12.1
SECTION 12
REFRIGERATION
Refrigeration Required to Cool an
Occupied Building
12.1
Determining the Displacement of a
Reciprocating Refrigeration
Compressor
12.4
Heat-Recovery Water-Heating from
Refrigeration Units
12.6
Computing Refrigerating Capacity
Needed for Air-Conditioning Loads
12.12
Water-Vapor Refrigeration-System
Analysis
12.15
Analyzing a Steam-Jet Refrigeration
System for Chilled-Water Service
12.17
Heat Pump and Cogeneration
Combination for Energy Savings
12.21
Comprehensive Design Analysis of an
Absorption Refrigerating System
12.28
Cycle Computation for a Conventional
Compression Refrigeration Plant
12.41

Comparison
12.72
Central Chilled-Water System Design to
Meet Chlorofluorocarbon (CFC) Issues
12.76
REFRIGERATION REQUIRED TO COOL AN
OCCUPIED BUILDING
The building in Fig. 1 is to be maintained at 75
Њ
F (23.9
Њ
C) dry bulb and 64.4
Њ
F
(18.0
Њ
C) wet-bulb temperatures. This building is situated between two similar units
which are not cooled. There is a second-floor office above and a basement below.
The south wall, containing 45 ft
2
(4.18 m
2
) of glass area, has a southern exposure.
On the north side of the building there are two show windows which are ventilated
to the outside and are at outside temperature conditions. Between the show windows
is a doorway. This doorway is normally closed but it is frequently opened and
allows an average of 600 ft
3
/min (16.98 m
3

C). The maximum outside
conditions for design purposes are 95
Њ
F (35.0
Њ
C) dry bulb and 78
Њ
F (25.6
Њ
C) wet-
bulb temperature. There is a 0.5 hp (373 W) fan motor in the interior of the building.
What is the refrigeration load for cooling this building?
Calculation Procedure:
1. Assemble the overall coefficients of heat transfer for the building materials
Using the ASHRAE handbook, find the U values, Btu/ft
2
⅐
h
⅐ Њ
F as follows: East
and west walls (24-in [60.96-cm] brick, plaster one side), 0.16; North partition
(1.25-in [3.18-cm] tongue-and-groove wood), 0.60; Plate-glass door, 1.0; South wall
(13-in [33-cm] brick, plaster one side), 0.25; Windows (single-thickness glass),
1.13; Floor (1-in [2.54-cm] wood, paper, 1-in [2.54-cm] wood over joists), 0.21;
Ceiling (2-in [5.08-cm] wood on joists, lath and plaster), 0.14. To determine the SI
overall coefficient, multiply the given value above by 5.68 to obtain the W/m
2
⅐
Њ
C. Thus, the values are: 0.908; 3.4; 5.68; 1.42; 6.42; 1.19; 0.79, respectively.

U(ft
2
[m
2
] surface area)(temperature difference).
For each of the surfaces in this building the heat leakage is computed thus: East
and west walls
ϭ
(0.16)(1440)(20)
ϭ
4610 Btu/h (1350.7 W); North partition
ϭ
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REFRIGERATION
REFRIGERATION
12.3
(0.6)(299)(20)
ϭ
3590 Btu / h (1051.9 W); Glass door
ϭ
(1.0)(35)(20)
ϭ
700 Btu
/h (205.1 W); South wall
ϭ
(0.25)(300)(20)
ϭ
1500 Btu/h (439.5 W); Windows

5. Find the air leakage heat load
Use the relation: Air leakage heat load, Btu / h (W)
ϭ
(air change, ft
3
/h)(specific
heat of air)(temperature difference)/(specific volume of air, ft
3
/lb). Or
(36,000)(0.24)(95
Ϫ
75)/13.70
ϭ
12,600 Btu/h (rounded off) (3691.8 W).
6. Calculate the sun effect heat load
For the glass on the south wall, the sun effect
ϭ
(45 ft
2
)(30 Btu / h ft
2
)
ϭ
1350
Btu/h (399.6 W). The sun effect on the south wall
ϭ
(300 ft
2
)(0.25)(120
Ϫ

F (35
Њ
C) 75
Њ
F (23.9
Њ
C)
Wet bulb 78
Њ
F (25.6
Њ
C) 64.4
Њ
F (18.0
Њ
C)
Percent relative humidity 47 55
Dew point 71
Њ
F (21.7
Њ
C) 58
Њ
F (14.4
Њ
C)
Grains per lb 114.4 (16,327.0 mg / kg) 71.9 (10,262.1 mg/ kg)
Specific volume — 13.7 cu ft/ lb (0.85 m
3
/kg)

ϭ
4.01
ϩ
1.67
ϭ
5.68 tons (19.98
kW).
Related Calculations. As you can see, if you are to perform repeated calcu-
lations for buildings and rooms, a form listing both the equations and items to be
computed will be helpful in saving you time. The procedure given here is useful
for the occasional computation of buildings of all types: residential, commercial,
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REFRIGERATION
12.4
PLANT AND FACILITIES ENGINEERING
industrial, etc. The same general procedure can be used for trucks, ships, aircraft
and other mobile applications.
DETERMINING THE DISPLACEMENT OF A
RECIPROCATING REFRIGERATION COMPRESSOR
What is the needed displacement of a reciprocating refrigeration compressor rated
at 50 tons (45.4 t) when operating with a refrigerant at 0
Њ
F(
Ϫ
17.8
Њ
C) in the evap-
orator expansion coils? At this temperature, the heat absorbed by the evaporation

2. Find the compressor bore
Let N
ϭ
the compressor speed, rpm; D
ϭ
compressor cylinder bore, ft (m); L
ϭ
compressor stroke, ft (m); V
ϭ
piston displacement, ft
3
(m
3
) per stroke. Then:
23
V
ϭ
0.785DL N
ϫ
V
ϭ
180 ft / min
3
180
ϭ
180/N
ϭ
⁄
180
ϭ

Related Calculations. With the phasing out of chlorofluorcarbons (CFC) be-
cause of environmental restrictions, engineers must be able to evaluate the perform-
ance of alternative refrigerants. The procedure given above shows exactly how to
perform this evaluation for any refrigerant whose thermodynamic and physical char-
acteristics are known by the engineer, or can be obtained from standard data ref-
erence.
While many liquids boil at temperatures low enough for refrigeration, few are
suitable for refrigeration purposes. Those liquids suitable for practical refrigeration
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REFRIGERATION
REFRIGERATION
12.5
(a)
(b)
SI Values
psi kPa F C
20 137.8 5.5 –14.7
185 1274.7 235 112.8
Insulation
Power saved
due to jacket
cooling
Superheat pickup
Controlled
superheat
Expansion valve
Low side High side
20 psi

raises its boiling point. Reducing the pressure on a refrigerant lowers its boiling
point. Refrigeration occurs when a refrigerant boils at a low temperature, permitting
heat flow from an item or area to be cooled to the refrigerant. Boiling of the
refrigerant takes place in the evaporator which is located in the area to be cooled,
Fig. 2a.
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REFRIGERATION
12.6
PLANT AND FACILITIES ENGINEERING
When the boiling refrigerant removes sensible heat from environment at a rate
equivalent to the melting of one ton (2000 lb; 980 kg) of water ice in 24 h, the
rate of heat removal is a ton (3.516 kW) of refrigeration. Since the heat of melting
(sublimation) of 1 lb (0.454 kg) of water ice is 144 Btu (151.9 kJ), a ton of
refrigeration is equivalent to 2000 lb (144 Btu/24h)
ϭ
12,000 Btu / h (3516 W, or
3,516 kW).
By comparison, the heat of sublimation (melting) of dry ice (CO
2
) is 275 Btu/lb
(640.8 kJ/ kg). To say that a refrigeration machine has a capacity of 10 tons (35.16
kW) is to say that the rate of refrigeration is 10
ϫ
200
ϭ
2000 Btu/min (35.16
kW). Note that 1 ton of refrigeration equals a rate of 200 Btu / min (3.516 kW).
To determine the amount of refrigerant that must be circulated, divide the re-

emissions because of reduced electrical power needed to run them. Payback time
for the new machines will be less than 2.5 years because the energy savings are so
significant. Several examples of typical piping arrangements for reciprocating re-
frigeration compressors and chillers are shown in Fig. 3 through Fig. 6.
Another example of using this procedure is substitution of natural-gas fueled
engine drives for refrigeration chillers using new refrigerants. In one department
store installation of such a chiller the estimated annual energy cost savings are
$54,875. Such drives reduce electric demand charges, are compact in size, are
environmentally friendly, and are used in hospitals, nursing homes, schools, col-
leges, office buildings, retail, and industrial / process facilities. Some of the newest
centrifugal chillers on the market report a required input of just 0.20 kW / ton when
operating at 60 percent load with entering condenser water at 55
Њ
F (12.8
Њ
C).
HEAT-RECOVERY WATER-HEATING FROM
REFRIGERATION UNITS
How much heat can be obtained from heating water for an apartment house having
150 apartment units served by two 200-ton (180 t) air-conditioning units if a 70
Њ
F
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REFRIGERATION
REFRIGERATION
12.7
FIGURE 3 Layout of suction and hot-gas lines for multiple-compressor
operation. (Carrier Corporation.)

ventional water-cooled condenser, can transfer 25 to 35 percent of the Btu (kJ)
rating of the refrigeration compressor, i.e., the air-conditioning unit’s rating. Using
the lower value in this range for these units gives, Heat Available
ϭ
0.25 (200)
(12,000 Btu/h/ton)
ϭ
0.25 (200)(12,000)
ϭ
600,000 Btu / h (633,000 kJ/h). With
35 percent, Heat Available
ϭ
0.35 (200)(12,000)
ϭ
840,000 Btu/ h (886,200 kJ/
h). With two refrigerating units the heat available would be double the computed
amount, or 1,200,000 Btu/h (1,266,000 kJ / h) and 1,680,000 Btu/h (1,772,400 kJ
/h).
2. Find the hourly water heating rate for the system
Water weighs 8.34 lb / gal (1.02 kg / L). To raise the temperature of one pound of
water (0.454 kg) 1
Њ
F (0.55
Њ
C) requires a heat input of 1 Btu (1.055 kJ). With the
specified 70
Њ
F (38.9
Њ
C) water-temperature rise required in this building, the rate of

3. Compute the daily total gallonage of hot water produced
Since air-conditioning refrigeration units operate varying numbers of hours per day,
depending on the outside weather conditions, the gallonage of hot water available
from heat recovery will vary. For the range of operating hours specified, we have,
per 200-ton (180-t) unit:
Gallonage available (L) per hours of operation
81216
25 percent 8232 (30,458) 12,348 (45,688) 16,464 (62,399)
35 percent 11,520 (42,624) 17,280 (65,491) 23,040 (87,322)
Again, we double these numbers for two units in the building.
Related Calculations. The normal discharge temperature of modern refriger-
ants makes heat recovery for domestic and/ or process water heating an attractive
option, especially in an environmentally conscious world. Further, heat recovery is
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REFRIGERATION
REFRIGERATION
12.11
FIGURE 8 Layout of heat-recovery heat-exchanger piping for refrigeration cycle.
such a simple design challenge that the investment is often recovered in fuel savings
in less than two years.
For maximum efficiency, the designer must try to match hot-water needs with
the operating time of the refrigeration unit. If there is a disparity between these
two variables, a sufficiently large water storage tank can be designed into the system
to store hot water during times when the refrigeration unit is not operating. Such
a storage tank is not expensive and it will have a long life if properly maintained.
Heat recovery for heating incoming cold water can be used in office buildings,
apartment houses, hotels, motels, factories, and commercial buildings wherever a
need for hot water exists and a refrigeration unit of some kind operates in the

85
Њ
F (29.4
Њ
C); wet-bulb temperature is 70
Њ
F (21.1
Њ
C). Condensate leaves the de-
humidifier at the same temperature as the outgoing air, which is at standard at-
mospheric pressure throughout the system. Air leaves the dehumidifier in the sat-
urated state. Determine the refrigerating load for this air-conditioned space.
Calculation Procedure:
1. Draws sketches of the system arrangement and a skeleton psychrometric
chart of the processes
Figures 9 and 10 show the system and the psychrometric chart with important state
points identified by number. Show the values of the enthalpies and humidities at
the various state points before attempting to make the air-conditioning design.
Thus, at point 3,
v
m
ϭ
13.39 ft
3
/lb of dry air (0.83 m
3
/kg). Then, the weight
flow rate of air, M
a
ϭ

ƒc
ϭ
51
Ϫ
32
ϭ
19 Btu/lb of water (44.2 kJ / kg).
3. Set up an energy balance about the dehumidifier
The energy balance for an adiabatic mixing process is given by M
1
(h
m1
)
ϩ
M
2
(h
m2
)
ϭ
M
3
(h
m3
), where the M values are the weight flow rates in the respective air-vapor
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REFRIGERATION
REFRIGERATION

Ϫ
(0.00436)(19)
ϭ
13.03 Btu/lb (30.36
kJ/kg). Then, the total heat removed at the dehumidifier
ϭ
(13.03)(746)
ϭ
9740
Btu/min (171.2 kW).
4. Find the refrigeration capacity required
Since 200 Btu/min
ϭ
1 ton of refrigeration, the refrigeration capacity required
ϭ
9740/200
ϭ
48.7 tons (43.6 t).
Related Calculations. This procedure could have been performed using the
pressure and humidity relations of the air streams. Approximately the same result
would have been obtained. By considering the enthalpy of the condensate the es-
timated refrigeration load is decreased by about 0.2 ton (0.18 t), which for this
installation is insignificant. The tonnage reduction will always be small when the
final dew point is low, and in most cases it can be neglected. Further, the true state
of which the condensate is removed is usually difficult to establish. Neglecting the
energy of the condensate will give a refrigeration capacity requirement that is on
the safe side, i.e., a larger value than would be arrived at with the exact solution.
Depending on the number of banks of spray nozzles, the direction of the spray,
and the air velocity, the percentage of untreated air passing through an air washer,
Fig. 12, may range from 5 to 35 percent for two banks and one bank of nozzles,

F (7.2
Њ
C). The makeup and recirculated water are at 60
Њ
F
(15.6
Њ
C), and the quality of the vapor leaving the evaporator and entering the com-
pressor is 0.97. Find: (a) the capacity of the machine, tons; (b) the pounds (kg) of
vapor to be removed from the evaporator per minute; (c) the volume of vapor to
be removed from the evaporator, ft
3
/min (m
3
/min) and ft
3
/(min
⅐
ton) (m
3
/t). Fur-
ther, consider that the machine in Fig. 13 is equipped with a centrifugal compressor
and mechanical vacuum pumps. The condenser pressure is 2 in Hg (5.1 cm) abs.
Compression efficiency is 0.65. Mechanical efficiency of the compressor is 98 per-
cent. The condensate leaves the condenser at 90
Њ
F (32.2
Њ
C). Power required to drive
the condensate and air pumps is 6 percent of the total power input. Find: (d) the

2087.5(15)
ϭ
31,312 Btu/min
ϭ
156.6 tons (140.9 t)
(b) Vapor removed, M
3
ϭ
31,312/1049.4
Ϫ
28.06
ϭ
30.66 lb/min (13.9 kg/ min)
(c) Vapor volume at compressor suction
ϭ
30.66(1975)
ϭ
60,550 ft
3
/min (1714
cu
3
/min) or 387 ft
3
/min/ton (12.2 m
3
/t)
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h
3
(276.3 kJ/kg)
h
Ј
4
Actual
⌬
h
ϭϭ
182.5 Btu/lb
ϭ
h
4
Ϫ
h
3
(425.2 kJ/kg)
118.6
0.65
h
4
ϭ
1049.4
ϩ
182.5
ϭ
1231.9 (2 in Hg and 380
Њ
F) (2870.3 kJ / kg) (5.08 cm

ϭϭ
5.17 (based on total hp)
200
0.914(42.4)
See Fig. 14 for a typical water-vapor refrigeration system using steam ejectors
and a surface condenser. Figure 15 shows a water-vapor refrigeration system using
a steam-jet (barometric condenser).
Related Calculations. The rather high coefficient of performance (COP) of this
machine is caused by the comparatively favorable temperature range through which
the machine operates. With today’s environmental concern over safer refrigerants,
water has a 0.00 ozone-depletion potential. Further, water also has a 0.00 immediate
global warming potential, an a 0.00 100-year global warming potential. And since
such refrigeration systems can be gas-fired, they have less potential atmospheric
pollution compared to coal or oil. Likewise, water has zero flammability.
Air cooling is one industrial application of refrigeration that does not ordinarily
require, in the refrigerating sense, low temperatures. Hence, the water-vapor refrig-
erating system is ideal for such applications. Further, water is a cheap refrigerant,
and it is truly a safe one, and as a result, it is finding greater use in industry today,
especially in view of the new, stricter environmental regulations that seem to be
imposed every year.
This procedure, and Figs. 13 through 15, are the work of Norman R. Sparks,
Professor and Head, Department of Mechanical Engineering, The Pennsylvania
State University, and Charles C. DiLio, Associate Professor of Mechanical Engi-
neering, The Pennsylvania State University. SI values were added by the handbook
editor.
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REFRIGERATION
REFRIGERATION

ϭ
90 percent. Entrainment efficiency
ϭ
65 percent. Diffuser efficiency
ϭ
75 percent. Steam consumption of the auxiliary ejectors is 6 percent of the total
steam requirement. Neglecting the power demands of the water pumps, find: (a)
Steam consumption of the main ejector; (b) the total steam consumption; (c) heat
rejected in the primary condenser.
Calculation Procedure:
1. Determine the pertinent steam enthalpies and quality
Referring to Fig. 16 for symbols, we have
h
A
ϭ
1203.5 at 140 lb/in
2
(abs), 370
Њ
F (2804 kJ/kg at 964.6 kPa, 187.8
Њ
C)
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REFRIGERATION
12.18
PLANT AND FACILITIES ENGINEERING
FIGURE 15 Steam-jet (barometric condenser), water-vapor refrigeration system.
(Ingersoll-Rand Co.)

x
b
ϭ
0.97; h
b
ϭ
1049.4 (2445.1 kJ / kg)
2. Compute M
a
and test its accuracy as a first trial
First trial:
Assume x
c
ϭ
0.87, for which h
c
ϭ
943 (2197.2 kJ / kg)
Then
ϭ
1043 (after isentropic compression to 2 in Hg) (2430.2 kJ/kg at 5.08h
Ј
d
cm Hg)
Ϫ
h
c
ϭ
100 Btu; h
d

d
; M
A
ϭ
h
Ϫ
h
db
h
Ϫ
h
Ad
1076.3
Ϫ
1049.4
M
ϭϭ
0.211 lb/ lb evaporator vapor (0.096 kg/ kg)
a
1203.5
Ϫ
1076.3
From e
e
M
a
ϭ
(M
a
ϩ

must be increased appreciably.
3. Make a second computation of M
a
Second trial:
Assume h
c
ϭ
1000 Btu/lb (2330 kJ / kg)
ϭ
1108;
Ϫ
h
c
ϭ
108; h
d
Ϫ
h
c
ϭϭ
144 (335.5 kJ / kg)
108
h
Ј
h
Ј
dd
0.75
h
d

c
and the value of M
a
derived from this second trial are considered correct.
4. Calculate the system variables
(a) Data, as necessary, have been taken from the previous procedure.
Evaporator vapor
ϭ
30.66 lb/min
Ejector steam consumption
ϭ
1.59(30.66)
ϭ
48.7 lb / in, or 2920 lb/h
(1325.7 kg/h)
Steam rate
ϭϭ
0.311 lb/(min
⅐
ton), or 18.7 lb/(h
⅐
ton) [9.43 kg/(h
48.7
156.6
⅐
t)]
(b) Total steam consumption
ϭ
2920/0.94
ϭ

,or,in
this instance, 1203.5
Ϫ
58
ϭ
1145.5 Btu/lb (2669 kJ / kg). The ratio of the refrig-
erating effect to the heat supplied to produce that effect is therefore 12,000 /
19.9(1145.5)
ϭ
0.526.
Related Calculations. Like the absorption system, the steam-jet water-vapor
cycle requires but a very small portion of mechanical energy for its operation. A
coefficient of performance (COP) may not, therefore, be truly expressed for this
type of machine. But a pseudocoefficient can be arbitrarily devised for the purpose
of comparison with other types of systems.
While there is no standard procedure for setting up a pseudocoefficient, if the
motive steam were used in, say, a turbine to drive a compressor, about 50 percent
of the energy available in expanding to condenser pressure might be considered as
the turbine work output and the compressor input. For this procedure, the energy
available per pound (kg) of motive steam between 140 lb/in
2
(abs) (964.6 kPa),
370
Њ
F (187.8
Њ
C) and 2 in (5.08 cm) Hg abs is 317.5 Btu/lb (739.8 kJ / kg). The
coefficient of performance would thus be, according to this standard, 12,000 / [19.9
(0.5)(317.5)]
ϭ

70° Air
70° Air
Condenser
68° Air
Compressor
Motor
Ground level
FIGURE 17 Schematic of heat pump used to heat a building.
est. This may be closely approximated by the application of the principles given
in this, and the preceding, calculation procedures.
This procedure is the work of Norman R. Sparks, Professor and Head, Depart-
ment of Mechanical Engineering, The Pennsylvania State University, and Charles
C. DiLio, Associate Professor of Mechanical Engineering, The Pennsylvania State
University. SI values were added by the handbook editor.
HEAT PUMP AND COGENERATION
COMBINATION FOR ENERGY SAVINGS
In the heat-pump system shown in Fig. 17, the building requires 100,000 Btu/ h
(29.3 kW) for heating. Calculate for this system: (a) The capacity of the heat pump,
tons (t) as a refrigerating machine; (b) the motor horsepower (kW); (c) the hourly
energy input to the motor; (d) the overall coefficient of performance; (e) the heat
received from the outside air; (ƒ) the electrical consumption per hour for direct
heating; (g) the coal consumption for direct heating with a 50 percent furnace
efficiency; (h) the oil consumption for direct heating with a 65 percent furnace
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REFRIGERATION
12.22
PLANT AND FACILITIES ENGINEERING
efficiency; (i) the natural-gas consumption for direct heating with a 70 percent

4970 Btu/(h
⅐
ton) (1.62 kW/t)
4325
0.87
The motor loss is therefore 4970
Ϫ
4325
ϭ
645 Btu / (h
⅐
ton) [0.21 kW / h
⅐
t)]
which may be considered as applied to direct heating.
Heat output of condenser (and cylinder jacket) per ton
ϭ
12,000
ϩ
4325
ϭ
16,325 Btu/h (4.8 kW)
Total heat output of machine per ton
ϭ
16,325
ϩ
645
ϭ
16,970 Btu/h (4.97
kW)

ϭ
100,000/3412
ϭ
29.3 kW
(g) Coal for direct heating
ϭ
100,000/0.5(14,000)
ϭ
14.3 lb/h (6.5 kg / h)
(h) Oil for direct heating
ϭ
100,000/0.65(19,000)
ϭ
8.1 lb / h or 8.1/0.88(8.33)
ϭ
1.1 gal/h (4.2 L/h)
(i) Natural gas for direct heating
ϭ
100,000/0.7(1100)
ϭ
130 ft
3
/h (12.1 m
3
/h)
(j) Manufactured gas for direct heating
ϭ
100,000/0.7(600)
ϭ
238 ft

Assuming that 80 percent of the waste can be recovered, a typically safe as-
sumption, the heat supplied to the building per ton (t) capacity is 16,325
ϩ
0.80
(10,210)
ϭ
24,495 Btu/h (7.17 kW), using data from the steps above in this pro-
cedure. Then:
Capacity required
ϭ
100,000/24,495
ϭ
4.08 tons (3.67 t)
Power required
ϭ
4.08(1.7)
ϭ
6.94 hp (5.2 kW)
Fuel required
ϭ
6.94(0.45)
ϭ
3.12 lb/h or 0.425 gal/h (1.6 L/h)
Coefficient of performance, based on shaft work
ϭ
100,000/6.94(2545)
ϭ
5.65
Related Calculations. Relative fuel and electrical energy costs can be obtained
for any given locality from the utility serving the area, and with the deregulation

x
and 0.1 g / bhp
⅐
h, to go into effect by 2004. The rare-earth
additive will increase fuel cost by 2 to 4 percent in the U.S.
For winter operation of this heat pump, a humidifier would be used. Provision
would be made for ventilation by admission of the required amount of fresh air to
be conditioned and mixed with that which is recirculated. For summer use, the
condenser and evaporator in Fig. 17 would be so connected that they would
exchange functions, the summer evaporator acting as an air cooler and dehumidifier.
Heat removed from the inside air would be rejected, along with the heat equivalent
of the compressor work, to the atmosphere or to the water which acts as the source
of heat for cold-weather operation, and as the cooling medium in warm weather.
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REFRIGERATION
12.24
PLANT AND FACILITIES ENGINEERING
FIGURE 18 Conventional unitary heat-pump system heats, cools, and dehumidifies at terminal
unitary heat pumps (UHPs). (Gershon Meckler, P.E., and HPAC magazine.). See procedure for
SI values.
Unitary heat pumps (UHPs) using closed-water loops are used extensively today
in office buildings because of their flexibility. As described by Gershon Meckler,
P.E., in HPAC magazine,* UHPs are small, individual air conditioners located and
controlled within each building zone, that dehumidify and cool or heat at the ter-
minal without activating a central chiller or other UHPs. Best of all, they can be
switched back and forth between heating and cooling, and some can be heating
while others are cooling. Figure 18 shows a conventional unitary heat-pump system.
To retain the benefits of UHPs, while making them more responsive to the new

Cooling tower
(380 nominal tons)
Typical
fire
hose
station
Fire annunciator
To UHPs and sprinkler heads
Plate-and-frame
heat exchangers
Desiccant enthalpy
exchange wheel
Run-around coil
Exhaust fan-1
Outside air
Exhaust
Primary AHU
Supply air Supply air
Dry
primary air
Ice
Supply fan-1
25,000-cfm
Run-around coil
Ice thermal storage/energy transfer subsystem
P-1A, B
P-3A, B
P-2A, B
P-4
4 in.