King Fahd University of Petroleum and Minerals Department of Mechanical Engineering. ME 431: Refrigeration. Laboratory Manual

King Fahd University of Petroleum and Minerals Department of Mechanical Engineering ME 431: Refrigeration Laboratory Manual February 2001

2 Table of Contents Contents Page # Safety Regulations 3 Lab Regulations and schedule 4 Experiment # 1 Basic vapor compression system: Expansion devices 6 Experiment # 2 Basic vapor compression system: Compressor analysis 11 Experiment # 3 Basic vapor compression system: Heat Exchanger 14 Experiment # 4 Basic vapor compression system: Two evaporatore 19 Experiment # 5 Computer lab # 1 22 Experiment # 6 Computer lab # 2 25 Experiment # 7 Computer linked vapor compression system: Demosntration 28 Experiment # 8 Computer linked vapor compression system: Energy balance 32 Experiment # 9 Computer linked vapor compression: Heat transfer coefficient 37 Experiment # 10 Experiment on cryogenics 42

3 Safety Regulations Users of Refrigeration & Air-conditioning Lab. must comply with the following safety instructions. 1 Smoking is not allowed inside the Lab. 2. Wear always pants and safety shoes when you operate any machine. Thobes, gutras and sandals are not allowed at all. 3. There should be no over-crowding. Only one person should operate one machine. 4. Watch for over-head hanging hoists and cranes. 5. Make sure that you stay away from hot exhaust lines and moving parts of machines. 6. Before operating any machine, you must be aware of the following a. Location of fire extinguishers, fire blanket and the outside exits. b. How the machine operate. Read instruction or manual of the machine before operating it. c. How to turn off the machine in case of damages. 7. When you hear or see a danger alarm from the machine that you using, stop the machine right away. 8. Make sure that there is no fuel or oil spill on the floor. 9. Do not run inside the lab and concentrate on the present task. 10. When moving heavy equipments or gas cylinders, use carts. 11. Always use the right tools for the given task. 12. Handle the tools and equipemnts with extreme care and return the tools to their proper places (Tool Cabinets). 13. For cleaning tools or equipments, use only the proper cleaner. Never use fuels such as gasoline or diesel for cleaning. 14. Handle fuels with extreme caution. a. Use the designated area for this purpose. b. Use the proper containers (safety cnas) to carry fuels. c. Make sure there is no electric spark present. d. Do not leave fuels in open containers. 15. Make sure that all gas cylinders are chained and well supported. 16. Before operating a machine, make sure that there is no fuel or gas leakage.

4 Lab Regulations 1. Lab grades covers 20% of the total course grade. The distribution is as follows. Lab reports: 8% Lab quiz: 10% Attendance: 2% 3. Grading Policy: The following is the grading policy for the lab report grades. A + = 10.0 A = 9.5 A - = 9.0 B + = 8.5 B = 8.0 B - = 7.5 C + = 7.5 C = 7.0 C - = 6.5 4. Make up Lab: Make up lab is only allowed in the case of valid excuse from the student affair department or permission from the department chairman. Students must have to attain their own sections. 5. Late Coming: Students should come on time in the class. Late coming will be panelized from the attendance grade. 7. Lab Exam: Lab exam will be during last week of classes. 8. Schedule: Following is the schedule for the semester.

5 Lab Schedule Week Exp. No Experiment Lab Report 1 0 Basic vapor compression refrigeration cycle (Introduction) and use of R12 enthalpy diagram. 2 1 Basic vapor compression refrigeration cycle ( Study of Expansion Devices) 3 2 Basic vapor compression refrigeration cycle (Compressor Analysis) 4 3 Basic vapor compression refrigeration cycle (Study of heat Exchanger) 5 4 Basic vapor compression refrigeration cycle (Working with two evaporators) No yes Yes Yes Yes 6 5 Computer Lab # 1 Yes 7 6 Computer Lab # 2 Yes 8 - Field Trip No 9 7 Computer linked vapor compression refrigeration cycle (Demonstration) 10 8 Computer linked vapor compression refrigeration cycle (Energy balance) 11 9 Computer linked vapor compression refrigeration cycle (Overall heat transfer coefficient) No Yes Yes 12 10 Experiment on Cryogenics No 13 - Rivision 14 - Final Lab Quiz

6 Experiment # 1 BASIC VAPOR COMPRESSION REFRIGERATION CYCLE ( Expansion Devices)

7 Experiment # 1 Mechanical Vapor Compression Refrigeration Cycle Object: Study the various refrigeration metering devices in a vapor compression refrigeration system. Equipment: Brodhead-Garrett Model 9501 Commercial Refrigeration Training Unit. Thermal Engineering Company, Toledo, Ohio, Serial # 662 Refrigerant 12. Components: 1. Reciprocating Compressor: Copeland Refrigeration Corporation Model KAN, Refrigerant 12 No. of Cylinders: 2, Bore: 11/8 inch, Stroke: 5/8 inch. RPM: 1750, Power: 1/2 HP. 2. Evaporator: Plate and tube type 24 pipes, 3/8 inches O.D, 63 plates, 8x3x0.006 in. thickness 3. Condenser: Similar to Evaporator. 4. Filter Dryer: It contains a screen mesh and grains of silica gel. The screen filter removes the dirt and other solid contamination. The silica gel absorbs water. 5. Flow Meter: It measures the liquid refrigerant as it passes through the system. It is designed for R12 (Sp. Gr. 1.26) and reads directly in lbm/min. 6. Moisture and liquid Indicator: This is a sight glass which contains a color dot inside. The dot will be colored green when the system is free from water, and yellow when there is excessive amount of moisture. The filter dryer must be changed when the color is yellow. Bubbles appearing in the indicator show that there is a shortage of R12 in the system. 7. Reserve Tank: It is made of Pyrex glass for viewing and its purpose is to supply additional refrigerant when needed. 8. Accumulator A: It is made of Pyrex glass and is used to check and catch any liquid refrigerant not completely vaporized in the evaporator. It will boil off from this tank and re-enter the main stream as vapor.

8 9. Accumulator B: It is made of Pyrex glass, and excess oil from the compressor pump may back up into this tank as an indication of possible oil in the system. Theory: Refrigeration works on the principle of heat absorption due to the evaporation of refrigerant. The refrigerant is evaporated by passing from a region of high pressure to a region of low pressure, thus reducing its saturation temperature below its actual temperature. The degree of cooling can then be controlled by controlling the amount of refrigerant passing into the low pressure region. The mechanism that controls the refrigerant flow into the low pressure region is called a metering device. Following three metering devices we will use in this experiment. 1.Capillary Tube (CTV): The most common metering device is used for household refrigeration systems is the simplest one, the capillary tube. The easiest way to create a high and low pressure difference is to put a restriction in the refrigerant flow path. The capillary tube is such a restriction. Its a copper tube 2 to 20 ft long with a inside diameter from 0.025 to 0.090 inches. The refrigerant begins to evaporate as soon as it reaches the evaporator but it is carried into the evaporator before it has completely evaporated. As the refrigerant evaporates, water from the air begins to freeze on the evaporator tubes. 2. Automatic Expansion Valve (AXV): Capillary tube system is insensitive to load changing condition. When load increased, the suction pressure went up causing inefficient cooling and strain on the compressor. Therefore this system can be used only when the load is approximately constant. For varying load condition, a device sensible to load changes must be used. Automatic expansive valve is one of these devices. AXV keeps the pressure in the evaporator a constant and thus load on the compressor remain same. When the load on the system increases, evaporation in the evaporator also increases and hence suction pressure rises. In order to keep the suction pressure constant, the automatic expansion valve restricts the refrigerant flow therefore slowing down the evaporation and bringing the suction pressure back to its original state. 3. Thermostatic expansion valve (TXV): The most popular type of expansion device for moderate sized refrigeration systems is the thermostatic expansion valve. Its forces a constant superheat in the evaporator. When there is an increase of load the refrigerant will evaporate faster in the evaporator. This in turn will cause a greater superheat at the evaporator outlet. This will cause the TXV to let more refrigerant pass and drive the superheat down. The net effect is then an increase in refrigeration when there is an increase in load. A thermostatic expansion valve is shown in Fig. 1. The sensing bulb contains fluid (normally it is the same refrigerant that is used for the refrigeration system) either in a vapor or a liquid form. This bulb is usually placed after the evaporation coil and near the suction line of the compressor. When the temperature of the refrigerant rises, it causes temperature of the fluid within the sensing bulb also to rise and forces the push rod downward which in turn moves the needle or the ball from its seat allowing liquid refrigerant to flow through and admit liquid to the evaporator coil. Because of this fresh charge the temperature of the refrigerant drops which is again sensed by the bulb and causing the pressure within it to drop. This reduces pressure against the diaphragm and the spring action from the bottom shuts the refrigerant passage. This opening and closing of the valve provides automatic regulation of refrigerant flow into the evaporator coil.

9 For further information, read chapter # 7 Refrigeration & Air-conditioning by W.F.Stoecker and Article # 3.1, 3.2, 3.3 from the textbook. Experimental Results: The temperature and pressure at different locations are indicated by the use of following thermocouples and pressure gauges. #1 Inlet to Evaporator #2 Outlet from Evaporator #9 Inlet to the Compressor #5 Outlet from the Compressor #6 Inlet to Condenser #7 Outlet from Condenser Lab Report: 1. Fill the attached table and for all expansion devices, 2. Plot the cycle on a p-h diagram. 3. Calculate the capacity of the evaporator in tons of refrigeration. 4. Calculate the amount of heat rejected from the condenser. 5. Calculate the work done on the compressor. 6. Calculate the COP of the cycle.

Mass flow of R-12 lb/min t 1 ( 0 C) t 1 ( 0 F) P 1 (psig) P 1 (psia) H 1 (Btu/lb) t 2 ( 0 C) t 2 ( 0 F) P 2 (psig) P 2 (psia) H 2 (Btu/lb) t 5 ( 0 C) t 5 ( 0 F) P 5 (psig) P 5 (psia) H 5 (Btu/lb) t 6 ( 0 C) t 6 ( 0 F) P 6 (psig) P 6 (psia) H 6 (Btu/lb) t 7 ( 0 C) t 7 ( 0 F) P 7 (psig) P 7 (psia) H 7 (Btu/lb) t 9 ( 0 C) t 9 ( 0 F) P 9 (psig) P 9 (psia) H 9 (Btu/lb) CTV AXV TXV

11 Experiment # 2 BASIC VAPOR COMPRESSION REFRIGERATION CYCLE ( Compressor Analysis )

12 Experiment # 2 Mechanical Vapor Compression Refrigeration Cycle Object: Study the performance of compressor in a vapor compression refrigeration system. Equipment: Brodhead-Garrett Model 9501 Commercial Refrigeration Training Unit. Thermal Engineering Company, Toledo, Ohio, Serial # 662 Refrigerant 12. Components: 1. Reciprocating Compressor: Copeland Refrigeration Corporation Model KAN, Refrigerant 12 No. of Cylinders: 2, Bore: 11/8 inch, Stroke: 5/8 inch. RPM: 1750, Power: 1/2 HP. 2. Evaporator: Plate and tube type 24 pipes, 3/8 inches O.D, 63 plates, 8x3x0.006 in. thickness 3. Condenser: Similar to Evaporator. Theory: The heart of the vapor compression system is the compressor. The three most common types of refrigeration compressors are the reciprocating, rotary and centrifugal. The reciprocating compressor consists of a piston moving back and forth in a cylinder with suction and discharge valves arranged to allow pumping to take place. The rotary and centrifugal compressors both have rotating members, but the rotary compressor is a positive-displacement machine, whereas the centrifugal compressor operates by centrifugal force. Compressor used in the lab is a Reciprocating Hermetically sealed compressor. A compressor whose crankshaft extended to the shaft is called an open-type compressor. A seal must be used where the shaft comes through the compressor housing to prevent refrigerant gas leaking out, or to prevent air from leaking it if the crankcase is at as pressure lower than atmospheric. Even though designers have continually developed better seals, the piercing of the housing always represents a source of leakage. To avoid leakage at the seal, the motor and compressor are often enclosed in the same housing. These type of compressors are Hermetically sealed compressors. Almost all small motor units used in household refrigerators and freezers are of the hermetic type. Also read from text book article # 4.5 and solve example 4.1 & 4.2 from the textbook.

13 Experimental Results: The temperature and pressure at different locations are indicated by the use of following thermocouples and pressure gauges. #1 Inlet to Evaporator #2 Outlet from Evaporator #9 Inlet to the Compressor #5 Outlet from the Compressor #6 Inlet to Condenser #7 Outlet from Condenser Lab Report: 1. Fill the attached table and do the followings, 2. Plot the cycle on a p-h diagram. 3. Calculate the capacity of the evaporator in tons of refrigeration. 4. Calculate the amount of heat rejected from the condenser. 5. Calculate the piston displacement of the compressor. 6. Calculate the work done on the compressor. 7. Calculate the compressor volumetric efficiency. 8. Calculate the Clearance Factor of the compressor. 9. Calculate the mechanical efficiency of the compressor. 10. Calculate the COP of the cycle. 11. Calculate the refrigeration efficiency of the system. No Thermocouple Location Temperature Pressure Enthalpy o C o F Psig Psia Btu/lb 1 Inlet to the Evaporator 2 Outlet from the Evaporator 5 Outlet from the Compressor 6 Inlet to the Condenser 7 Outlet form the Condenser 9 Inlet to the Compressor Mass flow rate of Refrigerant - 12 = Specific volume at the inlet to the compressor = Specific volume at the outlet to the compressor = lb/min ft 3 /lb ft 3 /lb

14 Experiment # 3 BASIC VAPOR COMPRESSION REFRIGERATION CYCLE ( With Heat Exchanger )

15 Experiment # 3 Mechanical Vapor Compression Refrigeration Cycle Object: Study the effect of sub-cooling and superheating in a vapor compression refrigeration cycle system. Equipment: Brodhead-Garrett Model 9501 Commercial Refrigeration Training Unit. Thermal Engineering Company, Toledo, Ohio, Serial # 662 Refrigerant 12. Components: 1. Reciprocating Compressor: Copeland Refrigeration Corporation Model KAN, Refrigerant 12 No. of Cylinders: 2, Bore: 11/8 inch, Stroke: 5/8 inch. RPM: 1750, Power: 1/2 HP. 2. Evaporator: Plate and tube type 24 pipes, 3/8 inches O.D, 63 plates, 8x3x0.006 in. thickness 3. Condenser: Similar to Evaporator. Theory: Some actual refrigeration systems utilize a liquid-to-suction heat exchanger. This heat exchanger sub-cools the liquid from condenser with suction vapor coming from the evaporator. The arrangement and the corresponding ph diagram are shown in the figures on last page of the handout. Effect of Superheating: 1. The compression work for the superheated cycle is more than that for the saturated cycle. 2. The temperature of the discharged vapors (exit of the compressor) is considerably higher for the superheated cycle. 3. Because of the higher outlet temperature of the compressed vapors, greater quantity of heat must be rejected in the condenser. It means load on the condenser increases. 4. Superheating of the vapors at the inlet to the compressor ensures no liquid enters in the compressor. Effect of sub-cooling: 1. Sub-cools liquid enters in the evaporator and hence increases the refrigeration effect of the cycle. Compared with the standard vapor compression cycle, the system using the heat exchanger may seen to have obvious advantages because of the increases refrigerating effect. But on the other hand, compressor power also increases because of the superheated vapors has to be compressed. Therefore, the COP of the cycle, which is ratio of the refrigeration effect to the compressor power of the cycle, not necessarily increases. The heat exchanger is definitely justified, however in situations where the vapor entering the compressor must be superheated to ensure that no liquid enters the compressor.

16 Another practical reason for using the heat exchanger is to sub-cool the liquid form the condenser to prevent bubbles of vapor from impending the flow of refrigerant through the expansion valve. Also read text book, Article #3.5 and Example #3.5 Page # 58, and solve problem #3.5. Experimental Results: The temperature and pressure at different locations are indicated by the use of following thermocouples and pressure gauges. #1 Inlet to Evaporator #2 Outlet from Evaporator #8 Inlet to the Heat Exchanger, Outlet form Evaporator #9 Outlet from the Heat Exchanger, Inlet to the Compressor #5 Outlet from the Compressor #6 Inlet to Condenser #7 Outlet from Condenser #10 Outlet form the Condenser, Inlet to the Heat Exchanger #11 Outlet form the Heat Exchanger, Inlet to the Evaporator Lab Report: 1. Fill the attached tables (#1 and #2) and the followings for both cycles. 2. Plot the cycle on a p-h diagram. 3. Calculate the power input to the compressor. 4. Calculate the capacity of the evaporator in tons of refrigeration. 5. Calculate the amount of heat rejected from the condenser. 6. Calculate the COP of the cycle. 7. Fill table # 3. Table # 1 Data without Heat exchanger No Thermocouple Location Temperature Pressure Enthalpy o C o F Psig Psia Btu/lb 1 Inlet to the Evaporator 2 Outlet from the Evaporator 5 Outlet from the Compressor 6 Inlet to the Condenser 7 Outlet form the Condenser 9 Inlet to the Compressor Mass flow rate of Refrigerant - 12 = lb/min

17 Table # 2 Data with Heat exchanger No Thermocouple Location Temperature Pressure Enthalpy o C o F Psig Psia Btu/lb 1 Inlet to the Evaporator 2 Outlet from the Evaporator 8 Inlet to the Heat exchanger Outlet from the Evaporator 9 Outlet form the Heat exchanger, Inlet to the compressor 5 Outlet from the Compressor 6 Inlet to the Condenser 7 Outlet form the Condenser 10 Outlet form the condenser, Inlet to the Heat exchanger 11 Outlet form the Heat exchanger, Inlet to the Evaporator Mass flow rate of Refrigerant - 12 = lb/min Table # 3 Comparison of results Compressor power Without Heat exchanger With Heat exchanger % Difference Refrigeration effect Condenser load COP

18

19 Experiment # 4 BASIC VAPOR COMPRESSION REFRIGERATION CYCLE ( With two Evaporators at same temperature)

20 Experiment # 4 Mechanical Vapor Compression Refrigeration Cycle with two evaporators Object: Study of a vapor compression refrigeration system with two evaporators in parallel operating at same temperatures. Equipment: Brodhead-Garrett Model 9501 Commercial Refrigeration Training Unit. Thermal Engineering Company, Toledo, Ohio, Serial # 662 Refrigerant 12. Components: 1. Reciprocating Compressor: Copeland Refrigeration Corporation Model KAN, Refrigerant 12 No. of Cylinders: 2, Bore: 11/8 inch, Stroke: 5/8 inch. PRM: 1750, Power: 1/2 HP. 2. Evaporator: Plate and tube type 24 pipes, 3/8 inches O.D, 63 plates, 8x3x0.006 in. thickness 3. Condenser: Similar to Evaporator. Theory: In many applications, more than one evaporator units are used in a single refrigeration system because of the necessity of refrigeration more than one spaces. The nature of application may require the evaporators to be maintained at the same or even at a different temperatures. For example, an industry which needs low-temperature refrigeration for a process and which must also provide air conditioning for some offices or a supermarket required air-conditioning for the shop and as well as refrigeration for the food storage. The temperature and pressure at the different locations are indicated by the use of the followings thermocouples and pressure gauges. #1 Inlet to the Evaporator #1 #2 Outlet from the Evaporator #1 #3 Inlet to the Evaporator #2 #4 Outlet form the Evaporator #2 #9 Inlet to the Compressor #5 Outlet form the Compressor #6 Inlet to the Condenser #7 Outlet from the Condenser

21 Lab Report: 1. Fill the attached table. 2. Plot the cycle on a p-h diagram. 3. Calculate the power input to the compressor and the COP of the cycle. 4. Solve problem # 3.8 from the textbook. No Thermocouple Location Temperature Pressure Enthalpy o C o F Psig Psia Btu/lb 1 Inlet to the Evaporator # 1 2 Outlet from the Evaporator # 1 3 Inlet to the Evaporator # 2 4 Outlet of the Evaporator # 2 5 Outlet from the Compressor 6 Inlet to the Condenser 7 Outlet from the Condenser 9 Inlet to the Compressor Mass flow rate of Refrigerant R-12 through evp # 1= Mass flow rate of Refrigerant R-12 through evp # 2= lb/min lb/min

22 Experiment # 5 Numerical Experiment # 1 SYSTEM SIMULATION OF BASIC VAPOR COMPRESSION REFRIGERATION CYCLE

23 Numerical Lab # 1 Objective: Analysis of a vapor compression refrigeration system for refrigerant R12 by using mathematical simulation system. Theory: The expected trends of refrigeration capacity and power requirements as function of the evaporating and condensing temperatures are given as follows. qe = C 1 + C 2 te + C 3 te 2 + C 4 tc + C 5 tc 2 + C 6 tetc + C 7 te 2 tc + C 8 tetc 2 + C 9 te 2 tc 2 (1) and P = D 1 + D 2 te + D 3 te 2 + D 4 tc + D 5 tc 2 + D 6 tetc + D 7 te 2 tc + D 8 tetc 2 + D 9 te 2 tc 2 (2) Rate of heat rejection from the condenser can be obtained for the following relation. qc = qe + P (3) The performance of the condenser can also be calculated form qc = F ( tc t amb ) (4) Where t amb is ambient temperature and F = 9.39 kw/k for air-cooled condenser. Lab Work: Assume t amb = 40, 35, 30, 25 o C. For each t amb, consider te = 10, 15, 20, 25 o C, calculate qe, P, qc and tc. Use Excel. You need to do iteration by using a macro. The procedure for making a macro in Exel is given as follows. How to make a Macro: To make a new macro, select tool and then record macro, Do the operation you want and then stop recording. Edit the macro and write for I = 1 to 50 in first line and next I in last line. To run the macro, select tool and macro and run the required macro. After calculating, plot characteristic curves 1/COP vs 1/qe. Constants for R-12 are given as follows. C 1 = 137.402 D 1 = 1.00618 C 2 = 4.60437 D 2 = -0.893222 C 3 = 0.061652 D 3 = -0.01426 C 4 = -1.118157 D 4 = 0.870024 C 5 = -0.001525 D 5 = -0.0063397 C 6 = -0.0109119 D 6 = 0.033889 C 7 = -0.00040148 D 7 = -0.00023875 C 8 = -0.00026682 D 8 = -0.00014746 C 9 = 0.000003873 D 9 = 0.0000067962

24 137.402 1.00618 Solution of the Lab 4.60437-0.893222 0.061652-0.01426-1.118157 0.870024-0.001525-0.0063397-0.0109119 0.033889-0.00040148-0.00023875-0.00026682-0.00014746 0.000003873 6.7962E-06 Ta Te Tc Qe P qc tc 1/COP 1/qe 40 10 55.16693 108.1082 34.30931 142.4175 55.16693 0.317361 0.00925 40 15 57.41366 126.2108 37.30348 163.5142 57.41366 0.295565 0.007923 40 20 59.84616 145.5353 40.82019 186.3555 59.84616 0.280483 0.006871 40 25 62.48035 165.9146 45.17584 211.0905 62.48035 0.272284 0.006027 35 10 50.75068 115.4857 32.41319 147.8989 50.75068 0.280668 0.008659 35 15 53.03702 134.4669 34.90068 169.3676 53.03702 0.259548 0.007437 35 20 55.50108 154.7023 37.80286 192.5051 55.50108 0.244359 0.006464 35 25 58.15619 176.0268 41.40984 217.4367 58.15619 0.235247 0.005681 30 10 46.29373 122.7807 30.21743 152.9981 46.29373 0.246109 0.008145 30 15 48.61806 142.6219 32.2017 174.8236 48.61806 0.225784 0.007012 30 20 51.11386 163.7546 34.50455 198.2592 51.11386 0.210709 0.006107 30 25 53.79169 186.0144 37.38955 223.404 53.79169 0.201004 0.005376 25 10 41.79493 129.9908 27.71361 157.7044 41.79493 0.213197 0.007693 25 15 44.15555 150.6728 29.1978 179.8706 44.15555 0.193783 0.006637 25 20 46.68327 172.6889 30.91694 203.6059 46.68327 0.179033 0.005791 25 25 49.38568 195.8738 33.10774 228.9815 49.38568 0.169026 0.005105 1/qe 1/COP 0.009249995 0.31736093 0.007923255 0.29556496 0.006871186 0.28048309 0.006027196 0.27228365 0.008659082 0.28066846 0.007436774 0.25954845 0.006464029 0.24435877 0.005680952 0.23524729 0.008144605 0.24610907 0.007011545 0.22578371 0.006106698 0.21070887 0.005375928 0.20100354 0.007692852 0.2131967 0.006636899 0.19378282 0.005790759 0.17903254 0.005105328 0.16902588

25 Experiment # 6 Numerical Experiment # 2 SYSTEM SIMULATION OF BASIC VAPOR COMPRESSION REFRIGERATION CYCLE

26 Numerical Lab # 2 Objective: Analysis of a vapor compression refrigeration system for refrigerant R-12 by using mathematical simulation system. Theory: This lab is an extension of numerical lab 1. In lab 1, the performance of the condenser is evaluated with the ambient while for the evaporator, it was assumed that the temperature is given. In practical applications, instead of evaporator temperature, the sourounding temperature of the evaporator is known and it effects the evaporator temperature. If twi is the sourounding temperature of the evaporator, the evaporator capacity qe with respect to twi, can be obtained as, qe = G ( twi te ) Assume G = 1 + 0.046 ( twi te ) for a given evaporator. If we include this in lab 1, the complete model will look like as, qe = C 1 + C 2 te + C 3 te 2 + C 4 tc + C 5 tc 2 + C 6 tetc + C 7 te 2 tc + C 8 tetc 2 + C 9 te 2 tc 2 (1) and P = D 1 + D 2 te + D 3 te 2 + D 4 tc + D 5 tc 2 + D 6 tetc + D 7 te 2 tc + D 8 tetc 2 + D 9 te 2 tc 2 (2) qc = qe + P (3) qc = F ( tc t amb ) or tc = t amb + ( qc/f ) (4) F = 9.39 kw/k for air-cooled condenser. qe = G ( twi te ) Where G = 1 + 0.046 ( twi te ) (5) This equation is used for te, te = twi ( qe/g ) (6) Lab Work: Assume t amb = 40, 35, 30, 25 o C. For each t amb, consider twi = 10, 15, 20, 25 o C, calculate qe, P, qc, tc and te. Use Excel. You need to do iteration by using a macro. The procedure for making a macro in Exel is given as follows. How to make a Macro: To make a new macro, select tool and then record macro, Do the operation you want and then stop recording. Edit the macro and write for I = 1 to 50 in first line and next I in last line. To run the macro, select tool and macro and run the required macro. After calculating, plot characteristic curves 1/qe vs 1/COP. Constants for R-12 are given as follows. C 1 = 137.402 D 1 = 1.00618 C 2 = 4.60437 D 2 = -0.893222 C 3 = 0.061652 D 3 = -0.01426 C 4 = -1.118157 D 4 = 0.870024 C 5 = -0.001525 D 5 = -0.0063397 C 6 = -0.0109119 D 6 = 0.033889 C 7 = -0.00040148 D 7 = -0.00023875 C 8 = -0.00026682 D 8 = -0.00014746 C 9 = 0.000003873 D 9 = 0.0000067962

27 137.4 1.01 Solution of the Lab 4.604-0.89 0.062-0.01-1.118 0.87-0.002-0.01-0.011 0.03-4E-04-0 -3E-04-0 4E-06 0 Ta tei te Tc qe P qc tc g te 1/qe 1/COP 40 10 0.805 51.487 78.51 29.36 107.87 51.487 1.423 0.8049 0.012738 0.37399 40 15 4.744 52.994 90.56 31.45 122.01 52.994 1.4718 4.7443 0.011042 0.347268 40 20 8.663 54.596 103.5 33.56 137.06 54.596 1.5215 8.6631 0.009662 0.324304 40 25 12.57 56.299 117.2 35.8 153.04 56.299 1.5719 12.568 0.008529 0.305336 35 10 0.395 46.83 83.09 27.99 111.08 46.83 1.4418 0.3953 0.012035 0.336875 35 15 4.315 48.357 95.62 29.8 125.42 48.357 1.4915 4.3151 0.010458 0.311625 35 20 8.215 49.976 109 31.58 140.62 49.976 1.5421 8.215 0.009171 0.289614 35 25 12.1 51.691 123.3 33.42 156.73 51.691 1.5933 12.102 0.00811 0.271076 30 10 0.002 42.14 87.58 26.41 113.99 42.14 1.4599 0.0016 0.011418 0.301566 30 15 3.904 43.683 100.6 27.93 128.49 43.683 1.5104 3.9039 0.009944 0.277741 30 20 7.787 45.318 114.4 29.38 143.83 45.318 1.5618 7.7866 0.008737 0.256748 30 25 11.66 47.045 129.2 30.85 160.06 47.045 1.6138 11.657 0.00774 0.2388 25 10-0.378 37.419 91.99 24.62 116.61 37.419 1.4774-0.378 0.010871 0.267626 25 15 3.509 38.975 105.4 25.84 131.22 38.975 1.5286 3.5092 0.009489 0.245156 25 20 7.376 40.622 119.7 26.96 146.69 40.622 1.5807 7.3763 0.008352 0.225219 25 25 11.23 42.361 134.9 28.07 163.02 42.361 1.6334 11.23 0.00741 0.207991 1/qe 1/COP 0.013 0.37 0.011 0.35 0.01 0.32 0.009 0.31 0.012 0.34 0.01 0.31 0.009 0.29 0.008 0.27 0.011 0.3 0.01 0.28 0.009 0.26 0.008 0.24 0.011 0.27 0.009 0.25 0.008 0.23 0.007 0.21

28 Experiment # 7 COMPUTER LINKED VAPOR COMPRESSION REFRIGERATION CYCLE ( Demonstration )

29 Experiment # 7 Computer Linked Mechanical Vapor Compression Refrigeration Cycle Object: Demonstration of the computer linked vapor compression refrigeration cycle and production of the cycle diagram under various conditions. Equipment: RC 712 Computer Linked Refrigeration Laboratory Unit ( Fig. 1 ). Introduction: The refrigeration cycle in various forms applications in countless industrial and domestic situations throughout the world. For example, the storage and transport of perishable foodstuff and drugs would be extremely difficult if not impossible without refrigeration. Similarly the efficient operation of offices and factories in many parts of the world would be impossible without the use of refrigeration plants in air conditioning systems. The most common type of refrigerator operates on the vapor compression cycle and require a work input. The Hilton Computer Linked Refrigeration Laboratory Unit RC 712 is a vapor Compression Refrigerator of this form and has been designed to enable students to safely study the cycle in detail. Description: Glass reinforced plastic panel houses a belt driven twin cylinder reciprocating compressor and electric motor. The motor is mounted on trunnions and connected via an arm to a load cell dynamometer. With a tachometer mounted on the compressor and instrumentation connected to the on board interface, it is possible to compute the power necessary to drive the compressor. Refrigerant 134 (a) vapor is drawn into the compressor from the electrically heated evaporator mounted on the front of the panel. Work is done on the gas and its pressure and temperature are raised. This hot gas at high pressure discharges from the compressor and flows into the water cooled condenser. Also mounted on the front of the panel. A measured and controllable flow of cooling water passes through a copper coil sealed inside the condenser cylinder. The hot gas desuperheats and then condenses on the surface of the cooing coil where it runs down to the bottom of the cylinder and is subcooled. A sight glass at the base of the condenser cylinder allows this liquid gas to be observed. The liquid flows through a refrigerant filter drier and an electronic paddle type flow-meter to the thermostatic expansion valve. Here it passes through a controlled orifice which allows its pressure to fall from that of the condenser to that of the evaporator. The liquid immediately starts to boil and takes in heat to accomplish this at low temperature. In order to allow control and measurement of the heat input at the evaporator, an electric heater element is used. This is rolled concentrically inside the copper tube carrying the low temperature liquid vapor mixture from the expansion valve. The power to the heaters is varied by a burst firing switch controlled by the electronic interface inside the panel. This in turn is controlled by the Hilton computer software. The relative power supplied to the evaporator is indicated visually by a panel mounted neon that lights when power is applied to the heaters. Hence at full power the light is on all of the time and at reducing power it flashes off for an increasing length of time unit at the no load condition it is off continuously. Measurements of the voltage, current, and hence power is carried out by voltage and current transformers also connected to the interface. The sensing bulb of the thermostatic expansion valve is mounted on the exit pipe from the evaporator and this detects the

30 degree of superheat of the gas leaving the evaporator and entering the compressor. If the superheat is low the valve will close and reduce the flow and if too high the reverse will occur. By this means stability is maintained under all condition of operation. measurements of condenser and evaporator pressures is achieved using two diaphragm type transducers connected directly to the interface. PROCEDURE: 1. Turn on the water supply to the RC 712 and, ensuring that the drain pipe is installed in a suitable drain, open the water control valve several turns to allow a moderate flow of water through the condenser coils. 2. Turn on the water supply to the RC 712 and depress the Main Switch on the panel. The compressor will start and the load Lamp, System Sampling and Interface Status Lamps should flash on momentarily and then go out. 3. Turn on the power supply to the Hilton Computer and switch on the Printer, Video and computer in that order. 4. After a short delay, the computer will issue an audible beep and the screen will display a warning message and the software version number. 5. After a further short delay the computer will issue four audible beeps and flash the system status light on and off at the same time. This indicated that both computer and interface are in communication. When this is completed the system status light remains on until the unit is either switch off, or power supply interrupted. Note that if either the cooling water flow is insufficient, or the computer is not turning for some reason, the unit will display a message to that effect until the fault is rectified. 6. The screen display changes to await input from the keyboard by the operator. Four entries are requested: (i) R12 Flowmeter Calibration Factor (found engraved on R12 Flowmeter body) (ii) Water Flowmeter Calibration Factor (found engraved on water flowmeter body) (iii) Compressor Friction Force: This is the motor load cell, or dynamo meter load when the compressor suction valve is closed and the compressor is doing no net work o the R12 gas. The typical value is 5 Newtons. (iv) atmospheric Pressure: this is the local atmospheric pressure in units of mm of Mercury e.g. typically 750 mm of Hg. When the last of these has been entered, Master Menu will be displayed as follows: No. 1. Display Schematic Diagram and System Parameters at 60 second internals with Optional Data Printout. 2. Display Transient Data - With Optional Data Printout. 3. Display Refrigeration Cycle Diagram and update at 60 second intervals With Optional Data Printout. 7. Select the above mentioned option one by one and study the different parameters of the vapor compression cycle.

31 Lab Report: 1. Give comments on the vapor compression refrigeration system.

32 Experiment # 8 COMPUTER LINKED VAPOR COMPRESSION REFRIGERATION CYCLE ( Energy Balance )

33 Experiment # 8 Computer Linked Mechanical Vapor Compression Refrigeration Cycle Object: Production of an energy balance for a vapor compression refrigeration cycle system. Equipment: RC 712 Computer Linked Refrigeration Laboratory Unit. Theory: The Ideal Vapor Compression Cycle: The idealized plant, T-S and P-H diagrams for the vapor compression cycle are shown in Fig 1. The cycle is as follows: 1-2 Isentropic compression of the vapor from the evaporating to the condensing pressure. 2-3 Condensation of the high pressure vapor during which heat is transferred to the high temperature region. 3-4 Adiabatic throttling of the condensed vapor from the condensing to the evaporating pressure. 4-1 Evaporation of the low pressure liquid during which heat is absorbed from the low temperature source. The energy transfers can readily be determined from the p-h diagram: W 1-2 = h 2 - h 1 q 2-3 = h 2 - h 3 h 3 = h 4 (Throttling process) q 1-4 = h 1 - h 4 The Practical Vapor Compression Cycle: The practical cycle differs from the idealized cycle in the following ways: (a) Due to friction, there will be a small pressure drop between the compressor discharge and expansion inlet, and between the expansion valve outlet and the compressor suction (b) The compression process is neither adiabatic nor reversible. There is usually a net heat loss from the compressor and there are frictional effects in piston rings, bearings,etc. (c) The vapor leaving the evaporator is usually slightly superheated. This makes possible automatic control of the expansion valve and improves compressor performance. (d) The liquid leaving the condenser is usually slightly subcooled, i.e., it is reduced to a temperature below the saturation temperature corresponding to its pressure. This improves the COP and reduces the possibility of the formation of vapor, due to pressure drop, in the line leading from the condenser to the expansion valve. (e) There may be small unwanted heat inputs from the surroundings to all parts of the cycle which operate below ambient temperature.

34 These effects are shown in the following Figures.

35 PROCEDURE: (1) Start the apparatus as explain in Exp # 7. (2) Enter program 1 and increase the evaporator load up to 45%. At this stage do not select hard copy of the data. (3) Allow the unit to stabilize. (4) Observe the four main components, i.e., evaporator, compressor, condenser and expansion valve, the connecting pipe work and the instrumentation. (5) Return to the menu, select program 1 and select a print out for the data. (6) Increase the evaporator load by 10% and repeat the steps from 3 to 5. Useful Data: Compressor: Two Cylinder Single Acting Bore = 40 mm, Stroke = 30 mm Swept Volume = 75.5 cm 3 per revolution Clearance Volume / Swept Volume = 0.025 Belt Pulley Ration (D/d) = 3.17 Typical Compressor Friction Force = 5 N Torque arm radius (Dynamometer) = 0.165 m Mean heat transfer area for Heat Exchanger = 0.75 m2 ENERGY BALANCE: Evaporator: Evaporator heat input Qe = Ve Ie Enthalpy change rate = m r ( h 1 - h 4 ) Heat leaks from/to evaporator = Difference of the above Condenser: Heat Transfer to cooling water = m w cp w ( t 6 - t 5 ) Enthalpy change rate = m r ( h 3 - h 2 ) Heat loss from condenser to ambient = Difference of the aobve Compressor: Shaft Power Ps = Tw Friction Power Pf = T f w Indicated Power Pi = Ps - Pf Enthalpy change rate = m r ( h 2 - h 1 ) Energy loss form compressor to ambient = Indicated power - enthalpy change rate. Mechanical Efficiency of the Compressor = ( Ps/Pi ) * 100

36 Lab Report: 1. When the heat input to the evaporator is increased by 10% what effect did you observe on the following parameters. ( a ) The evaporator pressure and temperature. ( b ) The refrigerant flow rate ( c ) The condenser pressure ( d ) The cooling water outlet temperature ( e ) The electric motor load cell force (torque) and power consumption. 2. Perform energy balances as given in hand out and hence calculate, for both load ( a ) Capacity of the evaporator in Tons of refrigeration. ( b ) COP of the system ( c ) Heat loss from the compressor to the ambient ( d ) Heat loss from the condenser to the ambient ( e ) Mechanical efficiency of the compressor 3. What effect does COP have with load and why?

37 Experiment # 9 COMPUTER LINKED VAPOR COMPRESSION REFRIGERATION CYCLE (Overall Heat Transfer Coefficient)

38 Experiment # 9 Computer Linked Mechanical Vapor Compression Refrigeration Cycle Object: Determination of the overall heat transfer coefficient for the condenser cooling coil. Equipment: RC 712 Computer Linked Refrigeration Laboratory Unit. Theory: Condenser receive superheated refrigerant from the compressor, remove the superheat, and then liquefy the refrigerant. Some abundant fluid such as air or water carries away the heat; this fluid characterizes the condenser as air-cooled or water-cooled. A water cooled condenser is used in RC 712 computer linked refrigeration laboratory unit. As an estimate, between 80 and 90% of the total heat transfer is occurring during the phase change. This takes place at constant temperature, t sat, corresponding to the pressure in the condenser. Therefore it is reasonable to use the temperature difference existing between the R134 saturation temperature at the condenser pressure and cooling water in order to determine the overall heat transfer coefficient. Log Mean Temperature Difference of the Condenser (LMTD): T 1 = t c - t 5 Tcondensing T 2 = t c - t 6 t 6 LMTD = T T T1 ln T 1 2 2 t 5 Heat Transfer from the condenser = m& ( h) UA( LMTD) LAB REPORT: r condensere = (1) Run the system for three different cooling water flow rates and calculate the followings. Heat Transfer in the condenser Log Mean Temperature Difference Overall Heat Transfer Coefficient Mean heat transfer area for Heat Exchanger A = 0.75 m 2 (2) Fill table # 1 and plot the Overall heat transfer coefficient with water flow rate and condensing temperature.

39 Table # 1 Water Flow Rate Q m 3 /sec Condensing Temperature Tc o C Overall Heat Transfer Coeff. U W/m 2 K

40

0 0 100 100 41

42 Experiment # 10 Introduction to Cryogenics

43 Experiment # 10 Ultralow-Temperature Refrigeration: Cryogenics Object: Introduction to Cryogenics system by visiting Cryogenics lab in Physics Department. Introduction: The lowest temperature achieved by refrigeratrion is about 250 o F (157 o C). This can be produced by cascade vapor-compression arrangement. It means to achieve absolute zero, a further 210 o F (116 o C) decrease in temperature is required. Such a low temprature can be achieved by liquification of gases. The term Crygenics is used to describe such applications. Applications: Most common use of this ultralow-temperatuer is to liquify the gases. Separation of atmospheric air is done by crygenics to obtain oxygen, nitrogen, argon, neon, xenon, and krypton. Hydrogen is obtained by sepration of coke oven gas and pure helium is commonly obtained by separation from helium-bearing natural gas. Problems: There are few problems in achieving such a low temperature. The first is the insulation problem to maintain the liquid gas at this low temperature. The second is the measurment of such low temperatures. Special considerations are required to ovide these two problems. Methods of Cryogenics: The ultralow-temperature can be produced by the expansion of gases. The expansion can be done by the following two ways. 1) When the gas performed work during expansion process, it produces cooling. 2) By throttling of a gas, it produces cooling. Cooling produces by throttling of a gas is of major importance. This will be discussed in detaile here. Following two quantities are important in throttling of a gas. Jule-Thomson Coefficient: It is defined as T µ = P For a perfect gas, Jule-Thomson coefficient is always zero. However for a real gas, it is not a constant and function of temperature and pressure. The value of µ is important to determine the trend of the temperature of a gas. If µ = 0, the temperature of the gas remains constant. If µ > 0, the temperature of the gas decreases. If µ < 0, the temperature of the gas increases. It means for producing cooling, a large positive value of µ is required. Inversion temperatue: The temperature for which µ = 0 for a real gas is called inversion temperature for the gas. h

44 Cycles for Liquification of Gases 1. Linde Air Liquefaction cylce: The block and T-S diagram for the eycle is shown in Fig. 1 and 2. The processes involved are explained as follows. 1-2 Isothermal compression (Discharge pressure is about 50 to 200 atmosphere and temperature is less than the inversion temperature) 2-3 Cooling at constant pressure in heat exchanger. 3-4 Throttling process in throttling valve. 4-5 Separation of liquid gas in separator. 4-6 Separation of gas in separator. 6-7 Heating at constant pressure in heat exchanger. FIG. 1 COMPONENT DIAGRAM

FIG. 2 T-S DIAGRAM 45

46 2. Claude Air Liquefaction Cycle: The components of a basic claude air liquefaction cycle are shown in Fig. 3 and T-S diagram in Fig. 4. The following are the processes in the cycle. 1-2 Isothermal compression. 2-3 Cooling at constant pressure in heat exchanger 1. 3-4 Cooling at constant pressure in heat exchanger 2. 3-8 Expansion in an expander (non-isentropic). 4-5 Throttling process in throttling valve. 5-6 Separation of liquid gas in separator. 5-7 Separation of gas in separator. 7-9 Mixing of gas coming from expander and separator. 9-10 Heating at constant pressure in heat exchanger 2. 10-11 Heating at constant pressure in heat exchanger 1. FIG. 3 COMPONENT DIAGRAM FIG. 4 T-S DIAGRAM