Tài liệu Handbook of Mechanical Engineering Calculations P12 pdf

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

Design of a Compound Compression-
Refrigeration Plant with Water-Cooled
Intercooler
12.45
Analysis of a Compound Compression-
Refrigeration Plant with a Water-
Cooled Intercooler and Liquid Flash
Cooler
12.47
Computation of Key Variables in a
Compression Refrigeration Cycle with
Both Water- and Flash-Intercooling
12.50
Refrigeration System Selection
12.54
Selection of a Refrigeration Unit for
Product Cooling
12.56
Energy Required for Steam-Jet
Refrigeration
12.62
Refrigeration Compressor Cycle Analysis
12.64
Reciprocating Refrigeration Compressor
Selection
12.68
Centrifugal Refrigeration Machine Load
Analysis
12.71
Heat Pump Cycle Analysis and

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

) of outside air to be admitted. Opening
of the door by customers will cause slightly more than two air changes per hour
in the building. The number of persons in the building is 35; lighting is 1100 watts
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
12.2
PLANT AND FACILITIES ENGINEERING
SI Values
ft m
48 14.6
15 4.6
23 7.0
Display
window
Display
window
N
48 ft 0 in.
23 ft 0 in.
15 ft 0 in. high
D R
Cloak room T T
Down
FIGURE 1 Plan of building cooled by refrigeration.
on a sunny day. Basement temperature is 80
Њ
F (26.7
Њ

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.

2. Compute the temperature differences for the walls, ceiling, and floor
For the walls and ceiling the temperature difference
ϭ
outside design temperature
Ϫ
indoor design temperature
ϭ
95
Ϫ
75
ϭ
20
Њ
F (36
Њ
C). For the floor, the temper-
ature difference
ϭ
80
Ϫ
75
ϭ
5
Њ
F(9
Њ
C).
3. Calculate the heat flow into the building
The heat leakage for any surface
ϭ

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

ϭ
(1.13)(91)(20)
ϭ
2060 (603.5 W); Floor
ϭ
(0.21)(1104)(5)
ϭ
1160 (339.9W);
Ceiling
ϭ
(0.14)(1104)(20)
ϭ
3090 (905.4 W). Summing these individual heat
leakages gives 16710 Btu/h (4896 W).
4. Determine the sensible heat load
Using the conventional heat loads for people, lights, motor hp, the sensible heat
load is: Occupants
ϭ
(35)(300)
ϭ
10,500 Btu/h (3076.5 W); Lights
ϭ
(1100)(3.413)
ϭ
3760 Btu/h (1101.7 W); Motor horsepower
ϭ
(0.5)(2546)
ϭ
1275
Btu/h (373.6 W).

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

95)
ϭ
1875 Btu/h (549.4 W).
7. Find the dry tons (W) of refrigeration required
Sum the heat gains computed above thus: 16,710
ϩ
10,500
ϩ
3760
ϩ
1275
ϩ
12,600
ϩ
1350
ϩ
1875
ϭ
48,070 Btu / h (14,084.5 W)
ϭ
grand total heat loss. The
dry tons (W) of refrigeration is then found from 48,070 / 12,000 Btu / ton
ϭ
4.01
tons (14.1 kW).
8. Evaluate the moisture latent heat load
List the air leakage conditions thus:
Conditions Outside air Inside air
Dry bulb 95
Њ

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)

First, determine the pounds (kg) of air per hour from (ft
3
/h)(outside air gr/lb
Ϫ
inside air gr / lb) / (specific volume of air)(7000 gr/lb). Or (36,000)(114.4
Ϫ
71.9)/
(13.7)(7000)
ϭ
15.95 lb/h (7.24 kg / h).
The, the air latent heat
ϭ
(lb/h)(latent heat of air, Btu/lb)
ϭ
(15.95)(1040)
ϭ
16,588 Btu / h (4860.3 W). Person load
ϭ
(35)(100)
ϭ
3500 Btu / h (1025.5 W).
Total latent heat
ϭ
16,588
ϩ
3500
ϭ
20,088 Btu/ h (5885.8 W), or 20,088/12,000
ϭ
1.67 tons. Then, the total refrigeration load

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

of 1 lb (0.45 kg) of the refrigerant is 500 Btu (527.5 kJ) refrigerating effect, and
the specific volume is 9 ft
3
/lb (0.56 m
3
/kg). The vapor enters the compressor in
the saturated state. If the compressor speed of rotation is 180 r/min, and the stroke
is 1.2
ϫ
bore, what is the bore and stroke of this single-acting compressor?
Calculation Procedure:
1. Determine heat absorbed and compressor displacement
The heat to be absorbed (tons of refrigeration)(heat equivalent of 1 ton of refrig-
eration, Btu / min). Or the heat absorbed
ϭ
50(200 Btu/min/ton refrigeration)
ϭ
10,000 Btu/min (10,550 kJ / min).
The compressor displacement
ϭ
(heat absorbed)(specific volume of the
refrigerant)/(refrigerating effect). Substituting for this refrigerant, displacement
ϭ
10,000(9)/(500)
ϭ
180 ft
3
(5.09 m
3
).

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

1ft
23
L
ϭ
1.2DV
ϭ
0.785D
ϫ
1.2D
ϭ
1ft
Rearranging and transposing, we see that D
ϭ
(1/1.2)(0.785)
0.333
ϭ
1.02 ft (0.31
m), or 12.24 in.
3. Find the compressor piston stroke
Using the relation in step 2, L
ϭ
1.2D
ϭ
1.2
ϫ
1.02
ϭ
1.224 ft (0.373 m), or
14.69 in. The bore and stroke as computed here are typical for a compressor of
this capacity.

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

5.5°F
185 psi
235°F
Evaporator Condenser
Compressor
(heat pump)
C
B
A
F
F'
D
B'
D'
E'
F
F'
E
B
A
DT
S
C
G
E or E'
D or D' 185 psi
96°F
FIGURE 2 (a) Typical reciprocating-compressor refrigeration cycle. (b) T-S plot of
refrigerant cycle.
applications are termed refrigerants. For any refrigerant, increased pressure on it

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-

frigerating effect of the refrigerant in But/lb (kg) into 200 Btu / min (W). Thus,
with a refrigerating effect 25 Btu/lb (58.3 kJ/kg), the quantity of refrigerant to be
circulated is 200/24
ϭ
4 lb/min (1.82 kg / min).
The work of compression is the amount of heat added to the refrigerant during
compression in the cylinder or rotary compressor. It is measured by subtracting the
heat content of 1 lb (0.454 kg) of refrigerant at the compressor suction conditions,
point F
Ј
, F, or A in Fig. 2b from the heat content of the same pound (kg) at the
compressor discharge conditions, point B or B
Ј
in Fig. 2b.
The theoretical horsepower (kW) requirements of a refrigeration compressor can
be found by multiplying the work compression in Btu/ lb (kJ / kg) by the pounds
(kg) of refrigerant circulated in one hour, and dividing this product by 2545 Btu/
hp-h, hp
t
ϭ
(work of compression)(refrigerant circulated) / 2545. Multiply by 0.746
to obtain theoretical kW input.
A good example of the practical value of this calculation is in the recent real-
life example of the upgrading of the HVAC system in a 400-unit apartment com-
plex. Two older refrigerating machines using CFC-refrigerants were replaced by
two new chillers using HCFC-123 refrigerant.
The older machines required an input of 0.81 kW per ton of cooling capacity
(refrigeration) while the newer machines require only 0.55 kW per ton. Annual
savings of more than 30 percent in energy costs for refrigeration are expected with
the new machines and refrigerant. The new machines also reduce greenhouse gas

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.)

FIGURE 4 Interconnecting piping for multiple condensing reciprocating refrigeration units.
(Carrier Corporation.)
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REFRIGERATION
12.8
PLANT AND FACILITIES ENGINEERING
FIGURE 5 Piping arrangement for two centrifugal coolers in series arrangement. (Carrier Cor-
poration.)
(38.9
Њ
C) temperature rise of the incoming cold water is required? Determine the
number of gallons (L) of water that can be heated per hour and the total gallonage
(L) of heated water that can be delivered with the air-conditioning units operating
8, 12, and 16 hours per day.
Calculation Procedure:
1. Determine the quantity of heat available
Heat is available from the high-pressure gas at the refrigeration compressor dis-
charge. This is valid regardless of the type of compressor used: reciprocating, rotary,
or centrifugal. The quantity of heat available from a specific compressor depends
on the outlet-gas temperature, gas flow rate, and the efficiency of the heat exchanger
used.
To recover heat from the hot gas, a heat exchanger, Fig. 7, is placed in the
compressor discharge line, ahead of the regularly used condenser. Cold water from
either the building’s outside water supply line, or from the building’s heated-water
storage tank, is pumped through the heat exchanger in the compressor discharge
line. Leaving the heat exchanger, the heated water returns to the hot-water storage
tank.
Experience shows that a typical well-designed heat exchanger, such as a con-

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

water heating will be: gal/h (L/h)
ϭ
(heat available)/(lb/gal)(temperature rise re-
quired).
With 25 percent heat transfer, we have, gph heated
ϭ
(600,000)/(8.33)(70)
ϭ
1029 gal / h (3900 L / h). And with 35 percent heat transfer, gph
ϭ
(840,000)/
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REFRIGERATION
12.9
FIGURE 6 (a) Schematic of bypass piping for absorption-type refrigerating unit used with a cooling tower. (b) Schematic of bypass piping
for absorption-type refrigerating unit with a central water source. (Carrier Corporation.)
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REFRIGERATION
12.10
PLANT AND FACILITIES ENGINEERING
FIGURE 7 Heat-recovery heat-exchanger location in a refriger-
ation cycle.
(8.33)(70)
ϭ
1440 gal / h (5458 L / h). Again, with two units the hourly heating rate
will be doubled, or 2058 gal/h (7800 L / h) and 2880 (10,915 L/h).

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

structure. Energy savings can range from 25 to 100 percent of the cost of heating
water for either domestic or process uses. A pump, Fig. 8, is used to force-circulate
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REFRIGERATION
12.12
PLANT AND FACILITIES ENGINEERING
FIGURE 9 Basic dehumidifier consisting of
cooler followed by heater.
water through the hot-gas heat exchanger while the refrigeration unit is operating.
This action recovers heat when it is available and stores the heated water for future
use.
A regulating valve, Fig. 8, controls water flow and the temperature of the water
leaving the heat exchanger. The valve ensures that maximum heat recovery will
occur. When designing the piping for heat recovery, be certain to arrange to have
the coldest water enter the heat exchanger. Then maximum benefit will be obtained
from the heat recovery.
COMPUTING REFRIGERATING CAPACITY
NEEDED FOR AIR-CONDITIONING LOADS
An air-conditioned space is supplied 10,000 ft
3
/min (283 m
3
/min) of air at 65
Њ
F
(18.3
Њ
C) and 60 percent relative humidity. Air is supplied to the dehumidifier at

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

flow rate, ft
3
/min/specific volume
ϭ
10,000/13.39
ϭ
746
lb dry air per min (338.7 kg/min).
2. Compute the quantity of moisture condensed and the air dew point
Moisture condensed
ϭ
(moisture content of incoming air
Ϫ
moisture content of
leaving air) / 7000 gr / lb
ϭ
(86
Ϫ
55.5)/7000
ϭ
0.00435 lb/lb dry air (0.00197
kg/kg). Figure 11 shows the psychrometric process. The dew point, from the chart,
t
2
ϭ
51
Њ
F (10.6
Њ
C); enthalpy h

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

12.13
FIGURE 10 Dehumidification process on skeletonized psychrometric chart.
mixtures, and the h values are the respective enthalpies. Then, the energy relation
for the cooler is Q
R
ϭ
(h
m1
Ϫ
h
m2
)
Ϫ
(W
v
1
Ϫ
W
v
2
)h
ƒc
where the symbols are as
given above.
Substituting, Q
R
ϭ
(33.96
Ϫ
20.85)

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,

respectively, according to ASHRAE. For a coil-type dehumidifier the percentage of
untreated air may range from 2 to 39 percent for coils eight rows to one row deep
operating with a face velocity of 300 ft/min (91.4 m/min). In general, the refrig-
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REFRIGERATION
12.14
PLANT AND FACILITIES ENGINEERING
FIGURE 11 Values for the dehumidification process.
FIGURE 12 Schematic of complete air-conditioning system for evaluating variables in design.
eration power requirements of an air-conditioning system are dependent upon the
heat-transfer characteristics of the dehumidifier used.
This procedure, and Figs. 9 through 12, 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
12.15
FIGURE 13 Schematic of water-vapor refrigeration system.
WATER-VAPOR REFRIGERATION-SYSTEM
ANALYSIS
A water-vapor refrigerating machine, Fig. 13, is to produce 250 gal/min (15.8 L/
s) of chilled water at 45
Њ

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

compressor hp (kW) and hp / ton (kW/t); (e) the heat rejected in the condenser;
Btu/min (kJ/min); (ƒ) the coefficient of performance.
Calculation Procedure:
1. Draw a sketch of the system showing the important state points
Figure 13 shows a schematic of the system and the state points.
2. Compute the rate of refrigerant flow
Using Fig. 13,
33
h
ϭ
28.06
ϭ
h ; h
ϭ
13.06; h
ϭ
1049.4;
v ϭ
1975 ft / lb (123.1 m / kg)
1 M 23 3
M
ϭ
250(8.35)
ϭ
2087.5 lb/ min (947.7 kg/min)
2
3. Calculate refrigerating effect, vapor removed, and vapor volume at the
compressor suction
(a) Refrigerating effect
ϭ

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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REFRIGERATION
12.16
PLANT AND FACILITIES ENGINEERING
4. Find the actual change in enthalpy
h
3
ϭ
1049.4; s
3
ϭ
2.0794; h
5
ϭ
58 (135.1 kJ / kg)
After isentropic compression to 2 in Hg,
ϭ
1168 (2721.4 kJ / kg)
h
Ј
4
Isentropic
⌬
h
ϭ
1168
Ϫ
1049.4
ϭ
118.6 Btu/lb
ϭϪ

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

Hg and 193.3
Њ
C)
5. Compute the compressor and total power input
ϭ
30.66(182.5)
ϭ
5595 Btu/min (5902.7 kJ / min)
W
i
J
Compressor hp
ϭϭ
134.6, or 0.859 hp/ton (0.71 kW/t)
5595
42.4(0.98)
Total hp
ϭϭ
143.2 or 0.914 hp/ton (0.758 kW/t)
134.6
0.94
6. Determine the condenser heat rejection and COP
Q
ϭ
30.66(1231.9
Ϫ
58)
ϭ
35,992 Btu/min, or 230 Btu/min/ton (269.6 kJ/t)
Coefficient of performance

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

12.17
FIGURE 14 Typical water-vapor refrigeration system using steam ejectors and surface condenser.
(Ingersoll-Rand Co.)
ANALYZING A STEAM-JET REFRIGERATION
SYSTEM FOR CHILLED-WATER SERVICE
A steam-jet water-vapor refrigerating machine is to produce 250 gal/ min (15.8 L/
s) of chilled water at 45
Њ
F (7.2
Њ
C). 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 ejector
is 0.97. Condenser pressure is 2 in (5.08 cm) Hg abs. Condensate leaves the con-
denser at 90
Њ
F (32.2
Њ
C). The motive steam is supplied at 140 lb/in
2
(abs) (946 kPa),
370
Њ
F (187.8
Њ
C). System manufacturer supplies the following efficiencies: Nozzle
efficiency

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.)

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REFRIGERATION
REFRIGERATION
12.19
FIGURE 16 Steam-jet compressor.
ϭ
803 (after isentropic expansion to 0.3 in Hg) (1871 kJ / kg) (0.76 cm Hg)h
Ј
a
h
A
Ϫϭ
400.5 Btu/lb motive steam (933.2 kJ/kg)h
Ј
a
h
A
Ϫ
h
a
ϭ
0.9(400.5)
ϭ
360.5 Btu/lb
ϭ
KE
a
(839.9 kJ/kg)

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

Ϫ
h
c
ϭϭ
133.3 Btu/lb (310.6 kJ / kg)
100
h
Ј
d
0.75
h
d
ϭ
943
ϩ
133.3
ϭ
1076.3 Btu/lb (2507.8 kJ / kg)
From M
A
h
A
ϩ
h
b
ϭ
(M
A
ϩ
1)h

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

1) e
e
M
a
KE
a
ϭ
(M
a
ϩ
1)KE
c
22
VV
ac
;
2gJ 2gJ
0.65(0.211) (360.5)

(0.211
ϩ
1) 133.3
49.5

161.5
Examination of this last equation and the results of this first trial indicate that
M
a
is too low. Therefore, h
c

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

ϭ
1000
ϩ
144
ϭ
1144 (2665.5 kJ / kg)
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REFRIGERATION
12.20
PLANT AND FACILITIES ENGINEERING
1144
Ϫ
1049.4
M
ϭϭ
1.59 lb/ lb evaporator vapor (0.72 kg/ kg)
a
1205.5
Ϫ
1144
0.65(1.59) (360.5)
ϭ
(2.59) (144)
373
ϭ
373
Since the entrainment energy equation shows a perfect check, the assumed
h

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

3110 lb/h, or 19.9 lb / (h
⅐
ton) [10.04
kg/(h
⅐
t)]
(c) Steam to primary condenser
ϭ
30.66
ϩ
48.7
ϭ
79.36 lb/min (36 kg / min)
Q
ϭ
79.36(1144
Ϫ
58)
ϭ
86,200 Btu/ min,
or 550 Btu / (min
⅐
ton) [644.7 Btu / (min
⅐
t)]
The heat required for the generation of the motive steam will be h
A
Ϫ
h
5

,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

3.80, based on actual steam consumption. The corresponding horse-
power per ton
ϭ
1.237 (1.025 kW / t).
As a practical problem, the determination of the minimum available energy of
the motive steam at which the ejector will operate is not of particular importance
since excessive steam consumption will render the operation unsatisfactory long
before this point is reached. However, the condition of the motive steam to be
supplied to an ejector for any given maximum allowable steam consumption, as
dictated, for example, by cooling-water considerations, may frequently be of inter-
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REFRIGERATION
REFRIGERATION
12.21
110° Air
SI Values
30 F air (–1.1 C)
70 F air (21.1 C)
68 F air (20.0 C)
18 F brine (–7.8 C)
110 F air (43.3 C)
25 F brine (–3.9 C)
Brine heater
30° Air
18° Brine
Brine
cooler
25° Brine

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

furnace efficiency; (j) the manufactured-gas consumption for direct heating with
70 percent furnace efficiency; (k) the energy-saving potential with a conventional
diesel engine driving the compressor in a cogeneration mode.
Calculation Procedure:
1. Specify the power and fuel constants to use in the design
For this system the following power and fuel constants will be sued; such constants
are readily available in standard reference works: Thus, the power consumption of
motor for the unit, considered as a standard refrigerating machine, will be 1.7 hp /
ton (1.49 kW/t). Motor efficiency is 87 percent. Heat of combustion of the fuels:
coal, 14,000 Btu/lb (32,620 kJ/kg); oil, 19,000 Btu / lb (44,270 kJ/kg); natural gas,
1100 Btu/ft
3
(41,007 kJ/m
3
); manufactured gas, 600 Btu/ft
3
(22,367 kJ/ m
3
). Spe-
cific gravity of the fuel oil
ϭ
0.88.
2. Find the total heat output of the heat pump
1.7 hp
ϭ
1.7(2544)
ϭ
4325 Btu/h (1.27 kW)
Motor input
ϭϭ

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)

In this instance, this is equivalent to 12,000
ϩ
4970.
3. Compute the system characteristics
(a) Capacity of unit, considered as a refrigerating machine
ϭ
100,000/16,970
ϭ
5.89 tons (6.5 t)
(b) Motor hp
ϭ
5.89(1.7)
ϭ
10.0 (7.46 kW)
(c) Motor input
ϭ
10(2544)/0.87
ϭ
29,240 Btu/h or 8.58 kW
(d) Coefficient of performance
ϭ
100,000/29,240
ϭ
3.42
(e) Heat received from outside air
ϭ
5.89(12,000)
ϭ
70,680 Btu/h (20.7 kW)
(ƒ) Electric power for direct heating

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

3
/h (22.1
m
3
/h)
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REFRIGERATION
REFRIGERATION
12.23
4. Determine the cogeneration potential for the installation
With a Diesel-engine drive for the compressor, the specific fuel consumption will
typically be 0.45 lb/bhp
⅐
h (0.27 kg/ kW
⅐
h). Then, the fuel required per ton (t)
refrigerating capacity will be 1.7 (0.45)
ϭ
0.765 lb/h (0.347 kg / h), representing
0.765 (19,000)
ϭ
14,535 Btu/h (4.26 kW). Of this, 4325 Btu / h (1.26 kW) leave
the engine in the form of work. The remainder, 10,210 Btu/h
⅐
ton (10,722 kJ / h
⅐
t), is rejected in cooling water, exhaust, and by radiation. Much of this loss can be
recovered and applied to direct heating.

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

taking place throughout the electrical power industry at this writing, competitive
costs may be obtained from two, or more, utilities. This means that even more
favorable rates, in general, will be obtained.
Cost calculations may indicate that, on the basis of fuel consumption alone, the
Diesel-driven heating unit would be more economical for the system above than
any other heating system, including direct heating with coal, in many localities. As
a heating unit alone, however, the initial cost of the heating cycle is greater than
that of any other system. But when a substantial proportion of the capacity of the
system may be applied to cooling in the summer, the combined heating and air-
conditioning first cost should compare favorably with that of a conventional heating
plant plus air-conditioning equipment. For this reason, air conditioning should pref-
erably be incorporated with the heating cycle.
Recently developed rare-earth additives for Diesel engines simultaneously cut
NO
x
emissions and eliminate more than 90 percent of particulates in the engine
exhaust. The system combines a cerium-based organic fluid and a conventional
ceramic particulate trap. Using such emission controls allows Diesel engines in
heavily populated areas where they formerly might not have been welcome. Thus,
Diesel-engine-drive of heat pumps in a cogeneration mode could be more popular
in the future. The additive and trap for Diesel engines results in NO
x
of 2.0 grams
per brake-horsepower-hour (2.68 g/kWh and 0.013 g/kWh), compared to 0.1 g/
bhp
⅐
h (0.13g/kWh) for best traps as of this writing. EPA has set emission limits
of 2.5 g / bhp
⅐
hNO

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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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

utility rate structures and more energy efficient, Gershon Meckler developed a UHP
system that reduces electric demand in several ways, amongst which are: (a) Shifts
the dehumidification load from the UHPs to a small, central ice plant that operates
off-peak (or, in an alternative version, to an efficient two-stage desiccant system);
(b) distributes dehumidification via a small quantity of very dry ventilation air (no
increase in primary air quantity over a conventional UHP system); (c) uses UHPs
for sensible heating / cooling only, with no condensation at terminals; (d) operates
UHPs in cooling mode at 55
Њ
F (12.8
Њ
C) refrigerant coil temperature (instead of at
40 to 45
Њ
F [4.4 to 7.2
Њ
C]), resulting in a 35 to 40 percent reduction in compressor
horsepower (kW); (e) deactivates the UHP compressors when winter cooling and
night heating are required.
As described by Gershon Meckler, P.E., Fig. 19 depicts schematically and Fig.
20 charts psychrometrically the ice dehumidification/unitary heat-pump system de-
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12.25
Compressor
DX DX
Compressor
DX DX

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.

75-ton chiller
Evaporative
condenser
Ice storage
(700 ton-hr)
Gas or
electric boiler
Primary duct
Unitary heat pump
(NE and NW perimeter zones)
UHP with chilled-water coil
(interior and SE/SW perimeter zones)
1500 gpm
8 in.
8 in.
8 in.
8 in.
Fire water
Fire alarm valve
Sprinkler heads
Sprinkler riser
Standpipe riser
FIGURE 19 Office-building UHP system with central plant ice dehumidification and UHP / water-coil sensible heating
and cooling (U.S. Patent 3918525 and other patents pending). (Gershon Meckler, P.E. and HPAC magazine.)
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REFRIGERATION