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 75ЊF (23.9ЊC) dry bulb and 64.4ЊF (18.0 ЊC) 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 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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 80ЊF (26.7 ЊC). The maximum outside conditions for design purposes are 95 ЊF (35.0ЊC) dry bulb and 78ЊF (25.6ЊC) 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 ЊF 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 20ЊF (36ЊC). For the floor, the temper- ature difference ϭ 80 Ϫ 75 ϭ 5ЊF(9ЊC). 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 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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 95ЊF (35ЊC) 75ЊF (23.9ЊC) Wet bulb 78 ЊF (25.6ЊC) 64.4ЊF (18.0ЊC) Percent relative humidity 47 55 Dew point 71 ЊF (21.7ЊC) 58ЊF (14.4ЊC) 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, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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 ЊF(Ϫ17.8ЊC) 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 Nϫ V ϭ 180 ft / min 3 180 ϭ 180/N ϭ ⁄180 ϭ 1ft 23 L ϭ 1.2DVϭ 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 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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 ЊF (12.8ЊC). 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 ЊF Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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.) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. REFRIGERATION 12.8 PLANT AND FACILITIES ENGINEERING FIGURE 5 Piping arrangement for two centrifugal coolers in series arrangement. (Carrier Cor- poration.) (38.9ЊC) 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 ЊF (0.55ЊC) requires a heat input of 1 Btu (1.055 kJ). With the specified 70 ЊF (38.9ЊC) 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)/ Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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.) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. 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 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. REFRIGERATION