Experimental Investigation of an Environmental Control Unit Utilizing Carbon Dioxide for Heating and Cooling

Transcription

1 University of Illinois at Urbana-Champaign Air Conditioning and Refrigeration Center A National Science Foundation/University Cooperative Research Center Experimental Investigation of an Environmental Control Unit Utilizing Carbon Dioxide for Heating and Cooling S. S. Wujek and P. S. Hrnjak ACRC TR-25 September 2006 For additional information: Air Conditioning and Refrigeration Center University of Illinois Department of Mechanical Science & Engineering 206 West Green Street Urbana, IL 680 Prepared as part of ACRC Project #6 Experimental and Modeling Investigation of Two Systems (2) P. S. Hrnjak, Principal Investigator

2 The Air Conditioning and Refrigeration Center was founded in 988 with a grant from the estate of Richard W. Kritzer, the founder of Peerless of America Inc. A State of Illinois Technology Challenge Grant helped build the laboratory facilities. The ACRC receives continuing support from the Richard W. Kritzer Endowment and the National Science Foundation. The following organizations have also become sponsors of the Center. Arçelik A. S. Behr GmbH and Co. Carrier Corporation Cerro Flow Products, Inc. Daikin Industries, Ltd. Danfoss A/S Delphi Thermal and Interior Embraco S. A. Emerson Climate Technologies, Inc. General Motors Corporation Hill PHOENIX Honeywell, Inc. Hydro Aluminum Precision Tubing Ingersoll-Rand/Climate Control Lennox International, Inc. LG Electronics, Inc. Manitowoc Ice, Inc. Matsushita Electric Industrial Co., Ltd. Modine Manufacturing Co. Novelis Global Technology Centre Parker Hannifin Corporation Peerless of America, Inc. Samsung Electronics Co., Ltd. Sanden Corporation Sanyo Electric Co., Ltd. Tecumseh Products Company Trane Visteon Automotive Systems Wieland-Werke, AG For additional information: Air Conditioning & Refrigeration Center Mechanical & Industrial Engineering Dept. University of Illinois 206 West Green Street Urbana, IL

3 Abstract This report details the tests performed on R22 box and R breadboard versions of the US Army Environmental Control Unit (ECU) with a nominal cooling capacity of 9,000 BTU/hr (0.5 Tons or 2.6 kw). An R22 ECU box currently used by the Army was tested and used as the baseline for comparison with the later R testing. This baseline system is generally used where portable air conditioning is needed in the field; the box is also equipped with electric resistance heaters. The baseline box was found to outperform its rated capacity. An R system was tested in a breadboard format, this system was comprised of components similar to what would be used in an ECU box if it were converted from R22. This breadboard system was capable of operating as both an air conditioner and as a heat pump. A cooling COP approximately equal to the box system was attained with the breadboard. The breadboard did provide a HPF greater than unity for all heat pump tests, making it is more efficient than the electric resistance heaters found in the box system. Detailed comparisons were made between several R heat exchangers in different orientations under varying operating conditions. s were tested in horizontal and vertical tube orientations. The flat top fin evaporator was found to work well in both horizontal and vertical orientations. The round top fin evaporator, which may have contained a manufacturing defect, performed much better with the tubes vertical, than with the tubes horizontal. Overall the best heat exchanger was the round top fin evaporator with the tubes vertical; however this heat exchanger performed the worst with the tubes horizontal. In air conditioning mode, both gas coolers performed similarly. In heat pump mode, the six-port tube heat gas cooler performed better in almost every regard in comparison to the four-port tube with the only exception being the air side pressure drop. iii

4 Table of Contents Page Abstract... iii List of Figures... v List of Symbols... vii Chapter. Introduction... Chapter 2. R22 ECU Box Baseline Tests Test setup Test results... Chapter 3. Breadboard Tests - Setup and Calibration Test setup Calibration... 3 Chapter. Breadboard System Results Cooling mode results Heating mode results Chapter 5. Heat Exchanger Performance in Breadboard Testing performance performance... 3 Chapter 6. Conclusions... Bibliography... 2 Appendix A. Raw air conditioning data... 3 Appendix B. Infrared Pictures... 2 Appendix C. EES code for air conditioning Appendix D. Raw heat pump data... 3 Appendix E. EES code for heat pump... 8 iv

5 List of Figures Page Figure. Schematic of R22 ECU box...3 Figure 2. Air Conditioning text matrix for R22 box tests...3 Figure 3. Schematic of air flow measurement setup... Figure. Air flow rate versus static pressure at evaporator exit... Figure 5. R22 box cooling capacity...5 Figure 6. R22 box COP...5 Figure. Flat top fins...6 Figure 8. Round top fins... Figure 9. Compressor... Figure 0. Insulation in outdoor chamber for heat pump tests...8 Figure. Laboratory...9 Figure 2. Schematic of system in air-conditioning mode...0 Figure 3. Schematic of system in heat pump mode... Figure. Indoor wind tunnel...2 Figure 5. Outdoor wind tunnel...2 Figure 6. Indoor differential air pressure versus voltage...5 Figure. Outdoor differential air pressure versus voltage...5 Figure 8. Absolute gas cooler outlet pressure versus voltage...6 Figure 9. Absolute evaporator outlet pressure versus voltage...6 Figure 20. Absolute compressor inlet pressure versus voltage... Figure 2. High side differential pressure versus voltage... Figure 22. Low side differential pressure versus voltage...8 Figure 23. Mass versus voltage...8 Figure 2. Data acquisition screenshot...9 Figure 25. Air side energy balance...20 Figure 26. Chamber side energy balance...20 Figure 2. Refrigerant side energy balance...2 Figure 28. Breadboard cooling mode test matrix...2 Figure 29. Breadboard heating mode test matrix...2 Figure 30. Capacity with varying high side pressures...23 Figure 3. COP with varying high side pressures...23 Figure 32. High side pressure control strategy...2 Figure 33. Cooling capacity...25 Figure 3. Coefficient of performance...25 Figure 35. Pressure-enthalpy diagram...26 Figure 36. Heating capacity from charge determination test...2 Figure 3. Heating capacity...28 Figure 38. Heating performance factor...28 v

6 Figure 39. Pressure-enthalpy diagram for heat pump testing...29 Figure 0. Indoor air discharge temperature...29 Figure. Air pressure drop across evaporator...30 Figure 2. air pressure drop versus volumetric flow rate...3 Figure 3. effectiveness for dry conditions...32 Figure. overall heat transfer coefficient...33 Figure 5. Infrared image of flat top fin evaporator with tubes horizontal...33 Figure 6. Infrared image of flat top fin evaporator with tubes vertical...3 Figure. Infrared image of round top fin evaporator with tubes horizontal...3 Figure 8. Infrared image of round top fin evaporator with tubes vertical...35 Figure 9. Water drainage rate for wet conditions...36 Figure 50. Percentage of water removed...36 Figure 5. Indoor heat exchanger approach temperature...3 Figure 52. approach temperature...38 Figure 53. air pressure drop versus volumetric flow rate...38 Figure 5. Correction Factor Plot (Bowman, Mueller, Nagle)...39 Figure 55. overall heat transfer coefficient...0 Figure 56. Outdoor heat exchanger effectiveness...0 vi

7 List of Symbols COP Coefficient of performance c pa Specific heat of air F Dimensionless correction factor h cpri Specific enthalpy of refrigerant at compressor inlet h cpro Specific enthalpy of refrigerant at compressor outlet h cpro,s Specific enthalpy of refrigerant at compressor outlet pressure with the same specific entropy as the compressor inlet h eai Specific enthalpy of dry air at evaporator inlet h eao Specific enthalpy of dry air at evaporator outlet h fg Latent heat of vaporization of water HPF Heating performance factor h steam Specific enthalpy of saturated steam at atmospheric pressure h water Specific enthalpy of water at evaporator air outlet temperature m Mass flow rate of dry air a m r Mass flow rate of refrigerant-oil mixture m Mass of water condensed w Q Heat transfer Q cooling Cooling capacity Q e,a Cooling capacity from air energy balance Q e,c Cooling capacity from chamber energy balance Q e,r Cooling capacity from refrigerant energy balance Q heating Heating capacity Q max Maximum theoretical heat transfer Q glycol Heat removed via chilled glycol Q trans Heat transmission through wall Q steam Latent load T cai Temperature of air at gas cooler inlet T cao Temperature of air at gas cooler outlet T cri Temperature of refrigerant at gas cooler inlet T cro Temperature of refrigerant at gas cooler outlet T eai Temperature of air at evaporator inlet T eao Temperature of air at evaporator outlet T eri Temperature of refrigerant at evaporator inlet T ero Temperature of refrigerant at evaporator outlet T wi Temperature, inside of chamber wall T wo Temperature, outside of chamber wall UA Overall heat transfer coefficient W blowers Electric power supplied to blowers W comp Compressor power W electric Total electric power supplied to chambers W heaters Electric power supplied to heaters W outlets Electric power supplied to outlets inside of chamber Mass concentration of oil in refrigerant x oil vii

8 Δh oil Δh r Δh R ΔT approach ΔT lmc ΔT lme ε hx η isen Change in specific enthalpy of pure oil Change in specific enthalpy of refrigerant oil mixture Change in enthalpy of pure R approach temperature Log mean temperature difference in gas cooler Log mean temperature difference in evaporator Heat exchanger effectiveness Isentropic efficiency of compressor viii

9 Chapter. Introduction This report details the tests performed on R22 Box and R Breadboard versions of the US Army Environmental Control Unit (ECU). The nominal cooling capacity of these units is 9,000 BTU/hr which is equivalent to 0.5 Tons or 2.6 kw. This work was done at the Air Conditioning and Refrigeration Center at the University of Illinois at Urbana-Champaign. The first section of this thesis deals with the testing of the R22 ECU box, which is used as the baseline for the later R testing. This baseline system is presently used by the US Army for numerous applications where mobile air conditioning is needed. Its primary function is as an air-conditioner; however, the box is also equipped with electric resistance heaters. The second section of this report details the testing methodology and calibration. Next, the system level testing of a breadboard, which is comprised of components similar to what would be used in an ECU box if it were converted from R22 to R, is discussed. Breadboard testing was conducted in both air-conditioning and heat-pump modes. The fourth section makes detailed comparisons between several R heat exchangers in several different orientations under varying conditions. Finally some conclusions from this experiment are recapitulated.

10 Chapter 2. R22 ECU Box Baseline Tests This section describes the test setup and results for the 9000 BTU/hr R22 box ECU. These tests were conducted in order to have adequate data from which comparisons to the R breadboard system can be made. 2. Test setup The main goal of the box testing was to determine the cooling capacity and coefficient of performance (COP) of the present, R22 system. The R22 box tests were conducted, using ASHRAE standard 6, in one environmental chamber which was partitioned by an interior wall into indoor and outdoor rooms. The air temperature and humidity on both sides of the wall could be independently controlled and measured. The sensible load could be added by means of electric resistance heaters and removed via heat exchange with a chilled glycolwater flow. The power to the electric resistance heaters, and all other electric power going into the chambers, was measured with watt transducers from Ohio Semitronics. The accuracy of these watt transducers is 0.02% of the reading. These devices were capable or recording up to 2 kw of electrical power, each. The electric power consumption of the box was used in the calculation of the COP. The electric power consumption of the evaporator and condenser blowers was measured. The blower power could be subtracted from the overall box consumption leaving the compressor power. The power going to the ECU was measured with a watt transducer from Ohio Semitronics which was accurate to within % of the full scale, 8 kw. The cooling from the glycol-water mixture was determined by multiplying the specific heat of the mixture by the mass flow rate of the mixture and the temperature difference between entering and exiting the chamber. The mass flow and density of the glycol-water mixture was measured with a Micromotion CMF-0 mass flow meter. The density measurement was used to determine the concentration of the glycol in solution. The latent load was determined by measuring the rate at which condensation fell into a bucket using a load cell. The heat transfer from both the indoor and outdoor rooms via conduction through the walls was determined. These tests were conducted by setting the duct blowers in both rooms to a fixed speed. As real blowers are inefficient, the room heated up. When the temperature remained constant, the power supplied to the blowers equaled the heat conduction through the walls. The heat leak from the indoor room to the outside of the chamber was 8.2 W/K. The heat leak from the outdoor room to the outside of the chamber was 8.9 W/K. Finally, the heat leak through the partition separating the indoor from outdoor rooms was 6.0 W/K. Knowing the electric heating, glycol cooling, latent load, and transmission through the walls allows the capacity to be determined by the chamber side energy balance. The temperature of the refrigerant before and after both the evaporator and gas cooler was measured with type-t immersion thermocouples. These thermocouples, from Omega, were covered in thermal paste in order to enhance conduction before being attached to the outside of the refrigerant piping. Insulation was placed around the thermocouples to decrease convection with the surrounding air. A schematic of the refrigerant circuit with temperature and power measurement locations labeled can be seen in Figure. No piping modifications were made to the box from the time when it was received. 2

11 Figure. Schematic of R22 ECU box The air temperature was measured at the inlet and exit of both the evaporator and condenser using type-t thermocouples which were joined in parallel in order to provide an average value. These thermocouples were arc welded in an Argon rich environment in order to decrease impurities in the metal. The humidity, before and after the evaporator, was measured with a chilled mirror dew point sensor from General Eastern. The air temperature and humidity were adjusted in order to meet the test conditions. The test conditions for the box tests can be seen in Figure 2. These conditions were selected to emphasis performance at high ambient conditions. No static pressure was added to the evaporator exit for the box testing. Thus, no duct work was attached to the air exit. Test Condition Outdoor Temp [ F/ C] Indoor Temp [ F/ C] Indoor Web Bulb [ F/ C] D 90/32.2 D2 0/3.3 DAS 20/8.9 Dry D3 25/5. W 90/ /32.2 W2 0/3.3 WAS 20/8.9 5/23.9 W3 25/5. Figure 2. Air Conditioning text matrix for R22 box tests Besides the capacity tests for the box ECU, it was also essential to measure the air flow rate in the R22 box in order that the same flow rate could be replicated in the R breadboard testing. Static pressure was added in varying degrees in order to produce a blower curve. The static pressure was measured with a water-filled U-tube manometer. The air flow rate was determined by connecting the air outlets of the evaporator and condenser to a box containing air flow rate nozzles and differential pressure transducers. This box was in turn connected to a blower 3

12 which could be adjusted via a variable frequency drive so as to compensate for the pressure drop caused by the nozzle box. A schematic of this setup can be seen in Figure 3. Figure 3. Schematic of air flow measurement setup 2.2 Test results The air flow rate of the box was measured using the schematic shown in Figure 3. This chart, known as a blower curve, shows the stiffness of the blower. Stiffness is a measure of how much the air flow rate is affected by changing pressure head on the blower. It was found that the air flow rate over the evaporator was 320 cubic feet per minute at room temperature with no static pressure. Figure shows the blower curve produced by increasing the static pressure at the evaporator exit. In this range, it can be seen that increasing static pressure causes a near linear decrease in air flow rate. In the field, an increase in static pressure in the evaporator air stream could be produced by such things as attaching duct work, water condensation, or, if below freezing, frosting. The condenser air flow rate was found to be 620 cubic feet per minute. A blower curve was not made for the condenser air flow since a duct is not attached to this stream in field operation nor does water condensation occur on this coil Air Flow Rate [CFM] Static Pressure [" water] Figure. Air flow rate versus static pressure at evaporator exit

13 The net cooling capacity for the R22 box can be seen in Figure 5. The net capacity is the cooling capacity of the evaporator minus the evaporator blower power, which was measured to be 2 W. The box exceeds the target capacity of 9000 BTU/hr (3/ tons or 2.6kW) at the rating point condition (W3) by about 30%. The net COP can be seen in Figure 6. The net COP is the net capacity divided by the total electrical energy supplied to the BOX. The condenser fan power was measured to be 23 W. In studying Figure 5 and Figure 6, it should be noted that the outdoor temperature is not equally spaced. The cooling capacity drops off quicker at the upper end of the outdoor temperature spectrum tested. 6 5 Net Cooling Capacity [kw] D D2 DAS D3 W W2 WAS W3 Figure 5. R22 box cooling capacity 2 Net COP [-] D D2 DAS D3 W W2 WAS W3 Figure 6. R22 box COP 5

14 Chapter 3. Breadboard Tests - Setup and Calibration This section details the test setup and calibration for the breadboard system. The purpose of this section is to provide information on how tests were designed and conducted. The information should allow the replication of the experiments. 3. Test setup The heat exchangers used for this testing were supplied by Modine. Part of the testing was to study the system effects caused by using different heat exchangers. Two different gas coolers were tested, the difference being that one had four-port tubes and the other had six-port tubes. Two different evaporators were tested. Both evaporators used six-port tubes. One evaporator had flat top fins, shown in Figure while the other had round top fins, shown in Figure 8. Both evaporators were tested in tubes horizontal and tubes vertical orientations. Figure and Figure 8 show the evaporators with the tubes running vertically. Typically, the manufacturer suggests using the evaporators so that the header is located at the bottom, with the flow vertically upward through the tubes. Figure. Flat top fins 6

15 Figure 8. Round top fins The compressor for this experiment was a prototype R compressor supplied by Tecumseh. The dimensions of the compressor were fourteen inches in both height and diameter. The compressor weighed approximately 23 pounds. The electrical power requirements for the compressor were 208 volts, 60 hertz, single phase, AC power. The compressor contained 00 grams of PAG oil. At standard conditions, the oil had a viscosity of 6 centistokes. The compressor discharge limits were MPa and 0ºC. The compressor can be seen in Figure 9. Figure 9. Compressor

16 The expansion valve used was a manual Swagelok needle valve. The pipe leading up to the expansion valve was insulated with polyethylene foam pipe insulation, and was thought of as being part of the adiabatic expansion mechanism. All piping connecting the components was thick walled 3/8 copper tube. /8 stainless steel tubing was used to connect to pressure gauges, which will be discussed later. Based on experience running an Army ECU with approximately double the capacity, it was decided to insulate the compressor, accumulator, and piping that were colder than ambient temperature in the outdoor chamber during heat pump testing. Closed-cell foam insulation was used on most of the pipes as well as the compressor and accumulator. Due to the high temperature of the refrigerant discharge lines, the pipe was insulated with fiberglass insulation commonly used on steam pipes (rated at 0.29 Btu/hr. x in./sq. ft.). Some of the insulation of these components for heat pump testing can be seen in Figure 0. Figure 0. Insulation in outdoor chamber for heat pump tests The environmental chamber used for the box testing was used in addition to another existing chamber which had been used to house the breadboard testing. An extensive amount of work was done in both chambers to update the electrical system in order to give more flexibility for future experiments. In order to minimize the effect of erroneous measurements on system results, three independent energy balances were utilized in this experiment for calculating cooling capacity. If there was close agreement, within 5%, between the three energy balances, then the data can be viewed as valid. All tests were conducted at steady state with the prescribed conditions. This procedure exceeds ASHRAE testing standards, which only call for two independent balances. A view of the lab showing the outdoor chamber, in addition to the computer and data logger can be seen in Figure. 8

17 Figure. Laboratory The first of these balances is the chamber side energy balance. For the experiment, there are two calorimetric chambers which are used to create the indoor and outdoor test conditions. The compressor, gas cooler, and an accumulator with an integrated suction-line heat exchanger are all located within the outdoor chamber. The evaporator is located in the indoor chamber. The expansion valve is just outside of the indoor chamber. A basic diagram of this test setup can be seen in Figure 2. The chamber is treated as a thermodynamic control volume, where the energy flows through the control volume are measured. Some of that energy transmission flows in the form of electrical power (W electrical ). Electrical power is used for a variety of sources which are summed in Equation. Electricity is used by blowers (W blowers ) to draw air through the heat exchanger and circulate air in order to make the room more homogeneous. Electric power is used in the PID-controlled resistance heaters (W heaters ) in order to control the room temperatures. Electrical power is also used for various small devices in the room which are plugged into outlets (W outlets ). W = W + W + W () electric blowers heaters outlets Steam is used to provide the latent load for the given test conditions. At steady state, the mass flow of steam equals the mass of water condensed ( m w ). The specific enthalpy of the steam (h steam ) entering the room is approximated by the fluid properties of saturated steam at atmospheric pressure. The specific enthalpy of water (h water ) is determined for atmospheric pressure and the refrigerant temperature at the evaporator exit. These terms are then summed to give the latent load (Q steam ) for the chamber balance as seen in Equation 2. Q steam w ( h h ) = m (2) steam water 9

18 The chambers can be forced to maintain a constant temperature by adding and removing heat. Heat can be removed by glycol (Q glycol ). The heat conduction through the walls of the chamber can be computed based on the temperatures on the interior and exterior faces of the wall (T wi and T wo, respectively) and the overall thermal conductivity (UA) of the chamber using Equation 3. The procedure for measuring the overall thermal conductivity can be found on Page. Q trans ( T ) = UA T (3) wi wo Summing all of the energy flows through the chamber walls allows for the calculation of heat exchanger capacity using only the chamber energy balance (Q e,c ) as shown in Equation. The form shown has positive values for all terms. Q = Q + W Q Q e, c steam electric trans glycol () For heat pump tests, all components in the outdoor chamber, including the refrigerant lines, were insulated, with the exception of the evaporator. A schematic of the system in heat pump mode can be seen in Figure 3. The electrical draw of the compressor goes both into compressing the refrigerant and to waste heat. Because of this, an independent chamber energy balance can not be computed for the outdoor chamber, as it requires overlapping information with the refrigerant balance. Figure 2. Schematic of system in air-conditioning mode 0

19 Figure 3. Schematic of system in heat pump mode The second energy balance is the air side energy balance, by which the change in the air s energy as it flows across the heat exchangers is measured. The indoor and outdoor wind tunnels can be seen in Figure andfigure 5 respectively. The air flow is directed through the wind tunnels, where it first passes through a grid of Type-T thermocouples. Next the air passes through the heat exchanger. Another thermocouple grid is located directly after the heat exchanger. These two grids allow for the calculation of the change in specific enthalpy of dry air by finding the difference of the specific enthalpy before (h eai ) and after (h eao ) the evaporator. The mass flow rate of the air ( m ) is calculated from differential static pressure measurements across air flow nozzles with known a diameters which are installed further downstream. Two nozzles, one with a three inch throat diameter and one with a two inch throat diameter, are used in the evaporator duct. One nozzle, with a six in throat diameter, is used in the gas cooler duct. The throat diameters of the nozzles were selected according to ASHRAE Standard Another set of thermocouples is located in the throat of the nozzles. The air flow rate is maintained at the test conditions by use of blowers that are controlled by variable frequency drives. For the indoor duct, General Eastern chilled mirror dew point sensors are located before and after the evaporator. This allows the change in the air s moisture content to be calculated. All of these measurements allow the calculation of heat exchanger capacity (Q e,a ) from air side measurements. The method of calculating capacity from air side measurements can be seen in Equation 5. Q = m ( h h ) + m h e, a a eai eao w fg (5)

20 Figure. Indoor wind tunnel Figure 5. Outdoor wind tunnel The third, and final, energy balance is the refrigerant side energy balance. One corriolis type mass flow meter from Micromotion is used to measure the mass flow rate of the refrigerant-oil mixture ( m ). The flow meter is located in the high pressure line between the suction line heat exchanger and the expansion valve. By this r 2

21 placement, any pressure drop can be seen as a continuation of the pressure drop which is experienced by the throttling in the expansion valve. Absolute and differential pressure sensors were used to determine the static pressure entering and leaving every major component. Type-T immersion thermocouples were placed in the center of the refrigerant stream before and after each component. The change in specific enthalpy, from component inlet to outlet, for both the refrigerant (Δh R ) and the oil (Δh oil ) can be determined from temperature and pressure measurements. Equation 6 shows this specific enthalpy difference. This equation shows that in the absence of enthalpy data for the R PAG oil mixture, the effects of oil were taken in account in two ways: One, by reducing the contribution from the refrigerant by the ratio of the oil in circulation. Two, by taking into account sensible heat contribution from the oil flow. This enthalpy change in the oil is calculated with a quadratic equation of state. The shortcoming of this method is that the heat of mixing was neglected. This methodology for taking into account the oil contribution was developed by Yin, et al. ( xoil ) ΔhR + xoilδhoil Δhr = (6) The locations for the pressure and temperature measurements, which can be used to find the specific enthalpies in Equation 6, are denoted by the dots in Figure 2 and Figure 3. Multiplying the refrigerant mass flow rate by the enthalpy difference across the heat exchangers yields the refrigerant side energy balance (Q e,r ), as seen in Equation. ( h ) Q e, r mr Δ e, r = () The cooling capacity (Q cooling ), which is presented throughout the paper, is the mean of the three independent chamber balances. The cooling capacity is calculated using Equation 8. Qe, a + Qe, c + Qe, r Qcooling = (8) 3 The coefficient of performance (COP) is method of calculating the system efficiency which relates cooling to the power requirement. It should be noted that for the breadboard tests, the coefficient of performance does take into account blower power. This can be seen in Equation 9, where the denominator is the compressor power (W comp ) and not the total power required to condition the air. Qcooling COP = (9) W comp Similarly, a heating efficiency called the Heating Performance Factor (HPF) can be determined using Equation 0. The theoretical minimum Heating Performance Factor is one. Qheating HPF = (0) W comp 3.2 Calibration In order to make the chamber balance, it is critical to know the heat conduction through the chamber walls (Q trans ). At steady state, when the temperature inside the chamber does not change relative to the temperature on the outside of the chamber, it is possible to determine the heat conduction through the walls. If the energy input to the environmental chambers comes exclusively from electrical power, then: 3

22 Q trans = W electric () In order to precisely determine the overall thermal conductivity of the chambers (UA), accurate wall temperatures measurements are required. The interior wall thermocouples in the chamber were replaced with new T-type thermocouples which were arc welded in an argon rich environment to ensure greater accuracy. Five of these thermocouples were attached to each interior face of the chamber; which yields 30 interior thermocouples per chamber. The thermocouples were all joined in parallel to give a mean temperature. Utilizing Equation, the thermal conductivity of the chambers was calculated by measuring the electrical power going into the chambers (W electric ) and the steady state difference in temperatures between the inside (T wi ) and outside (T wo ) of the chamber walls as seen in Equation2. ( T ) Welectric = UA Twi wo (2) All electric power going into the chambers for these calibrations was supplied to the blowers in order to create convection similar to what occurs during normal operation. Due to the insulation, which is roughly one foot thick, these tests take about three days to reach equilibrium. The outdoor chamber s thermocouple on the inside of the wall had a large noise, about / degree Celsius in amplitude, but by comparing it to other thermocouples in the outdoor chamber it was found to be statistically random noise and did not offset the value when averaged. The average UA value calculated for the indoor chamber was W/K. The average UA value for the outdoor chamber was 9. W/K. Note that these values are different than those from the box testing due to the change in layout. Besides determining the thermal conductivity of the walls, a great deal of instrument calibration had to be completed before tests could be conducted. To calculate the air side energy balance, the air flow rate must be calculated. Knowing the differential air pressure across the air flow nozzles in both ducts allows the air flow rate to be iteratively determined. The differential air pressure transducers were calibrated with a U-tube manometer which contained propanol (density=80 kg/m 3 ). The device was leveled and the offset measurement noted. The scale was labeled in hundredths of an inch. The blowers were raised and lowered in frequency between zero hertz and the frequency which yielded the upper limit of the transducers. The manometer reading and the output voltage were recorded and plotted. After the plot was complete, a linear least square curve fit was created. A plot of the indoor differential air pressure calibration can be seen in Figure 6. A similar plot for the outdoor wind tunnel can be seen in Figure. The curve fits shown in Figure 6 and Figure were then input into the data acquisition software to be utilized for future tests.

23 800 Differential pressure [Pa] y = x R 2 = Volts Figure 6. Indoor differential air pressure versus voltage 250 y = 9.08x R 2 = Differential pressure [Pa] Volts Figure. Outdoor differential air pressure versus voltage It is essential to know the refrigerant pressure at several locations in the thermodynamic cycle in order to determine the refrigerant side energy balance. The absolute refrigerant pressure transducers are used to determine the absolute pressure at the gas cooler outlet, the evaporator outlet, and the compressor inlet. The calibrations were conducted by increasing and decreasing the charge of nitrogen gas in the system. The pressure was measured using a Fluke 6 Pressure Calibrator, which measures the gauge pressure. To find absolute pressure, atmospheric 5

24 pressure was added to the gauge pressure. Figure 8 shows the absolute pressure at the gas cooler outlet versus voltage. Similarly Figure 9 and Figure 20 show the data for the evaporator outlet and the compressor inlet, respectively y = x R 2 = Pressure [kpa] Voltage Figure 8. Absolute gas cooler outlet pressure versus voltage y = x R 2 = Pressure [kpa] Voltage Figure 9. Absolute evaporator outlet pressure versus voltage 6

25 y = x R 2 = Pressure [kpa] Voltage Figure 20. Absolute compressor inlet pressure versus voltage In order to measure refrigerant pressure drop through the system components, differential pressure transducers were used. The refrigerant differential pressure transducers were calibrated in the same way as the absolute pressure transducers, except the low pressure side was open to the atmosphere. One differential pressure transducer was used on the high pressure side of the cycle, and one was used on the low pressure side. The differential pressure is plotted against the voltage for the high and low side differential pressure transducers in Figure 2 andfigure 22, respectively. 200 Differential Pressure [kpa] y = x R 2 = Voltage Figure 2. High side differential pressure versus voltage

26 200 Differential Pressure [kpa] y = x R 2 = Voltage Figure 22. Low side differential pressure versus voltage A load cell is used to measure the amount of water which is condensed out of the air by the evaporator. The load cell holds a bucket which collects the water. Several objects were weighed on an AND FP-6000 electronic balance. These objects were then placed in the bucket. The resulting voltages from the load cell, which correspond to different masses, were recorded. The masses of the measured items are plotted against the output voltage of the load cell in Figure 23. A least square curve fit, which is shown in the figure, creates a linear relation between mass and voltage y = x R 2 = Mass [g] Voltage Figure 23. Mass versus voltage 8

27 The data acquisition software, HPVEE, was programmed to show the most important variables, such as temperatures, pressure, and electric power on the screen while running tests. HPVEE also writes all measured data to Excel. Excel, using Refprex, shows an up to date pressure-enthalpy diagram based on the computer s second monitor using the data output by HPVEE. A screenshot of the HPVEE screen can be seen in Figure 2. Figure 2. Data acquisition screenshot Engineering Equation Solver (EES) was programmed to do the required data reduction for this system. EES contains the fluid property data for both the air and refrigerant. EES was programmed to provide the capacity from the three different energy balances, as well as other performance information. The error for the air, chamber, and refrigerant side energy balances, can be seen in Equations 3,, and 5. E E E a c r Qcooling Qe, a = 00% (3) Q colling Qcooling Qe, c = 00% () Q colling Qcooling Qe, r = 00% (5) Q colling 9

28 The air, chamber, and refrigerant side energy balances for all tests conducted are plotted against the mean of the energy balances (Q cooling ) in Figure 25, Figure 26, andfigure 2. As can be seen, most data is within 5% error, and almost all data points are within 0% error. Those tests points with large errors were re-run and the smaller error test points were utilized for the remainder of this report. Air side cooling capacity [kw] No Error 5% Error 0% Error Mean cooling capacity [kw] Figure 25. Air side energy balance Chamber side cooling capacity [kw] No Error 5% Error 0% Error Mean cooling capacity [kw] Figure 26. Chamber side energy balance 20

29 Refrigerant side cooling capacity [kw] No Error 5% Error 0% Error Mean Cooling Capacity [kw] Figure 2. Refrigerant side energy balance The test matrix used for cooling mode can be seen in Figure 28. All of the points used for breadboard testing were used in the box testing, but without the box test s DAS and WAS conditions. These test conditions were eliminated due to their closeness to conditions D3 and W3, respectively. The test matrix used for heating mode can be seen in Figure 29. The box did not have a heat pump mode, so this testing was unique to the breadboard. Test Condition Outdoor Temp [ F/ C] Indoor Temp [ F/ C] Indoor Wet Bulb [ F/ C] D 90/32.2 D2 0/3.3 D3 25/5. W 90/32.2 W2 0/3.3 W3 25/5. Figure 28. Breadboard cooling mode test matrix 90/32.2 Dry 5/23.9 (50%RH) Test Condition Outdoor Temp [ºF/ºC] Indoor Temp [ºF/ºC] HP 0 / -8 HP2 5 / -9 HP3 32 / 0 68/20 HP 50 / 0 Figure 29. Breadboard heating mode test matrix 2

30 Chapter. Breadboard System Results This chapter contains the overall system results for the breadboard tests. System results include such things as capacity and measures of system efficiency.. Cooling mode results The first set of tests which was run was the charge determination tests. These tests were run at test condition D3, which is the warmest test condition. The warmest of the outdoor temperatures should also require the largest charge. A properly functioning accumulator stores the excess charge for lower temperature conditions. A dry condition was chosen for the initial test condition because the condition can be more precisely controlled. Therefore condition D3 was used for the charge determination tests. Determining the ideal charge for the system is generally an iterative process. For this test however, the high side pressure was matched to the high side pressure equation found for the.5 ton ECU breadboard test. This was a reasonable starting place since there were many similarities between the systems. The charge which maximized the cooling capacity was found to be 300 g. The pressure which yields the maximum cooling capacity does not exist for ideal systems, however it exists for many real systems due to several secondary effects. The method used for increasing high side pressure in this system was closing the expansion valve. Closing the expansion valve decreases the mass flow rate of refrigerant. Besides the direct loss cost from decreasing the amount of refrigerant available to exchange enthalpy in a given time period, the lower flow rate also leads to more super-heat. The larger super-heat is responsible for lower volumetric efficiency in the compressor. Once the optimal charge was determined, the optimal high side pressure for this system was found. R systems are generally controlled by adjusting the high side pressure. Pressure is increased or decreased by closing and opening the expansion valve, respectively. As with the charge determination test, the tests were conducted with the flat top fin evaporator with the tubes running horizontal and the four-port tube gas cooler. A plot of capacity versus a range of pressures can be seen in Figure 30, in this figure the different test conditions are signified by different colors. It should be noted that at some conditions the high side pressure was limited by the high pressure limit of the system, which is MPa. The coefficient of performance (COP) for these same tests can be seen in Figure 3. This chart was then used in order to create a control strategy for the system. The high side pressure is controlled based on the refrigerant temperature leaving the gas cooler. The high side pressures which yielded the highest capacities at each test condition were plotted against the temperature of the refrigerant leaving the gas cooler in those tests. This plot, Figure 32, shows the high side pressure equation, which was made using a least squared curve fit through the points shown. This equation was used in all subsequent tests, and was further validated by those tests by varying the high side pressure around the pressure given by the equation. 22

31 Capacity [kw] D_300g_Pcro9500 D_300g_Pcro0300 D_300g_Pcro09 D_300g_Pcro00 Figure 30. Capacity with varying high side pressures D_300g_Pcro900 D2_300g_Pcro600 D2_300g_Pcro2200 D2_300g_Pcro2800 D2_300g_Pcro3500 D3_300g_Pcro300 D3_300g_Pcro00 W_300g_Pcro0200 W_300g_Pcro0800 W_300g_Pcro500 W_300g_Pcro2300 W2_300g_Pcro0600 W2_300g_Pcro00 W2_300g_Pcro2200 W2_300g_Pcro2300 W3_300g_Pcro3000 W3_300g_Pcro300 W3_300g_Pcro000 COP [-] D_300g_Pcro9500 D_300g_Pcro0300 D_300g_Pcro09 D_300g_Pcro00 Figure 3. COP with varying high side pressures D_300g_Pcro900 D2_300g_Pcro600 D2_300g_Pcro2200 D2_300g_Pcro2800 D2_300g_Pcro3500 D3_300g_Pcro300 D3_300g_Pcro00 W_300g_Pcro0200 W_300g_Pcro0800 W_300g_Pcro500 W_300g_Pcro2300 W2_300g_Pcro0600 W2_300g_Pcro00 W2_300g_Pcro2200 W2_300g_Pcro W3_300g_Pcro3000 W3_300g_Pcro300 W3_300g_Pcro000

32 Gas Cooler Outlet Pressure [kpa] y = x Gas Cooler Outlet Temperature [C] Figure 32. High side pressure control strategy The cooling capacity and COP of the system for the system in all configurations that were tested can be seen in Figure 33 and Figure 3. All points were run with the same quantity of refrigerant charge. The high side pressure was varied according to the empirically determined equation for this system which maximized cooling capacity. Due to size constraints of the charts, the points for each color are D, D2, D3, W, W2, and W3. The color coding system for Figure 33 and Figure 3 is as follows: Orange: Flat top fin evaporator w/ tubes horizontal; port tube gas cooler Green: Flat top fin evaporator w/ tubes vertical; port tube gas cooler Red: Round top fin evaporator w/ tubes vertical; port tube gas cooler Light Blue: Round top fin evaporator w/ tubes horizontal; port tube gas cooler Chartreuse: Round top fin evaporator w/ tubes vertical; 6 port tube gas cooler Blue: Flat top fin evaporator w/ tubes horizontal; 6 port tube gas cooler 2

33 Figure 33. Cooling capacity Figure 3. Coefficient of performance It was generally seen in the air conditioning tests, that the exit of the evaporator had a quality of about 95%. On average the suction line heat exchanger had an effectiveness of 5%. Effectiveness is the ratio of actual heat transfer to the maximum possible heat transfer. In calculating suction line heat exchanger effectiveness, it was assumed that both the high and low pressure streams had the same thermal capacitance. 25

34 Isentropic efficiency (η isen ) is a measure of how closely the compressor behaves to that of an ideal, or isentropic, compressor. In real compressors, the process is not reversible and adiabatic; therefore there is a second law inefficiency. As can be seen in Equation 6, isentropic efficiency is calculated from several specific enthalpies of the refrigerant, including at the compressor suction (h cpri ) and discharge (h cpro ). The specific enthalpy at the discharge pressure but the same specific entropy of the compressor inlet is given by the final term (h cpro,s ). hcpro, s hcpri η isen = (6) h h cpro cpri The isentropic efficiency of the compressor remained relatively constant throughout the air conditioning test matrix. The average isentropic efficiency of the compressor was 58%. The range of isentropic efficiencies measured was from 5% to 6%. The standard deviation of isentropic efficiencies through the entire test matrix was just 2%. An example pressure-enthalpy diagram for condition D2 can be seen in Figure 35. Pressure-enthalpy diagrams for all test conditions can be seen in Appendix A. Raw air conditioning data o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. Figure 35. Pressure-enthalpy diagram The refrigerant pressure drop through the high side of the suction line heat exchanger averaged 9.3 kpa. Through the low side of the suction line heat exchanger, which also functions as the accumulator, the pressure drop was 0.5 kpa..2 Heating mode results The testing procedure for heat pump tests was similar to that used for air conditioning tests. Based on previous heat pump experiences, generally the limiting factor for heat pump testing is the compressor discharge limit. The high temperature limit given for the compressor by its manufacturer is 0ºC. In order to allow a margin of safety, discharge temperatures were limited to 35ºC. The first test to be conducted was the charge determination test. The heating capacity at condition HP can be seen in Figure 36. From this plot it can be seen that 200g of 26

35 R gives the largest cooling capacity. The sharp increase seen between 50 and 200g was due in large part to a large increase in refrigerant mass flow rate. This sizeable increase was due to opening the expansion valve further in order to maintain discharge temperature and also due to increasing volumetric efficiency of the compressor due to greater density at the inlet Heating capacity [kw] g 050g 00g 50g 200g 250g 300g 350g Figure 36. Heating capacity from charge determination test Once the charge determination test was completed, the system was run at varying pressures. It was seen that for conditions HP2, HP3, and HP, the system attained the maximum capacity when the compressor discharge temperature was 35ºC. For the coldest condition, HP, the largest heating capacity occurred when the high side pressure was around 000kPa. The heating capacity of the system can be seen in Figure 3. The heating performance factor (HPF), which gives the ratio between heating capacity and compressor power can be seen in Figure 38. 2

36 3.5 Heating Capacity [kw] Port tube gas cooler Port tube gas cooler HP HP2 HP3 HP HP HP2 HP3 HP Figure 3. Heating capacity 3 6 Port tube gas cooler Port tube gas cooler 2 HPF [-] HP HP2 HP3 HP HP HP2 HP3 HP Figure 38. Heating performance factor A pressure enthalpy diagram can be produced for heat pump testing from refrigerant side measurements. Figure 39 depicts an example at condition HP. There is a great deal of enthalpy lost between the compressor discharge and the inlet of the indoor coil. This is most likely caused by the large temperature gradient between the discharge line and the ambient temperature. The length of the discharge pipe within the outdoor chamber is about 8 feet. An additional 20 feet of pipe are outside of the chamber. The temperature loss between these two points was anywhere from ºC to 38 ºC. This loss was in spite of the fiberglass insulation which was placed on the pipe. 28

37 The isentropic compressor efficiency was about 8%. The effectiveness of the suction line heat exchanger averaged 58%. The temperature of the air coming from the heat pump, sometimes called the quality of heat, is an important measure of comfort. A heat pump which is blowing air just warmer than room temperature can actually make a person feel colder, instead of warmer, due to an increase in convection. Figure 0 shows the air temperature after passing through the indoor coil o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. Figure 39. Pressure-enthalpy diagram for heat pump testing 5 Air temperature at exit of indoor coil [C] Port tube gas cooler Port tube gas cooler 0 HP HP2 HP3 HP HP HP2 HP3 HP Figure 0. Indoor air discharge temperature 29

38 Chapter 5. Heat Exchanger Performance in Breadboard Testing This section looks more in depth into the differences between the heat exchangers. The flat top fin and round top fin evaporators were compared in two different orientations. The four and six-port tube gas coolers were also compared. In these sections evaporator refers to the indoor coil and gas cooler refers to the outdoor coil. In this notation, the refrigerant evaporates in the gas cooler, and cools in the evaporator. 5. performance The two evaporators were compared using several different parameters. One of the areas of study was pressure drop of the air flow across both evaporators with both tube orientations. It can be seen in Figure that both evaporators had very similar air pressure drops for dry conditions no matter what the orientation. However, under wet conditions, the round top fin performed much better with the tubes vertical than horizontal. For the flat top fin, there was little difference between the two orientations. It should be noted that the air flow rate was the same for all tests. The values shown are averages Air deltap Evap [Pa] Flat top horizontal dry Flat top horizontal wet Flat top vertical dry Flat top Serpentine Serpentine Serpentine vertical wet vertical dry vertical wet horizontal dry Serpentine horizontal wet Figure. Air pressure drop across evaporator The pressure drop across the evaporator in dry conditions with the tubes horizontal is shown in Figure 2. air pressure drop versus volumetric flow rate It can be seen that the pressure drop is slightly higher for the flat top fin than for the serpentine fin for the same volumetric air flow rate. Also, the pressure drop across the both evaporators can be approximated by a quadratic function of volumetric air flow rate. 30

39 60 Differential air pressure [Pa] Flat top fin Round top fin y = x x -.09 y = x x Volumetric air flow rate [CFM] Figure 2. air pressure drop versus volumetric flow rate Another useful method of comparing evaporators is by studying their effectiveness. Effectiveness (ε) is a value between zero and one which describes the ratio of heat transferred (Q) to the maximum amount of heat that can theoretically be transferred (Q max ). The relationship can be seen in Equation. The maximum amount of heat that can theoretically be transferred in the evaporator is determined using Equation8. Q ε () Q hx = Q max = m c ( T T ) max a pa eai eri (8) Figure 3 shows the effectiveness for dry test conditions for both evaporators in both orientations. It can be seen that in general the flat top fin evaporator was more effective than the round top fin evaporator. Also, having the refrigerant flowing vertically upward from the headers made for greater effectiveness than horizontal flow in the tubes. 3

40 0.9 Flat top fin Tubes horizontal Flat top fin Tubes vertical Round top fin Tubes vertical Round top fin Tubes horizontal 0.8 Effectivness D D2 D3 D D2 D3 D D2 D3 D D2 D3 Figure 3. effectiveness for dry conditions Another method of quantifying the performance of a heat exchanger is studying its overall heat transfer coefficient. This overall heat transfer coefficient is the inverse of the thermal resistance. The usefulness of the overall heat transfer coefficient is that it allows the calculation of capacity based on the inlet and outlet temperatures of the fluid. For the evaporator, the overall heat transfer coefficient (UA) can be multiplied by the log mean temperature difference (ΔT lm ) to give the capacity (Q), as seen in Equation 9. The dimensionless correction factor (F) is used because the heat exchanger is cross counter flow, instead of simply counter flow. Since the temperature of the refrigerant in the evaporator is constant, the correction value equals one. The log mean temperature difference of the evaporator is given in Equation 20 as a function of the air temperatures at the inlet (T eai ) and outlet (T eao ) as well as the refrigerant temperatures at the inlet (T eri ) and outlet (T ero ). Q = F( UA) Δ (9) cooling T lm Teai + TTeri Tero Teao ΔTlm, e = (20) Teai Tero ln Teao Teri The overall heat transfer coefficient for both evaporators in both orientations can be seen in Figure. overall heat transfer coefficient It can be seen that the heat transfer coefficient is greater for the flat top fin than for the serpentine fin. Therefore the flat top fin is able to do more cooling with a given temperature difference. With the exception of the serpentine fin in the vertical orientation, the heat transfer coefficient tends to decrease for wet conditions. This is most likely due to the thermal resistance of the water which has condensed on the fin. For the serpentine fin in vertical orientation, the moisture may have increased the surface area of the evaporator. 32

41 0. Overall Heat Transfer Coeffecient [kw/k] Flat top horizontal dry Flat top horizontal wet Flat top vertical dry Flat top vertical wet Serpentine Serpentine Serpentine Serpentine vertical dry vertical wet horizontal dry horizontal wet Figure. overall heat transfer coefficient A useful tool which can be used to qualitatively study evaporators is an infrared camera. The camera sees temperatures, and can therefore help show distribution. Warm areas in the evaporator, which are seen as red in the images, are those which are getting little flow or contain only vapor. Cool areas, shown in blue, are receiving liquid refrigerant flow. Infrared pictures of the evaporator at the D2 condition are shown in Figure 5, Figure 6, Figure, and Figure 8. The infrared camera was looking at the evaporator from the air inlet, or refrigerant outlet, side. It can be seen that the round top fin evaporator is not receiving uniform flow, possibly due to a manufacturing defect. A more uniform refrigerant distribution is evident with the vertical tubes than with the horizontal tubes. Figure 5. Infrared image of flat top fin evaporator with tubes horizontal 33

42 Figure 6. Infrared image of flat top fin evaporator with tubes vertical Figure. Infrared image of round top fin evaporator with tubes horizontal 3

43 Figure 8. Infrared image of round top fin evaporator with tubes vertical Another qualitative technique employed was to take videos of water drainage off the evaporator coil when running wet conditions. The flat top fin evaporator had the normal trickle down type of drainage in both orientations. The round top fin evaporator also drained in this manner when it was oriented with is tubes vertical. However, the round top fin evaporator had poor drainage when it was oriented with tubes horizontal. In this case, water would build up in all air passages until they were full. Then the air pressure would eventually pull out all the water at once in a large sheet. It should be noted that the round top fin was specifically designed to drain with the tubes vertical, so this result was anticipated. Another method used to judge the ability for the evaporators to handle condensation is to look at the rate at which water condensates on, and fall off, the evaporator. The mass of the condensate accumulated per unit time was calculated based on the load cell readings during the test. At steady state, the rate that water condensates on the evaporator is equal to the rate at which the water drains from the coil. Figure 9 shows the rate of water drainage. Since humidity is difficult to control, only general trends should be garnered from the chart. Larger values mean more latent load is being removed from the air. It can be seen that the flat top fin evaporator removes water from the air at about the same rate in both orientations. The round top fin performed appreciably better with the tubes vertical. 35

44 Flat top fin Tubes horizontal Flat top fin Tubes vertical Round top fin Tubes vertical Round top fin Tubes horizontal 0. Water drainage [g/s] W W2 W3 W W2 W3 W W2 W3 W W2 W3 Figure 9. Water drainage rate for wet conditions Another way of presenting the latent cooling is by analyzing the percentage of the water removed from the air. This is done by dividing the water drainage rate, which is shown in Figure 50, by the mass of water being drawn into the evaporator Flat top fin Tubes horizontal Flat top fin Tubes vertical Round top fin Tubes vertical Round top fin Tubes horizontal 6 Water removal [%] W W2 W3 W W2 W3 W W2 W3 W W2 W3 Figure 50. Percentage of water removed When the system also operates as a heat pump the evaporator coil also functions as the gas cooler. For the heat pump testing, only the flat top fin evaporator in the tubes horizontal orientation was used. This was done 36

45 because the evaporator in this configuration is what could most readily be installed in an actual box system. The approach temperature for this heat exchanger is shown in Figure 5. The trend seen between HP2 and HP is due to a steadily opening expansion valve allowing more refrigerant flow rate. Since condition HP did not achieve its maximum capacity when by operating the system the compressor discharge temperature limit, it did not fit the trend as the point had a higher refrigerant flow rate Approach Temperature [C] HP HP2 HP3 HP Figure 5. Indoor heat exchanger approach temperature 5.2 performance Similar to the evaporator, the gas cooler was also compared on a number of different measures. The first method used in comparing the gas coolers was by measuring the approach temperatures. The approach temperature (ΔT approach ), given in Equation 2, is the difference between the air inlet (T cai ) and refrigerant outlet temperatures (T cro ). Δ Tapproach = Tcro Tcai (2) A smaller approach temperature is preferred because it means more enthalpy is being given up by the refrigerant to the air. An infinite heat exchanger, with an infinite amount of time, would have an approach temperature of zero. The approach temperatures for both gas coolers can be seen in Figure 52. It can be seen that the six-port gas cooler outperformed the four-port tube gas cooler in this regard. 3

46 3.5 Four Port Tube Gas Cooler Six Port Tube Gas Cooler 3 Approach Temperature [C] D D2 D3 W W2 W3 D D2 D3 W W2 W3 Figure 52. approach temperature The pressure drop across the gas cooler was also studied. The air pressure drop across the four-port tube gas cooler was 9.5 Pa, but was 0. Pa for the six-port tube gas cooler. The refrigerant pressure drop through the four-port tube gas cooler was 5. kpa; it was 3. kpa for the six-port tube while running at the test condition. These differences are small enough that there will be little effect on system performance due to the pressure drop across these two heat exchangers. The pressure drop across the gas cooler was also studied with respect to varying volumetric flow rates. The results of these tests, conducted at room temperature, can be seen in Figure 53. Once again, the six-port tube gas cooler is seen to have a higher pressure drop. The equations shown on the graph give a representation of the air pressure drop as a function of volumetric flow rate. 25 Differential air pressure [Pa] port tube 6 port tube y = E-05x x y = 8E-06x x Volumetric air flow rate [CFM] Figure 53. air pressure drop versus volumetric flow rate 38

47 Similar to the evaporator, the overall thermal conductivity (UA) of the gas cooler can be found. As this is a cross-counter flow heat exchanger, the correction factor (F) must be used. In the gas cooler, both the refrigerant and air temperatures change. Therefore, the correction factor must be found on correction factor charts. Figure 5. Correction Factor Plot (Bowman, Mueller, Nagle) shows the chart from Bowman which is most similar to the configuration seen in the gas cooler. The main difference is that the gas cooler used in our experiment has four passes, instead of two. This has the effect of causing further squaring off the curves, so that the correction factor approaches unity at higher values of P. A correction factor equal to 0.9 was estimated based on the dimensionless temperature ratios shown in Figure 5. Figure 5. Correction Factor Plot (Bowman, Mueller, Nagle) The formula used to determine the overall heat transfer coefficient can be seen in Equation 9, but with the cooling capacity replaced by the gas cooler capacity. The gas cooler capacity is determined from the refrigerant and air side energy balances. The log mean temperature difference of the gas cooler (ΔT lmc ), which can be calculated by utilizing Equation 22, is function of the air temperatures at the gas cooler inlet (T cai ) and outlet (T cao ) as well as the refrigerant temperatures at the inlet (T cri ) and outlet (T cro ). ΔT lmc T = cri + T cai T ln T cri cro T cro T T cao cai T cao The overall thermal conductivity can be seen in Figure 55. overall heat transfer coefficient The six-port tube gas cooler was able to transfer more heat given a certain temperature difference. This may be due to increased heat transfer area. (22) 39

48 Overall Heat Transfer Coeffecient (UA) [kw/k] port tube 6 port tube Figure 55. overall heat transfer coefficient In heat pump mode, this outdoor heat exchanger functions as an evaporator. Because the system was not designed with a defrost mechanism, the room was dehumidified before testing. Therefore, the dew point temperature had to be decreased to a point below the refrigerant temperature prior to running in order to prevent frost from forming on the coil. Frosting could be determined by monitoring the air pressure drop across the heat exchanger. Frost causes a larger drop in air pressure across the heat exchanger. Because of the required dehumidification the chamber doors could not be opened at any point the day of testing. This precluded the ability to take infrared pictures, as the camera needs to be periodically supplied with liquid nitrogen. When the outdoor coil functions as an evaporator, it is useful to look into heat exchanger effectiveness. The heat exchanger effectiveness for the outdoor coil in heat pump mode can be seen in Figure 56. As can be seen in this figure, the six-port tube gas cooler was far more effective than the four-port tube gas cooler Port tube gas cooler Port tube gas cooler 0. Effectivness [-] HP HP2 HP3 HP HP HP2 HP3 HP Figure 56. Outdoor heat exchanger effectiveness 0

49 Chapter 6. Conclusions Valuable knowledge was gained from this experiment. The baseline box outperformed its rated capacity, making it more challenging to beat its capacity with R. The R breadboard did perform well in both air conditioning and heap pump modes. Lessons learned in previous ECU work have yielded improvements. Due to an undersized compressor, the cooling capacity was not as large as that found on the R22 box. A COP approximately equal to the breadboard system was attained. The breadboard did provide a HPF greater than unity for all heat pump tests, making it is more efficient than the electric resistance heaters found in the baseline system. A great deal of knowledge was gathered from studying different evaporators and gas coolers in crosscounter flow orientation. s were tested in horizontal and vertical tube orientations. The flat top fin evaporator was found to work well in both horizontal and vertical orientations. The round top fin evaporator, which may have contained a manufacturing defect, performed much better with the tubes vertical, than with the tubes horizontal. Overall the best heat exchanger was the round top fin evaporator with the tubes vertical; however this heat exchanger performed the worst with the tubes horizontal. In air conditioning mode, both gas coolers performed similarly. In heat pump mode, the six-port tube heat exchanger performed better in almost every regard in comparison to the four-port tube heat exchanger. The only exception to this was air side pressure drop.

50 Bibliography Bowman, R. A., A. C. Mueller, and W. M. Nagle. Mean Temperature Difference in Design Transactions of the A.S.M.E., May 90. p Incropera, Frank P. and David P. DeWitt. Fundamentals of Heat and Mass Transfer, Fifth Edition, John Wiley and Sons, New York, Elbel S. and P. Hrnjak. Experimental and analytical validation of new approaches to improve transcritical CO 2 environmental control units. ACRC report CR-52. Urbana, Klein, S. A. Engineering Equation Solver V699-3D. F-Chart Software Yin J, Y. Park, D. Boewe, R McEnaney, A. Beaver, C Bullard, and P. Hrnjak. Experimental and model comparison of transcritical CO 2 versus R3a and R0A system performance, Proceedings of the third IIR- Gustav Lorentzen Conference on Natural Working Fluids, Oslo, 998. p

51 Appendix A. Raw air conditioning data This section contains the raw data for all air conditioning test points as well as their respective pressureenthalpy diagrams. The pressure-enthalpy diagrams were plotted using information calculated in the EES template. Many of these points were not directly used in determining system capacity, but were instead used in order to validate the high side pressure control strategy. The tests that were used as the representative points for their given setups and test conditions, as well as for all performance data, were numbers 9, 3, 6, 8, 23, 2, 28, 3, 32, 33, 3, 35, 36, 39, 3,, 5, 6,, 8, 5, 53, 5, 55, 6, 6, 65, 68, 0, 2,, 5, 6,, 8, and 9. 3

52 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 50g 3600kPa Dencg =.08 Pcpri = Teao =.35 Deneg =.08 Pcro = Tegi = 2.58 Denr =.9 Pero = Tego = dpca = 8. Tcai = 5.9 Tenao = 9.0 dpcna = 0.62 Tcao = 68.8 Teri = 3.0 dpcr = 55. Tcgi = -0.0 Tero = dpea = 30.3 Tcgo = Tewi = 3.3 dpena = 32.8 Tcnao =. Tewo = dper = 8.3 Tcpri = 6.58 Tslhx2ro = 5.09 dpslhx = 2.9 Tcpro = Tslhxri = dpslhx2 =.2 Tcro = Wcb = 0.2 Dslope = 0.00 Tcwi = 50.9 Wch =.82 Mcg = 2.05 Tcwo = 2.2 Wcomp =.80 Meg = 09.2 TDPeai =.2 Web = 0.5 Mr = 8. TDPeao =.2 Weh =.9 Mw = Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

53 Test number: 2 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 200g 3600kPa Dencg =.08 Pcpri = Teao =. Deneg =.08 Pcro = Tegi = 2.58 Denr =.9 Pero = Tego = 28. dpca = Tcai = Tenao = 9.29 dpcna =.62 Tcao = 69.2 Teri =.03 dpcr = 55.6 Tcgi = Tero = 6 dpea = Tcgo = 23. Tewi = 3.32 dpena = Tcnao =.6 Tewo = 20.8 dper = 55. Tcpri = 5.38 Tslhx2ro = 3.2 dpslhx = 3.6 Tcpro = 6 Tslhxri = 22.6 dpslhx2 =.6 Tcro = Wcb = 0.29 Dslope = 0.00 Tcwi = 5.08 Wch =.5 Mcg = 6.50 Tcwo = 2.3 Wcomp =.9 Meg = 0.3 TDPeai =.8 Web = 0.6 Mr = 9.32 TDPeao =.9 Weh =.96 Mw = -0.0 Teai = 32.0 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

54 Test number: 3 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 250g 300kPa Dencg =.08 Pcpri = 0 Teao = 6.96 Deneg =.08 Pcro = 360. Tegi = 2.56 Denr =.9 Pero = 50.2 Tego = 28.2 dpca = 8.6 Tcai = 5.83 Tenao = 9.20 dpcna =.32 Tcao = Teri =.66 dpcr = 5.2 Tcgi = Tero =.5 dpea = 30.2 Tcgo = 2.8 Tewi = 3.30 dpena = 32.5 Tcnao = 2.09 Tewo = 20.8 dper = 59.8 Tcpri = 3.5 Tslhx2ro =.02 dpslhx = 6.8 Tcpro = 50.8 Tslhxri =.9 dpslhx2 =.65 Tcro = Wcb = 0.29 Dslope = 0.00 Tcwi = 50.9 Wch =.50 Mcg = 62.6 Tcwo = 2.3 Wcomp =.8 Meg = TDPeai = 3.06 Web = 0.5 Mr = 9.96 TDPeao = 3.0 Weh =.9 Mw = -0.0 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

55 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 300g 300kPa Dencg =.08 Pcpri = Teao =.9 Deneg =.08 Pcro = 33.8 Tegi = 2.59 Denr =.9 Pero = 58.9 Tego = 28. dpca = 8.9 Tcai = 5. Tenao = 9.25 dpcna =.5 Tcao = Teri = 5.2 dpcr = 6.99 Tcgi = -.06 Tero = 5. dpea = 30.8 Tcgo = Tewi = 3. dpena = 320 Tcnao = 2.68 Tewo = dper = 65.8 Tcpri = 0.5 Tslhx2ro = 3.29 dpslhx = Tcpro =.2 Tslhxri = 5.33 dpslhx2 =.2 Tcro = Wcb = 0.30 Dslope = 0.00 Tcwi = 50.9 Wch =. Mcg = 2.3 Tcwo = 2.26 Wcomp =.83 Meg = 06.8 TDPeai = 3.0 Web = 0.6 Mr = 20.2 TDPeao = 3.5 Weh =.95 Mw = -0. Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

56 Test number: 5 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 9500kPa Dencg =.08 Pcpri = Teao = 9.82 Deneg =.08 Pcro = 3.05 Tegi = 2.83 Denr =.9 Pero = Tego = dpca = Tcai = 5.82 Tenao = dpcna = 0.8 Tcao = 0.65 Teri = 6.2 dpcr = 5.2 Tcgi = -.59 Tero = 6.88 dpea = 2. Tcgo = 8.30 Tewi = 3.58 dpena = Tcnao = 3.26 Tewo = 20.9 dper = 6.00 Tcpri = 6 Tslhx2ro = dpslhx = 8.90 Tcpro = 6.52 Tslhxri =.6 dpslhx2 = 5.3 Tcro = 5.20 Wcb = 0.29 Dslope = 0.00 Tcwi = Wch =.20 Mcg = 5.66 Tcwo = 2. Wcomp =.8 Meg = TDPeai = 20. Web = 0.6 Mr = 2.26 TDPeao = 8.9 Weh =.58 Mw = 0.8 Teai = 32. Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

57 Test number: 6 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 0300kPa Dencg =.08 Pcpri = 8.8 Teao =.30 Deneg =.08 Pcro = Tegi = 25.3 Denr =.9 Pero = 60.0 Tego = 26.3 dpca = 0.26 Tcai = 32.8 Tenao = 6.90 dpcna = 0.23 Tcao = 9.82 Teri =.8 dpcr = 59. Tcgi = -2.8 Tero = 2.3 dpea = Tcgo = 3.98 Tewi = dpena = 335. Tcnao = 5.6 Tewo = 20.9 dper = 2.9 Tcpri = 30.3 Tslhx2ro = dpslhx = 3.80 Tcpro = Tslhxri =.9 dpslhx2 = 9 Tcro = 38.3 Wcb = 0.2 Dslope = 0.00 Tcwi = 3.6 Wch =.2 Mcg = Tcwo = Wcomp =.29 Meg = TDPeai = 6.25 Web = 0.6 Mr = TDPeao = 6.39 Weh =.39 Mw = 0.6 Teai = 32.0 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

58 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 09kPa Dencg =.08 Pcpri = Teao = 3.20 Deneg =.08 Pcro = Tegi = 25.8 Denr =.9 Pero = Tego = dpca = 0.2 Tcai = 32.3 Tenao = 5.98 dpcna = 2.29 Tcao = Teri = 0.2 dpcr = Tcgi = -.83 Tero =.06 dpea = 30.9 Tcgo = 5.3 Tewi = 30.6 dpena = Tcnao = 52.8 Tewo = dper = 5.2 Tcpri = 25 Tslhx2ro = 2.8 dpslhx = 3.30 Tcpro = 09.2 Tslhxri = 0.6 dpslhx2 = 6.8 Tcro = 3.0 Wcb = 0.2 Dslope = 0.00 Tcwi = 3.92 Wch =.3 Mcg = Tcwo = 23.5 Wcomp =.39 Meg = 63. TDPeai = 6.85 Web = 0. Mr = 2.38 TDPeao = 6.98 Weh =. Mw = 0.6 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 50

59 Test number: 8 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 00kPa Dencg =.08 Pcpri = Teao = 2.86 Deneg =.08 Pcro = Tegi = 26. Denr =.9 Pero = 55 Tego = 2.6 dpca = 0.28 Tcai = 32. Tenao = 5.6 dpcna = 2.95 Tcao = 5.06 Teri = 9.65 dpcr = Tcgi = -.3 Tero = 0.6 dpea = Tcgo = 6 Tewi = dpena = Tcnao = 53.5 Tewo = dper = 5. Tcpri = 2.29 Tslhx2ro = 2 dpslhx = 3.0 Tcpro = 5.05 Tslhxri = 0.6 dpslhx2 = 5.6 Tcro = 35.2 Wcb = 0.2 Dslope = 0.00 Tcwi = 3.98 Wch =.03 Mcg = 96.2 Tcwo = Wcomp =.5 Meg = 66.3 TDPeai =.5 Web = 0. Mr = 20.6 TDPeao =.52 Weh =.3 Mw = 0.66 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

60 Test number: 9 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 900kPa Dencg =.08 Pcpri = 6.5 Teao = 2.85 Deneg =.08 Pcro = Tegi = Denr =.9 Pero = 53.5 Tego = 2. dpca = 0.25 Tcai = 3 Tenao = 5.68 dpcna = 3.06 Tcao = 2 Teri = 9.3 dpcr = 8.62 Tcgi = Tero = 0.9 dpea = Tcgo = 6.23 Tewi = 30.9 dpena = Tcnao = Tewo = dper =.2 Tcpri = 26.8 Tslhx2ro = 2.99 dpslhx =. Tcpro = Tslhxri = 0.33 dpslhx2 =.93 Tcro = 35.3 Wcb = 0.2 Dslope = 0.00 Tcwi = Wch =.6 Mcg = 39.3 Tcwo = Wcomp =.52 Meg = TDPeai =.96 Web = 0.6 Mr = 20. TDPeao = 8.02 Weh =.8 Mw = 0.65 Teai = 3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 52

61 Test number: 0 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 600kPa Dencg =.08 Pcpri = 60.6 Teao = 2.90 Deneg =.08 Pcro = 92. Tegi = 26.5 Denr =.8 Pero = Tego = 2.53 dpca = 0.2 Tcai = 39 Tenao = dpcna = 3.5 Tcao = 5.6 Teri = 9.3 dpcr = 8. Tcgi = -0.0 Tero = 0.50 dpea = 30.9 Tcgo = 6. Tewi = 30.9 dpena = 33.5 Tcnao = 53.8 Tewo = 20.9 dper = 2.5 Tcpri = 26. Tslhx2ro = 2.96 dpslhx = 0.60 Tcpro = 25. Tslhxri =.39 dpslhx2 =.60 Tcro = 3.5 Wcb = 0.2 Dslope = 0.00 Tcwi = Wch =. Mcg = 26.3 Tcwo = 23.0 Wcomp =.58 Meg = 69.9 TDPeai = 8.2 Web = 0. Mr = 9.8 TDPeao = 8.23 Weh =.2 Mw = 0.6 Teai = 33 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 53

62 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 2200kPa Dencg =.08 Pcpri = Teao = 6.59 Deneg =.08 Pcro = Tegi = 28. Denr =.9 Pero = Tego = 29.3 dpca = 9.38 Tcai = 3.52 Tenao = 8.65 dpcna = Tcao = Teri = 3.80 dpcr = Tcgi = Tero =.2 dpea = 3.25 Tcgo = 5.0 Tewi = 3.80 dpena = Tcnao = 63.3 Tewo = 2.0 dper = 6.09 Tcpri = Tslhx2ro = dpslhx =.5 Tcpro = 22.8 Tslhxri = 3.8 dpslhx2 = 6.65 Tcro =.03 Wcb = 0.2 Dslope = 0.00 Tcwi = 2. Wch = 3.2 Mcg = 2.88 Tcwo = 23.5 Wcomp =.5 Meg = 8.2 TDPeai = 6.25 Web = 0.6 Mr = 22.0 TDPeao = 6.39 Weh = 2.23 Mw =.5 Teai = 3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

63 Test number: 2 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 2800kPa Dencg =.08 Pcpri = Teao = 6.9 Deneg =.08 Pcro = Tegi = 28. Denr =.9 Pero = Tego = dpca = 9. Tcai = 3. Tenao = 8. dpcna = 6.3 Tcao = 6.3 Teri = 3.05 dpcr = 50.2 Tcgi = -.93 Tero = 3.6 dpea = 3.86 Tcgo =.52 Tewi = 3.2 dpena = Tcnao = Tewo = 2.0 dper = 5. Tcpri = 35. Tslhx2ro = 33.8 dpslhx = 5.35 Tcpro = 29.8 Tslhxri = 3.20 dpslhx2 = 8 Tcro = 6. Wcb = 0.28 Dslope = 0.00 Tcwi = 2.9 Wch = 3.02 Mcg =.6 Tcwo = Wcomp =.62 Meg = 8.89 TDPeai = 6.85 Web = 0.6 Mr = 2. TDPeao = 6.98 Weh = 2.9 Mw =.5 Teai = 32.9 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 55

64 Test number: 3 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 3500kPa Dencg =.08 Pcpri = Teao = 6.02 Deneg =.08 Pcro = Tegi = 28.9 Denr =.9 Pero = Tego = 29.3 dpca = 9. Tcai = 3.5 Tenao = 8.2 dpcna = 5.69 Tcao = 6.50 Teri = 2. dpcr =. Tcgi = -.8 Tero = 3. dpea = 3 Tcgo =. Tewi = 3. dpena = Tcnao = Tewo = 2.0 dper = Tcpri = 3.50 Tslhx2ro = 3.52 dpslhx = 5.6 Tcpro = Tslhxri = 2.0 dpslhx2 =.5 Tcro = 5.60 Wcb = 0.28 Dslope = 0.00 Tcwi = 2.8 Wch = 2.8 Mcg = 2.29 Tcwo = 23.8 Wcomp =.69 Meg = 2 TDPeai =.5 Web = 0.6 Mr = TDPeao =.52 Weh = 2.2 Mw =.5 Teai = 32.5 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 56

65 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 300kPa Dencg =.08 Pcpri = Teao = 6.35 Deneg =.08 Pcro = Tegi = 28.2 Denr =.9 Pero = Tego = dpca = 9. Tcai = 3.9 Tenao = 8.06 dpcna = 6. Tcao = 6.86 Teri = 2.0 dpcr = 5.2 Tcgi = -.8 Tero = 2.9 dpea = 35.8 Tcgo =.88 Tewi = 3. dpena = Tcnao = 6.09 Tewo = 2. dper =.5 Tcpri = 33.9 Tslhx2ro = 3.22 dpslhx =. Tcpro = 2. Tslhxri = 2 dpslhx2 = 3.8 Tcro = 5.32 Wcb = 0.28 Dslope = 0.00 Tcwi = 2.83 Wch = 2.69 Mcg = Tcwo = Wcomp =. Meg = 8.0 TDPeai = 8.02 Web = 0. Mr = 9. TDPeao =.5 Weh = 2.26 Mw =.50 Teai = 32.8 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

66 Test number: 5 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 300g 00kPa Dencg =.08 Pcpri = Teao =.5 Deneg =.08 Pcro = Tegi = Denr =.9 Pero = Tego = 29.2 dpca = 8.6 Tcai = 5.8 Tenao = 9.5 dpcna =.8 Tcao = 69.6 Teri = 5. dpcr = 9.36 Tcgi = -.5 Tero = 6.25 dpea = 30.0 Tcgo =.08 Tewi = 3.6 dpena = Tcnao = 2.2 Tewo = 2.3 dper = 6. Tcpri =.63 Tslhx2ro = 39.5 dpslhx =.5 Tcpro = 3. Tslhxri = 5.8 dpslhx2 = 5. Tcro = 5.29 Wcb = 0.2 Dslope = 0.00 Tcwi = 5.02 Wch = 3.69 Mcg = 3.63 Tcwo = 23.9 Wcomp =. Meg = 9.52 TDPeai = 6.85 Web = 0.6 Mr = 2.30 TDPeao = 6.98 Weh =.9 Mw =.8 Teai = 32.6 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 58

67 Test number: 6 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 300g 0200kPa Dencg =.08 Pcpri = 528 Teao =.3 Deneg =.08 Pcro = Tegi = Denr =.9 Pero = Tego = 29. dpca = 8.6 Tcai = 5.9 Tenao = 9.2 dpcna =.66 Tcao = 69.8 Teri = 5.2 dpcr =.28 Tcgi = -.2 Tero = 5. dpea = Tcgo =. Tewi = 3.85 dpena = Tcnao = 2.3 Tewo = 2.0 dper = Tcpri =.2 Tslhx2ro = 3. dpslhx = 9.0 Tcpro = 9.0 Tslhxri = 5.2 dpslhx2 =.28 Tcro = 53.6 Wcb = 0.2 Dslope = 0.00 Tcwi = Wch = 3.95 Mcg = 3.69 Tcwo = Wcomp =.8 Meg = TDPeai =.5 Web = 0.6 Mr = TDPeao =.52 Weh = 2.05 Mw =.9 Teai = 36 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 59

68 Test number: Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 0800kPa Dencg =.08 Pcpri = 98.2 Teao = 9.38 Deneg =.08 Pcro = Tegi = 2.3 Denr =.9 Pero = 895 Tego = 2 dpca = 0.3 Tcai = 32.6 Tenao = 9.2 dpcna = 2.9 Tcao = Teri = 3.08 dpcr = Tcgi = -.06 Tero = 8.88 dpea =.5 Tcgo = 20. Tewi = dpena = Tcnao = Tewo = 20.9 dper = 2. Tcpri = 32.0 Tslhx2ro = 3.8 dpslhx = 3.03 Tcpro = 08.0 Tslhxri = 9.0 dpslhx2 = 9.6 Tcro = 3.62 Wcb = 0.2 Dslope = 0.00 Tcwi = 3.92 Wch = 5.3 Mcg = Tcwo = 2.62 Wcomp =.38 Meg = 0. TDPeai = 20.3 Web = 0. Mr = TDPeao =.38 Weh = 0.52 Mw = 0.35 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 60

69 Test number: 8 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 500kPa Dencg =.08 Pcpri = 90.9 Teao = 9.5 Deneg =.08 Pcro = Tegi = 2.23 Denr =.9 Pero = Tego = 28.0 dpca = 0.59 Tcai = 32.6 Tenao = 9.8 dpcna = 9.03 Tcao = 52.0 Teri = 2.2 dpcr = 6. Tcgi = -6. Tero = 20. dpea = 5.9 Tcgo = 2.00 Tewi = 30.2 dpena = Tcnao = 5.9 Tewo = 2.0 dper = Tcpri = 3.5 Tslhx2ro = 3.22 dpslhx = 2.32 Tcpro = 3.83 Tslhxri = dpslhx2 = 8.69 Tcro = 36.6 Wcb = 0.28 Dslope = 0.00 Tcwi = 32.0 Wch = 5. Mcg = -.0 Tcwo = 2.6 Wcomp =.5 Meg = 2.33 TDPeai = 20.8 Web = 0.6 Mr = 22.0 TDPeao =. Weh = 0.62 Mw =.3 Teai = 32.0 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

70 Test number: 9 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 2300kPa Dencg =.08 Pcpri = Teao = 9.6 Deneg =.08 Pcro = Tegi = 2.35 Denr =.9 Pero = Tego = 28.6 dpca = 0.55 Tcai = 32.6 Tenao = 9.52 dpcna = 9. Tcao = 52.9 Teri =.3 dpcr = Tcgi = Tero = 2.29 dpea = 6. Tcgo = 2.0 Tewi = 30.0 dpena = 3.29 Tcnao = 5.32 Tewo = 20.9 dper = Tcpri = Tslhx2ro = 29.9 dpslhx =.58 Tcpro = 2.2 Tslhxri = dpslhx2 =.8 Tcro = Wcb = 0.28 Dslope = 0.00 Tcwi = 3.98 Wch = 5.33 Mcg = Tcwo = 2.59 Wcomp =.5 Meg = 2.30 TDPeai = Web = 0.6 Mr = 2. TDPeao = 6.98 Weh = 0.9 Mw = 0.5 Teai = 32.0 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 62

71 Test number: 20 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 0600kPa Dencg =.08 Pcpri = Teao = 9.2 Deneg =.08 Pcro = Tegi = 2.3 Denr =.9 Pero = 5.89 Tego = 2 dpca = 0.30 Tcai = Tenao = 9.52 dpcna = 3.3 Tcao = 52.6 Teri =.50 dpcr = 6.66 Tcgi = Tero = dpea = 5.83 Tcgo = 2.02 Tewi = 30.3 dpena = 3.35 Tcnao = 5.82 Tewo = 20.9 dper = 6.8 Tcpri = 30.6 Tslhx2ro = dpslhx =.3 Tcpro = 2.95 Tslhxri = dpslhx2 = 6.53 Tcro = 3.6 Wcb = 0.2 Dslope = 0.00 Tcwi = 3.96 Wch = 5.25 Mcg = -2.9 Tcwo = 2.58 Wcomp =.63 Meg =.6 TDPeai = Web = 0.6 Mr = 20.9 TDPeao = 6.9 Weh = 0.8 Mw =.62 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 63

72 Test number: 2 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W2 300g 00kPa Dencg =.08 Pcpri = Teao = 20.6 Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = Tego = 28.6 dpca = 9.2 Tcai = 3.8 Tenao = 2.23 dpcna =.3 Tcao = 60.3 Teri =.05 dpcr = Tcgi = -.05 Tero = 8.8 dpea = 5.9 Tcgo = 20 Tewi = 3.9 dpena = Tcnao = 62.2 Tewo = 20.9 dper = Tcpri = 39.0 Tslhx2ro = 0.3 dpslhx = 0.22 Tcpro =.62 Tslhxri = dpslhx2 = Tcro =.5 Wcb = 0.2 Dslope = 0.00 Tcwi = 2.80 Wch = 3.35 Mcg = -3.9 Tcwo = 2.56 Wcomp =. Meg = 82.9 TDPeai = Web = 0. Mr = 25.9 TDPeao = 8.98 Weh =.5 Mw = 0.2 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

73 Test number: 22 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W2 300g 2200kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = 392. Tegi = 2.2 Denr =.9 Pero = Tego = 28.3 dpca = 9.5 Tcai = 3.56 Tenao = 20.0 dpcna = 3.8 Tcao = 6. Teri = 5. dpcr = 6. Tcgi = Tero = 8.26 dpea = 6.30 Tcgo = 2.82 Tewi = 3.6 dpena = Tcnao = Tewo = dper = 9.06 Tcpri = Tslhx2ro = dpslhx = 5.26 Tcpro = 20.2 Tslhxri = 8.6 dpslhx2 = 8.8 Tcro =.8 Wcb = 0.2 Dslope = 0.00 Tcwi = 2.69 Wch = 3.2 Mcg = -.26 Tcwo = 2.5 Wcomp =.52 Meg = 8. TDPeai = 20. Web = 0. Mr = TDPeao = 8.60 Weh =.86 Mw = 2.0 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 65

74 Test number: 23 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W2 300g 2300kPa Dencg =.08 Pcpri = Teao = 20.3 Deneg =.08 Pcro = Tegi = 28.0 Denr =.9 Pero = Tego = 29.0 dpca = 9.2 Tcai = 3. Tenao = dpcna =.00 Tcao = 62.8 Teri =.96 dpcr = 59.9 Tcgi = -3.8 Tero = 8.29 dpea = 8.2 Tcgo = Tewi = 3.68 dpena = 36 Tcnao = 6.9 Tewo = dper = Tcpri = 38.3 Tslhx2ro = 3.5 dpslhx =.8 Tcpro = 28.8 Tslhxri = 8.80 dpslhx2 =. Tcro = 6.8 Wcb = 0.28 Dslope = 0.00 Tcwi = 2. Wch = 3. Mcg = Tcwo = 2.5 Wcomp =.6 Meg = 8.32 TDPeai = 20.9 Web = 0. Mr = 22.0 TDPeao = 8.20 Weh =.69 Mw =.0 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 66

75 Test number: 2 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W2 300g 3000kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = Tegi = 2.96 Denr =.9 Pero = 53.8 Tego = dpca = 9.2 Tcai = 3.55 Tenao = 20.5 dpcna =.3 Tcao = 62.8 Teri =.8 dpcr = Tcgi = Tero = 9.6 dpea =.55 Tcgo = Tewi = 3.6 dpena = Tcnao = 6.5 Tewo = 20.9 dper = Tcpri = 38.9 Tslhx2ro = 3.56 dpslhx = 3.96 Tcpro = Tslhxri = 9.65 dpslhx2 =.6 Tcro = 6.8 Wcb = 0.28 Dslope = 0.00 Tcwi = 2.8 Wch = 3.3 Mcg = 2. Tcwo = 2.55 Wcomp =.62 Meg = 8. TDPeai = 20.8 Web = 0. Mr = 22. TDPeao = 8.3 Weh =.63 Mw = 0.5 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

76 Test number: 25 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 300kPa Dencg =.08 Pcpri = 52.9 Teao = 20. Deneg =.08 Pcro = Tegi = Denr =.9 Pero = Tego = dpca = 8.60 Tcai = 5.85 Tenao = 2.28 dpcna = 2. Tcao = 69.5 Teri = 6.83 dpcr = Tcgi = -.3 Tero = 8.2 dpea = 5.2 Tcgo = 23.9 Tewi = 3.6 dpena = 32.3 Tcnao = 2.02 Tewo = dper = Tcpri = 3.23 Tslhx2ro = 2.22 dpslhx =.0 Tcpro = 3.0 Tslhxri = 9.0 dpslhx2 = 6.86 Tcro = Wcb = 0.2 Dslope = 0.00 Tcwi = 50.6 Wch = 5.0 Mcg = -.6 Tcwo = 2.60 Wcomp =.3 Meg = 9.9 TDPeai = 20.6 Web = 0. Mr = TDPeao = 9.9 Weh =.52 Mw = 2.29 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 68

77 Test number: 26 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 000kPa Dencg =.08 Pcpri = Teao = 20. Deneg =.08 Pcro = Tegi = Denr =.9 Pero = Tego = dpca = 9 Tcai = 5.92 Tenao = 2.26 dpcna = 3.02 Tcao = 0.36 Teri = 6.56 dpcr = Tcgi = Tero = 8.6 dpea = 6. Tcgo = 2.20 Tewi = 3.63 dpena = Tcnao = 2.89 Tewo = dper = 6.55 Tcpri = 3.29 Tslhx2ro =.22 dpslhx =.09 Tcpro =.8 Tslhxri = 9.26 dpslhx2 = 5.88 Tcro = 5.30 Wcb = 0.2 Dslope = 0.00 Tcwi = 5.0 Wch =.68 Mcg =.3 Tcwo = 2.58 Wcomp =.82 Meg = TDPeai = 20.8 Web = 0. Mr = 5 TDPeao = 9.06 Weh =.5 Mw = 0.22 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 69

78 Test number: 2 Flat top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 500kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = Tegi = Denr =.9 Pero = Tego = dpca = 8.62 Tcai = 5.93 Tenao = 2.0 dpcna = 2.96 Tcao = 0. Teri = 6.28 dpcr = Tcgi = Tero = 8.28 dpea = 6.3 Tcgo = 23.9 Tewi = 3.60 dpena = Tcnao = 3.3 Tewo = dper = 69.3 Tcpri = 3.2 Tslhx2ro = 0.33 dpslhx = 0.3 Tcpro = 8.68 Tslhxri = 9.03 dpslhx2 = 5.36 Tcro = 5.05 Wcb = 0.2 Dslope = 0.00 Tcwi = 5.00 Wch =.62 Mcg = Tcwo = 2.60 Wcomp =.8 Meg = TDPeai = Web = 0.8 Mr = 2.08 TDPeao = 8.83 Weh =.50 Mw = 0.53 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 0

79 Test number: 28 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D 300g 200kPa Dencg =.08 Pcpri = 68.6 Teao = 3. Deneg =.08 Pcro = Tegi = 26. Denr =.9 Pero = Tego = 2.6 dpca = 0.3 Tcai = Tenao = 6.53 dpcna =.2 Tcao = 5. Teri = 0.03 dpcr = Tcgi = -.09 Tero =.65 dpea = 3.32 Tcgo = 20.3 Tewi = 3.05 dpena = Tcnao = 53.8 Tewo = dper = 53.5 Tcpri = 25.9 Tslhx2ro = dpslhx = 0.3 Tcpro =.0 Tslhxri =.22 dpslhx2 = 5.63 Tcro = 35.6 Wcb = 0.30 Dslope = 0.00 Tcwi = 3.6 Wch = 5.29 Mcg = 2.25 Tcwo = Wcomp =.5 Meg = 0.62 TDPeai =.02 Web = 0. Mr = 2.2 TDPeao =.5 Weh = 0.60 Mw = Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

80 Test number: 29 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D 300g 3000kPa Dencg =.08 Pcpri = Teao = 3.2 Deneg =.08 Pcro = Tegi = 2.5 Denr =.9 Pero = 69 Tego = dpca = 0.33 Tcai = 32.2 Tenao = 6.5 dpcna =.28 Tcao = 5.93 Teri = 9.83 dpcr = 50 Tcgi = -.6 Tero =.58 dpea = 3.3 Tcgo = 20.3 Tewi = 3.30 dpena = 33.8 Tcnao = 5. Tewo = dper = 6.32 Tcpri = 25.2 Tslhx2ro = 23. dpslhx = 0.6 Tcpro = Tslhxri =.5 dpslhx2 = 5.8 Tcro = 35.3 Wcb = 0.30 Dslope = 0.00 Tcwi = 3.69 Wch = 5.3 Mcg = 95.6 Tcwo = 23.3 Wcomp =.62 Meg = TDPeai = 5.89 Web = 0. Mr = TDPeao = 5.9 Weh = 0.2 Mw = Teai = 3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 2

81 Test number: 30 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D2 300g 3300kPa Dencg =.08 Pcpri = 53.0 Teao = 5.9 Deneg =.08 Pcro = 30.9 Tegi = 2.8 Denr =.9 Pero = Tego = dpca = 9.2 Tcai = 3.6 Tenao = 8.6 dpcna =.95 Tcao = 62.9 Teri = 3.3 dpcr = 55. Tcgi = Tero =.58 dpea = 3. Tcgo = Tewi = 3.52 dpena = 3.0 Tcnao = 65.2 Tewo = dper = 63.8 Tcpri = Tslhx2ro = dpslhx =.6 Tcpro = 33.9 Tslhxri =.5 dpslhx2 = 5.00 Tcro = 6.2 Wcb = 0.29 Dslope = 0.00 Tcwi = 3.2 Wch = 3.59 Mcg = 6.2 Tcwo = Wcomp =. Meg = TDPeai = 6.6 Web = 0. Mr = 2.06 TDPeao = 6.83 Weh = 0.89 Mw = Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 3

82 Test number: 3 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D2 300g 000kPa Dencg =.08 Pcpri = Teao = 5. Deneg =.08 Pcro = Tegi = 2.63 Denr =.9 Pero = Tego = 28. dpca = 9.23 Tcai = 3.9 Tenao = 8 dpcna = 2. Tcao = 62.8 Teri = 3.2 dpcr = 55.0 Tcgi = -6 Tero =.52 dpea = 3.8 Tcgo = 9.2 Tewi = 3.0 dpena = 3.6 Tcnao = Tewo = dper = 58.2 Tcpri = 33.5 Tslhx2ro = 30.8 dpslhx = 2.63 Tcpro = Tslhxri =.0 dpslhx2 =.52 Tcro = 6.26 Wcb = 0.30 Dslope = 0.00 Tcwi = 2.9 Wch = 3.3 Mcg =.8 Tcwo = Wcomp =. Meg = 96. TDPeai = 6.50 Web = 0. Mr = TDPeao = 6.5 Weh =.38 Mw = -0.0 Teai = 32. Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

83 Test number: 32 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D3 300g 300kPa Dencg =.08 Pcpri = Teao =.58 Deneg =.08 Pcro = Tegi = 2.86 Denr =.9 Pero = Tego = dpca = 0 Tcai = 6 Tenao = 9.93 dpcna = 0.66 Tcao = 0.2 Teri = 3 dpcr = Tcgi = -. Tero = 6.52 dpea = 3.06 Tcgo = 8.6 Tewi = 3.56 dpena = Tcnao = 2.66 Tewo = 20.8 dper =.8 Tcpri = Tslhx2ro = 35.9 dpslhx =.9 Tcpro = 3.93 Tslhxri = 6.09 dpslhx2 =.69 Tcro = 53.9 Wcb = 0.29 Dslope = 0.00 Tcwi = 50.6 Wch =.6 Mcg = 0.23 Tcwo = Wcomp =.88 Meg = 9. TDPeai =. Web = 0. Mr = 2.20 TDPeao =.20 Weh = 0.2 Mw = Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

84 Test number: 33 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W 300g 2600kPa Dencg =.08 Pcpri = Teao =.80 Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = 90.3 Tego = 28.6 dpca = 0.2 Tcai = Tenao = 9.8 dpcna = 6. Tcao = 52.6 Teri = 2. dpcr = 0.6 Tcgi = -0.8 Tero = 6. dpea = Tcgo = 3.5 Tewi = 3.60 dpena = Tcnao = 5.6 Tewo = dper = 6.0 Tcpri = 29.3 Tslhx2ro = 2 dpslhx = 5.28 Tcpro =.8 Tslhxri = 6.83 dpslhx2 = 8.0 Tcro = Wcb = 0.3 Dslope = 0.00 Tcwi = 3.69 Wch = 3.8 Mcg = 9.52 Tcwo = Wcomp =.53 Meg = 9.99 TDPeai = 20. Web = 0. Mr = 2.6 TDPeao = 8.33 Weh = 0.69 Mw =. Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

85 Test number: 3 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W2 300g 3900kPa Dencg =.08 Pcpri = Teao = 9.26 Deneg =.08 Pcro = Tegi = 2.69 Denr =.9 Pero = Tego = dpca = 9.22 Tcai = 3.8 Tenao = 2.09 dpcna = 3.3 Tcao = Teri = 5.3 dpcr = Tcgi = -.92 Tero = 6.26 dpea = Tcgo = 2.8 Tewi = 3.68 dpena = Tcnao = 65.6 Tewo = dper = 66.6 Tcpri = Tslhx2ro = 3 dpslhx = 6. Tcpro = 2.9 Tslhxri = 6. dpslhx2 =. Tcro =.5 Wcb = 0.30 Dslope = 0.00 Tcwi = 3.8 Wch = 2.86 Mcg = 02.0 Tcwo = Wcomp =.68 Meg = 98.0 TDPeai = Web = 0.6 Mr = 2.95 TDPeao = 9.23 Weh = 2.03 Mw =.69 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

86 Test number: 35 Flat top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W3 300g 500kPa Dencg =.08 Pcpri = Teao = 9.60 Deneg =.08 Pcro = 390. Tegi = 28.0 Denr =.9 Pero = 52.6 Tego = dpca = 2 Tcai = 9 Tenao = 0 dpcna = 2.30 Tcao = 0. Teri = 6.5 dpcr = Tcgi = -3.0 Tero =.39 dpea = 6.55 Tcgo = 5.89 Tewi = 3.0 dpena = Tcnao = 3.23 Tewo = dper =.9 Tcpri =.02 Tslhx2ro = 38.0 dpslhx = 0. Tcpro = Tslhxri = 6.95 dpslhx2 = Tcro = 5.36 Wcb = 0.30 Dslope = 0.00 Tcwi = Wch = 3.83 Mcg = Tcwo = 23.2 Wcomp =.86 Meg = 98.3 TDPeai = 20.5 Web = 0.8 Mr = 2.2 TDPeao = 20.9 Weh =.6 Mw = 0.3 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

87 Test number: 36 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D 300g 2260kPa Dencg =.08 Pcpri = 9.3 Teao = 6.38 Deneg =.08 Pcro = 6. Tegi = Denr =.9 Pero = 29.2 Tego = 2.26 dpca = 0.5 Tcai = 32. Tenao = 8.25 dpcna =.23 Tcao = 50.6 Teri = dpcr = 2.2 Tcgi = Tero = 9.8 dpea = Tcgo = 2. Tewi = 3. dpena = 35.6 Tcnao = Tewo = 9.28 dper = Tcpri = Tslhx2ro = 2.25 dpslhx = 3.66 Tcpro = Tslhxri = 9.0 dpslhx2 = 0.56 Tcro = 3.92 Wcb = 0.30 Dslope = 0.00 Tcwi = 3. Wch = 3.92 Mcg = 66.2 Tcwo = Wcomp =.52 Meg =.52 TDPeai =.53 Web = 0.8 Mr = 8.0 TDPeao =.53 Weh =.8 Mw = Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

88 Test number: 3 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D 300g 2620kPa Dencg =.08 Pcpri = Teao = 6. Deneg =.08 Pcro = Tegi = 2.39 Denr =.9 Pero = 399. Tego = 28. dpca = 0.23 Tcai = Tenao = 8.06 dpcna = 5.09 Tcao = 50. Teri = 8.2 dpcr = 36.0 Tcgi = Tero = 9. dpea = Tcgo = Tewi = 3.83 dpena = 36.3 Tcnao = Tewo = 9. dper = 8.0 Tcpri = Tslhx2ro = dpslhx = 2. Tcpro = 28. Tslhxri = 8.3 dpslhx2 = 0.23 Tcro = 3.2 Wcb = 0.30 Dslope = 0.00 Tcwi = 3.89 Wch = 2. Mcg = 6.09 Tcwo = Wcomp =.60 Meg = 0.85 TDPeai =.8 Web = 0. Mr =.96 TDPeao =.8 Weh = 2.62 Mw = 2.0 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 80

89 Test number: 38 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D2 300g 2680kPa Dencg =.08 Pcpri = Teao =.23 Deneg =.08 Pcro = Tegi = 2.32 Denr =.9 Pero = 99.6 Tego = 28.2 dpca = 9.39 Tcai = 3.55 Tenao = 9.00 dpcna =.2 Tcao = 6.30 Teri = 2.0 dpcr = Tcgi = Tero = dpea = Tcgo = 22.8 Tewi = 3.82 dpena = Tcnao = 63. Tewo = 9.62 dper = 86.3 Tcpri = 32.3 Tslhx2ro = 29.0 dpslhx = 3.5 Tcpro = 3.22 Tslhxri = 2. dpslhx2 = 2. Tcro = 5.0 Wcb = 0.3 Dslope = 0.00 Tcwi = 2.65 Wch =.59 Mcg = 9.96 Tcwo = 23. Wcomp =.66 Meg =.02 TDPeai =.80 Web = 0.5 Mr = 9.25 TDPeao = 8.3 Weh = 2. Mw =. Teai = 32. Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

90 Test number: 39 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D2 300g 3520kPa Dencg =.08 Pcpri = Teao =.65 Deneg =.08 Pcro = Tegi = 26.9 Denr =.9 Pero = Tego = 28.2 dpca = 9. Tcai = 3.5 Tenao = 9.09 dpcna =.59 Tcao = 6.56 Teri = 2.6 dpcr = 58.3 Tcgi = Tero = 3.02 dpea = Tcgo = 2.05 Tewi = 3.6 dpena = Tcnao = 63.6 Tewo = dper = 8. Tcpri = Tslhx2ro = 29.6 dpslhx = 3.8 Tcpro = Tslhxri = 2.62 dpslhx2 = 2.5 Tcro = 6.5 Wcb = 0.30 Dslope = 0.00 Tcwi = 2.80 Wch = 5.9 Mcg = Tcwo = Wcomp =.68 Meg = 9.29 TDPeai = 0.02 Web = 0.5 Mr = 9.3 TDPeao = 0. Weh = 2.30 Mw =.92 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 82

91 Test number: 0 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D2 300g 2950kPa Dencg =.08 Pcpri = Teao =.52 Deneg =.08 Pcro = Tegi = 2.3 Denr =.9 Pero = Tego = 28.8 dpca = 9. Tcai = 3.0 Tenao = 8.98 dpcna =.2 Tcao = 6.5 Teri = 2.20 dpcr =.2 Tcgi = - Tero = 2.0 dpea = Tcgo = 22.9 Tewi = 3.5 dpena = Tcnao = Tewo = dper = 9. Tcpri = 3.82 Tslhx2ro = dpslhx = 2.86 Tcpro =.63 Tslhxri = 2.23 dpslhx2 = 0.68 Tcro = 5. Wcb = 0.30 Dslope = 0.00 Tcwi = 2.6 Wch = 5.02 Mcg = 2.23 Tcwo = Wcomp =. Meg = 8.8 TDPeai = 0.80 Web = 0.5 Mr = 8.6 TDPeao = 0.95 Weh = 2.3 Mw =.9 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 83

92 Test number: Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D3 300g 3520kPa Dencg =.08 Pcpri = Teao = 9.9 Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = Tego = dpca = 8.2 Tcai = 5.8 Tenao = dpcna = 0 Tcao = Teri = 6.6 dpcr = Tcgi = Tero = 6. dpea = Tcgo = 2. Tewi = 3.85 dpena = Tcnao =.56 Tewo = 9.86 dper = Tcpri = Tslhx2ro = 3. dpslhx = 5.22 Tcpro = 3.6 Tslhxri = 5.0 dpslhx2 =.08 Tcro = 5.35 Wcb = 0.29 Dslope = 0.00 Tcwi = 50.8 Wch = 5.98 Mcg = 3.09 Tcwo = 2.0 Wcomp =. Meg = 9.99 TDPeai = 3.06 Web = 0. Mr = 20.9 TDPeao = 3. Weh =.92 Mw =.8 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

93 Test number: 2 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D3 300g 020kPa Dencg =.08 Pcpri = Teao =.52 Deneg =.08 Pcro = Tegi = 2.3 Denr =.9 Pero = Tego = 28.8 dpca = 9. Tcai = 3.0 Tenao = 8.98 dpcna =.2 Tcao = 6.5 Teri = 2.20 dpcr =.2 Tcgi = - Tero = 2.0 dpea = Tcgo = 22.9 Tewi = 3.5 dpena = Tcnao = Tewo = dper = 9. Tcpri = 3.82 Tslhx2ro = dpslhx = 2.86 Tcpro =.63 Tslhxri = 2.23 dpslhx2 = 0.68 Tcro = 5. Wcb = 0.30 Dslope = 0.00 Tcwi = 2.6 Wch = 5.02 Mcg = 2.23 Tcwo = Wcomp =. Meg = 8.8 TDPeai = 0.80 Web = 0.5 Mr = 8.6 TDPeao = 0.95 Weh = 2.3 Mw =.9 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 85

94 Test number: 3 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: D3 300g 0980kPa Dencg =.08 Pcpri = Teao = 8.93 Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = Tego = dpca = 8.3 Tcai = Tenao = dpcna = 3.09 Tcao = 0.09 Teri = 5.06 dpcr = 62.3 Tcgi = -6.9 Tero = 5.32 dpea = Tcgo = 23.6 Tewi = 3.69 dpena = 330. Tcnao = 2.2 Tewo = 9.92 dper = 86.5 Tcpri = 38. Tslhx2ro = 3.8 dpslhx = 3.39 Tcpro = 8.2 Tslhxri =.90 dpslhx2 =. Tcro = 53.0 Wcb = 0.29 Dslope = 0.00 Tcwi = Wch = 5.9 Mcg = Tcwo = Wcomp =.86 Meg = TDPeai = 2. Web = 0. Mr = 9.52 TDPeao = 2. Weh =.98 Mw =.85 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 86

95 Test number: Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W 300g 290kPa Dencg =.08 Pcpri = 8.22 Teao = 20.8 Deneg =.08 Pcro = Tegi = 26. Denr =.9 Pero = Tego = dpca = 0.6 Tcai = 32.8 Tenao = dpcna = 80 Tcao = 5.8 Teri =.95 dpcr = 58.0 Tcgi = -2.8 Tero = 5.8 dpea = 36.5 Tcgo = 9.3 Tewi = 3.33 dpena = Tcnao = 53.8 Tewo = 9.3 dper = Tcpri = Tslhx2ro = 2.6 dpslhx =.3 Tcpro =.53 Tslhxri = 6.38 dpslhx2 = 5.22 Tcro = Wcb = 0.3 Dslope = 0.00 Tcwi = 3.2 Wch = 5.9 Mcg = 28.6 Tcwo = 23. Wcomp =. Meg = TDPeai = 20.6 Web = 0. Mr = 20. TDPeao = Weh =.80 Mw =.2 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

96 Test number: 5 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W2 300g 360kPa Dencg =.08 Pcpri = Teao = 9.82 Deneg =.08 Pcro = Tegi = 2.52 Denr =.9 Pero = Tego = dpca = 9.58 Tcai = 3.65 Tenao = dpcna = 80.5 Tcao = Teri =.5 dpcr = 65.6 Tcgi = -5.8 Tero =.38 dpea = 35.9 Tcgo =.39 Tewi = 3.5 dpena = Tcnao = 6.36 Tewo = 9.09 dper = Tcpri = Tslhx2ro = 32. dpslhx = 5.3 Tcpro = Tslhxri = 3.9 dpslhx2 =.39 Tcro = 5.85 Wcb = 0.3 Dslope = 0.00 Tcwi = 2.5 Wch = 5.32 Mcg = Tcwo = Wcomp =.66 Meg = 33. TDPeai = Web = 0. Mr = 20.2 TDPeao = 9.58 Weh =.6 Mw =.65 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 88

97 Test number: 6 Round top fin evaporator with tubes vertical Four port tube gas cooler Test condition: W3 300g 0800kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = 36.2 Tegi = 2.68 Denr =.9 Pero = Tego = dpca = 8.92 Tcai = 5.6 Tenao = 9 dpcna = Tcao = 69.6 Teri = 5.95 dpcr = 69.9 Tcgi = Tero = 6.0 dpea = 3.3 Tcgo = 9.8 Tewi = 3.6 dpena = Tcnao = 2.0 Tewo = 9. dper = Tcpri = 0.6 Tslhx2ro = 3. dpslhx =.6 Tcpro = 5. Tslhxri = 5.68 dpslhx2 = 9.29 Tcro = Wcb = 0.3 Dslope = 0.00 Tcwi = 50.0 Wch = 5. Mcg = Tcwo = Wcomp =.83 Meg = 20. TDPeai = Web = 0.6 Mr = TDPeao = 9.96 Weh =.66 Mw =.0 Teai = 38 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 89

98 Test number: Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 230kPa Dencg =.08 Pcpri = Teao = 8.0 Deneg =.08 Pcro = Tegi = 26.6 Denr =.9 Pero = 3. Tego = 28. dpca = 0.8 Tcai = Tenao = 8.35 dpcna = Tcao = 5 Teri = 5.3 dpcr = 39.3 Tcgi = -3.8 Tero = 6.25 dpea = Tcgo =.5 Tewi = 3.96 dpena = Tcnao = Tewo = 8. dper = 6.20 Tcpri = 22. Tslhx2ro = 9.6 dpslhx = 2.9 Tcpro = 22.0 Tslhxri = 5.8 dpslhx2 = 0.69 Tcro = Wcb = 0.29 Dslope = 0.00 Tcwi = 3.2 Wch =.96 Mcg = 9.28 Tcwo = Wcomp =.6 Meg = 00 TDPeai = 9.3 Web = 0.5 Mr =.2 TDPeao = 9.0 Weh =.23 Mw =. Teai = 36 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 90

99 Test number: 8 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 230kPa Dencg =.08 Pcpri = 5.23 Teao =.08 Deneg =.08 Pcro = Tegi = 25.0 Denr = Pero = Tego = 2.0 dpca = 9. Tcai = 3. Tenao = 8.66 dpcna = 85.9 Tcao = Teri = 2.5 dpcr = 9.3 Tcgi = Tero = 2.0 dpea = 2.32 Tcgo = 9.53 Tewi = 3.50 dpena = Tcnao = 62. Tewo = 8. dper = 6.88 Tcpri = 3.5 Tslhx2ro = dpslhx = 3.0 Tcpro = 3.50 Tslhxri =.65 dpslhx2 =.25 Tcro = 5.00 Wcb = 0.29 Dslope = 0.00 Tcwi = 2.28 Wch = 3.0 Mcg = 65.9 Tcwo = 23. Wcomp =.62 Meg = TDPeai = 2.0 Web = 0.5 Mr = 9.0 TDPeao = 2.85 Weh = 2.6 Mw =.60 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

100 Test number: 9 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 2620kPa Dencg =.08 Pcpri = Teao = 9. Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = Tego = 28. dpca = 9.8 Tcai = 3.58 Tenao = 9.5 dpcna = 8.36 Tcao = Teri = 9.39 dpcr = Tcgi = -.62 Tero = 0.0 dpea = 36.0 Tcgo = 23.6 Tewi = 3.60 dpena = Tcnao = Tewo = 8.0 dper =.0 Tcpri = 3.5 Tslhx2ro = 2. dpslhx = 2.9 Tcpro = 3.55 Tslhxri = 9.60 dpslhx2 = 2.05 Tcro = 5.28 Wcb = 0.29 Dslope = 0.00 Tcwi = 2.5 Wch = 3.85 Mcg = 80.8 Tcwo = 22.3 Wcomp =.6 Meg = 99.9 TDPeai =.9 Web = 0.5 Mr = 8.03 TDPeao =.3 Weh = 0. Mw =.5 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 92

101 Test number: 50 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D2 300g 3960kPa Dencg =.08 Pcpri = Teao = 6.93 Deneg =.08 Pcro = Tegi = 26. Denr = Pero = 60.3 Tego = 2.82 dpca = 9.8 Tcai = 3.03 Tenao = 3 dpcna = 85.0 Tcao = Teri = dpcr = 0 Tcgi = -8.2 Tero =.92 dpea = 2.3 Tcgo = 9.2 Tewi = 3.30 dpena = Tcnao = 62. Tewo = 8.8 dper =.2 Tcpri = Tslhx2ro = 2.3 dpslhx = 3.0 Tcpro = Tslhxri =.6 dpslhx2 = 0. Tcro =. Wcb = 0.29 Dslope = 0.00 Tcwi = 2.30 Wch = 3.53 Mcg = 8. Tcwo = 23.2 Wcomp =.65 Meg = 0.26 TDPeai = 3. Web = 0. Mr = 8.0 TDPeao = 3.2 Weh = 2.30 Mw =.6 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 93

102 Test number: 5 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D3 300g 020kPa Dencg =.08 Pcpri = Teao = 8.6 Deneg =.08 Pcro = Tegi = 2. Denr =.9 Pero = 55.9 Tego = 29.3 dpca = 9.03 Tcai = 0 Tenao = dpcna = 8.9 Tcao = Teri =.8 dpcr = 59.6 Tcgi = Tero = 5.32 dpea = 28.6 Tcgo = 2.23 Tewi = 32.3 dpena = 33.3 Tcnao =.96 Tewo = 8. dper = Tcpri = 3.55 Tslhx2ro = dpslhx = 3.6 Tcpro =.2 Tslhxri =.89 dpslhx2 = 3.5 Tcro = 53.2 Wcb = 0.28 Dslope = 0.00 Tcwi = Wch =.3 Mcg = Tcwo = 2 Wcomp =.89 Meg = TDPeai = 5.00 Web = 0.5 Mr = 9.96 TDPeao = 5.9 Weh = 2.3 Mw =. Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

103 Test number: 52 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 00kPa Dencg =.08 Pcpri = Teao = 2.96 Deneg =.08 Pcro = Tegi = 2.6 Denr = Pero = Tego = dpca = 0.6 Tcai = 32. Tenao = 2.6 dpcna = Tcao = Teri = 2.69 dpcr = Tcgi = -.29 Tero =. dpea = 5.82 Tcgo =.3 Tewi = 3.6 dpena = Tcnao = 5.2 Tewo = 8.90 dper = 3.3 Tcpri = 29. Tslhx2ro = 28.0 dpslhx = 8.22 Tcpro = 2.95 Tslhxri =.56 dpslhx2 = 8 Tcro = Wcb = 0.29 Dslope = 0.00 Tcwi = 32.0 Wch =.6 Mcg = Tcwo = Wcomp =. Meg = 38.2 TDPeai = 20.2 Web = 0. Mr = 0 TDPeao = 23.8 Weh = 0. Mw =. Teai = 3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 95

104 Test number: 53 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W 300g 2580kPa Dencg =.08 Pcpri = 5.88 Teao = 20.5 Deneg =.08 Pcro = Tegi = 2.3 Denr = Pero = 63.9 Tego = dpca = 0. Tcai = 32.2 Tenao = 9.83 dpcna = 8.08 Tcao = 5.60 Teri = 2.60 dpcr = Tcgi = Tero =.8 dpea = 5.9 Tcgo =.08 Tewi = 3.50 dpena = Tcnao = Tewo = dper = Tcpri = 26.3 Tslhx2ro = 2. dpslhx =.3 Tcpro =.3 Tslhxri =.38 dpslhx2 = 2.39 Tcro = 3.9 Wcb = 0.29 Dslope = 0.00 Tcwi = 3. Wch = 2.8 Mcg = 20. Tcwo = 23.9 Wcomp =. Meg = 6.83 TDPeai = Web = 0. Mr = 20. TDPeao = 8.25 Weh =.80 Mw =. Teai = 32.9 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 96

105 Test number: 5 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W2 300g 30kPa Dencg =.08 Pcpri = Teao = 20.9 Deneg =.08 Pcro = Tegi = 2.6 Denr = 88.6 Pero = Tego = 29.0 dpca = 9.6 Tcai = 3.5 Tenao = 20.6 dpcna = 83.5 Tcao = 6.8 Teri =.2 dpcr = 6.8 Tcgi = -8.0 Tero =.68 dpea = Tcgo = 8.0 Tewi = 3.52 dpena = 332 Tcnao = 6. Tewo = 8.9 dper = 96.0 Tcpri = 3.0 Tslhx2ro = dpslhx = 5.0 Tcpro = 30.6 Tslhxri =.30 dpslhx2 =.65 Tcro = 5.80 Wcb = 0.29 Dslope = 0.00 Tcwi = 2.85 Wch = 3.85 Mcg = Tcwo = 23.6 Wcomp =.6 Meg = 5.9 TDPeai = Web = 0.8 Mr = TDPeao = 8.99 Weh =.62 Mw =.89 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

106 Test number: 55 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 3980kPa Dencg =.08 Pcpri = Teao = 2.86 Deneg =.08 Pcro = Tegi = 2.52 Denr =.2 Pero = Tego = dpca = 8.85 Tcai = Tenao = 2 dpcna = 8.0 Tcao = 69.9 Teri = 6.23 dpcr = 2.6 Tcgi = Tero = 6.5 dpea = 56.8 Tcgo = Tewi = 3.83 dpena = Tcnao = 2.2 Tewo = 8. dper = 9. Tcpri = 0.2 Tslhx2ro = 3.00 dpslhx =.2 Tcpro = 3.25 Tslhxri = 6. dpslhx2 = 9.86 Tcro = Wcb = 0.28 Dslope = 0.00 Tcwi = 5.0 Wch =.23 Mcg = 99.5 Tcwo = 23.5 Wcomp =.83 Meg =.36 TDPeai = 20.6 Web = 0.6 Mr = 20.8 TDPeao = 9.90 Weh =.83 Mw = 0.95 Teai = 3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 98

107 Test number: 56 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: W3 300g 060kPa Dencg =.08 Pcpri = Teao = 2.63 Deneg =.08 Pcro = Tegi = 2. Denr = 6.2 Pero = Tego = dpca = 8.8 Tcai = 5.92 Tenao = 2.3 dpcna = 9.6 Tcao = 69.5 Teri = 6.0 dpcr = Tcgi = Tero = 6.0 dpea = Tcgo = Tewi = 3.6 dpena = Tcnao = 2. Tewo = 8. dper = 88.8 Tcpri = Tslhx2ro = 36.3 dpslhx =.28 Tcpro = 5.00 Tslhxri = 5.98 dpslhx2 =.8 Tcro = 53.5 Wcb = 0.28 Dslope = 0.00 Tcwi = 5.3 Wch =.05 Mcg = 8.6 Tcwo = 23.0 Wcomp =.83 Meg = 0.3 TDPeai = 20.0 Web = 0.6 Mr = TDPeao = 9.69 Weh =. Mw =.38 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 99

108 Test number: 5 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 500kPa Dencg =.08 Pcpri = Teao = 9.90 Deneg =.08 Pcro = Tegi = Denr = Pero = Tego = 26.6 dpca = 0. Tcai = 32.2 Tenao = 8.33 dpcna = 8. Tcao = 8.0 Teri = 6.6 dpcr = Tcgi = -9.0 Tero = 6.08 dpea =.30 Tcgo = 3.9 Tewi = 3.6 dpena = Tcnao = 50. Tewo = 8.3 dper = 0.3 Tcpri = 22.8 Tslhx2ro = 9.2 dpslhx = 2. Tcpro = Tslhxri = 5.8 dpslhx2 = Tcro = 3.02 Wcb = 0.29 Dslope = 0.00 Tcwi = 3. Wch = Mcg = Tcwo = Wcomp =.0 Meg = 0.99 TDPeai =.0 Web = 0.5 Mr = 6.8 TDPeao =.03 Weh =.9 Mw = -0.0 Teai = 32.5 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 00

109 Test number: 58 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 2500kPa Dencg =.08 Pcpri = 0.33 Teao = 9. Deneg =.08 Pcro = 6.6 Tegi = Denr = 88.9 Pero = 09.5 Tego = 2.98 dpca = 0. Tcai = Tenao =.82 dpcna = 8.9 Tcao = 8.3 Teri = 6.9 dpcr = 30.5 Tcgi = Tero = 5.86 dpea = 0.5 Tcgo = 3.0 Tewi = 3.0 dpena = Tcnao = Tewo = 8.0 dper = Tcpri = 2.6 Tslhx2ro = 8.9 dpslhx =.3 Tcpro = Tslhxri = 9 dpslhx2 = Tcro = 33.5 Wcb = 0.29 Dslope = 0.00 Tcwi = 3.3 Wch = 2. Mcg = 83.5 Tcwo = 23.0 Wcomp =.8 Meg = TDPeai = 0.29 Web = 0.6 Mr = 6.2 TDPeao = 0. Weh =. Mw = 0. Teai = 32. Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 0

110 Test number: 59 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 030kPa Dencg =.08 Pcpri = Teao = 9.5 Deneg =.08 Pcro = Tegi = Denr = 89.3 Pero = Tego = 28.9 dpca = 0. Tcai = 32.2 Tenao =.9 dpcna = 85.2 Tcao = 8.2 Teri = 6.63 dpcr = 2.06 Tcgi = -9.2 Tero = 6.0 dpea = 0.3 Tcgo = 3.88 Tewi = 3.0 dpena = Tcnao = Tewo = 8. dper = Tcpri = 2.06 Tslhx2ro =.9 dpslhx = 0.65 Tcpro = 3.2 Tslhxri = 5.9 dpslhx2 = 2.35 Tcro = Wcb = 0.29 Dslope = 0.00 Tcwi = 3. Wch = 2.39 Mcg = Tcwo = 23. Wcomp =.59 Meg = TDPeai = 0.2 Web = 0.5 Mr = 5.8 TDPeao = 0.6 Weh =.08 Mw = 0.6 Teai = 32.6 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 02

111 Test number: 60 Round top fin evaporator with tubes horizontal Four port tube gas cooler Test condition: D 300g 0960kPa Dencg =.08 Pcpri = Teao = 9.35 Deneg =.08 Pcro = Tegi = 2.3 Denr = Pero = Tego = 28.9 dpca = 0.3 Tcai = 32.9 Tenao =.85 dpcna = Tcao = 8.3 Teri = 6.3 dpcr = 3.2 Tcgi = Tero = 5.66 dpea = Tcgo = 3.38 Tewi = 3.05 dpena = Tcnao = 9.99 Tewo = 8.3 dper = Tcpri = 2.90 Tslhx2ro = 8.9 dpslhx =.85 Tcpro = 2.03 Tslhxri = 5.30 dpslhx2 = Tcro = 33.2 Wcb = 0.29 Dslope = 0.00 Tcwi = 3.6 Wch = 2.0 Mcg =.66 Tcwo = 23. Wcomp =. Meg = 66.0 TDPeai = 0.06 Web = 0.5 Mr = 6.2 TDPeao = 9.8 Weh =.02 Mw = 0.2 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 03

112 Test number: 6 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: D 300g 0990kPa Dencg =.08 Pcpri = Teao =.85 Deneg =.08 Pcro = Tegi = 2.05 Denr = 823. Pero = 8.3 Tego = 28. dpca =.5 Tcai = 32.0 Tenao = 6.03 dpcna = Tcao = 50.3 Teri = 6 dpcr = 28 Tcgi = Tero = 9.9 dpea = 2.8 Tcgo = 6.23 Tewi = 3. dpena = 33 Tcnao = 50.5 Tewo = 8. dper = 9.62 Tcpri = Tslhx2ro = 2.5 dpslhx =.89 Tcpro = 3.82 Tslhxri = 9.35 dpslhx2 =.8 Tcro = 3.50 Wcb = 0.0 Dslope = 0.00 Tcwi = 32.2 Wch =.2 Mcg = 3. Tcwo = 2.8 Wcomp =.6 Meg = 28.9 TDPeai =.62 Web = 0.5 Mr = 9.5 TDPeao =.20 Weh =. Mw = Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 0

113 Test number: 62 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: D 300g 2550kPa Dencg =.08 Pcpri = 58.8 Teao =.6 Deneg =.08 Pcro = Tegi = Denr = Pero = 5.22 Tego = dpca =.56 Tcai = 3 Tenao = 5.88 dpcna = 8.82 Tcao = 5.2 Teri = 8.85 dpcr = 2.35 Tcgi = -0.6 Tero = 9.92 dpea =. Tcgo = 5.8 Tewi = dpena = 3.9 Tcnao = 50.8 Tewo = 8.6 dper = 8.33 Tcpri = 25.3 Tslhx2ro = 23.0 dpslhx = 5.08 Tcpro = 3.63 Tslhxri = 9.5 dpslhx2 = -.90 Tcro = 3.93 Wcb = 0.39 Dslope = 0.00 Tcwi = 32.0 Wch = 5.38 Mcg = Tcwo = 23.0 Wcomp =.6 Meg = 8.89 TDPeai =.89 Web = 0. Mr = 9.55 TDPeao =.3 Weh =.25 Mw = 0.3 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 05

114 Test number: 63 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: D2 300g 250kPa Dencg =.08 Pcpri = Teao = 6.6 Deneg =.08 Pcro = Tegi = 26. Denr = 9 Pero = Tego = dpca = 0.6 Tcai = 3.59 Tenao =.66 dpcna = 83. Tcao = 6.23 Teri =.86 dpcr = 32 Tcgi = -.9 Tero = 2. dpea = 0.9 Tcgo = 0.36 Tewi = 3.3 dpena = 3.5 Tcnao = 60.9 Tewo = 8.32 dper = 6. Tcpri = 33.5 Tslhx2ro = 30.8 dpslhx =.3 Tcpro = 3.50 Tslhxri = 2. dpslhx2 = 3.5 Tcro = 5.8 Wcb = 0.38 Dslope = 0.00 Tcwi = 3.05 Wch = 5.30 Mcg = Tcwo = 22.5 Wcomp =.68 Meg = TDPeai =.30 Web = 0.5 Mr = 9.80 TDPeao =.06 Weh = 2.2 Mw = 0.3 Teai = 32.0 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 06

115 Test number: 6 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: D2 300g 380kPa Dencg =.08 Pcpri = Teao = 6.0 Deneg =.08 Pcro = 2 Tegi = Denr = Pero = Tego = dpca = 0.5 Tcai = 3.50 Tenao = 8.0 dpcna = 83.8 Tcao = 6.28 Teri = dpcr = 36.2 Tcgi = Tero = 3.0 dpea =.5 Tcgo = 0.53 Tewi = 3.9 dpena = Tcnao = 6.08 Tewo = 8.32 dper = 8.00 Tcpri = 3. Tslhx2ro = 3.9 dpslhx = 5.05 Tcpro = 3.3 Tslhxri = 2.65 dpslhx2 = 5.0 Tcro = 5.39 Wcb = 0.39 Dslope = 0.00 Tcwi = 3.02 Wch = 5.2 Mcg = Tcwo = 25 Wcomp =.69 Meg = 20.5 TDPeai =.35 Web = 0.6 Mr = 9.9 TDPeao =.3 Weh = 2.6 Mw = 0.30 Teai = 33 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 0

116 Test number: 65 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: D3 300g 9230kPa Dencg =.08 Pcpri = Teao = 8.23 Deneg =.08 Pcro = Tegi = Denr = 5.9 Pero = 5.28 Tego = 2 dpca = 9.89 Tcai = Tenao = 9.62 dpcna = Tcao = Teri =.6 dpcr = Tcgi = -.55 Tero = 5.33 dpea = 0.8 Tcgo = 8.36 Tewi = 3.3 dpena = 33.6 Tcnao = Tewo = 8.3 dper =.9 Tcpri = 0.23 Tslhx2ro = dpslhx = 3.3 Tcpro = 5.28 Tslhxri =.98 dpslhx2 = 6.88 Tcro = Wcb = 0.38 Dslope = 0.00 Tcwi = Wch = 5.06 Mcg = 53.2 Tcwo = Wcomp =.8 Meg = 29.5 TDPeai = 6.35 Web = 0.5 Mr = 9.9 TDPeao = 6.5 Weh = Mw = 0.3 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 08

117 Test number: 66 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W 300g 0230kPa Dencg =.08 Pcpri = Teao = 9. Deneg =.08 Pcro = Tegi = 2.06 Denr = Pero = Tego = dpca =.66 Tcai = 32. Tenao = 20.6 dpcna = 82.2 Tcao = 9.98 Teri =.0 dpcr = 2 Tcgi = -.29 Tero =.2 dpea = 55.3 Tcgo =.56 Tewi = 3.68 dpena = 3.2 Tcnao = 9.8 Tewo = 8.2 dper = 9.56 Tcpri = Tslhx2ro = 33.9 dpslhx = 22.8 Tcpro = 95 Tslhxri =.62 dpslhx2 =.8 Tcro = 38. Wcb = 0.39 Dslope = 0.00 Tcwi = 3.9 Wch = 3.3 Mcg = Tcwo = 23.9 Wcomp =.23 Meg = 2.08 TDPeai = Web = 0.6 Mr = 2.82 TDPeao = 8.60 Weh = 0. Mw = 2.00 Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 09

118 Test number: 6 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W 300g 080kPa Dencg =.08 Pcpri = 9 Teao = 9.06 Deneg =.08 Pcro = Tegi = 26.8 Denr = 52.6 Pero = Tego = dpca =.6 Tcai = 32. Tenao = 9.6 dpcna = 82.0 Tcao = 5.9 Teri = 2.25 dpcr =.5 Tcgi = -6.9 Tero = 8.95 dpea = Tcgo = 5. Tewi = 3.5 dpena = 38. Tcnao = 0 Tewo = 8.2 dper =.60 Tcpri = 3. Tslhx2ro = dpslhx = 9.8 Tcpro = 0. Tslhxri = 9.30 dpslhx2 = 3.00 Tcro = Wcb = 0.39 Dslope = 0.00 Tcwi = 3.9 Wch = 3.2 Mcg = 29.6 Tcwo = 23.3 Wcomp =.38 Meg = TDPeai = 20.8 Web = 0.6 Mr = TDPeao =.65 Weh = 0.8 Mw =.26 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 0

119 Test number: 68 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W 300g 2320kPa Dencg =.08 Pcpri = 8. Teao = 8.96 Deneg =.08 Pcro = 09.3 Tegi = 26.6 Denr =.9 Pero = Tego = 28.6 dpca =.60 Tcai = 32.2 Tenao = 9.66 dpcna = 82.3 Tcao = Teri =.8 dpcr = 3.85 Tcgi = Tero = dpea = 5 Tcgo = 5.38 Tewi = 3. dpena = 38.9 Tcnao = 5.88 Tewo = 8.3 dper = 68.9 Tcpri = 30.6 Tslhx2ro = 30. dpslhx = 8. Tcpro =. Tslhxri = 20.6 dpslhx2 = 3.3 Tcro = Wcb = 0.39 Dslope = 0.00 Tcwi = Wch = 3. Mcg = 8. Tcwo = Wcomp =.6 Meg = 2.36 TDPeai = Web = 0. Mr = 2.3 TDPeao =.32 Weh = 0.96 Mw = 0.5 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

120 Test number: 69 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W2 300g 2620kPa Dencg =.08 Pcpri = 52.9 Teao = 9.82 Deneg =.08 Pcro = Tegi = 2.63 Denr = 9.8 Pero = Tego = 29.0 dpca = 0.65 Tcai = 3.52 Tenao = 20.9 dpcna = Tcao = Teri =.3 dpcr = 3.59 Tcgi = Tero =.62 dpea = 55.9 Tcgo =.65 Tewi = 3. dpena = Tcnao = 62.3 Tewo = dper = Tcpri = 3.56 Tslhx2ro = 36.0 dpslhx =.3 Tcpro = 2.2 Tslhxri = 8. dpslhx2 =.90 Tcro = 6.09 Wcb = 0.39 Dslope = 0.00 Tcwi = 3.02 Wch =.2 Mcg = 5.2 Tcwo = 2.00 Wcomp =.65 Meg = 66.3 TDPeai = Web = 0.6 Mr = 2. TDPeao = 8.3 Weh =.80 Mw =.6 Teai = 3.96 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 2

121 Test number: 0 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W2 300g 3060kPa Dencg =.08 Pcpri = 56.3 Teao = 9.88 Deneg =.08 Pcro = Tegi = 2.0 Denr = 5.6 Pero = Tego = dpca = 0.63 Tcai = 3.2 Tenao = 20.5 dpcna = 82. Tcao = 62.6 Teri =.35 dpcr = Tcgi = -5.0 Tero = 8.36 dpea = Tcgo =.29 Tewi = 3.80 dpena = 33 Tcnao = 68 Tewo = dper = Tcpri = 3.63 Tslhx2ro = 36.0 dpslhx = 6.9 Tcpro = 30.3 Tslhxri = 9.2 dpslhx2 = 6.90 Tcro = 5. Wcb = 0.38 Dslope = 0.00 Tcwi = 2.99 Wch =.02 Mcg = 8.38 Tcwo = 2.02 Wcomp =.0 Meg = 63.8 TDPeai = Web = 0. Mr = 2.3 TDPeao = Weh =.93 Mw =.38 Teai = 32.5 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 3

122 Test number: Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W2 300g 3000kPa Dencg =.08 Pcpri = 520. Teao = 9.69 Deneg =.08 Pcro = Tegi = 2.8 Denr =.2 Pero = Tego = dpca = 0.62 Tcai = 3. Tenao = 20.3 dpcna = Tcao = 62.3 Teri = 3.9 dpcr = 3.0 Tcgi = Tero = 8.3 dpea = Tcgo =.3 Tewi = 3.53 dpena = Tcnao = Tewo = dper = Tcpri = 3.3 Tslhx2ro = dpslhx = 6.29 Tcpro = 3.2 Tslhxri = 9.2 dpslhx2 =.55 Tcro = 5.30 Wcb = 0.38 Dslope = 0.00 Tcwi = 3.00 Wch = 3.99 Mcg = Tcwo = 23.9 Wcomp =.6 Meg = 62.2 TDPeai = 20.5 Web = 0.6 Mr = 20.6 TDPeao = 5 Weh =.55 Mw =.9 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

123 Test number: 2 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W3 300g 3820kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = Tegi = 2.92 Denr =.32 Pero = Tego = dpca = 9.96 Tcai = 5.6 Tenao = 2.22 dpcna = 8.05 Tcao = Teri = 6.3 dpcr = 3.0 Tcgi = Tero =.05 dpea = 5.33 Tcgo = 6.83 Tewi = 3.9 dpena = Tcnao = Tewo = 9 dper = Tcpri = Tslhx2ro =.52 dpslhx = 6.36 Tcpro = 3.28 Tslhxri =.89 dpslhx2 =.5 Tcro = 5.02 Wcb = 0.39 Dslope = 0.00 Tcwi = 5.03 Wch =.63 Mcg = 60.6 Tcwo = Wcomp =. Meg = TDPeai = Web = 0.8 Mr = 2.66 TDPeao = 9.53 Weh =.8 Mw =. Teai = 32.8 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 5

124 Test number: 3 Flat top fin evaporator with tubes horizontal Six port tube gas cooler Test condition: W3 300g 0920kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = Tegi = 2.28 Denr =.88 Pero = Tego = 28.0 dpca = 9.92 Tcai = 5.9 Tenao = 2.0 dpcna = 82.8 Tcao = 0.03 Teri = 5.93 dpcr = 2.8 Tcgi = Tero = 6.98 dpea = Tcgo = 8.66 Tewi = 3.68 dpena = Tcnao = 0.35 Tewo = 8.2 dper = 5.93 Tcpri = 2.8 Tslhx2ro = 0.0 dpslhx = 5.60 Tcpro = 2.95 Tslhxri = 8.8 dpslhx2 = 0.23 Tcro = 53. Wcb = 0.38 Dslope = 0.00 Tcwi = 5.6 Wch =.89 Mcg = -.98 Tcwo = 23.2 Wcomp =.8 Meg = TDPeai = 20.5 Web = 0.6 Mr = TDPeao = 9.53 Weh =.53 Mw = 0.8 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 6

125 Test number: Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: D 300g 220kPa Dencg =.09 Pcpri = 39.0 Teao = 5.9 Deneg =.08 Pcro = Tegi = 2.50 Denr = Pero = Tego = 26.0 dpca =.5 Tcai = Tenao =.33 dpcna = 82.5 Tcao = 50.3 Teri =.2 dpcr = 2.53 Tcgi = Tero = dpea = Tcgo =.86 Tewi = dpena = Tcnao = 9. Tewo = 8.35 dper = 89.5 Tcpri = 23.2 Tslhx2ro = 20.6 dpslhx =.26 Tcpro =.29 Tslhxri = 8.05 dpslhx2 =.06 Tcro = 3.3 Wcb = 0.0 Dslope = 0.00 Tcwi = 32. Wch = 2.66 Mcg = 6.9 Tcwo = 2.0 Wcomp =. Meg = TDPeai = 8.98 Web = 0. Mr = 8.92 TDPeao = 9.0 Weh =.59 Mw = Teai = 32.8 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0.

126 Test number: 5 Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: D2 300g 3690kPa Dencg =.08 Pcpri = Teao = 6.88 Deneg =.08 Pcro = Tegi = Denr = Pero = Tego = 2.32 dpca = 0.3 Tcai = 2.9 Tenao = 8.8 dpcna = 8.32 Tcao = Teri =.58 dpcr = 36 Tcgi = -5.2 Tero = 2.3 dpea = Tcgo = 9.6 Tewi = 30. dpena = Tcnao = 60. Tewo = 8.36 dper = Tcpri = 3.25 Tslhx2ro = 28.2 dpslhx =.20 Tcpro = 29.9 Tslhxri =.59 dpslhx2 = 2.6 Tcro =.83 Wcb = 0.0 Dslope = 0.00 Tcwi = Wch =.5 Mcg = 80.9 Tcwo = 2.33 Wcomp =.66 Meg = 92.8 TDPeai = 0.00 Web = 0. Mr = 9.6 TDPeao = 0.0 Weh =.09 Mw = Teai = 32.3 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 8

127 Test number: 6 Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: D3 300g 0990kPa Dencg =.08 Pcpri = 55 Teao = 8. Deneg =.08 Pcro = Tegi = Denr = 95.3 Pero = Tego = 2.56 dpca = 9.86 Tcai = 5. Tenao = 9. dpcna = Tcao = Teri =.66 dpcr = 39.2 Tcgi = Tero = 5.03 dpea = Tcgo =.35 Tewi = 30.8 dpena = Tcnao = Tewo = 8.3 dper = Tcpri = 3.3 Tslhx2ro = dpslhx = 3.88 Tcpro =.0 Tslhxri =.50 dpslhx2 = 5.25 Tcro = Wcb = 0.39 Dslope = 0.00 Tcwi = 50.9 Wch =.62 Mcg = Tcwo = 2.62 Wcomp =.8 Meg = TDPeai = 0.56 Web = 0. Mr = TDPeao = 0.6 Weh = 0. Mw = Teai = 32. Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 9

128 Test number: Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: W 300g 20kPa Dencg =.08 Pcpri = 82.6 Teao = 8.89 Deneg =.08 Pcro = Tegi = Denr = Pero =.5 Tego = 2.93 dpca =.2 Tcai = Tenao = 20.2 dpcna = Tcao = Teri =.5 dpcr = 3.0 Tcgi = Tero = 2.23 dpea = Tcgo = 2.00 Tewi = 3.5 dpena = 30.3 Tcnao = 5.95 Tewo = 8.8 dper = 2.25 Tcpri = Tslhx2ro = 23.8 dpslhx =.6 Tcpro = Tslhxri =.9 dpslhx2 = 2.3 Tcro = 35.2 Wcb = 0.0 Dslope = 0.00 Tcwi = 32.8 Wch =.32 Mcg = Tcwo = 2.93 Wcomp =.9 Meg = 29.9 TDPeai = Web = 0.5 Mr = 2.93 TDPeao = 8.66 Weh = 2.0 Mw =. Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 20

129 Test number: 8 Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: W2 300g 3820kPa Dencg =.09 Pcpri = 52. Teao = 9. Deneg =.08 Pcro = 23.9 Tegi = Denr = 80.3 Pero = Tego = 28. dpca = 0.5 Tcai = 3.58 Tenao = dpcna = 8 Tcao = Teri =.39 dpcr = 6.88 Tcgi = Tero =.83 dpea = 3.0 Tcgo = 9.9 Tewi = 3.63 dpena = Tcnao = Tewo = 8.93 dper = 03.9 Tcpri = 32.8 Tslhx2ro = 30.6 dpslhx = 6.29 Tcpro = 2.2 Tslhxri =. dpslhx2 = 6.5 Tcro = 5.92 Wcb = 0.0 Dslope = 0.00 Tcwi = 3.23 Wch =.03 Mcg = 65.6 Tcwo = 22.0 Wcomp =.68 Meg = 2.05 TDPeai = Web = 0.5 Mr = 2.80 TDPeao = 9.9 Weh =.8 Mw =.28 Teai = 32.2 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 2

130 Test number: 9 Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: W3 300g 390kPa Dencg =.08 Pcpri = Teao = 20.6 Deneg =.08 Pcro = Tegi = 26.8 Denr = 93.2 Pero = 52. Tego = dpca = 0.0 Tcai = 5.3 Tenao = 2. dpcna = 8.3 Tcao = Teri = 5. dpcr =.22 Tcgi = -.52 Tero = 6.6 dpea = 36. Tcgo = 3.2 Tewi = 3.2 dpena = Tcnao = 69. Tewo = 9. dper = Tcpri = 3.5 Tslhx2ro = 3.3 dpslhx =.63 Tcpro = Tslhxri = 5.0 dpslhx2 = 5.65 Tcro = Wcb = 0.39 Dslope = 0.00 Tcwi = 50.9 Wch = 5. Mcg = 22.0 Tcwo = 2 Wcomp =.86 Meg = TDPeai = Web = 0.5 Mr = 2. TDPeao = 9.6 Weh =.6 Mw =.53 Teai = Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 22

131 Test number: 80 Round top fin evaporator with tubes vertical Six port tube gas cooler Test condition: W3 300g 060kPa Dencg =.08 Pcpri = Teao = Deneg =.08 Pcro = Tegi = 2.6 Denr = Pero = Tego = 28. dpca = 9.93 Tcai = 5.89 Tenao = 2.80 dpcna = 8.3 Tcao = 0. Teri = 6.5 dpcr = 6.05 Tcgi = -. Tero = 6.80 dpea = 36. Tcgo = 8.38 Tewi = dpena = Tcnao = 0.60 Tewo = 0 dper = Tcpri = 38.2 Tslhx2ro = 35. dpslhx =.92 Tcpro = 39.0 Tslhxri = 6.32 dpslhx2 =.5 Tcro = Wcb = 0.39 Dslope = 0.00 Tcwi = 5.8 Wch =.5 Mcg = 5.6 Tcwo = 22. Wcomp =.8 Meg = 8.00 TDPeai = Web = 0.6 Mr = 2.8 TDPeao = 20.5 Weh =.8 Mw = 0.9 Teai = 32.6 Wel = o C kj/kg-k 0. kj/kg-k. kj/kg-k.3 kj/kg 0.5 kj/kg-k x=0. x= x=0. 23

132 Appendix B. Infrared Pictures This section contains infrared pictures for dry test conditions and for both evaporators in both orientations. The camera was located on the air inlet (refrigerant outlet) side. Flat top fin, tubes horizontal, D 2

133 Flat top fin, tubes horizontal, D2 Flat top fin, tubes horizontal, D3 Flat top fin, tubes vertical, D3 25

134 Round top fin, tubes vertical, D2 Round top fin, tubes vertical, D3 26

135 Round top fin, tubes horizontal, D Round top fin, tubes horizontal, D2 2