Steam Jet Refrigeration

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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, without the written permission of the Vice President, Engineering Services, Saudi Aramco.

Chapter : Process File Reference: CHE21004

For additional information on this subject, contact R.A. Al-Husseini on 874-2792

Engineering Encyclopedia

Process Steam Jet Refrigeration

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 ADDENDUM ........................................................................................................................................... 26

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LIST OF FIGURES 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

Figure 12.

Enthalpy Values for Water (Btu/lb)

22

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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 lowpressure. 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.

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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.

Figure 2. Approximate Range of Operation of Ejectors

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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.

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TEMPERATURE °C °F 0 32 1.11 34 2.22 36 3.33 38 4.44 40 5.55 42 6.67 44 7.78 46 8.89 48 10.0 50 11.11 52 12.22 54 13.33 56 14.44 58 15.56 60 21.11 70 26.67 80 32.22 90 37.78 100 43.33 110 48.89 120 Source:

VAPOR PRESSURE kPa psia 0.61 0.089 0.66 0.096 0.72 0.104 0.77 0.112 0.84 0.122 0.90 0.131 0.98 0.142 1.05 0.153 1.14 0.165 1.23 0.178 1.32 0.192 1.42 0.206 1.53 0.222 1.64 0.238 1.76 0.256 2.50 0.363 3.50 0.507 4.81 0.698 6.54 0.949 8.79 1.275 11.67 1.693

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.

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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.

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Source:

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

Figure 5. Steam Jet Refrigeration Unit

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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 m 3 (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.

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Nozzle throat Steam nozzle

STEAM

Steam inlet Steam chest Source:

Suction chamber Diffuser throat

Discharge

Mixture of Steam and Refrigerant Vapor

Diffuser

Nozzle Suction plate

Refrigerant Vapor

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.

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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.

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Estimating the sizes of steam jet refrigeration systems 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:

wr = Chilled-water rate, (lb/hr) where:

Qr

=

Qr houtput −hinput

Refrigeration load, Btu/hr

houtput

=

Enthalpy of chilled water from the load Btu/lb

hinput

=

Enthalpy of chilled water to the load, Btu/lb

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(EQN A)

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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:

wvc −wr = where:

(houtput − hinput)

hvapor − h makeup

wvc

=

Water vaporized for cooling, lb/hr

wr

=

Chilled-water rate, lb/hr

houtput

=

Enthalpy of chilled water from load, Bu/lb

hinput

=

Enthalpy of chilled water to load, Btu/lb

hvapor

=

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

hmakeup

=

Enthalpy of makeup water

(EQN B)

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: ws = Qrn where:

ws

=

Steam required for the ejector, lb/hr

Qr

=

Specified refrigeration load, tons

n

=

Required steam, lb/hr per ton of refrigeration

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(EQN C)

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The cooling water requirement of the condenser is then calculated as follows: 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: Qc – (hs – hl)ws + (hvapor – hl )wvc where:

(EQN D)

Qc

=

Condenser duty, Btu/hr

hs

=

Enthalpy of the steam from Figure 12, Btu/lb

hl

=

Liquid enthalpy at the condenser temperature (H makeup), Btu/lb

ws

=

Steam rate, lb/hr

hvapor =

Vapor enthalpy at the chilled temperature, Btu/lb

wvc

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:

wc = where:

Qc (hcwout − hcwin)

wc

=

Cooling-water rate, Btu/hr

Qc

=

Condenser duty, Btu/lb

(EQN E)

hcwout =

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

hcwin

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.

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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.

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WORK AID 1: PROCEDURE FOR ESTIMATING THE SIZES OF STEAM JET 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.

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ENTHALPY TEMPERATURE °F 32 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110

Liquid Btu/lb -0.01 3.00 8.02 13.04 18.06 23.07 28.08 33.09 38.09 43.09 48.09 53.08 58.07 63.06 68.05 73.03 78.02

Vapor Btu/lb 1075.4 1076.7 1078.9 1081.1 1083.3 1085.5 1087.7 1089.9 1092.0 1094.2 1096.4 1098.6 1100.7 1102.9 1105.0 1107.2 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.

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

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7.

Use the following equation to determine the required chilled water flow rate.

wr =

where:

Qr houtput − hinput

(EQN A)

wr

=

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

Qr

=

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

houtput

=

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)

hinput

=

Enthalpy of the chilled water that is entering the refrigeration load unit, Btu/lb, obtain from Figure 7

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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 vc = wr

where:

9.

houtput − hinput hvapor − hmakeup

(EQN B)

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)

houtput

=

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)

hinput

=

Enthalpy of the chilled water that is entering the refrigeration load unit, Btu/lb, obtain from Figure 7

hvapor

=

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

hmakeup

=

Enthalpy of liquid makeup from condenser

Note that the water rate that is required for makeup is the same as the rate that is vaporized (W VC = WM)

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Based on the condenser temperature, select the appropriate figure from among Figures 8, 9, 10, and 11. 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.

40

45

50

110

55

CHILLED WATER TEMPERATURE °F 60

CONDENSER TEMPERATURE °F

10.

105 100 95 90 85 80 0

10

20

30

40

50

POUNDS STEAM PER TON HOUR 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

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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

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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

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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

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11.

Use the following formula and the steam requirement that is found in Step 10 to calculate the steam requirement for the refrigeration unit.

w s = Qr n where:

ws

=

Qr = n 12.

=

(EQN C)

Estimated steam requirement to provide the refrigeration, lb/hr (to convert lb/hr to kg/hr, multiply by 0.453 592) 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) Required steam, lb/hr per ton of refrigeration, that was determined in Step 10

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

ENTHALPY PRESSURE psia 25 50 75 100 125 Source:

TEMPERATURE °F 240.08 281.03 307.63 327.86 344.39

Liquid Btu/lb 208.52 250.24 277.61 298.61 315.90

Vapor Btu/lb 1160.7 1174.4 1182.4 1187.8 1191.8

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.

Figure 12. Enthalpy Values for Water (Btu/lb)

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13.

Use the following equation to determine the heat being absorbed in the condenser from the ejector stream.

(

)

Q c = (hs − hl )ws + hvapor − h l wvc

where:

(EQN D)

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)

hvapor

=

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

wvc

=

Vapor rate from the evaporator, lb/hr

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14.

Use the following equation to calculate the required cooling water flow rate. Assume cooling water temperature out is equal to condensing temperature. (This allows use of only one exchanger shell.)

wc

where:

15.

=

Qc

(hcwout

−

hcwin )

(EQN E)

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)

hcwout

=

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

hcwin

=

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

Determine the net condensate that is produced. The rate is equal to the steam rate to ejector.

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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.

thermocompressor

Another name for ejector.

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ADDENDUM A: EQUATIONS USED IN ChE 210.04

wr = where:

wr =

Required chilled-water flow rate

Qr =

Refrigeration load, Btu/hr

(EQN A)

houtput =

Enthalpy of the chilled water that is leaving the refrigeration load unit

hinput =

Enthalpy of the chilled water that is entering the refrigeration load unit

wvc = wr where:

Qr houtput − hinput

wvc = wr =

houtput − hinput hvapor − hmakeup

Required evaporation rate to chill the returning chilled- water stream Required chilled-water flow rate

houtput =

Enthalpy of the chilled water that is leaving the refrigeration load unit

hinput =

Enthalpy of the chilled water that is entering the refrigeration load unit

hvapor =

Enthalpy of the vapor that is being removed from the evaporator

hmakeup =

(EQN B)

Enthalpy of liquid makeup from condenser

Saudi Aramco DeskTop Standards

26

Engineering Encyclopedia

Process Steam Jet Refrigeration

ADDENDUM A: EQUATIONS USED IN ChE 210.04, CONT'D

ws = Qrn where:

(EQN C)

ws =

Estimated steam requirement to provide the refrigeration

Qr =

Specified refrigeration load

n

=

Required steam, lb/hr per ton of refrigeration

(

)

Q c = (hs − hl )ws + hvapor − h l wvc

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

hvapor = wvc

=

Enthalpy of the vapor that is being removed from the evaporator Vapor rate from the evaporator

wc

where:

=

Qc

(hcwout

−h cwin

)

wc =

Required cooling-water flow

Qc =

Heat to be removed from steam stream by the condenser

hcwout = hcwin =

(EQN D)

(EQN E)

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

Saudi Aramco DeskTop Standards

27

Engineering Encyclopedia

Process Steam Jet Refrigeration

ADDENDUM B: SYMBOLS USED IN ChE 210.04 hcwin = hcwout = hinput = hl

=

hmakeup = houtput = hs = hvapor = n

=

Enthalpy of the cooling water that is entering the condenser Enthalpy of the cooling water at outlet temperature Enthalpy of the chilled water that is entering the refrigeration load unit Enthalpy of the water that is leaving the condenser Enthalpy of liquid makeup from condenser Enthalpy of the chilled water that is leaving the refrigeration load unit Enthalpy of the steam Enthalpy of the vapor that is being removed from the evaporator Required steam

Qc =

Heat to be removed from ejector stream by the condenser

Qr =

Specified load

wc =

Required cooling-water flow,

wr =

Required chilled-water flow rate

ws =

Estimated steam requirement to provide the refrigeration

wvc =

Vapor rate from the evaporator

Saudi Aramco DeskTop Standards

28