Steam Jet Refrigeration

Note: The source of the technical material in this volume is the Professional

Engineering Development Program (PEDP) of Engineering Services.

Warning: The material contained in this document was developed for Saudi

Aramco and is intended for the exclusive use of Saudi Aramco’s

employees. Any material contained in this document which is not

already in the public domain may not be copied, reproduced, sold, given,

or disclosed to third parties, or otherwise used in whole, or in part,

Saudi Aramco DeskTop Standards

CONTENTS PAGES

IDENTIFYING WHEN TO USE STEAM JET REFRIGERATION SYSTEMS ...1

Process Description ...1

Advantages, Disadvantages, Features, and Limitations ...9

Advantages ...9

Disadvantages ...9

Features...9

Limitations...9

When to Use Steam Jet Refrigeration Systems...9

ESTIMATING THE SIZES OF STEAM JET REFRIGERATION SYSTEMS ... 10

WORK AID 1: PROCEDURE FOR ESTIMATING THE SIZES OF STEAM JET REFRIGERATION SYSTEMS, GIVEN SAUDI ARAMCO REQUIREMENTS, VENDOR INFORMATION, AND APPLICABLE FORMULAS ... 14

GLOSSARY... 25

Figure 1. Steam Jet Refrigeration System 2

Figure 2. Approximate Range of Operation of Ejectors 2

Figure 3. Common Pressure Equivalents 3

Figure 4. Water Vapor Pressure 4

Figure 5. Steam Jet Refrigeration Unit 6

Figure 6. Basic Steam Jet Ejector or Booster Assembly 8

Figure 7. Enthalpy Values for Water (Btu/lb) 15

Figure 8. 689 kPa (ga) (100 psig) Steam Demand for Steam Jet

Refrigeration System 18

Figure 9. Effect of Steam Pressure on Steam Demand at 32°C (90°F) Condenser 19 Figure 10. Effect of Steam Pressure on Steam Demand at 38°C (100°F) Condenser 20 Figure 11. Effect of Steam Pressure on Steam Demand at 43°C (110°F) Condenser 21

Saudi Aramco DeskTop Standards 1

IDENTIFYING When to use steam jet refrigeration systems

This section discusses steam jet refrigeration systems. The section covers the following items: • Process description

• Advantages and features

• When to use steam jet refrigeration systems

Process Description

The steam jet refrigeration process is based on evaporated water. When water is exposed to a vacuum, evaporation occurs. If the vacuum is constant and the water vapor is removed at a constant rate, a constant cooling of the unevaporated water occurs. In a steam jet refrigeration system, the cooled water is the refrigerant.

Figure 1 shows a simplified diagram of a steam jet refrigeration system. Steam is supplied to a nozzle in the steam jet ejector (also called exhausters, thermocompressors, and eductors). As the steam travels through the nozzle opening, the steam accelerates to high (usually supersonic) velocity and causes an extremely low-pressure. This low-pressure draws the vapor out of the evaporator and lowers the vapor pressure inside the evaporator. The mix of water vapor and steam is condensed in the condenser and returned via the condensate pump to the evaporator through spray nozzles. Some of the water that enters the evaporator through the spray nozzles and some of the water in the reservoir evaporates and heat is removed from the liquid that remains. The chilled water from the reservoir in the evaporator is then pumped by the chilled water pump to the refrigeration chamber where the chilled water provides cooling. After the water is heated in the cooling process, the water flows back to the evaporator where the water is sprayed into the evaporator, and the cycle is repeated.

Figure 1. Steam Jet Refrigeration System

When multiple steam jet ejectors are employed in a series, relatively large quantities of gas or vapor can be evacuated at very low absolute pressures. The approximate pressure range for multi-stage ejector systems is provided in Figure 2.

NUMBER OF ABSOLUTE PRESSURE

STAGES mm of Hg kPa* psia*

1 760 - 75 101.3 - 10.0 14.7 - 1.45

2 100 - 10 13.3 - 1.3 1.93 - 0.193

3 30 - 3 4 - 0.4 0.58 - 0.058

4 5 - 0.3 0.67 - 0.04 0.097 - 0.0058

5 0.35 - 0.01 0.047 - 0.0013 0.0068 - 0.00019

* kPa and psia values calculated from the mm of Hg values given in the source document.

Source: Air Conditioning Refrigerating Data Book, 10th Edition, American Society of Refrigerating Engineers, 1957, p.

23-09, table 1.

Saudi Aramco DeskTop Standards 3 Figure 3 shows some common pressure equivalents that are used in low-pressure calculations.

1 inch Hg = 25.4 mm Hg @ 0°C 1 mm Hg = .03937 inch Hg @ 0°C 1 psi = 2.036 inch Hg @ 0°C 1 psi = 51.715 mm Hg @ 0°C 1 inch H2O @ 39.2°F = 1.868 mm Hg @ 0°C 1 kPa = 7.5006 mm Hg @ 0°C

Figure 3. Common Pressure Equivalents

To obtain water in the evaporator that is at a specific temperature, the vapor pressure in the evaporator must be maintained at the vapor pressure of water for that temperature. Figure 4 provides vapor pressure data for water at some selected temperatures. For example, assume that refrigerated water at 5.5°C (42°F) is required. Figure 4 indicates that a vapor pressure of .90 kPa (.131 psia) is required, and Figure 2 indicates that a three-stage ejector is the minimum required.

TEMPERATURE VAPOR PRESSURE °C °F kPa psia 0 32 0.61 0.089 1.11 34 0.66 0.096 2.22 36 0.72 0.104 3.33 38 0.77 0.112 4.44 40 0.84 0.122 5.55 42 0.90 0.131 6.67 44 0.98 0.142 7.78 46 1.05 0.153 8.89 48 1.14 0.165 10.0 50 1.23 0.178 11.11 52 1.32 0.192 12.22 54 1.42 0.206 13.33 56 1.53 0.222 14.44 58 1.64 0.238 15.56 60 1.76 0.256 21.11 70 2.50 0.363 26.67 80 3.50 0.507 32.22 90 4.81 0.698 37.78 100 6.54 0.949 43.33 110 8.79 1.275 48.89 120 11.67 1.693

Source: Steam Tables, Joseph H. Keenan, Frederick G. Keyes, Philip G. Hill, and Joan G. Moore, John Wiley & Sons,

Inc., New York, NY, 1969, pp. 2, table 1.

Figure 4. Water Vapor Pressure

Although Figure 2 indicates that three stages are required to obtain and maintain the required vapor pressure, the compression to condenser pressure occurs in only one stage, the first stage. This stage is usually called the main stream ejector or booster. The other two stages compress the noncondensables to atmospheric pressure and remove them from the system. The ejectors for the other two stages are small when they are compared to the main stream ejector.

Saudi Aramco DeskTop Standards 5 Figure 5 illustrates a typical steam jet refrigeration unit. The steam is supplied to the steam jet ejector. After traveling through the ejector and picking up water vapor from the evaporator, the steam enters the condenser where it is cooled to the recovery temperature. In this process, the vapor stream must be compressed from the vapor pressure that is required at the evaporator to the pressure that is required of the barometric condenser. This compression occurs in the outlet tube of the ejector. The ejector compression ratio is the ratio between the condenser pressure and the evaporator pressure. The ejector efficiency depends on the compression ratio. A lower ejector compression ratio results in a higher ejector efficiency. Therefore, it is desirable to operate with the condenser temperature as low as possible and the evaporator temperature as high as possible. The

evaporator temperature is established by the cooling requirements and the condenser temperature is established by the available cooling water.

Source: Air Conditioning Refrigerating Data Book, 10th Edition, American Society of Refrigerating Engineers, 1957, p. 23-10, Figure 9.

Saudi Aramco DeskTop Standards 7 The evaporator is a large vessel in which the water to be chilled is allowed to flash down (depressure) to the evaporator temperature that is established by the main stream ejector pressure. The water that enters the evaporator must be dispersed in small drops or thin sheets to facilitate evaporation, and the design must allow the incoming water to reach the equilibrium temperature before reaching the bottom of the evaporator. There must be large openings to allow the water vapor to enter the ejector, because large volumes of vapor must be removed and very little pressure drop can occur between the evaporator and the ejector. A 0.18 kPa (.026 psi) pressure difference is equivalent to an evaporator temperature increase from 4.4°C (40°F) to 7.2°C (45°F). A kg (1 lb) of water vapor at 4.4°C (40°F) has a volume of approximately 69.2 m3 (2,445 ft3). Water must be added to replace the water that evaporates during the cooling process.

The condenser may be either a surface condenser (Figure 1) or barometric condenser (shown in Figure 5). Because a surface condenser is more compact, the recovery of the condensate for use as boiler feedwater is practical. However, a barometric condenser is less expensive to construct and more thoroughly cools the discharge water, due to the direct contact between the condensing water and the vapor stream. Unless condensate recovery or condenser height are important, the barometric condenser is usually preferred. A basic steam jet ejector is illustrated in Figure 6. In an injector, high-pressure steam enters the steam chest. Steam from the steam chest enters the steam nozzle with essentially zero initial velocity. The steam expands in the converging section of the steam nozzle. The velocity increases to a maximum at the nozzle steam throat, which is the velocity of sound. In the diverging section of the steam nozzle, the steam expands further and the pressure decreases. For example, if saturated steam enters the steam chest at essentially zero velocity at 689, kPa (ga) (100 psig), then the exit velocity at the diffuse end of the steam nozzle has a velocity on the order of 1,200 m/s (4,000 ft/s). The high velocity steam travels across the suction chamber and entrains refrigerant vapor. The mixture of steam and refrigerant vapor is compressed to the condenser pressure in the diffuser. The mixing of the steam stream and the refrigerant vapor is a constant momentum process that generates heat through a decrease in kinetic energy. In addition, because the steam stream is usually supersonic, as the mixture becomes subsonic somewhere within the diffuser, preferably in the diffuser throat, a shock wave is created that results in an increase in pressure. In the diverging segment of the diffuser, the stream velocity decreases until the pressure reaches the condenser pressure.

Nozzle throat Steam nozzle Suction chamber Diffuser throat Diffuser Discharge Suction Nozzle plate Steam chest Steam inlet Refrigerant Vapor Mixture of Steam and Refrigerant Vapor STEAM

Source: Air Conditioning Refrigerating Data Book, 10th Edition, American Society of Refrigerating Engineers, 1957, p.

23-12, Figure 10.

Figure 6. Basic Steam Jet Ejector or Booster Assembly

As illustrated in Figure 1, a steam jet refrigeration system will have auxiliary equipment that consists of a chilled-water circulating pump, heat exchanger, and a condensate water pump. In addition, the system will have gauges, valves, and other auxiliary equipment to provide control of the system. This auxiliary equipment typically requires utilities to operate that can add to the overall cost of the system. However, as a percentage of the total, costs for utility usage are negligible.

Saudi Aramco DeskTop Standards 9

Advantages, Disadvantages, Features, and Limitations

Advantages

In a steam jet refrigeration unit, the lack of moving parts yields a system free of vibration. Maintenance requirements are low. Waste steam can be used, and the refrigerant is a low-pressure liquid that can be transported in small lines. Thus, the system is inexpensive to build and operate. Water is used as the refrigerant, and water is neither toxic nor flammable.

Disadvantages

Systems that use water as a coolant are practical only where the required coolant temperatures are above freezing, 0°C (32°F). The low operating pressures of the evaporator and condenser require the use of large equipment.

Features

In locations where steam is inexpensive and cooling water is plentiful, operating costs for steam jet refrigeration systems are relatively low.

Limitations

The maximum capacity of a steam jet system depends upon the following: • Amount and pressure of steam available

• Amount and temperature of cooling water available • Cooling temperature required

• Physical size of the unit that can be constructed

When to Use Steam Jet Refrigeration Systems

Steam jet systems should be used when the following conditions apply: • Cheap steam is available

• Adequate cooling water is available

• The level of refrigeration that is required allows the use of water as the refrigerant

As always, investment, operating costs, and schedule availability should be considered when selecting a system.

Sizing steam jet refrigeration systems requires calculation of the material and energy requirements of the system. The refrigeration load and temperature level are known, and the following must be calculated:

• Chilled-water flow rate • Makeup-water rate • Steam requirements • Cooling-water rate • Net condensate produced

The method provided below and covered in Work Aid 1 will allow these quantities to be calculated for

screening purposes. Accurate values should be obtained from the ejector vendor for the model that is proposed, because some models may be more efficient than those that are used for this course.

The curves in Work Aid 1 are based on Tons of refrigeration. This number is obtained by the following: Tons Refrigeration = Refrigeration Load (Btu/hr)/ 12,000

The chilled-water return temperature must be selected. The temperature selected will affect the size of the installation. Several temperatures should be tried in an effort to minimize the investment and operating cost. The temperature of the chilled water from the refrigeration unit is usually specified by the refrigeration user. The exit and return temperatures are used to obtain the enthalpy values from steam tables (Figure 7) and to calculate the chilled-water rate as follows:

Chilled-water rate, (lb/hr)

w

=

Q

h

−

h

(EQN A) where: Q_r = Refrigeration load, Btu/hr

h_output = Enthalpy of chilled water from the load Btu/lb h_input = Enthalpy of chilled water to the load, Btu/lb

Saudi Aramco DeskTop Standards 11 A condenser temperature must be selected. The temperature selected will affect the equipment size and cooling water rate. Several temperatures should be tried in an effort to minimize investment and operating cost. The rate at which water must be vaporized, in the evaporator, to provide the required cooling is then calculated as follows:

w

−

w

(

h

−

h

)

h

−

h

(EQN B) where: w_{vc =} Water vaporized for cooling, lb/hr

w_r = Chilled-water rate, lb/hr

h_output = Enthalpy of chilled water from load, Bu/lb h_input = Enthalpy of chilled water to load, Btu/lb

h_vapor = Enthalpy of vapor at chilled water supply temperature, (chilled temp), Btu/lb

h_makeup = Enthalpy of makeup water The water-makeup rate is the same as the amount vaporized.

With steam jet units, the amount of steam that is required to generate one kW of refrigeration can vary widely even in well designed systems. Steam pressure, condenser temperature, and evaporator temperature all affect the amount of steam that is required. Data from several manufacturers were combined and averaged to arrive at the data shown in Figure 8 in Work Aid 1. This figure provides information for a constant steam pressure of 689 kPa (ga) (100 psig) with variation in condenser temperature at specific evaporator (chilled water) temperatures. If the refrigeration system uses steam at this pressure, the amount of steam required, per ton of refrigeration, can be obtained directly from Figure 8. The total steam requirement is then obtained by:

w_s = Q_rn (EQN C)

where: w_s = Steam required for the ejector, lb/hr Q_r = Specified refrigeration load, tons

Obtain the enthalpy of the steam used in the ejector from steam tables or Figure 12. Calculate the condenser heat duty with the following equation:

Q_c – (h_s – h_l)w_s + (h_{vapor – hl)wvc} (EQN D)

where: Q_{c =} Condenser duty, Btu/hr

h_s = Enthalpy of the steam from Figure 12, Btu/lb

h_l = _{Liquid enthalpy at the condenser temperature (Hmakeup), Btu/lb} w_s = Steam rate, lb/hr

h_vapor = Vapor enthalpy at the chilled temperature, Btu/lb w_vc = Vapor rate from the evaporator, lb/hr

Calculate the cooling water rate as follows:

Assume the cooling water temperature out is the same as the condensing temperature. This will allow use of only a single exchanger shell. The rate is:

w

=

Q

h

−

h

(

)

where: w_{c = Cooling-water rate, Btu/hr} Q_c = Condenser duty, Btu/lb

h_cwout = Enthalpy of cooling water at outlet temperature, Btu/lb (Use Figure 7) h_cwin = Enthalpy of cooling water at inlet temperature, Btu/lb (Figure 7)

The last item to calculate in sizing the system is the net condensate that is produced. The net condensate is equal to the steam rate to the ejector.

Saudi Aramco DeskTop Standards 13 As discussed earlier in this module, when the condenser temperature is lowered, the amount of steam that is required is decreased, because the compression ratio in the ejector is lowered and the efficiency is increased. As the temperature of the chilled water in the evaporator is lowered, the amount of steam that is required increases because the compression ratio in the ejector increases. Therefore the efficiency decreases. Higher steam pressures require less steam for a given amount of refrigeration. Figures 9, 10 and 11 have been

provided to allow estimation of steam rates with various steam pressures. Each figure is for a different specific condenser temperature.

Examination of Figures 9, 10, and 11 in Work Aid 1 show, that while less steam is required as steam pressure increases, there is a point where increased steam pressure no longer has significant impact on the steam requirement. Furthermore, comparison of Figures 9, 10, and 11 reveal that as condenser temperature rises and chilled-water temperature decreases, the point where increased steam pressure no longer has significant impact on the steam requirement also rises.

REFRIGERATION SYSTEMS, GIVEN SAUDI ARAMCO

REQUIREMENTS, VENDOR INFORMATION, AND APPLICABLE

FORMULAS

First, sketch a simple diagram of a steam jet refrigeration system, similar to Figure 1. Use this to record given information and track calculations.

1. Obtain the amount and temperature of refrigeration that is required and calculate tons of refrigeration that is;

Tons

=

Btu / hr

12,000

2. Obtain or determine the temperature of the cooling water available.

3. Obtain or determine the pressure of the steam that is available or that is to be generated for refrigeration. 4. Determine the condenser temperature.

5. Obtain or determine the temperature of the chilled water that is returning to the evaporator. (This water is typically 8.3°C (15°F) warmer than the chilled water temperature.)

6. Use Figure 7 to determine the enthalpy of the water that is used for refrigeration (chilled water) at its input and output points to the unit that is being cooled.

Saudi Aramco DeskTop Standards 15 ENTHALPY

TEMPERATURE Liquid Vapor

°F Btu/lb Btu/lb 32 -0.01 1075.4 35 3.00 1076.7 40 8.02 1078.9 45 13.04 1081.1 50 18.06 1083.3 55 23.07 1085.5 60 28.08 1087.7 65 33.09 1089.9 70 38.09 1092.0 75 43.09 1094.2 80 48.09 1096.4 85 53.08 1098.6 90 58.07 1100.7 95 63.06 1102.9 100 68.05 1105.0 105 73.03 1107.2 110 78.02 1109.3

Source: Steam Tables, Joseph H. Keenan, Frederick G. Keyes, Philip G. Hill, and Joan G. Moore, John Wiley & Sons, Inc., New York, NY, 1969, pp. 2-3, table 1.

w

=

Q

h

−

h

(EQN A)

where: w_r = Required chilled-water flow rate, lb/hr (to convert lb/hr to kg/hr, multiply by 0.453 592)

Q_r = Refrigeration load, Btu/hr (to convert kW to Btu/hr, multiply by 3,415; to convert tons to Btu/hr, multiply by 12,000)

h_output = Enthalpy of the chilled water that is leaving the refrigeration load unit, Btu/lb, which is obtained from Figure 7 (to convert kJ/kg to Btu/lb, multiply by 0.430) h_input = Enthalpy of the chilled water that is entering the refrigeration load unit, Btu/lb,

Saudi Aramco DeskTop Standards 17 8. Use the following equation to determine the rate at which water must be vaporized in the evaporator to

chill the returning chilled water.

w

=

w

h

−

h

h

−

h

(EQN B)

where: _{wvc =} Required evaporation rate to chill the returning chilled- water stream, lb/hr (to convert lb/hr to kg/hr, multiply by 0.453 592)

wr = Required chilled-water flow rate, lb/hr (to convert kg/hr to lb/hr, multiply by 2.205)

h_output = Enthalpy of the chilled water that is leaving the refrigeration load unit, Btu/lb, which is obtained from Figure 7 (to convert kJ/kg to Btu/lb, multiply by 0.430) h_input = Enthalpy of the chilled water that is entering the refrigeration load unit, Btu/lb,

obtain from Figure 7

h_vapor = Enthalpy of the vapor that is being removed from the evaporator, which is obtained from Figure 7

h_makeup = Enthalpy of liquid makeup from condenser

Use the selected figure to determine the approximate steam requirement per ton based on the steam pressure and chilled-water temperature. Figure 8 is for 100 psig steam pressure. Figures 9, 10, and 11 are used for other steam pressures at condenser temperatures of 90°F, 100°F, and 110°F, respectively.

110 105 100 95 90 85 80

POUNDS STEAM PER TON HOUR

CONDENSER TEMPERATURE °F

0 10 20 30 40 50 CHILLED WATER TEMPERATURE °F

60 55 50 45 40

Source: Air Conditioning Refrigerating Data Book 10th Edition, The American Society of Refrigerating Engineers, 1957, p. 23-13, Fig. 11.

Figure 8. 689 kPa (GA) (100 psig) Steam Demand for Steam Jet

Refrigeration System

Saudi Aramco DeskTop Standards 19

Source: Air Conditioning Refrigerating Data Book 10th Edition, The American Society of Refrigerating Engineers,

1957, p. 23-13, Fig. 12.

Figure 9. Effect of Steam Pressure on Steam Demand at 32°C (90°F)

Condenser

Source: Air Conditioning Refrigerating Data Book 10th Edition, The American Society of Refrigerating Engineers,

1957, p. 23-13, Fig. 13.

Figure 10. Effect of Steam Pressure on Steam Demand at 38°C (100°F)

Condenser

Saudi Aramco DeskTop Standards 21

Source: Air Conditioning Refrigerating Data Book 10th Edition, The American Society of Refrigerating Engineers,

1957, p. 23-13, Fig. 14.

Figure 11. Effect of Steam Pressure on Steam Demand at 43°C (110°F)

Condenser

requirement for the refrigeration unit.

w

=

Q

n

where: _{ws = Estimated steam requirement to provide the refrigeration, lb/hr (to convert lb/hr to} kg/hr, multiply by 0.453 592)

Qr = Specified refrigeration load, tons (to convert kW to tons, multiply by 0.284; to convert Btu/hr to tons, multiply by 0.000 083 33)

n = Required steam, lb/hr per ton of refrigeration, that was determined in Step 10

12. Use Figure 12 to determine the enthalpy of the steam that is being supplied to the refrigeration system.

ENTHALPY

PRESSURE TEMPERATURE Liquid Vapor

psia °F Btu/lb Btu/lb

25 240.08 208.52 1160.7

50 281.03 250.24 1174.4

75 307.63 277.61 1182.4

100 327.86 298.61 1187.8

125 344.39 315.90 1191.8

Source: Steam Tables, Joseph H. Keenan, Frederick G. Keyes, Philip G. Hill, and Joan G. Moore, John Wiley & Sons,

Inc., New York, NY, 1969, pp. 9-11, table 2.

Saudi Aramco DeskTop Standards 23 13. Use the following equation to determine the heat being absorbed in the condenser from the ejector

stream.

Q

=

(

h

−

h

)

w

+

(

h

−

h

)

w

(EQN D)

where: _{Qc =} Heat to be removed from ejector stream by the condenser, Btu/hr (to convert Btu/hr to mW, multiply by 0.2931)

hs = Enthalpy of the steam, which was obtained in Step 12, Btu/lb (to convert kJ/kg to Btu/lb, multiply by 0.430)

hl = Enthalpy of the water that is leaving the condenser (h_MAKEUP), obtain from Figure 7, Btu/lb

ws = Estimated steam requirement to provide the refrigeration, lb/hr (to convert kg/hr to lb/hr, multiply by 2.205)

h_vapor = Enthalpy of the vapor that is being removed from the evaporator, obtain from Figure 7

temperature out is equal to condensing temperature. (This allows use of only one exchanger shell.)

w

Q

h

cwout −

h

(

)

where: _{wc =} Required cooling-water flow, lb/hr (to convert lb/hr to kg/hr, multiply by 0.453 592; to convert lb/hr to L/hr, multiply by 0.453 592; to convert lb/hr to gal/min @ 60°F, multiply by 0.002)

Qc = Heat to be removed from steam stream by the condenser, Btu/hr (to convert MW to MBtu/hr, multiply by 3.412)

h_cwout = Enthalpy of the cooling water at outlet temperature, Btu/lb, obtain from Figure 7, Btu/lb

h_cwin = Enthalpy of the cooling water that is entering the condenser, obtain from Figure 7, Btu/lb

Saudi Aramco DeskTop Standards 25 GLOSSARY

booster Main stream ejector.

condenser The component in which the steam and entrained refrigerant vapor are cooled and condensed.

eductor Another name for ejector.

ejector Jet pump used to withdraw fluids from a space.

entrainment The capturing of particles by a moving stream.

evaporator The container in which the cooling agent gets cooled in a jet steam refrigeration process.

exhauster Another name for ejector.

flash The evolution of vapor from a bubble-point liquid when the pressure is

reduced.

main stream ejector The ejector that actually entrains the vapor from the evaporator.

w

=

Q

h

−

h

(EQN A) where: w_r = Required chilled-water flow rate

Q_r = Refrigeration load, Btu/hr

h_output = Enthalpy of the chilled water that is leaving the refrigeration load unit h_input = Enthalpy of the chilled water that is entering the refrigeration load unit

w

=

w

h

−

h

h

−

h

(EQN B) where: w_vc = Required evaporation rate to chill the returning chilled- water stream

w_r = Required chilled-water flow rate

h_output = Enthalpy of the chilled water that is leaving the refrigeration load unit h_input = Enthalpy of the chilled water that is entering the refrigeration load unit h_vapor = Enthalpy of the vapor that is being removed from the evaporator h_makeup = Enthalpy of liquid makeup from condenser

Saudi Aramco DeskTop Standards 27 ADDENDUM A: EQUATIONS USED IN ChE 210.04, CONT'D

w

=

Q

n

where: w_s = Estimated steam requirement to provide the refrigeration Q_r = Specified refrigeration load

n = Required steam, lb/hr per ton of refrigeration

Q

=

(

h

−

h

)

w

+

(

h

−

h

)

w

(EQN D)

where: _{Qc = Heat to be removed from ejector stream by the condenser} hs = Enthalpy of the steam

hl = Enthalpy of the water that is leaving the condenser ws = Estimated steam requirement to provide the refrigeration h_vapor = Enthalpy of the vapor that is being removed from the evaporator

wvc = Vapor rate from the evaporator

w

Q

h

cwout −

h

(

)

where: _{wc = Required cooling-water flow}

Qc = Heat to be removed from steam stream by the condenser h_cwout = Enthalpy of the cooling water at outlet temperature

h_cwin = Enthalpy of the cooling water that is entering the condenser h_cwout = Enthalpy of the cooling water at outlet temperature

h_input = Enthalpy of the chilled water that is entering the refrigeration load unit h_l = Enthalpy of the water that is leaving the condenser

h_makeup = Enthalpy of liquid makeup from condenser

h_output = Enthalpy of the chilled water that is leaving the refrigeration load unit h_s = Enthalpy of the steam

h_vapor = Enthalpy of the vapor that is being removed from the evaporator n = Required steam

Q_c = Heat to be removed from ejector stream by the condenser Q_r = Specified load

w_c = Required cooling-water flow, w_r = Required chilled-water flow rate

w_s = Estimated steam requirement to provide the refrigeration w_vc = Vapor rate from the evaporator