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MECHANICAL ENGINEERING
HVAC & REFRIGERATION
STUDY PROBLEMS
PSYCHROMETRICS & BASIC HVAC
CALCULATIONS
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How to use this book
The exam specifications in effect since April 2017 state that approximately 8 of the 32 problems (25%)
from the “Psychrometrics” topic will be in the morning “Principles” portion of your HVAC&R exam.
Reviewing all the problems in this book will prepare you for all these problems in the morning
“Principles” portion. Additionally, the problems in this book provide additional review for the
“Energy/Mass Balances” topic, which is expected to contain 5 of the 32 problems (16%) of the
“Principles” portion.
The exam specifications also state that the “Heating/Cooling Loads” topic comprise 8 of the 48
problems (17%) in the “Applications” portion of the exam. The problems in section 8 of this book are
designed to help you review this topic. Furthermore, cooling towers are listed as one of the relevant
examples of “Equipment and Components” in the afternoon “Applications” portion. The problems in
section 7 of this ebook are designed to help you cover this topic.
How it works
This study problems book works on what we call the “principle of progressive overload”. With this
technique you start with very easy problems and smoothly progress towards more complex problems.
A good example of progressive overload is the story of the famous wrestler Milo of Croton in ancient
Greece. This extraordinarily strong man was allegedly capable of carrying a fully grown bull on his
shoulders. He was reported to have achieved this tremendous strength by walking around town with a
new born calf on his shoulders every single day. As the calf grew, so did the man's strength.
We recommend you work the problems in this book in the order they are presented. Within each
section of the book, the first problems will feel “light”, like carrying that baby calf – you might even be
tempted to skip them. We strongly urge you to resist this temptation. As you progress, the problems
become harder, but the work you've been putting in with all the previous problems will bear fruit. You
will be pleasantly surprised at how relatively easy those “hard” problems will seem. You will soon be
carrying intellectual bulls on your shoulders! The problems that are considered “exam-level
difficult” are denoted with an asterisk.
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This book is comprised of the following sections:
Sections
Page
01: Properties of Moist Air
5
02: The Psychrometric Chart
13
03: Simple Cooling and Heating
19
04: Adiabatic Mixing
23
05: Cooling with Dehumidification and Heating with Humidification
27
06: Evaporative Cooling
33
07: Wet Cooling Towers
35
08: Basic HVAC System Calculations
41
09: Answers
51
For the most part, these sections are not independent and build from the previous ones. We recommend
you go through them in the order presented, and be sure to review them all. Each section begins with a
brief discussion of the relevant concepts and equations. These discussions are laser-focused on the
aspects that are relevant to the P.E. exam and do not go into derivations with academic rigor.
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SECTION 01: Properties of Moist Air
Psychrometrics is the science involving thermodynamic properties of moist air and the effect of
atmospheric moisture on materials and human comfort.
Air is a mixture of nitrogen, oxygen, and small amounts of some other gases. Air in the atmosphere
normally contains some water vapor (or moisture) and is referred to as atmospheric air. Conversely,
air that contains no water vapor is called dry air. At the relatively low atmospheric temperature we can
treat dry air as an ideal gas with constant specific heat. The water vapor in atmospheric air is typically
at very low pressures (i.e., the partial pressure of water in the mixture 1) and thus can be treated as an
ideal gas without loss of accuracy (even if it is saturated vapor).
Consider a container with 10 pounds of dry air and 0.1 pounds of water vapor, thoroughly mixed, at
atmospheric pressure (14.7 psia) and 75°F. The humidity ratio ω (also known as specific humidity)
is the ratio of the mass of water vapor to mass of dry air:
ω=
mv
ma
(1-1)
where the subscript “v” denotes water vapor, and “a” denotes dry air. In our example,
ω =0.1/10=0.01 lbm H2 O /lbm dry air . The mass of water vapor is sometimes expressed in grains,
which is a unit of mass such that 7,000 grains = 1 pound-mass. Therefore, in our example,
ω =70 grains H 2 O/lbm dry air . The humidity ratio is not to be confused with the water vapor mass
fraction: mf v=mv /(mv +ma )= ω /(1+ ω )
In our example, the moles of dry air N a=m a / M a=10 lbm/ 28.97(lbm/lbmol )=0.3452 lbmol and the
moles of water vapor N v =mv / M a=0.1 lbm /18(lbm /lbmol)=0.0056 lbmol . Therefore, the mole
fraction of water vapor y v=0.0056 /(0.0056+0.3452)=0.01584 . We can can thus obtain the partial
pressure of the water vapor (or vapor pressure) from its definition:
(1-2)
pv= yv p
thus p v =0.01584×14.7=0.2328 psia in our example. At this pressure, the saturation temperature is
57.3°F, so the vapor in the mixture is actually superheated vapor with a superheat of 75 – 57.3 =
17.7°F.
1 For a review of the concept of partial pressure and the modeling of mixtures of ideal gases consult our
“Thermodynamics and Energy Balances” e-book
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We can repeat these calculations for a mixture of 10 pounds of dry air and this time with 0.15 pounds
of water (i.e, we are increasing the amount of water in the mixture by 50%, so now
ω =0.015 lbm H 2 O /lbm dry air ). In this case;
N a=m a / M a=10 lbm/ 28.97(lbm/lbmol )=0.3452 lbmol
N v =m v / M a =0.15 lbm /18(lbm/lbmol )=0.00833 lbmol
y v=0.00833/(0.00833+0.3452)=0.02357
p v =0.02357×14.7=0.3465 psia
We see that for this mixture (which contains more water) the vapor pressure is higher, and the
saturation temperature corresponding to it is lower. For 0.3465 psia, T sat =68.6 °F and thus the
superheat is reduced to 75 – 68.6 = 6.4°F. If the mass of water is further increased again, this time to
0.1873 pounds, the vapor pressure will be 0.43017 psia – the saturation pressure at 75°F – thus the
water vapor in the mixture will be saturated vapor.
This example shows that, as we increase the mass of vapor in the mixture the closer that vapor comes
to the saturated vapor state. We denote with mg the mass of water in the mixture such that the water
vapor is saturated (i.e., not superheated, that is, when the vapor pressure corresponds to the saturation
pressure at the mixture temperature). In our example, mg =0.1873 lbm . Saturated air is a mixture of
dry air and saturated water vapor. The amount of water vapor in moist air varies from zero in dry air to
a maximum value mg (which depends on the pressure and temperature) in saturated air (when the vapor
is saturated.) The relative humidity ϕ (sometimes also referred to as “rh”) is defined as the ratio of the
mass of water in the air to the mass of water in saturated air at the same temperature and pressure:
ϕ=
mv
mg
(1-3)
In our example for the container with 10 pounds of dry air and 0.1 pounds of water, the relative
humidity would be ϕ =0.1/0.1873=0.534=53.4 % . When the mass of water vapor is increased to 0.15
pounds, the relative humidity increases to ϕ =0.15/0.1873=0.8=80 % .
Note that, since mg depends on temperature, we can change the relative humidity of the mixture by
changing the temperature, without adding or removing any water. Consider our example of 10 pounds
of dry air and 0.1 pounds of water vapor ( ω =0.01 lbm /lbm ) at 14.7 psia and 75°F. We saw earlier
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that N a=0.3452 lbmol ; N v =0.0056 lbmol and y v=0.01584 . Consider what happens when we cool the
mixture down to 65°F, without adding (or removing) any water vapor: At 65°F, the saturation pressure
for water is 0.306 psia. We would like to determine mg , the mass of water vapor that would be required
to make the water in the mixture a saturated vapor. A vapor pressure of .306 psia means that the mole
fraction of water is: y v= p v / p=0.306 /14.7=0.02081 . It can be shown that this mole fraction is
obtained when the mass of water vapor is 0.13212 pounds. That is, for a mixture temperature of 65°F,
mg =0.13212 lbm if ma=10 lbm . So, when the mixture temperature is decreased to 65°F,
ϕ =0.1/0.13212=0.7569=75.7 % .
So, humidity ratio provides a measure of the concentration of water vapor in atmospheric air and
relative humidity provides a measure of how close the vapor in atmospheric air is to being saturated
vapor. The relative humidity of air changes with temperature even when its humidity ratio remains
constant. The amount of moisture in the air has a definite effect on how comfortable people feel in an
environment. However, the comfort level depends more on how close the water vapor is to being
saturated; that is, the comfort level is more related to the relative humidity.
We can re-cast equation (1-1) by inserting the ideal gas law:
pv V
m
R T R p
p
ω = v = v = a v =0.622 v
ma pa V R v p a
pa
Ra T
Now use Dalton's law of additive pressures: p= pa+ p v ,
ω =0.622
(
pv
p− pv
)
(1-4)
which can be rearranged as:
ω
( ω +0.622 ) p
pv=
(1-5)
Similarly, we can use the ideal gas law with equation (1-3):
pv V
m
R T
p
ϕ = v= v = v
m g p g V pg
Rv T
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(1-6)
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where p g = psat @ T is the saturation pressure at the mixture temperature. Combining (1-6) with (1-5) we
obtain:
ϕ=
ω
p
( ω +0.622 ) p
(1-7)
g
or alternatively:
ω=
0.622 ϕ p g
p− ϕ p g
(1-8)
T he humidity ratio at saturation ω s – which is the amount of water vapor in saturated air at a
specified temperature and pressure – can be determined from equation (1-4) by replacing p v by p g , the
saturation pressure of water at that temperature, or by setting ϕ =1 in equation (1-8).
The degree of saturation μ is the ratio of the actual humidity ratio ω to the humidity ratio ω s of
saturated air at the same temperature and pressure:
μ =ω / ω s
(1-9)
Relative humidity and specific humidity are frequently used in engineering and atmospheric sciences,
and it is desirable to relate them to easily measurable quantities such as temperature and pressure. The
term dry-bulb temperature T (sometimes simply referred to as the temperature) is the temperature
measured by an ordinary thermometer placed in atmospheric air. The wet-bulb temperature T wb , is
read from an ordinary liquid-in-glass thermometer whose bulb is enclosed by a wick moistened with
water. When unsaturated air passes over the wet wick, some of the water in the wick evaporates. As a
result, the temperature of the water drops, creating a temperature difference (which is the driving force
for heat transfer) between the air and the water. After a while, the heat loss from the water by
evaporation equals the heat gain from the air, and the water temperature stabilizes. The thermometer
reading at this point is the wet-bulb temperature. The wet-bulb temperature is an excellent
approximation to another concept known as the adiabatic saturation temperature T as . It can be shown
that relative and absolute humidity can be obtained if T as , T , and p are all known.
Consider a mass of moist air at a known temperature T and pressure p . Equation (1-7) indicates that
these two variables are not enough to determine ϕ and ω . The additional relationship involves ω and
T as (and therefore T wb because T as≈T wb ). In customary US units, this relationship is:
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ω=
( 1,093−0.556T w.b. ) ω s−0.24 ( T −T w.b. )
1,093+0.444 T −T w.b.
(1-10)
where temperatures are in °F, and ω *s is the humidity ratio at saturation for the wet bulb temperature.
The condensation of part of the water vapor in atmospheric air when the temperature is reduced is an
important aspect of the behavior of moist air. This is encountered in the condensation of vapor on
window panes and on cold pipes, as well as condensation of water vapor from cool flue gases. The
dew-point temperature T dp is defined as the temperature at which condensation begins when moist air
is cooled at constant pressure. In other words, T dp is the saturation temperature of water corresponding
to the vapor pressure, T dp =T sat ( pv ) .
In most practical applications, the amount of dry air in the air–water-vapor mixture remains constant,
but the amount of water vapor changes. Therefore, the enthalpy of atmospheric air is expressed per
unit mass of dry air instead of per unit mass of the air–water vapor mixture. The enthalpy of a mixture
of perfect gases is equal to the sum of the individual partial enthalpies of the components. The enthalpy
of moist air is then:
h=ha + ω h v
where ha is the specific enthalpy for dry air and hv is the specific enthalpy for water vapor at the
temperature of the mixture. For most HVAC applications, hv≈h g where h g is the enthalpy of saturated
vapor at the mixture temperature, thus:
h=ha+ ω h g
(1-11)
For most HVAC applications, the following correlation in the form of equation (1-11) works quite
well:
h=0.24T + ω ( 1,061+0.444 T )
(1-12)
which yields the enthalpy in Btu/(lbm of dry air) as long as T is in °F and ω is in (lbm H2O)/(lbm dry
air). The specific volume v of a moist air mixture is expressed in terms of a unit mass of dry air:
pa V =ma R a T
Ra T
V
⇒ =
ma
pa
Ra T
⇒ v=
pa
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and using Dalton's law, we get:
v=
=
Ra T
p− pv
Ra T
(
p 1−
pv
p
)
which can be combined with equation (1-5):
Ra T
v=
( (ω ω ))
p 1−
+0.622
thus yielding:
v=
(
Ra T
ω
1+
p
0.622
)
(1-13)
where
{
53.34 ft⋅lbf / lbm⋅°R
3
0.3704 psia⋅ft /lbm⋅°R
Ra = 640.08 psia⋅in 3 /lbm⋅°R
0.0686 Btu /lbm⋅°R
0.287 kJ / kg⋅K
}
is the gas constant for dry air.
PROBLEMS
01-01. A sample of air at 14.7 psia and 75°F has a humidity ratio of 0.015 lbm H 2 O/lbm dry air .
Determine:
a) the partial pressure of water vapor, in psi.
b) the dew point temperature, in °F.
c) the relative humidity.
d) the degree of saturation.
e) the enthalpy in Btu/lbm dry air
f) the specific volume in ft3/lbm dry air
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01-02. A sample of air at 14.7 psia and 75°F has a dew point of 52°F. Determine the humidity ratio and
the relative humidity.
01-03. A sample of air at 14.7 psia and 75°F has a wet bulb temperature of 55°F. Determine the
humidity ratio and the relative humidity.
01-04. A sample of air is at 70°F, 50% rh, and 14.7 psia. What is the lowest temperature to which the
air can be cooled at constant pressure if condensation is to be avoided?
01-05*. The conditions at the intake of an air conditioning system are 95°F, 30% rh, and 14.7 psia. The
air flows at a rate of 1,000 cubic feet per minute (cfm). The rate (lbm/hour) at which water vapor enters
the system is most nearly:
(A) 0.74
(B) 10.5
(C) 44.5
(D) 55.4
01-06*. The air conditions at the intake of an air compressor are 70°F, 50% rh, and 14.7 psia. The air is
compressed to 50 psia, then sent to an intercooler. If condensation of water vapor from the air is to be
prevented, the lowest temperature (°F) to which the air can be cooled in the intercooler is most nearly:
(A) 31
(B) 46
(C) 51
(D) 86
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SECTION 02: The Psychrometric Chart
Many of the properties of moist air discussed in the previous section can be estimated from a
psychrometric chart, a plot of dry-bulb temperature (horizontal axis) and humidity ratio (vertical axis).
In other words, the chart is a graphical representation of all the equations in the previous section.
Psychrometric charts for sea level altitude are available for free download from the “Free Resources”
section of www.SlayThePE.com. Lines of constant humidity ratio are horizontal on the graph. Lines of
constant (dry-bulb) temperature exhibit a slight tilt to the left of vertical, with the degree of tilt
increasing with lower temperature. The left top portion of the plot is terminated at the saturation line,
which represents both the 100% relative humidity curve and the plot for dew-point temperature. Lines
of constant enthalpy appear as straight lines that slope down as the temperature increases. Lines of
constant wet-bulb temperature are nearly parallel to lines of constant enthalpy. Lines of constant
specific volume also slope down as temperature increases but at a much greater angle compared to
enthalpy and wet-bulb lines. Lines of constant relative humidity curve up as temperature increases.
Processes performed with air can be plotted on the chart for quick visualization, as well as for
determining changes in significant properties such as temperature, humidity ratio, and enthalpy for the
process.
To demonstrate the use of the chart, obtain one for sea level altitude and locate the point corresponding
to a dry-bulb temperature of 80°F and a wet-bulb temperature of 67°F:
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67°F (wb)
80°F
Using this condition as a starting point on the chart, make sure you can verify that:
• Humidity ratio—Move right horizontally to the axis and read, ω ≈0.0112 lbm /lbm (78 grains/lbm)
67°F (wb)
80°F
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• Relative humidity—Interpolate between the ϕ =50 % and ϕ =60 % lines, ϕ ≈52 % .
67°F (wb)
80°F
• Enthalpy—Follow a constant enthalpy line and read from the enthalpy scale, h≈31.5 Btu / lbm :
67°F (wb)
80°F
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• Dew point—Move left horizontally to the saturation curve ( ϕ =100 % ) and read T dp≈60.5 °F
67°F (wb)
80°F
• Specific volume—Interpolate between the 13.5 and 14.0 lines, to get v=13.85ft 3 /lbm
67°F (wb)
80°F
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PROBLEMS
02-01. A sample of air at 14.7 psia and 75°F has a humidity ratio of 0.015 lbm H 2 O/lbm dry air . Use a
psychrometric chart to determine:
a) the dew point temperature, in °F.
b) the relative humidity.
c) the degree of saturation.
d) the enthalpy in Btu/lbm dry air
e) the specific volume in ft3/lbm dry air
02-02. Use a sea-level psychrometric chart to complete the following table:
Dry Bulb
°F
Wet Bulb
°F
70
55
Dew Point
°F
100
Humidity Ratio
lbm/lbm
R.H.
%
Enthalpy
Btu/lbm
40
40
70
0.01
60
13.8
40
40
50
60
85
80
Specific Volume
ft3/lbm
30
0.012
80
02-03*. The conditions at the intake of an air conditioning system are 95°F, 30% rh, and 14.7 psia. The
air flows at a rate of 1,000 cubic feet per minute (CFM). The rate (lbm/hour) at which water vapor
enters the system is most nearly:
(A) 0.74
(B) 10.5
(C) 44.5
(D) 55.4
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SECTION 03: Simple Cooling and Heating
The air in heating systems is heated by circulating it through a duct that contains a heat exchanger (for
the hot combustion gases) or electric resistance wires. The amount of moisture in the air remains
constant during this process since no moisture is added to or removed from the air. That is, the specific
humidity of the air ω remains constant during a heating (or cooling) process with no humidification or
dehumidification. Such a heating process proceeds in the direction of increasing dry-bulb temperature
following a line of constant specific humidity on the psychrometric chart, which appears as a horizontal
line.
Notice that the relative humidity of air decreases
during a heating process even though the specific
humidity remains constant. Therefore, the relative
Simple Heating
ϕ 2< ϕ 1
T 2>T 1
h
2
humidity of heated air may be well below
comfortable levels, causing dry skin, respiratory
h1
ϕ2
ϕ1
difficulties, and an increase in static electricity.
As air flowing at a rate ṁa is heated from 1 to 2, an
ω =const
1
2
amount of heat is added at a rate Q̇ . This could occur
in the heating section of an HVAC device. This
could also occur as cold air warms as it flows from
T1
T2
the supply register(s) in a room towards the return
register(s). An energy balance reveals that:
Q̇= ṁa ( h2 −h1 )
(heating)
where h1 and h2 are enthalpies per unit mass of dry air at the inlet and the exit of the heating section,
respectively.
A cooling process at constant specific humidity is similar to the heating process discussed above,
except the dry-bulb temperature decreases and the relative humidity increases during such a process.
Cooling can be accomplished by passing the air over some coils through which a refrigerant or a
chilled coolant flows. The heat transfer rate for a simple cooling process can be obtained from an
energy balance on the cooling section as:
Q̇= ṁa ( h1−h 2 )
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Since the specific volume of air varies with temperature, all calculations should be made with the mass
of air instead of the volume. Nevertheless, volume values are required when selecting coils, fans, ducts,
and other components. One method of using volume while still including mass is to use volume values
based on measurement at standard air conditions. ASHRAE defines one standard condition as dry air at
20°C and 101.325 kPa (68°F and 14.7 psia). Under that condition the density of dry air is about 1.204
kg/m3 (0.075 lbm/ft3) and the specific volume is 0.83 m 3/kg (13.3 ft3/lbm). Saturated air at 15°C
(59.5°F) has about the same density or specific volume. Thus, in the range at which air usually passes
through the coils, fans, ducts, and other equipment, its density is close to standard and is not likely to
require correction.
The heat transfer to or from the air in simple heating and cooling processes is known as sensible heat
which corresponds to the change in dry-bulb temperature for a given airflow (standard conditions). The
sensible heat in Btu/h as a result of a difference in temperature ΔT in °F between the incoming air and
leaving air flowing at ASHRAE standard conditions is:
Q̇sens = ṁa Δ h= ṁa c p , a Δ T
Which can be written in terms of volumetric flow, and combined with equation (1-12):
Q̇sens = ρ a V̇ a ( 0.24+0.444 ω ) ΔT [ °F ]
Now, use ρ a=0.075lbm /ft 3 , the density of dry air at standard conditions and define a new variable,
CFM which is the volumetric flow rate in cubic feet per minute. Then the following equation yields the
sensible heat in Btu/h:
Q̇ sens [ Btu /h ] =0.075
∣
∣
lbm
60 min
×CFM×
( 0.24+0.444 ω ) Δ T [ °F ]
3
1h
ft
Since ω ≈0.01 in many air-conditioning problems, the sensible heat is typically approximated by:
Q̇sens [ Btu /h ] =1.1×CFM×ΔT [ °F ]
(3-1)
If standard conditions are specified as 70°F and 50% RH, replace 1.1 with 1.08 in equation (3-1). For
the purposes of the P.E. exam, equation (3-1) yields results that are accurate enough. However, you
must exercise caution if the air experiences a very large temperature difference across the heating or
cooling section (say, 30°F or greater) in which case it is advisable to use the mass flow based
equations, Q̇= ṁa ( Δ h ) .
If the actual volumetric airflow is needed at any particular condition or point, the corresponding
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specific volume is obtained from the psychrometric chart (or equation 1-13) and the volume at standard
conditions is multiplied by the ratio of the actual specific volume to the standard value of 13.3. For
example, assume the outdoor airflow rate at ASHRAE standard conditions is 1,000 cfm. The actual
outdoor air condition is 95°F dry bulb, and 75°F wet bulb [v = 14.3 ft 3/lb]. The actual volume flow rate
at this condition would be 1,000(14.3/13.3) = 1,080 cfm.
PROBLEMS
03-01. Air is heated to 80°F without adding water, from 60°F dry-bulb and 50°F wet-bulb temperature.
Use a sea-level psychrometric chart to find:
(a) relative humidity of the original mixture, (b) original dew-point temperature, (c) original humidity
ratio, (d) initial enthalpy, (e) final enthalpy, (f) the heat added, and (g) final relative humidity.
03-02. Air at 14.7 psia is cooled down to 55°F, 90% rh from 90°F. Use a sea-level psychrometric chart
to find the enthalpy change in Btu per pound of dry air.
03-03.* A gas furnace produces 60,000 Btu/h with an airflow of 2,800 cfm heated air with an inlet
condition of 65°F, 45% rh. The relative humidity of the outlet air is most nearly:
(A) 23
(B) 46
(C) 62
(D) 85
03-04.* A stream of 500 cfm of saturated air at 50°F and 14.7 psia is heated without adding or
removing water until its relative humidity drops to 40%. The heat input (Btu/h) required of the heater is
most nearly:
(A)
245
(B) 9,200
(C) 14,540
(D) 25,500
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03-05.* A heating section consists of a 15-in.-diameter duct that houses an electric resistance heater.
Air enters the heating section at 14.7 psia, 50°F, and 40% rh at a velocity of 25 ft/s and is discharged
with a relative humidity of 31 percent. The heater input (kW) is most nearly:
(A) 4.3
(B) 6.2
(C) 80.3
(D) 14,000
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SECTION 04: Adiabatic Mixing of Air Streams
Mixing of two moist air streams is a common air-conditioning process. The point of the
thermodynamic analysis of such processes is typically to determine the flow rate and state of the
exiting stream for given flow rates and states of each of the two inlet streams. It is typically assumed
that the heat lost to the environment surrounding the mixing chamber is negligibly small, hence the
adjective “adiabatic”. As shown in Figure 4-1, two streams at states 1 and 2 enter the chamber (at
steady state) and one stream leaves at state 3.
ṁ a1 , T 1, ω 1
ṁ a2 , T 2, ω 2
1
2
Mixing Chamber
3
ṁ a3 , T 3, ω 3
Perfect Insulation
Figure 4-1: Adiabatic mixing of two moist air streams.
The mass balances for dry air and water vapor are, respectively:
(4-1)
ṁa1 + ṁa2= ṁa3
ṁv1+ ṁv2= ṁ v3
and since ṁv= ω ṁa , this water vapor mass balance becomes:
ω 1 ṁa1+ ω 2 ṁa2 =ω 3 ṁa3
(4-2)
With the assumption of no heat transfer leaving the mixing box, the energy balance is:
ṁa1 ( ha1+ ω 1 hg1 ) + ṁa2 ( ha2 +ω 2 h g2 )= ṁa3 ( ha3+ ω 3 hg3 )
(4-3)
where we have used equation (1-11) for the enthalpy of the mixture. If the inlet conditions of both
inputs are known, then equations (4-1) through (4-3) can be solved to determine ṁa3 , ω 3 ,and
ha3+ ω 3 hg3 – which can be used to determine T 3 . It can be shown that the solutions to these equations
are:
T 3=
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T 2+T 1 ( ṁa1 / ṁa2 )
1+( ṁa1 / ṁa2 )
23
(4-4)
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h3=
ω 3=
h2 +h1 ( ṁa1 / ṁa2 )
(4-5)
1+( ṁa1 / ṁa2 )
ω 2+ω 1 ( ṁa1 / ṁa2 )
1+( ṁa1 / ṁa2 )
(4-6)
and if v 2≈v1 we get:
T 3≈
T 2+T 1 ( CFM1 /CFM 2 )
1+( CFM 1 /CFM 2 )
; h3≈
h2 +h1 ( CFM 1 /CFM 2 )
1+ ( CFM 1 /CFM 2 )
; ω 3≈
ω 2+ ω 1 ( CFM1 /CFM 2 )
1+( CFM 1 /CFM 2 )
(4-7)
Another way to quickly determine the resulting conditions of the mixing of two moist air streams is the
graphical approach using a psychrometric chart. Locate points 1 and 2 on a psychrometric chart, and
draw a straight line between them (as shown in Figure 4-2).
1
L12
L13
3
2
Figure 4-2: Adiabatic mixing of two moist air streams – graphical representation on psychrometric chart
Then, use a ruler to measure the distance between points 1 and 2, L12 . Then, it can be shown that point
3 will be on the line connecting 1 and 2, and the distance between 1 and 3 is given by:
L13=L 12
(
) (
ma2
CFM 2
≈ L12
ma1+ma2
CFM1 +CFM 2
)
(4-8)
Note that in the limiting case when CFM 2 ≫CFM 1 ⇒ L 13 → L12 which means point 3 will be very close
to point 2. In the other extreme, if CFM 2 ≪CFM 1 ⇒ L 13 → 0 which means point 3 will be very close to
point 1.
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PROBLEMS:
04-01. Air at 40°F dry bulb and 35°F wet bulb is mixed with air at 100°F dry bulb and 77°F wet bulb in
the ratio of 2 lb of cool air to 1 lb of warm air. Determine the humidity ratio of the mixed air.
04-02. During an air-conditioning process, 900 cfm of conditioned air at 65°F and 30 percent relative
humidity is mixed adiabatically with 300 cfm of outdoor air at 80°F and 90 percent relative humidity at
a pressure of 1 atm. Determine (a) the temperature, (b) the specific humidity, and (c) the relative
humidity of the mixture.
04-03.* In a mixing process of two streams of air, 10,000 cfm of air at 75°F and 45% rh is mixed with
hot air at 98°F and 40% rh. The desired humidity ratio of the resulting mixture is 0.01 lbm/lbm. Under
these conditions, the flow rate (cfm) of hot air into the mixing box must be most nearly:
(A) 1,200
(C) 7,500
(B) 3,300
(D) 30,000
04-04.* In a mixing process of two streams of air, 10,000 cfm of air at 75°F and 50% rh mix with 4,000
cfm of air at 98°F dry-bulb and 78°F wet-bulb temperature. The dew point temperature (°F) of the
resulting mixture is most nearly:
(A) 58
(C) 72
(B) 61
(D) 80
04-05.* A room is being maintained at 75°F and 50% rh. The outdoor air conditions are 40°F and 50%
rh at this time. Return air from the room is cooled and dehumidified by mixing it with fresh ventilation
air from the outside. The total air flow to the room is 60% outdoor and 40% return air by mass. The
humidity ratio (lbm-H20/lbm-air) of the mixed air going to the room is most nearly:
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(A) 0.0025
(C) 0.0067
(B) 0.0052
(D) 0.014
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SECTION 05: Cooling with Dehumidification and Heating with Humidification
Cooling with Dehumidification:
The cooling process with dehumidifying is illustrated schematically and on the psychrometric chart in
Fig. 5-1.
h1
h2
ADP
x
Refrigerant or chilled fluid
ω =const
Warm
humid air
1
ṁ a
1
2
Cooling coil
2
Cooled & dehumidified
air
Condensate line
ṁ cond
T1
T adp T 2
Figure 5-1: Cooling with dehumidification
Warm, humid air enters the cooling section at state 1 at a mass flow rate ṁa . As it passes through the
cooling coils, its temperature decreases and its relative humidity increases at constant specific
humidity. If the cooling section is sufficiently long, air reaches its dew point (state x, saturated air).
Further cooling of air results in the condensation of part of the moisture in the air. The condensate
leaves at a mass flow rate ṁcond . Air remains saturated during the entire condensation process, which
follows a line of 100 percent relative humidity until the final state is reached. In a “perfect”
refrigeration coil, the final state is saturated air at the apparatus dew point temperature, T adp which is
the average surface temperature on the outside of the cooling coil. In reality, some air passes through
without ever contacting the fins of the coil. The actual exit state, 2, is the result of the adiabatic mixing
of the air at that bypasses the coil (and thus is at condition 1) and the air that is cooled to T adp . See the
left side of Figure 5-1. The bypass factor (BF) is defined as the equivalent fraction of air that bypasses
the cooling coil while the remaining portion of the air is completely cooled to the apparatus dew-point
temperature. It can be shown that:
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BF=
T 2−T adp
T 1−T adp
(5-1)
The location of the outlet condition, 2, can be found by drawing a line from the inlet condition, point 1,
on the psychrometric chart to ADP, as shown in Figure 5-1, and then calculating the outlet dry bulb
temperature by rearranging Equation 5-1 to solve for T 2 using Equation 5-2. Point 2 will be located on
the line connecting 1 and ADP.
T 2=T adp+BF ( T 1 −T adp )
(5-2)
The total heat removed from the air (and absorbed by the fluid in the coil) is known as the coil load
Q̇ coil and is obtained from an energy balance on the cooling section:
Q̇ coil = ṁa ( h1 −h2 ) − ṁcond hcond
(5-3)
where it is typically assumed that the condensate leaves the cooling section at the apparatus dew point
temperature, hence hcond =h f (T adp) and from a water mass balance:
ṁcond = ṁa ( ω 1 − ω 2)
(5-4)
In HVAC calculations, it is typical to neglect the enthalpy of the condensate in equation (5-3) so the
coil load is obtained as:
Q̇ coil≈ ṁa ( h1−h 2 )
60 min
CFM×
1h
≈
( h1−h2 )
3
v [ft /lbm]
∣
∣
and using the specific volume at standard conditions (dry air at 68°F and 14.7 psia) of 13.3 ft 3/lbm the
above equation becomes:
Q̇ coil≈4.5 CFM ( h1 −h2 )
(5-5)
(Note: If standard conditions are specified as 70°F and 50% RH, replace 4.5 with 4.44.)
Point 2 has a lower temperature than point 1 but also a lower moisture content. As discussed in section
3, the heat transfer associated with the change in dry bulb temperature is known as the sensible heat.
The latent heat is the heat transfer associated with the phase change of the water.
Q̇ lat = ṁcond h fg
which in view of equation (5-4) becomes:
Q̇ lat = ṁa ( ω 1− ω 2) h fg
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and using the specific volume at standard conditions (dry air at 68°F and 14.7 psia) of 13.3 ft 3/lbm the
above equation becomes:
∣601min
h ∣
h
CFM×
fg ( ω 1 − ω 2 )
3
v [ft /lbm ]
CFM ( 60 )
≈
( 1,076 ) ( ω 1− ω 2 )
13.33
Q̇ lat≈
In the equation above, 1,076 (Btu/lbm) is the approximate energy content of the superheated water
vapor at 75°F (1094.7 Btu/lb), less the energy content of water 50°F (18.07 Btu/lb). This difference is
rounded up to 1,076. A temperature of 75°F is a common design condition for an occupied space and
50°F is normal condensate temperature from cooling and dehumidifying coils. Combining the
constants, the latent heat gain is:
Q̇ lat≈4,840 CFM ( ω 1− ω 2 )
(5-6)
(Note: If standard conditions are specified as 70°F and 50% RH, replace 4,840 with 4,680.)
The sensible heat factor (SHF), also called the sensible heat ratio (SHR), is the ratio of the sensible
heat for a process to the summation of the sensible and latent heat for the process:
SHR=
Q̇ sens
Q̇sens
=
Q̇sens +Q̇ lat Q̇ total
(5-7)
For a cooling coil, the sensible heat ratio can be found by applying equation (3-1) to find sensible
cooling, equation (5-5) to find the total cooling, and equation (5-7) to find SHR.
There are a couple of ways to graphically determine the SHR with a psychrometric chart. The first one
is to “break down” the cooling process into two imaginary processes as shown in Figure 5-2: process 11' in which there is no cooling but only dehumidification followed by process 1'-2 in which there is no
dehumidification but only cooling. The enthalpies of states 1, 1' and 2 are read from the enthalpy scale.
The sensible heat is (h1'−h2 ) the latent heat is (h1−h1' ) , and the total coil heat is (h1−h 2) . Therefore
SHR is calculated as SHR=(h1'−h2 )/(h1−h2) . The other graphical way to determine SHR is to use the
SHR protractor that some psychrometric charts have on the upper left corner. You draw a line parallel
to the process line 1-2 through the center of the protractor and use the SHR scale on the protractor to
read the value of SHR.
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h1
Parallel to 1-2
h1'
SHR scale
h2
total
sens.
1
1'
2
Figure 5-2: Calculation of Sensible Heat Ratio (SHR)
Heating with Humidification:
Injecting steam into a moist air stream to raise the humidity ratio of the moist air is a frequent airconditioning process (Figure 5-3). This is typically done in a humidification step after simple heating.
If the mixing is adiabatic, the following equations apply:
ṁ a1 , T 1, ω 1
1
Mixing
Chamber
2
ṁ a2 , T 2, ω 2
Humidifier
Perfect Insulation
ṁ steam , h steam
Figure 5-3: Adiabatic mixing of moist air with steam.
When steam is introduced in the humidification section, this will result in humidification with heating (
T 2>T 1 , ω 2> ω 1 ), as shown in Figure 5-4.
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2
1
Figure 5-4: Adiabatic mixing of moist air with steam – graphical representation on psychrometric chart.
The mass flow of steam injected into the air is obtained from a mass balance for water:
ṁsteam = ṁa ( ω 2 − ω 1 )
PROBLEMS
05-01. In a cooling and dehumidification coil, 71,000 cfm of air at 80°F dry bulb, 60% rh, and standard
atmospheric pressure, are conditioned to 57°F dry bulb and 90% relative humidity. Calculate the
following:
(a) cooling capacity of the air-conditioning unit, in Btu/h and in tons of refrigeration
(b) rate of water (condensate) removal from the unit
(c) coil sensible heat load, in Btu/h
(d) coil latent heat load, in Btu/h
(e) the dew point of the air leaving the conditioner
(f) the apparatus dew point
(g) the coil bypass factor
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05-02*. Air enters a 1-ft-diameter cooling section at 14.7 psia, 90°F, and 60% rh at 600 ft/min. The air
is cooled by passing it over a cooling coil through which chilled water flows. The chilled water
experiences a temperature rise of 14°F and the air leaves the cooling section saturated at 70°F. The
required flow rate of chilled water (gpm) is most nearly:
(A) 1.5
Chilled water, Δ T =14 °F
(B) 2.2
Cooling coil
(C) 12
Air, 600 fpm
90°F
60% r.h.
(D) 120
D=1 ft
70°F
saturated
1
2
05-03*. A quantity of 1,600 cfm of air at 14.7 psia, 80°F dry-bulb and 67°F wet-bulb flows through a
cooling coil with a 0.12 bypass factor and a 45°F apparatus dew point. Under these conditions, the wet
bulb temperature (°F) of the air leaving the coil is most nearly:
(A) 47
(C) 49
(B) 48
(D) 50
05-04*. Air flowing at a rate of 1,600 cfm at 78°F, 65°F w.b. enters a cooling unit with a total capacity
of 60,000 Btu/h and a sensible heat ratio of 0.75. The amount of water (gallons per hour) expected to
drain through the condensate line is most nearly:
(A) 2
(C) 6
(B) 4
(D) 8
05-05*. An air-conditioning system is to take in outdoor air at 50°F and 30 percent relative humidity at
a steady rate of 1,600 cfm and to condition it to 77°F and 60 percent relative humidity. The outdoor air
is first heated to 72°F in the heating section and then humidified by the injection of hot steam in the
humidifying section. Assuming the entire process takes place at a pressure of 14.7 psia, the mass flow
rate of the steam (pounds per hour) required in the humidifying section is most nearly:
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(A) 17
(C) 89
(B) 72
(D) 95
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SECTION 06: Evaporative Cooling
Cooling in a hot dry climate (e.g., the southwest United States) can be accomplished by evaporative
cooling. This consists of either spraying liquid water into air, or forcing air through a soaked pad that is
kept replenished with water. This is the principle of operation of the so-called swamp coolers popular
in the southwestern states. In extremely dry climates, evaporative cooling of air has the added benefit
of conditioning the air with more moisture. Because of the low humidity of the air entering the
evaporative cooler, part of the injected water evaporates. The energy required for this phase change of
the water is taken from the air, which manifests in a reduction of air temperature. It can be shown that
the evaporative cooling process takes place at nearly constant wet-bulb temperature. Therefore, during
evaporative cooling, the wet-bulb temperature of the air remains constant but the dry-bulb temperature
drops as the humidity rises. Since the constant wet-bulb temperature lines almost coincide with the
constant-enthalpy lines, the enthalpy of the airstream can also be assumed to remain constant.
Therefore, when air is cooled evaporatively all the way to saturation, it is said to have undergone
adiabatic saturation.
Evaporative coolers have a continuously recirculated water stream. It can be shown that during steady
state operation, this water reaches an equilibrium temperature that equals the entering air wet-bulb
temperature. The evaporative cooling process is schematically shown on a psychrometric chart in
Figure 6-1. The process 1-2 follows a line of constant wet-bulb temperature.
2'
2
T w.b., 1
1
T2
T1
T 2'=T w.b., 1
Figure 6-1: Evaporative Cooling
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Note from Figure 6-1 that the lowest air temperature that can be theoretically achieved corresponds to
state 2', that is when the air at the exit of the cooler is saturated air. The lowest temperature
theoretically achievable by evaporative cooling is therefore the wet-bulb temperature of the incoming
air.
PROBLEMS
06-01. Air enters an evaporative (or swamp) cooler at 14.7 psi, 95°F, and 20 percent relative humidity,
and it exits at 80 percent relative humidity. Determine the exit temperature of the air.
06-02. What is the lowest temperature that air can attain in an evaporative cooler if it enters at 1 atm,
95°F, and 40 percent relative humidity?
06-03*. Saturated air at 40°F is first preheated and then saturated adiabatically. This saturated air is
then heated to a final condition of 105°F and 28% rh. The temperature (°F) to which the air must
initially be heated in the preheat coil is most nearly:
(A) 65
(C) 101
(B) 95
(D) 105
06-04*. Air at 1 atm, 60°F, and 60 percent relative humidity is first heated to 85°F in a heating section
and then passed through an evaporative cooler where its temperature drops to 77°F. The exit relative
humidity (%) is most nearly:
(A) 26
(C) 60
(B) 42
(D) 100
06-05*. Air enters an evaporative cooler at 14.7 psia, 90°F, and 20 percent relative humidity at a rate of
9,800 cfm, and it leaves with a relative humidity of 90 percent. The required rate of water supply to the
evaporative cooler (gallons per minute) is most nearly:
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(A) 0.067
(C) 4.2
(B) 0.5
(D) 8.5
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SECTION 07: Wet Cooling Towers
Large-scale applications such as power plants, large commercial air-conditioning systems, and some
industrial sites generate large quantities of waste heat that is often rejected to cooling water from
nearby lakes or rivers. In some cases, however, the cooling water supply is limited or excessive thermal
pollution to a water body is undesirable. In such cases, the waste heat must be rejected to the
atmosphere, with cooling water recirculating and serving as a means of transport for heat between the
source and the sink (the atmosphere). Cooling towers provide a common way to achieve this.
A wet cooling tower is essentially a semi-enclosed evaporative cooler. An induced-draft wet cooling
tower is shown schematically in Figure 7-1.
Air out
Warm water
from process
Air in
Basin
Cooled water
to process
Make-up
water in
Pump
Figure 7-1: Induced-Draft Wet Cooling Tower
Note that – unlike in an evaporative cooler – the water in a cooling tower is supplied at a temperature
higher than the dry-bulb temperature of the incoming air. Air is drawn into the tower from the sides
and leaves through the top. Warm water is pumped to the top of the tower and is sprayed into this
airstream. The purpose of spraying is to expose a large surface area of water to the air. As the water
droplets fall under the influence of gravity, a small fraction of water (usually a few percent) evaporates
and cools the remaining water. The temperature and the moisture content of the air increase during this
process. Inside the tower, some designs use a series of solid structures known as packing or fill to
promote both the maximum contact surface and the maximum contact time between air and water. The
cooled water collects at the bottom of the tower in the basin and is pumped back to the process.
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Makeup water must be added to the cycle to replace the water lost by evaporation and drift, which is
circulating water lost from the tower as liquid droplets entrained in the exhaust air stream. To minimize
water carried away by the air, drift eliminators (baffles or similar elements designed to remove
entrained liquid water from the exhaust air) are installed in the wet cooling towers above the spray
section. The amount of makeup water is essentially equal to the moisture gained by the air; that is, air
mass flow rate times the increase in specific humidity.
The air circulation in the cooling tower described above is provided by a fan, and therefore it is
classified as a mechanical-draft cooling tower. The schematic of Figure 7-1 shows the fan located in
the exiting air stream drawing air through the tower; this is called induced draft. Conversely, if the fan
is located in the ambient air stream entering the tower thus blowing the air through the tower, this is
known as forced draft. Another type of cooling tower is the natural-draft cooling tower. The air in
the tower has a high water-vapor content, and thus it is lighter than the outside air. Therefore, the light
air in the tower rises, and the heavier outside air fills the vacant space, creating an airflow from the
bottom of the tower to the top – a “stack” effect. The flow rate of air is controlled by the conditions of
the atmospheric air. Natural-draft cooling towers do not require any external power to induce the air,
but they cost more to build than mechanical-draft cooling towers and are significantly taller. The
natural-draft cooling towers are hyperbolic in profile but this profile is for structural integrity reasons,
not for any thermodynamic or fluid dynamics reason.
The cooling tower range is the difference between the hot water temperature and the cold water
temperature. Similarly as evaporative coolers, the lowest temperature to which water can be cooled by
evaporation is the wet-bulb temperature of the incoming air. The approach is the difference between
the cold water temperature and the wet-bulb temperature of the incoming air. The tower cooling
efficiency is based on the water temperatures. It is defined as the ratio of the actual water temperature
drop over the theoretical maximum water temperature drop:
η CT =
T w,in −T w,out
T w,in −T wet bulb, in
(7-1)
The heat load is the total heat to be removed from the water by the cooling tower per unit time:
Q̇ CT = ṁ w c p ,w ( T w,in −T w,out )
(7-2)
Cooling tower problems are typically solved by performing energy and mass balances on the tower.
Referring to the tower in Figure 7-2, we can write down a mass balance for water as follows:
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86 °F
air, out 100% r.h.
2
3
Warm
water in 95 °F
68 °F
60% r.h.
1 air, in
4
5
Cool
water out
Make-up
water
Figure 7-2. Schematic of a wet cooling tower
of water = Rate of water
[ Rate
mass flow in ] [ mass flow out ]
(7-3)
ṁ3+ ṁ 4+ ṁair ω 1= ṁair ω 2 + ṁ5
Therefore, the rate of make-up water ṁ 4 required can be obtained as:
ṁmakeup = ṁ4 = ṁ air ( ω 2 − ω 1 )+ ṁ5 − ṁ3 = ṁair ( ω 2− ω 1 )
(7-4)
Now we will consider the energy balance. Although there is some mechanical work (power) consumed
by the fan(s), this energy interaction is typically neglected when performing an energy balance on
cooling towers, so we have:
at which = Rate at which
[ Rate
energy enters ] [ energy exits ]
ṁair h1 + ṁ3 h3 + ṁ4 h4= ṁair h2+ ṁ5 h5
Now, assuming the make-up water is at a temperature not too different than that of the cooled water,
we set h 4 =h 5 . Therefore: ṁair (h2 −h1 )= ṁ3 h3−( ṁ5− ṁ4 ) h5 . Combining this expression with the water
mass balance of equation (7-4) we obtain:
ṁair [(h2−h1 )−( ω 2− ω 1)h5 ]= ṁ3 (h3 −h5)
= ṁw c p ,w ( T w,in −T w,out )
(7-5)
and we recognize the right hand side of equation (7-5) as the tower heat load.
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PROBLEMS
07-01*. A cooling tower cools water by passing it through a stream of 1,400 cfm of air at 85°F dry bulb
and 78°F wet bulb entering the tower and leaving it saturated at 95°F. The water enters the tower at
110°F with a flow rate of 10 gpm. Under these conditions, the temperature (°F) of the water leaving the
tower is most nearly:
(A) 78
(B) 86
(C) 95
(D) 99
07-02*. The cooling water from the condenser of a power plant enters a wet cooling tower at 110°F at a
rate of 720 gpm. Water is cooled to 80°F in the cooling tower by air that enters the tower at 1 atm,
76°F, and 60 percent relative humidity and leaves saturated at 95°F. Under these conditions, the
volume flow rate of air (cfm) entering the tower is most nearly:
(A) 66,900
(B) 71,000
(C) 78,600
(D) 87,400
07-03*. A cooling tower with a cooling efficiency of 57% is being considered for the heat rejection
needs of a large commercial HVAC system. The water must leave the tower at 77°F. The air flow
through the tower is 263,000 cfm entering at 68°F, 50% rh and exiting at 90°F, 98% rh. Under these
conditions, the water flow (gpm) entering the tower is most nearly:
(A)
475
(B)
950
(C) 2,000
(D) 2,600
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07-04*. A cooling tower operates with a heat load of 600,000 Btu/hour. The air will enter the tower at
68°F, 30% rh and will be discharged as saturated air at 95°F, while the water will be cooled down to
70°F. Under these conditions, the water flow (gallons per hour) required as makeup water is most
nearly:
(A)
58
(B) 230
(C) 480
(D) 3,300
07-05.* Water enters a cooling tower at a rate of 30 gpm and 95°F. The cooling efficiency of the tower
is 75%. The air enters the tower at 60°F, 30% rh. Under these conditions, the cooling tower heat load
(Btu per hour) is most nearly:
(A)
9,250
(B) 66,600
(C) 395,000
(D) 555,100
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SECTION 08: Basic HVAC System Calculations
A single conditioner serving a single temperature control zone is the simplest form of an all-air HVAC
system. The unit may be installed remote from, or within, the space it serves, and it may operate either
with or without distributing ductwork. Well-designed systems can maintain temperature and humidity
closely and efficiently.
A single-zone system is one which responds to only one set of space conditions. Its use is limited to
situations where variations occur approximately uniformly throughout the zone served or where the
load is stable. Single-zone systems are used in such applications as small department stores, small
individual shops in a shopping center, individual classrooms for a small school, and computer rooms.
A rooftop unit complete with a refrigeration system serving an individual space is an example of a
single-zone system. The refrigeration system, however, may be remote and serving several single zone
units in a larger installation.
A schematic of the single-zone central unit is shown in Figure 8-1.
Return air
Exhaust air
Return Fan
Return air
(from room)
Optional section
Supply air
(to room)
Outside air
Supply Fan
Pre-heat coil (optional)
Cooling
coil
Humidifier (optional)
Reheat coil (optional)
Figure 8-1: Single-Duct System
Several strategies of control can be employed on a single-zone system. These strategies include on-off
operation, varying the quantity of cooling medium providing reheat, using face and bypass dampers, or
a combination of these. A single-duct system with a reheat coil satisfies variations in load by providing
independent sources of heating and cooling. Humidity control completely responsive to space needs is
possible when a humidifier is included in the system. Since control is directly from space temperature
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and humidity, close regulation of the system conditions may be achieved. Generally, a return air fan is
needed if the total resistance of the return air system (grilles and ductwork) exceeds about 0.25 in.
water gage.
The basic equations for individual rooms (zones) are the same for all single-path systems. Air supplied
to each room must be adequate to take care of each room’s peak load conditions whether or not it
occurs simultaneously in all rooms. The peak may be governed by sensible or latent room cooling
loads, heating loads, outdoor air requirements, air motion, or exhaust.
If qsens, summer is the room's peak summer sensible load, then the summer room supply air volume
required to satisfy this peak sensible load is:
CFM sens, summer =
qsens, summer
1.1×( T r −T s )
(8-1)
where T r is the room (or return) air temperature and T s is the supply air temperature required to satisfy
the summer peak load.
If qlat, summer is the room's peak summer latent load, then the summer room supply air volume required to
satisfy this peak latent load is:
CFM lat, summer =
qlat, summer
4,840×( ω r − ω s )
(8-2)
where ω r is the room (or return) air humidity ratio and ω s is the supply air humidity ratio required to
satisfy the summer peak load.
If qsens, winter is the room's peak winter sensible load (less any auxiliary heat), then the winter room
supply air volume required to satisfy this peak sensible load is:
q
CFM sens, winter = sens, winter
1.1×( T s −T r )
(8-3)
where T r is the room (or return) air temperature and T s is the supply air temperature required to satisfy
the winter peak load.
If qlat, winter is the room's peak winter latent load, then the winter room supply air volume required to
satisfy this peak latent load is:
CFM lat, winter =
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qlat, winter
4,840× ( ω s −ω r )
42
(8-4)
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where ω r is the room (or return) air humidity ratio and ω s is the supply air humidity ratio required to
satisfy the winter peak load.
In some instances, the air volume required for ventilation purposes might be larger than that required
by sensible or latent loads. There are three basic categories of ventilation air volume requirements:
1. Outdoor air requirements dictate the supply volume to a room as determined by an actual
quantity of outdoor air required CFM oa req and the ratio of the system's total outdoor air to its
total supply air, X o :
CFM ventilation =
CFM oa req
Xo
2. Exhaust air requirements in applications in which the air discharged to the outside (exhaust air)
exceeds the volume required to satisfy any loads. Then the supply air must match the exhaust
air requirement and serve as makeup for exhaust flow out of the room.
CFM ventilation =CFM exhaust
3. Air movement requirements in some applications might be larger than the volume required to
satisfy any loads. In these cases, the supply air volume has to be chosen to satisfy these air
movement requirements, which are expressed either in terms of air changes per hour (ACH) or
as proportional to the room floor area.
3
CFM ventilation =(Room volume in ft )×ACH/60 , or CFM ventilation=K (Room floor area )
Both ACH and K are empirical values that vary according to designers’ experiences and local
building code requirements.
EXAMPLE 8-1:
Consider a space which is designed to have a summer inside temperature of 75°F and relative humidity
of 50% and a winter inside temperature of 72°F and relative humidity of 25%. The summer supply air
conditions are 55°F, 90% rh, while the winter supply air temperature is 110°F with a humidity ratio of
0.0065 lb/lb. The summer design loads are qsens, summer =18,000 Btu /h and qlat, summer =3,000 Btu /h and
the winter design loads are qsens, winter =20,000 Btu / h and qlat, winter =2,500 Btu / h .The outdoor air
requirement is 120 cfm and the ratio of outdoor air to total supply air is 0.35. Determine the required
supply air in CFM to satisfy summer, winter, and ventilation conditions.
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Solution:
CFM sens, summer =
CFM lat, summer =
CFM ventilation=
18,000
=818cfm
1.1× ( 75−55 )
CFM sens, winter =
20,000
=479 cfm
1.1×( 110−72 )
3,000
2,500
=620 cfm CFM lat, winter =
=225 cfm
4,840×( 0.0093−0.0083 )
4,840× ( 0.0065−0.0042 )
CFM oa req 120
=
=343cfm
Xo
0.35
In order to satisfy all design parameters, the design volume flow rate should be selected as the
maximum flow requirement of 818 cfm. This is what is required for the summer sensible design load.
All of the other design parameters will be satisfied with this volume flow but would require some form
of control to maintain temperature, relative humidity, and outside ventilation air quality.
The operation of a single path system is illustrated in Figure 8-2. Each state point is shown with
corresponding nomenclature in the cycle diagram and in the summer and winter representations on the
psychrometric chart of Figure 8-3. In Figure 8-2, the red arrows represent heat loads. Each change in
temperature or humidity ratio is a result of sensible or latent heat loss or gain. Here we describe the
flow paths in a typical single-duct, single-zone system. In this illustration all return air is assumed to
pass from the room through a hung-ceiling return air plenum.
Roof
CFMEXH
CFMR
Tr
CFMr
Return Fan
Tr
CFMR
TRP
Trd
CFMS
CFMO
TO
CFMs
Tm
CFMS
TCC
THC
TSF
Supply Fan
Cooling
coil
plenum
lights
Ts
CFMR
Troom
window
CFMexf
room
occupants
Heating coil
equipment
Floor
Figure 8-2: Single-Duct, Single-Zone Cycle Diagram
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In Figure 8-2, supply air leaves the supply fan discharge at a rate CFM s and at temperature T sf in the
summer mode. This air absorbs transmitted supply duct heat and supply air fan velocity pressure
energy, thus raising the temperature to T s . The air enters the room where it absorbs room sensible and
latent heat qsens and qlat along the room sensible heat factor (SHF) line s-R, thus reaching the desired
room state, T room and ω room . Room (internal) sensible loads which determine CFM s consist of (for
example): (1) ceiling transmission from the hung ceiling above the room, and floor transmission from
the floor deck below, (2) direct light heat emissions to the room, (3) transmissions from other surfaces
such as walls, windows, etc., (4) appliance heat and occupancy heat, and (5) infiltration loads (not
shown). Additional loads in the plenum space move the conditions from “room” to “RP” and the
volume of air in the return system CFM R is CFM s minus the volume lost by exfiltration CFM exf . The
return air picks up return duct transmissions and return fan heat gains so it reaches the intake plenum
entrance at T r . Mixing of outdoor air, CFM o at state “o”, with final return air occurs along process line
r-o to mixture state “m”. Total system air flowing at rate CFM s passes through the cooling coil (process
line m-cc), terminating in state cc. The temperature rise through the supply fan results in T sf which
completes the cycle. Note that there may be more loads and flow rates (for example duct leakage) not
mentioned in this discussion.
o
summer
R
cc
m
rp r
sf s
R
m
hc
r
s sf
o
Figure 8-3: Single-Duct, Single-Zone Psychrometric Chart
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Consider the simplified single-duct system of Figure 8-4 shown operating in cooling (summer) mode.
With respect to the control volume defined by the imaginary red dashed line we notice the following
heat transfer interactions (shown as red arrows): the coil load, which is heat (sensible and latent)
removed from the supply air by the cooling coil, and the space load which is heat (sensible and latent)
absorbed by the air as it flows through the conditioned space.
x
Q̇ space
r
ṁ a, x
Conditioner
r
Fan
o
ṁ a, o
ṁ w, space
s
m
Space
c
Q̇ coil
ṁ cond
Condensate
Figure 8-4: Thermal and Moisture Loads for a Single-Duct, Single-Zone System
Also, the water mass flow rates (blue arrows) are: the condensate removed by dehumidification in the
conditioner and the space moisture load, which is water vapor added to the air as it flows through the
conditioned space.
The outdoor air at state “o” enters the control volume at a mass flow rate ṁa , o and is discharged at the
exhaust at state “x” (which is assumed for simplicity to be identical to “r”) at a mass flow rate ṁa , x . If
we neglect any building infiltration and exfiltration, the steady-state mass balance for air in the red
control volume requires that: ṁa , o= ṁa , x .
The energy balance requires that the rate at which energy enters the control volume equals the rate at
which energy leaves the control volume. Thus:
ṁ a , o ho + Q̇ space = ṁ a , x h r +Q̇ coil + ṁ cond hcond
⇒ Q̇ coil =Q̇ space + ṁ a , o ( ho−hr ) − ṁ cond hcond
and if we neglect the condensate enthalpy,
Q̇ coil =Q̇space + ṁa , o ( h o−hr )
(8-5)
We note then, that the coil size has to exceed the space load by at least the term underlined in equation
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8-5. This underlined term is the amount of energy that is required to take the outdoor air from the
outdoor condition “o” to the room condition, “r” and it is known as the load due to outdoor air.
Outdoor air that flows through a building is often used to dilute and remove indoor air contaminants.
However, the energy required to condition this outdoor air can be a significant portion of the total
space-conditioning load. The magnitude of the outdoor airflow into the building must be known for
proper sizing of the HVAC equipment and evaluation of energy consumption.
Air exchange of outdoor air with the air already in a building can be divided into two broad
classifications: ventilation and infiltration. Ventilation is the intentional introduction of air from the
outside into a building; it is further subdivided into natural ventilation and forced ventilation. Natural
ventilation is the intentional flow of air through open windows, doors, grilles, and other planned
building envelope penetrations, and it is driven by natural and/or artificially produced pressure
differentials. Forced ventilation is the intentional movement of air into and out of a building using
fans and intake and exhaust vents; it is also called mechanical ventilation. Infiltration is the
uncontrolled flow of outdoor air into a building through cracks and other unintentional openings and
through the normal use of exterior doors for entrance and egress. Infiltration is also known as air
leakage into a building. Exfiltration is the leakage of indoor air out of a building. Like natural
ventilation, infiltration and exfiltration are driven by natural and/or artificial pressure differences.
PROBLEMS
08-01: Consider an air-conditioned space with a summer sensible design load of 100,000 Btu/h, and a
summer design latent load of 20,000 Btu/h. The space is maintained at 75°F and 55% rh. Conditioned
air leaves the apparatus (and enters the room) at 58°F. The outdoor air is at 96°F, 77°F wet bulb, and is
20% of total flow to the conditioning apparatus.
Complete the following table where o, r, m, and s refer to outdoor, return, mix (r+o) and supply.
Point
T(°F)
o
96
r
75
ϕ
T wb (°F)
h (Btu/lbm)
ω (lbm/lbm) v a (ft3/h)
ṁ (lbm/h)
CFM
SCFM
77
55
m
s
58
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08-02. Determine the size of the required cooling unit in Btu/h and tons for the system of problem 0801.
08-03. For the system of problem 08-01; what percent of the required cooling is for sensible cooling,
and what percent is for dehumidification?
08-04. For the system of problem 08-01; what percent of the required cooling is due to the outdoor air
load?
08-05*. A room is to be maintained at 76°F and 40% rh. Supply air at 39°F is to absorb 100,000 Btu of
sensible heat and 35 lbm of moisture per hour. Assume the moisture has an enthalpy of 1100 Btu/lbm.
The required mass flow rate of air (lbm of dry air per hour) is most nearly:
(A) 1,230
(B) 2,457
(C) 11,100
(D) 16,500
08-06*. Moist air enters a chamber at 40°F dry-bulb and 36°F wet-bulb temperature at a rate of 3,000
scfm. In passing through the chamber, the air absorbs sensible heat at a rate of 116,000 Btu/h and picks
up 83 lbm/h of saturated steam at 230°F. The relative humidity (%) of the leaving air is most nearly:
(A) 5
(B) 10
(C) 30
(D) 50
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08-07*. In an auditorium, a sensible-heat load of 350,000 Btu and 1,000,000 grains of moisture must be
removed per hour. The room must be maintained at a temperature not to exceed 77°F, and a relative
humidity not to exceed 55%,. Air is supplied to the auditorium at 67°F. Under these conditions, the
dew-point (°F) of the supplied air is most nearly:
(A) 55
(B) 59
(C) 65
(D) 67
08-08*. A flow rate of 30,000 lbm/h of conditioned air at 60°F and 85% rh is added to a space that has
a sensible load of 120,000 Btu/h and a latent load of 30,000 Btu/h. A mixture of 50% return air and
50% outdoor air at 98°F dry bulb and 77°F wet bulb enters the air conditioner. The required size (tons)
of the cooling coil is most nearly:
(A) 10
(B) 12.5
(C) 17
(D) 27
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SECTION 09: Answers
01-01.
a) Use equation (1-5):
ω
( ω +0.622
)p
0.1
=(
×14.7 psia
0.1+0.622 )
pv=
=0.3462 psia
b) The dew-point is the saturation temperature corresponding to p v =0.3462 psia , so from the steam
tables, T dp =T sat (0.3462 psia)=68.6 °F
c) Use equation (1-6):
ϕ=
pv
pg
where p g = psat @ T =0.4302 psia . Thus, ϕ =0.3462/0.4302=0.805=80.5% .
d) Use equation (1-9) μ = ω / ω s where ω s is found by setting ϕ =1 in equation (1-8):
ω s=
0.622 p g 0.622×0.4302 psia
=
=0.01875
p− p g
( 14.7−0.4302 ) psia
Therefore μ = ω / ω s =0.015/0.01875=0.8=80 % .
e) Use equation (1-12):
h=0.24T + ω ( 1,061+0.444T )=34.42 Btu /lbm
f) Use equation (1-13):
(
)
Ra T
ω
1+
p
0.622
3
psia ft
0.3704
×( 75+459.67 ) °R
3
lbm °R
0.015
ft
=
× 1+
=13.8
14.7 psia
0.622
lbm
v=
(
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01-02. Use the definition of dew point to obtain the water vapor pressure:
T dp =T sat ( pv )=52°F ⇒ pv =0.192 psia
Now use equation (1-4) to find ω :
ω =0.622
(
)
(
)
pv
0.192
lbm
=0.622
=0.008232
p− pv
14.7−0.192
lbm
Use equation (1-6) to find ϕ :
ϕ=
pv
pg
where p g = psat @ T =0.4302 psia . Thus, ϕ =0.192/0.4302=0.4463=44.6 % .
01-03. We can use equation (1-10) to find the humidity ratio, but first we need ω *s (the humidity ratio
at saturation if at the wet bulb temperature) by setting ϕ =1 in equation (1-8), with
p g = psat @ T =0.2141 psia :
w.b.
ω *s =
0.622×0.2141
lbm
=0.0092
14.7−0.2141
lbm
Therefore,
ω=
( 1,093−0.556T w.b. ) ω s−0.24 ( T −T w.b. )
1,093+0.444 T −T w.b.
=0.00463
lbm
lbm
and
ω
(ω +0.622
)p p
0.00463
14.7
=(
×
=0.252=25.2 %
0.00463+0.622 ) 0.4302
ϕ=
sat @ T
01-04. If condensation is to be avoided, then the lowest temperature to which the air can be cooled is
the dew point temperature, T dp =T sat ( pv ) where p v is the vapor pressure. We can find p v from equation
(1-6)
p v = ϕ psat @ T =0.5×0.36336=0.18168 psia
Therefore,
T dp =T sat (0.18168 psia)=50.5°F
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01-05*. The correct answer is (C)
The flow rate of water vapor ṁv will be the mass flow rate of dry air ṁa , times the humidity ratio ω .
We can use equation (1-8) to find ω with p g = psat @ 95 °F=0.81642 psia :
ω=
0.622 ϕ p g 0.622×0.3×0.81642
lbm
=
=0.01054
p− ϕ p g
14.7−0.3×0.81642
lbm
The mass flow rate is obtained from the given volumetric flow rate ṁa=V̇ a /v a where va is the specific
volume of air – calculated with equation (1-13):
(
)
Ra T
ω
1+
p
0.622
3
psia ft
0.3704
×( 95+459.67 ) °R
3
lbm °R
0.01054
ft
v=
× 1+
=14.21
14.7 psia
0.622
lbm
v=
(
ṁv= ω ( V̇ a / v a )
3
)
∣
ft
60 min
1,000
×
lbm H 2 O
min
1h
=0.01054
×
3
lbm air
ft
14.21
lbm
∣ = 44.5 lbm
h
01-06*. The correct answer is (D)
The moisture content of the air remains constant through the process. We can use equation (1-8) to find
ω at the compressor inlet use( p g = psat @ 70 °F =0.36336 psia ):
ω=
0.622 ϕ p g 0.622×0.5×0.36336
lbm
=
=0.007784
p− ϕ p g
14.7−0.5×0.36336
lbm
For this moisture content, we can find the vapor pressure with equation (1-5) when the pressure is
raised to 50 psia:
ω
( ω +0.622
)p
0.007784
=(
×50 psia=0.618 psia
0.007784+0.622 )
pv=
Therefore,
T dp =T sat (0.618 psia)=86.1 °F
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02-01.
a) T dp≈68.5 °F
0.015 lbm/lbm
=105 grains/lbm
75°F
b) ϕ ≈80 %
c) Use equation (1-9) μ = ω / ω s where ω s is the humidity ratio if the air were saturated at the given
temperature. To find ω s in the chart, locate 75°F on the saturation curve and move horizontally to find
the corresponding value of the humidity ratio. This yields ω s≈131 grains/lbm
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0.015 lbm/lbm
=131 grains/lbm
0.015 lbm/lbm
=105 grains/lbm
75°F
Therefore μ = ω / ω s =105/131=0.8=80 % .
d) h≈34.5 Btu / lbm .
3
e) v≈13.8
ft
lbm
02-02.
Dry Bulb
°F
Wet Bulb
°F
Dew Point
°F
Humidity Ratio
lbm/lbm
R.H.
%
Enthalpy
Btu/lbm
Specific Volume
ft3/lbm
70
55
42.7
0.0058
37
23.1
13.5
100
77.9
70
0.0158
38
41.4
14.5
97.1
76.5
68.8
0.0151
40
40
14.4
79
65
57.3
0.01
47.2
29.9
13.8
85.6
60
40
0.0052
20
26.3
13.9
40
33.6
24
0.0026
50
12.4
12.6
74.7
65
60
0.0110
60.2
30
13.7
85
69.6
62.6
0.012
47
33.7
14.0
80
80
80
0.022
100
43.6
14.1
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02-03*. The correct answer is (C)
The flow rate of water vapor ṁv will be the mass flow rate of dry air ṁa , times the humidity ratio ω .
From the psychrometric chart:
ω ≈0.0105
lbm
lbm
The mass flow rate is obtained from the given volumetric flow rate ṁa=V̇ a /v a where va is the specific
volume of air, obtained from the chart:
3
v≈14.21
ft
lbm
Hence:
ṁv= ω ( V̇ a / va )
3
∣
ft
60 min
1,000
×
lbm H 2 O
min
1h
=0.0105
×
3
lbm air
ft
14.2
lbm
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56
∣ = 44.4 lbm
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03-01.
ω =const
50°F (wb)
60°F
80°F
(a) ϕ 1 ≈49 % , (b) T dp,1 ≈41 °F , (c) ω 1≈0.0054 lbm /lbm , (d) h1≈20.2 Btu /lbm , (e) h2 ≈25 Btu /lbm ,
(f) h2−h1≈4.8 Btu /lbm , and (g) ϕ 2≈25 % .
03-02.
90%
ω =const
55°F
90°F
h2≈30.7 Btu /lbm , h1≈22.2 Btu /lbm ; h2 −h1≈8.5 Btu /lbm
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03-03.* The correct answer is (A)
For the exit condition, the specific humidity will be the same as for the inlet. Use a psychrometric chart
to get ω 1=0.00588 lbm /lbm=ω 2 . We can find the exit temperature with equation (3-1):
Q̇ sens [ Btu/ h ] =1.1×CFM×Δ T [ °F ]
Q̇sens
⇒T 2≈T 1 +
1.1×CFM
60,000
≈65+
=85°F
1.1×2,800
Now, use a psychrometric chart to find ϕ 2≈23 %
03-04.* The correct answer is (C)
If we label the inlet to the heater as state (1) and the exit as state (2), the energy balance yields:
Q̇= ṁa ( h2−h1 )
where
ṁa=( V̇ a /v a )
500
=
3
∣
ft
60 min
×
min
1h
13.0
3
∣ = 2,307.7 lbm
h
ft
lbm
and we have used the value of specific volume at the inlet conditions (50°F, 100% rh) from the
psychrometric chart. Also from the chart: h1≈20.3 Btu /lbm , h2≈26.6 Btu /lbm ; h2−h1≈6.3 Btu / lbm
Q̇= ṁa ( h2−h1 )
lbm
Btu
Btu
=2,307.7
×6.3
=14,538.5
h
lbm
h
Alternative solution: Use the psychrometric chart to get the exit temperature, T 2 ≈76 °F and the
approximate equation based on volumetric flow rate, equation (3-1):
Q̇sens [ Btu /h ] =1.1×CFM×ΔT [ °F ]
=1.1×500×26
Btu
=14,300
h
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03-05.*
D =15 in
14.7 psia
50ºF
40% r.h.
25 ft/s
.
Q=?
31% r.h.
2
1
Heater
An energy balance on the heater yields:
Q̇= ṁa (h2−h1)
The enthalpy at the inlet can be read directly from the chart, h1≈15.3 Btu /lbm . To obtain h 2 we can
use the fact that the heating occurs at constant moisture content, ω 2 =ω 1 to locate state 2 on a
psychrometric chart. Then we would read h2 from the chart. Doing this, we find h2≈17 Btu /lbm ,
T 2 ≈57 °F .
The mass flow rate of air is:
ft
AV
s
lbm
ṁa= υ =
=2.38
3
s
12.9 ft /lbm
2
1.227 ft ×25
therefore:
Q̇= ṁa (h2−h1 )
lbm
Btu 3,600 s
=2.38
× ( 17−15.3 )
×
×
s
lbm
1h
1 kW
=4.28 kW
∣ ∣3,412.14
Btu
∣
∣
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04-01.
We can use equation (4-6) to find the resulting humidity ratio, but for that we need the humidity ratio
for both input streams. Use a psychrometric chart to find ω 1=ω ( 40 °F , 35°F wb )≈0.0031 lbm/lbm
and ω 2= ω ( 100 °F , 77 °F wb )≈0.0147 lbm /lbm . We have labeled the cool air with 1, therefore the
mass ratio is ṁa1 / ṁa2 =2
ω 3=
ω 2+ω 1 ( ṁa1 / ṁa2 ) 0.0147+2×0.0031
lbm
=
≈0.007
1+2
lbm
1+( ṁa1 / ṁa2 )
04-02.
Label the cold stream as “1”, so that ω 1=ω ( 65°F , 30 % rh )≈0.0039 lbm /lbm ; CFM 1 =900 cfm ;
ω 2= ω ( 80 °F , 90 % rh )≈0.020 lbm / lbm and CFM 2=300 cfm
The temperature is obtained directly with the temperature version of equation (4-7):
T 3≈
T 2+T 1 ( CFM1 /CFM 2 )
1+( CFM 1 /CFM 2 )
=
80+3×65
≈68.8 °F
1+3
The specific humidity is obtained directly with the specific humidity version of equation (4-7):
ω 3=
ω 2+ω 1 ( CFM 1 /CFM 2 ) 0.02+3×0.0039
lbm
=
≈0.008
1+3
lbm
1+( CFM 1 /CFM 2 )
Now use a psychrometric chart to obtain ϕ 3≈53 %
04-03.* The correct answer is (B).
Label the cold stream as “1”, so that ω 1=ω ( 75°F , 45 % rh )≈0.008 lbm /lbm ; CFM 1=10,000 cfm ;
ω 2= ω ( 98 °F , 40 % rh )≈0.016 lbm /lbm and we need to determine CFM 2 so that ω 3≈0.01 lbm /lbm
We use the specific humidity version of equation (4-7):
ω 2 + ω 1 ( CFM1 /CFM 2 )
1+( CFM 1 /CFM 2 )
ω −ω
0.01−0.008
⇒ CFM 2 =CFM 1 ω 3− ω 1 =10,000 cfm×
=3,333 cfm
2
3
0.016−0.01
ω 3≈
(
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04-04.* The correct answer is (B).
The graph below shows the graphical approach with a psychrometric chart. The resulting mixture is
labeled “mix”, then the dew point temperature is read by moving to the left up to the saturation curve.
We obtain a dew point temperature of approximately 61°F.
78°F wb
mix
50% rh
75°F
98°F
04-05.* The correct answer is (B).
Label the cold stream as “1”, so that:
ω 1=ω ( 40 °F , 50 % rh )≈0.0026 lbm /lbm ; ω 2= ω ( 75 °F ,50 % rh )≈0.0092 lbm /lbm
The mass ratio as given is ṁ1 /( ṁ1+ ṁ2 )=0.6 which can be re-arranged as ṁ1 / ṁ2 =1.5 . Now use
equation (4-6):
ω 3=
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ω 2+ω 1 ( ṁa1 / ṁa2 ) 0.0092+1.5×0.0026
lbm
=
≈0.00524
1+1.5
lbm
1+( ṁa1 / ṁa2 )
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05-01.
(a) cooling capacity of the air-conditioning unit, in Btu/h and in tons of refrigeration
From a psychrometric chart, enthalpy values are h1=33.6 Btu /lbm and h2 =23.3 Btu /lbm at the inlet
and outlet, respectively. Also, v1=13.9 ft 3 /lbm , therefore the total cooling capacity is obtained with
equation (5-3):
Q̇ coil = ṁa ( h1−h2 ) − ṁcond hcond
and we will neglect the enthalpy of the condensate:
3
71,000
Q̇ coil = ṁa ( h1−h2 ) =
∣
ft
60 min
×
min
1h
3
∣×(33.6−23.3 ) Btu
ft
lbm
lbm
Btu
=306,474.8
×( 33.6−23.3 )
h
lbm
Btu
=3,156,690
=263 tons
h
13.9
lbm
Alternatively, we can use the approximate equation, (5-5)
Q̇ coil≈4.5 CFM ( h1 −h2 )≈3,290,850 Btu / h=274 tons
(b) rate of water (condensate) removal from the unit.
The water removed is equal to the drop in moisture content of the air as it flows through the coil. From
a psychrometric chart, specific humidity values are ω 1=0.013 lbm /lbm and ω 2=0.0089 lbm /lbm at
the inlet and outlet, respectively. Therefore:
ṁcond = ṁa ( ω 1 −ω 2 )
lbm
lbm
=306,474.8
× ( 0.013−0.0089 )
h
lbm
lbm
=1,256.5
h
(c) coil sensible heat load, and (d) coil latent load, in Btu/h
The sensible load is associated with the change in dry bulb temperature, and the latent load is
associated with the change in specific humidity. We can determine these with the representation of the
process in a psychrometric chart: h1=34.6 Btu /lbm , h2 =23.3 Btu /lbm , and h1' =29 Btu /lbm
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h1
h 1'
60% r.h.
h2
1
2
1'
57°F
80°F
Q̇ lat = ṁa ( h1 −h1 ' )
lbm
Btu
× ( 33.6−29 )
h
lbm
Btu
=1,409,784
h
=306,474.8
Q̇sens =Q̇ coil −Q̇ lat
=( 3,156,690−1,409,784 )
Btu
Btu
=1,746,906
h
h
(e) the dew point of the air leaving the conditioner
From the psychrometric chart, state 2 has a dew point of approximately 54°F
(f) the apparatus dew point is determined by extending the 1-2 line until it reaches the saturation curve.
This is shown below, and we find that T adp ≈49°F .
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h1
h 1'
60% r.h.
h2
1
2
1'
ADP
57°F
80°F
(g) the coil bypass factor is found with equation (5-1):
BF=
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T 2−T adp 57−49
=
=0.258=25.8%
T 1−T adp 80−49
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05-02*. The correct answer is (B)
Chilled water, Δ T =14 °F
Cooling coil
Air, 600 fpm
90°F
60% r.h.
D=1 ft
70°F
saturated
1
2
The heat “picked up” by the chilled water is the heat lost by the air in the cooling process, or, the coil
load. Neglecting the enthalpy of the condensate,
Q̇ coil = ṁa ( h1−h2 )
where the mass flow rate of air is
∣
∣
π D 2 V π 1 ft 2 ×600 ft × 60 min
V̇ a 4
4
min
1h
lbm
ṁa= =
=
=1,982
3
va
va
h
ft
14.26
lbm
Now, use a psychrometric chart to obtain the air inlet and outlet values of enthalpy:
Q̇ coil = ṁa ( h1−h2 )
lbm
Btu
Btu
=1,982
×( 41.7−34 )
=15,267
h
lbm
h
This same heat is transferred into the chilled water, which causes its temperature rise:
Q̇= ṁc.w. c p Δ T
Q̇
15,267 Btu / h
lbm
⇒ ṁc.w.=
=
=1,090.5
c p ΔT
Btu
h
1
×14°F
lbm °F
So, the volumetric flow rate is:
lbm
1,090.5
ṁc.w.
h
7.48052 gal
1h
V̇ c.w.= ρ =
×
×
=2.18gpm
3
c.w.
lbm
60
min
1
ft
62.4 3
ft
∣
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∣∣
65
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05-03*. The correct answer is (B)
On a psychrometric chart, draw a line from the inlet condition of 80°F, 67°F w.b., and the ADP. We
know the coil exit condition is somewhere on this line.
80°F,
67°F w.b.
ADP
The temperature of the exit condition is found with equation (5-2)
T 2 =T adp +BF ( T 1 −T adp ) =45+0.12× ( 80−45 )=49.2 °F
The point on the line for which the dry bulb temperature is approximately 49.2°F has been indicated
with a green dot. For this point, we see that the wet bulb temperature is approximately 48°F.
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05-04*. The correct answer is (A)
On a psychrometric chart, locate the inlet condition. Then on the protractor draw the line with a slope
corresponding to SHR=0.7 (the green line below). Draw a line parallel to this one through the inlet
condition (the red line below). This will be the process line, so the outlet condition will be somewhere
on this red line.
For the inlet, we find ω 1 =0.01 lbm /lbm and h1=30 Btu /lbm .
With the given coil load, we can use equation (5-3) to find the enthalpy change:
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Q̇ coil≈ ṁa ( h1−h2 )
Q̇
⇒ h2 =h1 − coil
V̇
v
( )
⇒ h2 =30
Btu
−
lbm
60,000
(
Btu
h
∣
3
1,600
ft
1
60 min
×
×
3
min
1h
ft
13.77
lbm
∣)
=21.4
Btu
lbm
Therefore, the outlet condition will be the intersection of the constant enthalpy line corresponding to
21.4 Btu/lbm (the purple line on the chart above) and the process (red) line we drew on the
psychrometric chart. We thus locate the outlet and find that ω 2=0.0077 lbm /lbm . The mass flow rate
of condensate is then:
ṁcond = ṁa ( ω 1 −ω 2 )
(
3
= 1,600
=15.9
ft
×
min
∣
∣)
1
60 min
lbm
×
× ( 0.01−0.0077 )
3
1
h
lbm
ft
13.77
lbm
lbm
h
So, the volumetric flow rate is:
ṁcond
V̇ cond = ρ
=
cond
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lbm
h
7.48052 gal
gallons
×
=1.9
3
lbm
hour
1 ft
62.4 3
ft
15.9
∣
68
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05-05*. The correct answer is (B)
Humidifying
Section
Heating
Section
ṁ a
1
50 °F
30 %rh
2 ṁ steam
72 °F
3
77 °F
60 %rh
The process is shown in the psychrometric chart below. The humidity ratio stays constant during the
simple heating of 1-2, thus ω 2= ω 1≈0.0023 lbm/lbm . Also, from the chart ω 3 ≈0.012 lbm/lbm .
3
2
1
From a mass balance of water:
ṁsteam = ṁa ( ω 3− ω 2 )
(
3
ft
= 1,600
×
min
=72.2
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∣
∣)
1
60 min
lbm
×
×( 0.012−0.0023 )
3
1h
lbm
ft
12.9
lbm
lbm
h
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06-01. Locate the inlet state (95°F, 20% r.h.) on a psychrometric chart. The evaporative cooling
process follows a constant wet-bulb-temperature line up to the 80% r.h. Line. The exit temperature is
read as 70°F
2
1
06-02. The lowest temperature that can be achieved in an evaporative cooler is the wet-bulbtemperature of the incoming air. For the conditions given, this corresponds to 75°F
06-03. The correct answer is (C)
Heating
Section
Humidifying
Section
Pre-Heating
ṁ a
1
40°F
100 %rh
3
2
100 %rh
ṁ water
4
105 °F
28 % rh
The problem is asking for the temperature at the end of the pre-heating section, or T 2 . First locate the
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inlet state (40°F, 100% r.h.). State 2 will be somewhere to the right of 1 on a horizontal line (1-2 is
simple heating, which is at constant ω ). Now locate state 4 (105°F, 28% r.h.) and 3 will be on the
saturation line with the same specific humidity as 4. Since the 2-3 process occurs along a constant wetbulb line (or what is approximately the same thing; a constant enthalpy line) we can locate state 2:
3
4
1
2
Thus, we find T 2≈101 °F
06-04*. The correct answer is (B).
The solution is quickly found graphically with a psychrometric chart as shown below. We find
ϕ 3≈42 %
3
1
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2
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06-05*. The correct answer is (B)
1
2
ṁ a
90 %rh
90 °F
20 % rh
ṁ water
A water mass balance can be performed on the cooler, to show that:
ṁwater= ṁa ( ω 2− ω 1 )
where ω 1≈0.006 lbm/ lbm . Point 2 is found on a psychrometric chart by drawing a constant wet-bulbtemperature line through point 1 up to the 90% relative humidity line. Doing this, we find
ω 2≈0.012 lbm /lbm .
ṁwater= ṁ a ( ω 2− ω 1 )
( )
3
ft
min
lbm
=
× ( 0.012−0.006 )
3
lbm
ft
14
lbm
lbm
=4.2
min
9,800
So, the volumetric flow rate is:
ṁwater
V̇ water = ρ
=
water
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lbm
min
=0.5 gpm
lbm
8.34
gallon
4.2
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07-01*. The correct answer is (B).
Use Figure 7-2 as a reference, and use the energy balance – equation (7-5) – to find the enthalpy of
water leaving the tower:
ṁair [(h2−h1 )−( ω 2− ω 1)h5 ]= ṁ3 (h3−h5)
ṁ3 h3− ṁair (h 2−h1 )
⇒ h5 =
ṁ3 − ṁair ( ω 2− ω 1)
h1=41.4 Btu /lbm , ω 1 =0.019 lbm /lbm , h2=63.2 Btu /lbm , ω 2=0.037 lbm/lbm are obtained from
the chart, and h3≈h f (110°F)=78.1 Btu /lbm from the steam tables. Also, the mass flow rates of air and
water are:
3
ft
min
lbm
ṁair =
3 =98.6
min
ft
14.2
lbm
1,400
ṁ3 =10
gal
lbm
lbm
×8.34
=83.4
min
gal
min
Therefore,
lbm
Btu
lbm
Btu
×78.1
−98.6
×( 63.2−41.4 )
min
lbm
min
lbm
Btu
≈53.5
lbm
lbm
lbm
lbm
83.4
−98.6
× ( 0.037−0.019 )
min
min
lbm
83.4
h5=
So now we scan the saturated water table to find the temperature for which the enthalpy of saturated
liquid is 53.5 Btu/lbm. We find that T w,out≈86 °F .
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07-02*. The correct answer is (C).
Use Figure 7-2 as a reference, and use the energy balance – equation (7-5) – to find the mass flow of
air:
ṁ3 c p ,w ( T 3−T 5 )
ṁ3 (h3−h5)
ṁair =
=
(h 2−h1 )−( ω 2−ω 1) h5 (h2−h1 )−( ω 2− ω 1 )h5
where
ṁ3 =720
gal
lbm
lbm
×8.34
=6,005
min
gal
min
and h1=30.8 Btu / lbm , ω 1=0.012 lbm / lbm , h2 =63.2 Btu /lbm , ω 2 =0.037 lbm/lbm can be obtained
from the psychrometric chart and h5≈h f (80°F)=48.1 Btu /lbm from the steam tables . Plugging in
numerical values, we obtain:
lbm
Btu
×1
×( 110−80 ) °F
min
lbm °F
lbm
=
= 5,712.8
Btu
lbm
Btu
min
( 63.2−30.8 )
−( 0.037−0.019 )
×48.1
lbm
lbm
lbm
6,005
ṁair
So, the volumetric flow rate is:
3
V̇ air= ṁair v air=5,712.8
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lbm
ft
×13.75
=78,551 cfm
min
lbm
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07-03*. The correct answer is (D).
Use Figure 7-2 as a reference, and use the energy balance – equation (7-5) – to find the mass flow of
water entering the tower:
ṁ3 = ṁair
[(h2−h1)−(ω 2− ω 1 )h5 ]
(h3−h5 )
= ṁair
[ (h2−h1)−(ω 2− ω 1) h5 ]
c p ,w ( T 3−T 5 )
where h1=24.2 Btu /lbm , ω 1=0.007 lbm /lbm , h2 =55.1 Btu /lbm , ω 2=0.03 lbm /lbm come from the
psychrometric chart and h5≈h f (77°F)=45.1 Btu /lbm from the steam tables. Also, the mass flow rate
of air is:
3
ft
min
lbm
ṁair =
=19,481.5
3
min
ft
13.5
lbm
263,000
So we only need the incoming water temperature. The problem statement does not provide this value,
but we can figure it out from the definition of cooling efficiency, equation (7-1):
T −T w,out
η CT = w,in
T w,in −T wet bulb, in
⇒ ( T w,in −T w,out ) = η CT ( T w,in −T wet bulb, in )
⇒T w,in ( 1− η CT ) =T w,out− η CT T wet bulb, in
T
− η CT T wet bulb, in
⇒T w,in= w,out
1− η CT
77°F−0.57×56.7 °F
=
=103.9°F
1−0.57
Now plug in values into the previously obtained expression for water mass flow rate:
ṁ3 = ṁair
[(h2−h1)−(ω 2− ω 1 )h5 ]
=19,481.5
c p ,w ( T 3 −T 5)
lbm
×
min
Btu
lbm
Btu
−( 0.03−0.007 )
×45.1
lbm
lbm
lbm
Btu
1
× ( 103.9−77 ) °F
lbm °F
( 55.1−24.2 )
= 21,627.1
lbm
min
Therefore the volume flow is (21,627.1 lbm /min)÷(8.34 lbm /gal)=2,593.2 gpm
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07-04*. The correct answer is (A).
See Figure 7-2 for reference. The mass flow of makeup water is used to match the water lost by
evaporation, as is shown by equation (7-4):
ṁmakeup= ṁair ( ω 2− ω 1 )
where ω 2 =0.037 lbm/lbm and ω 1=0.0043 lbm/lbm are obtained from the psychrometric chart. To
obtain the mass flow rate of air, use the energy balance, equation (7-5) bearing in mind that the right
hand side of that equation is the tower heat load (which is a given parameter in this problem):
ṁair [(h2−h1 )−( ω 2− ω 1)h5 ]= Q̇CT
Q̇ CT
⇒ ṁ air=
(h2 −h1)−( ω 2 − ω 1) h5
where h1=21.1 Btu /lbm , h2=63.2 Btu /lbm , can be obtained from the psychrometric chart and
h5≈h f (70 °F)=38.1 Btu /lbm from the steam tables.
Q̇CT
ṁair =
=
(h 2−h1 )−( ω 2−ω 1 ) h5
600,000
( 63.2−21.1 )
Btu
h
Btu
lbm
Btu
− ( 0.037−0.0043 )
×38.1
lbm
lbm
lbm
=14,686.4
lbm
h
Therefore
ṁmakeup=14,686.4
lbm
lbm
lbm
×( 0.037−0.0043 )
=480.3
h
lbm
h
So, the volume flow rate is (458.9 lbm/ h)÷(8.34 lbm /gal)≈58 gal /h
07-05*. The correct answer is (D).
The cooling tower load is given by equation (7-2):
Q̇ CT= ṁ w c p ,w ( T w,in −T w,out )
which can be combined with equation (7-1) yielding:
Q̇ CT= ṁw c p ,w η CT ( T w,in −T wet bulb, in )
gal
lbm
60 min
Btu
Btu
= 30
×8.34
×
×1
×0.75× ( 95−45.7 ) °F=555,069
min
gal
1h
lbm °F
h
(
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08-01.
Here's a schematic flow diagram for the system:
r
Space
Conditioner
r
Fan
s
o
m
Condensate
c
For states o and r, the psychrometric properties can be read directly from the psychrometric chart. The
values in bold were provided in the problem statement.
Point
T(°F)
ϕ
o
96
43
77
40.3
r
75
55
64
m
79
54
s
58
89
T wb (°F)
ω
h (Btu/lbm)
(lbm/lbm)
v a (ft3/h)
ṁ (lbm/h)
CFM
SCFM
0.0157
14.35
4,822
1,153
1071
29.1
0.0102
13.7
19,286
4,404
4285
67
31.4
0.0113
13.84
24,108
5,557
5,356
56.1
24
0.0094
13.24
24,108
5,320
5,356
We know that the outdoor air mass flow is 20% of the supply air flow. Therefore, first we find the
supply air mass flow:
Q̇ sens= ṁ s c p , a ( T r −T s )
⇒ ṁs =
Q̇ sens
=
c p ,a ( T r −T s )
100,000
0.244
Btu
h
Btu
× ( 75−58 ) °F
lbm °F
=24,108
lbm
h
thus,
ṁo=0.2× ṁs =4,822
lbm
h
Now the volumetric flow rate is found as:
3
V̇ o= ṁo v o=4,822
∣
∣
lbm
ft
1h
×14.35
×
=1,153cfm
h
lbm 60 min
and to find the scfm we use the specific volume at standard conditions:
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3
V̇ o= ṁo vstd =4,822
∣
∣
lbm
ft
1h
×13.33
×
=1,071 scfm
h
lbm 60 min
The mass flow rate of return air mixed with outdoor air is:
ṁr = ṁs − ṁo=24,108−4,822=19,286
lbm
h
and the cfm and scfm for return air are found the same way we just did for outdoor air.
Now for the mixed air. With an energy balance on the mixing box we can get the enthalpy, and with a
moisture mass balance on the mixing box we can get the specific humidity.
Energy balance:
ṁr h r + ṁo ho= ṁ m h m
ṁr
ṁo
Btu
⇒ hm =
h r+
ho=0.8×29.2+0.2×40.2=31.4
ṁm
ṁm
lbm
(this is the same as using equation 4-5).
Moisture mass balance:
ṁr ω r + ṁo ω o= ṁm ω m
ṁr
ṁ
lbm
⇒ ω m=
ω r + o ω o=0.8×0.0102+0.2×0.0157=0.0113
ṁm
ṁm
lbm
(this is the same as using equation 4-6). We can now use a psychrometric chart to read all the other
properties for state m.
For the supply air, we need one more property since we only have the dry bulb temperature. A moisture
mass balance on the space will give us the humidity ratio at s: Since the latent load is 20,000 Btu/h we
can estimate the amount of water absorbed by the air in the space if we assume the typical enthalpy of
vaporization of 1,076 Btu/lbm (re-visit the discussion of the derivation of equation 5-6 on pages 28 –
29)
ṁwater =
Q̇ sens
20,000 Btu / h
=
=18.59 lbm /h
1,076 Btu /lbm 1,076 Btu /lbm
So, the moisture mass balance of the conditioned space is:
ṁs ω s + ṁwater = ṁr ω r
ṁwater
18.59
⇒ ω s =ω r −
=0.0102−
=0.00943 lbm /lbm
ṁ s
24,108
With ω s and T s use the chart to get all other properties.
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08-02. To find the coil load, use equation 5-3
Q̇ coil = ṁs ( h m −hs )+ ṁcond h cond
where it is typically assumed that the condensate leaves the cooling section at the apparatus dew point
temperature, hence hcond =h f (T adp ) and from a water mass balance for the coil:
ṁcond = ṁa ( ω m− ω s )
In HVAC calculations, it is typical to neglect the enthalpy of the condensate when calculating the coil
load. Nevertheless we will not do that here just for illustrative purposes. First, we need to find the
apparatus dew point. We do that with a psychrometric chart, by locating points m and s, and extending
the line that joins them all the way to the saturation line:
m
ADP
s
We thus find T adp ≈53°F hence hcond =h f (53°F)≈21.1 Btu /lbm from a steam table. Therefore,
Q̇ coil = ṁs ( h m −hs )− ṁcond h cond
lbm
Btu
lbm
lbm water
Btu
( 31.4−24 )
( 0.0113−0.00094 )
=24,108
−24,108
×21.1
h
lbm
h
lbm
lbm water
Btu
=177,432
=14.8 tons
h
If we had neglected the effect of the condensate in the calculation above, we would have obtained
178,400 Btu/h or 14.9 tons – a difference of less than 1%. We could also use equation 5-5:
Q̇ coil =4.8×5,356×( 31.4−24 )=178,355
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08-03. The percentage of the coil load that is required for sensible cooling is the sensible heat ratio
which can be determined on the chart by using the protractor. Alternatively we can calculate the
sensible load as (see Section 03):
Q̇sens = ṁs c p , a Δ T =24,108
lbm
Btu
Btu
×0.244
× ( 79−58 ) =123,529
h
lbm °F
h
therefore the percent used for sensible cooling is 123,529/178,355 ≈ 0.69 or 69%. The percent used for
latent cooling is then 31%.
08-04.
As discussed in connection with equation 8-5, the load due to outside air is:
Q̇ oa= ṁa ,o ( ho−h r )=4,822
lbm
Btu
Btu
× ( 40.3−29.1 )
=54,006
h
lbm
h
which is approximately 30% of the coil load of 177,432 Btu/h.
08-05*. The correct answer is (C).
Use equation 8-1 to find the cfm at standard conditions (scfm) to satisfy the sensible load:
CFM sens, summer =
qsens, summer
100,000
=
=2,457 scfm
1.1×( T r −T s ) 1.1×( 76−39 )
Since we are asked for mass flow rate, use the density at standard condition to convert scfm to lbm/h:
3
ft
2,457
V̇ a
min
60 min
lbm
ṁa= =
×
=11,084
3
v std
1
h
h
ft
13.3
lbm
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08-06*. The correct answer is (D)
Air
Chamber
1
2
3
Steam
116,000 Btu/h
The energy balance for the chamber is:
ṁa h1 +Q̇sens + ṁsteam h steam= ṁa h2
For the inlet conditions we can obtain the following from the psychrometric chart:
h1=13.4
Btu
lbm
ω 1=0.0035
lbm
lbm
also hsteam =1,156.9 Btu /lbm is the enthalpy of saturated steam at 230°F. So, we can obtain h2 from the
energy balance:
Q̇sens + ṁsteam hsteam
h2=h1 +
ṁa
Btu
=13.4
+
lbm
=29.1
Btu
lbm
Btu
+ 83
×1,156.9
h
h
lbm
3
ft
1 lbm
60 min
3,000
×
×
min 13.3 ft 3
1h
116,000
)∣
(
Btu
lbm
∣
We can obtain the resulting humidity from a moisture mass balance:
ṁa ω 1+ ṁ steam= ṁa ω 2
ṁsteam
⇒ ω 2= ω 1 +
ṁa
lbm
=0.0035
+
lbm
=0.0096
lbm
h
3
ft
1 lbm
60 min
3,000
×
×
min 13.3 ft3
1h
83
(
lbm
lbm
)∣
∣
With the humidity and enthalpy, we can locate state 2 on the psychrometric chart:
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s
so we find ϕ 2≈48 % .
08-07*. The correct answer is (A).
For the supply air, we only have the dry-bulb temperature. We need one more property to fully define
the state of the supply air and thus being able to determine the dew point. We can find the humidity
ratio of the supply air from a moisture mass balance on the auditorium:
ṁa ω s + ṁwater= ṁa ω r
where
ṁwater=106
∣
∣
grains
1 lbm
lbm
×
=142.9
h
7,000 grains
h
ω r =ω (77°F , 55 % rh)=0.0109
lbm
lbm
and the scfm of air can be estimated from equation 8-1:
ṁa
3
13.3
ft
lbm
=
350,000
lbm
=31,818.2 scfm ⇒ ṁa=423,181.8
h
1.1× ( 77−67 )
Now, return to the moisture mass balance:
ω s= ω r −
ṁwater
142.9
lbm
=0.0109−
=0.00943 lbm/lbm=0.0106
ṁa
423,181.8
lbm
With this humidity ratio and the given value of T s=67 °F we can use a psychrometric chart to
determine the dew point of the supply air as T dp , s ≈59°F
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08-08*. The correct answer is (D)
To determine the coil size we need the mass flow rate, and the supply air conditions (all of this is
given) but we also need the conditions of the air entering the coil, which is the mixture of outdoor air
and return air. We therefore need to find the conditions of return air.
r
Space
Conditioner
r
Fan
o
s
m
c
Condensate
The given mass flow rate of supply air of 30,000 lbm/h is equivalent to 6,650 scfm. Use equation 8-1 to
find the return air temperature:
120,000
( T r −T s )= 1.1×6,650 =16.4 °F ⇒ T r =( 60+16.4 ) °F=76.4 °F
Now, use a psychrometric chart to determine the humidity ratio of supply air:
ω s= ω (60 °F , 80 % rh)=0.0088
lbm
lbm
and equation 8-2 to determine the return air humidity ratio:
( ω r−ω s )=
30,000
lbm
lbm
.
=0.000932
⇒ ω r =( 0.0088+0.000932 ) °F=0.00973
4,840×6,650
lbm
lbm
Since its a 50/50 mixture, the temperature of mixed air will be the simple average of the temperature of
return air and the temperature of outdoor air:
T m =0.5 T r +0.5 T o=.5×76.4+0.5×98=87.2°F
Now, use a psychrometric chart to determine the humidity ratio of outdoor air:
ω o= ω (98 °F ,77 °F wb)=0.0152
lbm
lbm
so, the humidity ratio of mixed air is:
ω m =0.5 ω r +0.5 ω o=0.0125
lbm
lbm
Then the sensible coil load is calculated with equation 8-1:
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Q̇ coil, sens =1.1×6,650× ( 87.2−60 )=198,968
Btu
h
and the latent coil load with equation 8-2:
Q̇ coil, lat=4,840×6,650× ( 0.0125−0.0088 )=119,088
Btu
h
The total coil load is then 318,056 Btu/h or 26.5 tons
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