg/min
e) the rate of heat transfer to the air passing through the heating unit in kW.
Plot all state points on a psychrometric chart.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-6 The total (sensible plus latent) air conditioning load for a building is 7.3 kW when the
outdoor conditions are at 35°C dry-bulb, 25°C wet-bulb and the indoor space is
maintained at 24°C dry-bulb, 50% relative humidity. For ventilation purposes, outdoor
air is mixed with recirculated air in a 1 to 4 proportion. The mixed air is cooled and
dehumidified as it flows through a cooling cool. The air exiting the cooling coil is
saturated. The conditioned air is then blown into the space to supply the air conditioning
load. The cooling coil heat transfer effectiveness, e, is approximately 0.70, defined as

Tair ,in  Tair ,out
Tair ,in  Tsupply
where Tair ,in is the dry-bulb temperature of the mixed air entering the cooling coil, Tair ,out
is the dry-bulb temperature of the air leaving the cooling coil and Tsupply is the
temperature of the chilled water entering the coil.
a.) Determine the temperature and relative humidity of the air entering the cooling coil.
b.) Calculate and plot the sensible heat ratio provided by the conditioned air as a function
of Tsupply for Tsupply values between 2°C and 12°C.
c.) Calculate and plot the required volumetric flowrate of conditioned air needed to meet
the total load as a function of Tsupply for Tsupply values between 2°C and 12°C.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-7 Supermarkets have unusual cooling needs during the summer. Because the freezer cases
are often left uncovered, frost accumulates on the frozen food products making them
unsightly. The frost also reduces the efficiency of the frozen food cases. To reduce this
problem, the humidity in the store can be lowered. A cooling system designed for this
purpose is shown in Figure 12.B-7.
Exhaust
Return air
6
Supermarket
Bypass
Control
Valve
Ventilation
Air
1
Reheat Coil
Coiling Coil
2
3
4
5
Figure 12.B-7: Supermarket refrigeration system
4000 cfm of air at 62°F, 55% relative humidity is supplied at state 5 to maintain comfort
conditions in the supermarket. Air returns from the supermarket at 74°F, 54% relative
humidity at state 6. 15% of the return air (mass basis) is exhausted and replaced with
outdoor ventilation air at 82°F, 48% relative humidity (state 1). Some of the recirculated
air is mixed with the ventilation air (state 2) and passed through the cooling coil from
which it emerges at saturated conditions (state 3). The rest of the recirculated air
bypasses the cooling coil, mixes with the cooling coil outlet air and then enters the reheat
coil at state 4. Calculate and plot the following quantities as a function of the fraction of
the recirculation air that bypasses the cooling coil. What is the largest value of the
bypass that will allow the air at state 3 to remain above the freezing temperature?
a.) the required heat transfer rate in the cooling coil
b.) the cooling coil outlet air temperature
c.) the rate of heat transfer in the reheat coil
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-8 A cooling tower operates with a flow of condenser water of 50,000 lbm/hr that enters the
tower at 95°F. The ambient air used to cool the water is at a dry bulb temperature of
70°F and a relative humidity of 50 %. The air flow rate is 10,000 cfm. The performance
of a cooling tower can be estimated in terms of a cooling tower effectiveness defined as:

m a  hair ,out  hair ,in 
Q

Q max m a hsat ,T  hair ,in
w ,in


where m a is the dry air mass flowrate, m w,in is the inlet water flowrate, hair ,in and hair ,out
are the specific enthalpies of the air entering and leaving the cooling tower and hsat ,Tw ,in is
the specific enthalpy of air assuming it is saturated at the water inlet temperature. Air
exits the tower at a relative humidity of 0.98 at all effectiveness values.
a.) Plot the rate of the water outlet temperature and required rate of water replacement as
a function of the cooling tower effectiveness for values between 0.2 and 0.8.
b.) Plot the cooling tower Range and Approach parameters s a function of the cooling
tower effectiveness for values between 0.2 and 0.8.
c.). Determine the heat transfer rate that would result for the same effectiveness and air
and water flowrates if a ‘dry’ counterflow heat exchanger were used instead of a
cooling tower.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-9 Shown in Figure 12.B-9 is a schematic of an indirect/direct evaporative cooling system.
Air at 26°C, 40% relative humidity (state 1) entering the equipment is split into two flow
streams. One stream passes through an evaporative cooler that lowers the temperature of
the air to T2. This cooled air is then used as the cold stream in a countercurrent heat
exchanger to lower the temperature of the other ambient air stream to state 4. The heat
exchanger effectiveness is 0.85 based on the minimum capacitance rate. The air exiting
the heat exchanger at state 4 then passes through a second evaporative cooler to emerge
at the final conditioned state.
Evaporative Cooler
2
conditioned
air
5
26°C
40% rel.hum.
3
exhaust
Evaporative Cooler
1
Heat Exchanger
4
Figure 12.B-9: Indirect/direct evaporative cooling system
a.) Assume that the effectivenesses of the evaporative coolers are both 0.75 at any air
flow rate. Calculate and plot the energy rate at which conditioned air (state 5) is
provided (relative to state 1) as function of the ratio of the mass flow rates at state 1
and 3.
b.) If the effectiveness of one of the evaporative coolers could be increased to 0.85 while
the other remained at 0.75, which evaporative cooler would you select to have the
higher effectiveness?
c.) Indicate advantages and disadvantages of the system in Figure 12.B-9 compared to a
traditional evaporative cooling system.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-10 The purpose of this problem is to estimate the dry air mass flow rate and cooling capacity
of the cooling coil necessary to maintain indoor design conditions of 24°C, 50% relative
humidity for a the system shown in Figure 12.B-10. Ventilation air from outdoors is
supplied to the building 32°C, 45% relative humidity at a rate of 0.25 kg/s. A portion of
the air exiting the building is mixed with the ventilation air and the rest is exhausted.
Conditioned air is provided to the building at 12°C. The sensible heat gain occurring in
the building due to lights, people and equipment is 10 kW. The moisture gain in the
space is not known, so generate your results for sensible heat ratios over the expected
range of 0.7 to 0.9. Neglect the fan energy. The result of your analysis should be plots of
the mass flow rate at state 3, the relative humidity at state 3 and the required cooling coil
capacity as a function of the sensible heat ratio.
32°C, 45% rh
0.25 kg/s
1
Mixing
Section
Exhaust
4
2
Conditioned Space
24°C. 50% rel. hum
12°C
10 kW generation
SHR 0.7 to 0.9
3
Supply
Fan
Figure 12.B-10: Building air-conditioning system
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-11 An air conditioning system is to be designed to deliver 15,000 ft3/min of air at 60°F and a
relative humidity than is less that 50% for all outdoor conditions. A schematic of the
system is shown in Figure 12.B-11. Hourly values of the outdoor air temperature and
relative humidity are provided for a 12 hour period in Table 12.B-11. Write a program
that will calculate the total cooling provided and total condensate removed in the cooling
coil and the total heating that must be supplied in the reheat coil for the 12 hour period.
Outdoor Air
15000 cfm
1
60°F
<50%
Cooling
Coil
2
Reheat
Coil
3
condensate
Figure 12.B-11: Air-conditioning system
Table 12.B-11: Outdoor temperature and relative humidity for 12 hours
Time
7:00
8:00
9:00
10:00
11:00
12:00
Outdoor
Temp., °F
60
70
75
80
85
90
Relative
Humidity
90
80
75
70
65
50
Time
13:00
14:00
15:00
16:00
17:00
18:00
Outdoor
Temp., °F
95
95
90
80
70
65
Relative
Humidity
55
60
65
75
80
85
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-12 Figure 12.B-12 shows a novel air conditioning system. It is similar to a conventional air
conditioning system, but it includes an additional heat exchanger. Air at 28°C, 45%
relative humidity (state 1) is cooled to state 2 without condensation) in the heat
exchanger. The air then passes through a conventional cooling coil where cooling and
condensation both occur. The pinch point occurs at state 3 where the air leaving the
cooling coil is 10°C warmer than the evaporating refrigerant. The air exiting the cooling
cool passes through the heat exchanger as the cold stream (without mixing with stream 12) and then to the reheat coil where heat is provided so that air exits 18°C, 40% relative
humidity.
Cooling
Coil
Heat
Exchanger
Reheat Coil
Refrigerant
out
18°C, 40% rh
5
4
3
28°C, 45% rh
1
2
Refrigerant
in
Condensate
Figure 12.B-12: Schematic of a novel air-conditioning system
The purpose of this problem is to determine the benefit of the heat exchanger. In your
analysis, please do the following:
a.) determine the maximum value of the heat exchanger effectiveness that will cool the
air at state 2 without condensation occurring.
b.) determine the required rates of cooling and reheat per kg of dry air for effectiveness
values between 0 and the value you determined in part a).
c.) indicate the benefits (if any) and disadvantages of the proposed system.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-13 An air-conditioning system for a supermarket uses two cooling coils as shown in Figure
12.B-13. The cooling coil in the outdoor air duct cools and dehumidifies the ventilation
air so that air exiting this coil is at 40°F and a humidity ratio of 0.005. The return air
cooling coil operates at higher temperature and does not provide any condensation. The
amount of cooling and the air flows for each cooling coil are controlled so as to provide
supply air at 55°F, with a wet bulb temperature of 52°F. The return air taken from the
supermarket is maintained at 75°F, 50% relative humidity. The total cooling load for the
supermarket is 25 tons when the outdoor conditions are 95°F, 50% relative humidity.
Return
Air
Exhaust
5
Supermarket
Outdoor Air
95°F, 50% rh
1
75°F, 50% rh
Cooling
Coil
3
2
4
condensate
Figure 12.B-13: Supermarket cooling system with two cooling coils
Determine:
a.) The supply dry air flow rate
b.) The dry air flow rates through each cooling coil.
c.) The cooling load for each cooling coil.
d.) The operating cost in $/hr if the COP of the outdoor cooling coil is 2.8, the COP of
the return air cooling coil is 3.6 and electricity is purchased at $0.15/kwhr. (Neglect
the cost to operate the fans.)
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-14 In the air-conditioning system shown in Figure 12.B-14, 200 lbm dry air/min at 95°F,
55% relative humidity (state 1) is cooled to 48°F (state 2) in a cooling coil. Water enters
the cooling coil at 55°F and exits at 62°F. The cooled air is then mixed with 1000 cfm of
outdoor air to produce the conditions at state 3. The air is then heat exchanged with
water that enters at 140°F and exits at 125°F. The water flow rate is controlled so that the
mixture exits at 68°F (State 4).
1000 cfm
Outdoor Air
95°F, =55%
200 lb m/min
Cooling
Coil
125°F
68°F
=?
48°F
1
2
62°F
140°F
3
Heating
Coil
4
55°F
condensate
48°F
Figure 12.B-14: Schematic of an air conditioning system
a.) Determine the rate of condensate formation.
b.) Determine the tons of cooling that must be provided in the cooling coil and the
required chilled water flow rate.
c.) What are the temperature and relative humidity at state 3?
d.) What is the rate of rate that hot water is provided to the heating coil?
e.) What is the relative humidity of the air delivered at state 4?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-15 One method of controlling the amount of cooling required for a coil is to use a bypass
method in which only part of the air passes through the coil, and the other bypasses the
coil completely as shown in Figure 12.B-15. For a specific situation, the inlet air flow
rate is 10,000 lbm/hr at a temperature of 90F and a relative humidity of 40%. The
desired supply state is 65F with a maximum relative humidity of 60%. The air leaving
the cooling coil (state 2) can be assumed to be saturated.
cooling
coil
1
10,000 lbm/hr
90°F, =40%
2
bypass
damper
3
4
65°F,
=60%
Figure 12.B-15: Air conditioning unit with bypass damper
a.) Plot the required rate of cooling, the required rate of heating, and the required
saturation temperature at state 2 as a function of the bypass flow, f, where f=0
indicates that none of the air passes through the cooling coil and f=1 indicates that all
of the flow passes through the cooling coil.
b.) Based on your plot, what is your recommendation for the bypass damper setting?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-16 The air-conditioning system shown in Figure 12.B-16 is intended to provide conditioned
air for a greenhouse. Air enters the system at 24°C, 20% relative humidity at 0.32 m3/s.
This air flow is split into two equal mass flows. One stream passes through an adiabatic
evaporative cooler that is supplied with water at 25°C. The effectiveness of the
evaporative cooler is 0.90. The two streams are then combined in a mixing chamber and
provided to the greenhouse. The greenhouse is maintained at 27°C, 60% relative
humidity.
Figure 12.B-16 Air conditioning system for a greenhouse
a.) Determine the temperature and relative humidity at state 5.
b.) Determine the total cooling load for the greenhouse and the sensible heat ratio of the
load .
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-17 A cooling tower is shown in Figure 12.B-17 with operating conditions indicated.
air
4
water from condenser
Vw  6000 gal/hr
Tw ,in  95F
1
3
water to condenser
2
5
air
Tamb  74F
amb  0.5
Va  10,000 cfm
makeup water
Tmuw = 80°F
Figure 12.B-17: Cooling tower.
The cooling tower is designed to cool water with flow rate Vw = 6,000 gallons per hour
that enters the tower with temperature Tw,in = 95°F (35°C) from the condenser of a large
air conditioning system. Outdoor air at Tamb = 74°F (23.3°C) and amb = 50% relative
humidity is forced through the tower at a rate of Va = 10,000 cfm (4.72 m3/s). Makeup
water is provided at Tmuw = 80°F (26.7°C). The temperature of the air leaving the
cooling tower at state 4 is Ta,out = 89°F (31.7°C) and the relative humidity of the air
leaving at state 4 is a,out = 0.94. Use the Psychrometric Chart to do this problem. If
you need properties of pure water use the table in your text.
a.) Determine the rate of makeup water (in gal/hr) that must be provided to the cooling
tower.
b.) Determine the temperature of the water leaving the cooling tower at state 2.
c.) Determine the rate of cooling that is provided to the condenser using the cooling
tower (in ton).
d.) Clearly indicate states 3 and 4 on the Psychrometric chart. Note that one of the
reasons that the cooling tower works so well is that the air passing through the
tower is not only heated, it is also humidified. Therefore, both its temperature and
humidity ratio are increased. What fraction of the total enthalpy change of the air is
due to it temperature change only (i.e., what fraction of the enthalpy change of the
air would be achieved if its humidity ratio did not change but it left at the same
temperature, Ta,out)? This is referred to as the sensible heat ratio of the device.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.B-18 Figure 12.B-18 illustrates a building that is being cooled using an earth tube system.
outdoor air
Tout = 30°C
Q  5.8 kW
building
out = 70%
infiltration
3
V3  0.08 m /s
exhaust at
3
indoor air
3
V1  0.51 m /s
4 conditions
2
1
Tg = 8°C
ground tube
Figure 12.B-18: Earth tube system.
Outdoor air at out = 70% and Tout = 30°C is drawn into tubes at state 1 with volumetric
flow rate V1 = 0.51 m3/s. The tubes are buried in the ground. The ground temperature
is Tg = 8°C and the air leaving the tubes and entering the building at state 2 has been
cooled to within Tg = 3°C of the ground temperature. There is an infiltration of
outdoor air directly into the building at state 3. The volumetric flow rate of outdoor air
that directly enters the building is V3 = 0.08 m3/s. The building experiences a heat
transfer of Q = 5.8 kW. Assume that no moisture is added to the air as it passes
through the building. Indoor air leaves the building through the exhaust duct at state 4.
Assume that the pressure of the air is atmospheric throughout the system and solve this
problem using the Psychrometric chart.
a.) Is there any condensation within the ground tube? If so then determine the
volumetric flow rate of condensation draining into the ground in the ground tube (in
gallons/day).
b.) Determine the temperature and relative humidity of the indoor air.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-1 Consider a summer day with outdoor conditions at 27C and 40% relative humidity. An
air flow is split into two (possibly unequal) streams as shown in Figure 12.C-1. One
stream is passed through an evaporative cooler that cools that has an effectiveness of 1.
The two streams then enter a device that uses the temperature difference between the
streams to produce power. The work producing device may be adiabatic and the streams
do not mix as shown in the figure. Assuming that power be produced in this manner:
Water, 27°C
Air
27°C, 40%rh
1
Evaporative
Cooler
2
3
Work
Producing
Device
4
Figure 12.C-1 Power production from an evaporative cooler
a.) Explain why this power production scheme does not violate the Kelvin-Planck
statement of the Second Law which states that “It is impossible for any device that
operates continuously to receive heat from a single reservoir and produce a net
amount of work”.
b.) How would you define the efficiency of this work producing process? What is the
upper bound for this efficiency?
c.) Estimate the maximum possible work per kg of processed air that can be produced if
the two streams have the same air flow rate.
d.) What fraction of the air passing through the evaporative cooler will result in the
larges power?
e.) What do you see as practical limitations that prevent this work production method
from becoming commercially successful? What improvements could be made?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-2 An auditorium is to be maintained at 25°C dry-bulb temperature and 40% relative
humidity when the outdoor conditions are 32°C dry-bulb and 25°C wet bulb. The space
will be conditioned by the system shown in the figure. Outdoor air enters the duct at state
5 at 0.25 m3/s. The air exiting the cooling coil (state 7) is saturated at 7°C. Air is blown
into the space (state 1) at 0.90 m3/s. Pressure losses in the ducts and fan power are
negligible. As shown in Figure 12.C-2, the system provides a bypass damper that
controls how much of the recirculated air passes through the cooling coil. Unless
otherwise specified, assume that the bypass damper is positioned such that the 50% of the
recirculated air at state 3 passes through the cooling coil.
Figure 12.C-2: HVAC system for an auditorium
a. Determine the temperature, humidity ratio, relative humidity and mass flow rate at all
points.
b. Determine the rate of condensate removal in the cooling coil.
c. Determine the sensible heat ratio defined as the ratio of the sensible to the total
cooling load.
d. Determine the cost to condition this space assuming that the cooling coil is served by
a cooling system that has a system COP of 3.5 at these conditions for the conditions
of part a.
e. Vary the bypass damper position between 0 and 1, where 0 indicates that all of the air
passes through the cooling coil and 1 indicates that only the outdoor air that enters at
state 5 passes through the cooling coil. Prepare a plot of the total load and operation
cost as a function of the bypass damper position. Explain the results you see in the
plot.
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-3 Figure 12.C-3(a) illustrates a simple Rankine cycle power plant. The condenser rejects
heat from the plant to the ambient. In some cases it is possible to run lake water or river
water through the condenser in order to accomplish this heat rejection.
Qb
TH
boiler
W p
turbine
pump
Wt
condenser
Qcond
water from lake
water returned to lake
Figure 12.C-3(a): Rankine cycle power plant with a water-cooled condenser.
However, often a river or lake is not available (or maybe is available, but is protected by
environmental regulation). In this case, the power plant must reject heat to the ambient
air. Air is not a great heat transfer fluid and therefore it is not usually attractive to replace
the water flow with an air flow; the heat exchanger would have to be too large and
expensive. Instead, a cooling tower is often used, as shown in Figure 12.C-3(b).
A flow of water is pumped through the condenser. The water is sprayed into the cooling
tower as a mist so that it is in direct contact with air that is pulled through the tower by
fans. There is both heat transfer as well as evaporation of water into the air, leading to
very high rates of heat transfer. Because the heat transfer takes place directly from the
water droplets to the air, there is no need for expensive heat exchanger surfaces. Cooling
tower are used extensively for thermal-fluid systems; the very large distinctive towers
that you see near nuclear power plants (for example see Figure 12.C-3(c)) are actually
large, natural draft cooling towers that service the plant (the stuff coming out of the top is
actually just clouds of water vapor). You can see cooling towers on top of many
buildings, serving as heat rejection devices for the condensers used in the building HVAC
systems (for example, see Figure 12.C-3(d)).
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
Qb
TH
boiler
W p
pump
Wt
turbine
condenser
4
Qcond
air/water mixture
T4 = 35°C
P4 = 1 atm
4 = 1.0
1
water, m w
T1  40C
P1  4 atm
pump
cooling tower
3
2
water, m w
T2  20C
P2  1 atm
5
ambient air/water mixture
T3 = 18°C
3 = 0.3
P3 = 1 atm
makeup water
T5 = 18°C
P5 = 1 atm
Figure 12.C-3(b): Rankine cycle power plant with a cooling tower.
(a)
(b)
Figure 12.C-3 (c) and (d): Cooling towers used to provide heat rejection for a (c) power plant and (d)
building HVAC system.
The power plant shown in Figure 12.C-3(b) provides Wnet = 1000 MW (1 GW) of power
and has efficiency  = 0.35.
a.) What is the rate of heat transfer in the condenser, Q cond ?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
Ambient air at temperature T3 = 18°C and pressure P3 = 1 atm is drawn into the bottom of
the tower by large fans (you may neglect the fan power for this problem). The relative
humidity of the ambient air is 3 = 0.3. Warm pure liquid water flows from the
condenser at T1 = 40°C and P1 = 4 atm; the water is sprayed from nozzles at the top of the
tower in order to break it into small droplets that mix with the upward stream of air.
Therefore, the air leaving the top of the cooling tower is saturated with water and has
temperature T4 = 35°C, relative humidity 4 = 1.0, and pressure P4 = 1.0 atm. Cooled
water collects at the bottom of the tower at temperature T2 = 20°C and pressure P2 = 1
atm where it is pumped back to the condenser. Because some water is lost to
evaporation, makeup water at temperature T5 = 18°C and pressure P5 = 1 atm is added at
the bottom of the tower. You can neglect the pump work and assume that the system is at
steady state. The cooling tower is adiabatic. Use the internal functions in EES and the
substances AirH2O and Water to model air-water mixtures and pure water, respectively.
b.) What is the mass flow rate of water pumped through the condenser ( m w in Figure
12.C-3(b))?
c.) What is the mass flow rate of makeup water required at state 5 ( m muw ) and the total
volumetric flow rate of air drawn in at state 3 by the fans ( V )?
3
d.) What is the rate of entropy generation in the cooling tower?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-4 In hot dry climates, air conditioning can be achieved simply by evaporative cooling.
Water is sprayed into the dry air and subsequently evaporates with a resulting decrease in
the temperature of the exit air stream. The evaporative cooler design is characterized by
an effectiveness that is defined as
T T
  in out
Tin  Twb
where
Tin the entering air temperature
Tout is the exiting air temperature
Twb is the wet bulb temperature
In a particular case, outdoor air at 35°C, 20% relative humidity enters an evaporative
cooler at 6.0 m3/sec and is evaporatively cooled with liquid water at 35°C. Calculate and
plot the following quantities as a function of the evaporator cooler effectiveness.
a.) the outlet air temperature
b.) the rate of exergy destruction
c.) the Second-law efficiency
Is the Second-law efficiency of this process equal to one when the effectiveness is one?
If not, explain why. What do you see as the advantages and disadvantages of evaporative
cooling?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-5 Many electric utilities rely on gas turbine generators to help supply their peak electrical
demand during the summer months. Gas turbines operate less efficiently than the large
Rankine power cycle and this is one reason why electrical rates are higher in summer
than in winter. Other disadvantages of gas turbines are that they operate with a nearly
constant volumetric flow rate of inlet air. The reduced density of air on hot summer days
results in reduced power output at the times in which the power is most needed. One way
to increase the capacity and efficiency of a stationary gas turbine power system is to cool
the inlet air with an evaporative cooler, as shown in Figure 12.C-5. Cooling the air in this
manner lowers its temperature at state 1 which increases the density of the air and thus
the mass flow rate of working fluid, assuming a constant volumetric flow rate. Also, the
added water in the air at state 1, which acts as an inert component in the air, also
increases the mass flow rate, resulting in increased capacity. One purpose of this
problem is to compare the efficiency and capacity of the gas turbine power system with
and without the evaporative cooler.
Liquid water
at 25°C
6
2
Evaporative
Cooler
Compressor
Outdoor air
35°C, 35% rh 0
1
Regenerator 3
Combustion 4
chamber
Turbine
5
Generator
Fuel
46,000 kJ/kg
Figure 12.C-5: Gas turbine system using evaporative cooler to cool inlet air
Assume the isentropic efficiencies of the compressor and turbine to be 0.72 and 0.84,
respectively. The evaporative cooler has an effectiveness of 0.95. The maximum
temperature at state 4 is 1150C. Pressure losses can be neglected in this analysis. The
regenerator effectiveness is 0.30. The heating value of the fuel is 46,000 kJ/kg. Treat
the combustion process as a heat input so the composition of the gas at state 4 is the
same as at state 3. Consider at peak day in which the ambient air (state 0) is at 35C,
101.3 [kPa] and 35% relative humidity.
Calculate and compare the First-Law efficiencies of the gas turbine system without and
with the evaporative cooler unit. In each case, determine the pressure ratio that
maximizes the power output. Estimate the increase in power output resulting from the
evaporative cooler considering that the mass flow though the turbine is increased. What
is your assessment of evaporative cooler based on your calculations?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-6 A standard residential dehumidifier consists of a refrigeration unit and a small fan. Air
from the space to be dehumidified is blown over the evaporator coil where it is cooled
below its dewpoint temperature. Condensate drips off the evaporator coil into a drain
pan. This cool dry air is then blown past the condenser coil and returned to the room
warmer and dryer than it was initially. Figure 11.C-6 below shows a schematic of a
proposed new product which we call the regenerative dehumidifier unit. It is similar to a
conventional dehumidifier but in includes an additional cross-flow air to air heat
exchanger. The purpose of this problem is to determine the performance of the
regenerative dehumidifier as a function of the effectiveness of the heat exchanger which
can be approximated to be
T T
 HX  1 2
T1  T3
Condenser
8
Evaporator
9
4
5
3
7
24°C
55% rh
1
2
Heat
Exchanger
6
Compressor
Condensate
Figure 12.C-6: Schematic of a regenerative dehumidifier
During typical operation, air at 0.030 kg/sec, 24°C, and 55% relative humidity (state 1) is
cooled to state 2 in the heat exchanger. (The heat exchanger is not designed to provide
condensation.) The air then passes through a cooling coil that both cools and
dehumidifies the air to state 3. Air leaving the cooling coil at state 3 is saturated at a
temperature 10°C above that of the evaporating refrigerant.. Air leaving the evaporator
at state 3 reenters the heat exchanger as the cooler fluid and exits at state 4. (Streams 1-2
and 3-4 do not mix.) The air is then blown past the condenser coil to remove the heat of
condensation and returns to the room at state 5. The saturation temperature in the
condenser is 10°C higher than the temperature of the air state 4. The fans in the system
draw at total of 40 W.
The refrigeration cycles uses R134a and it standard in all respects. The compressor is
adiabatic with an efficiency of 0.65. The compressor is sized such that the volumetric
flowrate at the compressor inlet is 0.00040 m3/s. Assume that saturated vapor exits the
evaporator and saturated liquid exits the condenser.
Dehumidifier capacity is measured in pints of condensate per day. Calculate and
plot the the capacity and pints per kW-hr of the regenerative dehumidifier as a
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
function of the effectiveness of the heat exchanger. How does it compare to a
non-regenerative design?
© S.A. Klein and G.F. Nellis
Cambridge University Press, 2011
12.C-7 The purpose of this problem is to compare the three evaporative cooling systems shown
in Figures 12.C-7a, b, and c for a particular case in which the outdoor conditions are
85°F, 30% rel, humidity. (The water supplies to the evaporative coolers are not shown.)
The indoor temperature is to be maintained at 75°F requiring a sensible cooling load of 3
tons (36,000 Btu/hr) for these conditions. Assume that there is no water generation in the
conditioned space. Since the conditioned air is blown directly into the space, the
humidity ratio in the building is approximately equal to that of the conditioned air. The
basic design calls for evaporative coolers having an effectiveness of 0.75. Model the
three evaporative cooler systems and use the results of your models to answer the
following equations.
Figure 12.C-7a, b, and c. Alternative evaporative cooling systems
a.) What effect does the regenerator have on the required volumetric air flow rate needed
to meet the cooling load. (A plot of volumetric air flow rate versus regenerator
effectiveness would be helpful in answering this question.
b.) How important is the evaporative cooler on the exhaust side of the building in Figure
12.C-7c?
c.) What relative humidity is maintained in the conditioned space with the three systems?
Is it acceptable?