Here is a spread sheet that will help plot single stage cycles for several refrigerants. It does work on Office Mobile, but won't work on Google Sheets. The tool uses lookup functions that Sheets doesn't seem to have, Enjoy! Find any errors ley me know, and the sheet isn't protected so any change you make is changed. stick to only changing the blue cells and the sheet should be fine,
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With a little system information, it possible to estimate the refrigerant charge required using the pressure enthalpy diagram. Below, I plotted a R22 system cycle based for split system airconditioner operating at a saturated coil of 45 degrees Fahrenheit, and a saturated condenser operating at 120 degrees Fahrenheit. The table is the data used to make the plot. The temperature lines are in 10 degree increments. This is a 3ton split system with a 3/8” liquid line and a ¾” suction line 50foot long. The discharge line is ¾” and 2foot long. There are no other components that add volume to the system. (There is a volume associated with suction gas cooled compressors but it relatively small compared to system charge.) The net refrigeration effect, or evaporator cooling, is the difference in specific enthalpy of the evaporator, or, 175.41 109.3 = 66.11 Btu per pound of refrigerant. Three tons of cooling is 600 Btu/minute. Divide 600 by 66.11 and you get the refrigerant flow of 600 / 66.11 = 9 pounds per minute of refrigerant flow required to meet the capacity. Now, the dynamic condition of the refrigerant isn’t all liquid of nine pounds. The refrigerant is moving at a given velocity so that 9 pounds gets through evaporator and the condenser in one minute, not all at once. let’s examine where the liquid is. Below is a sketch of the system, and marked in red are the points in the system that are 100% liquid. Notice all the 100% liquid is in the last part of the condenser and the liquid line running to the expansion valve. The rest of the system is either a mixture of liquid and gas, or 100% gas. The evaporator gets a stream of liquid from the expansion valve at a temperature of 44.507 degrees F and a quality of .23, which means, the fluid is 23% vapor and 77% liquid. When the refrigerant leaves the evaporator, it will be a superheated gas with no liquid remaining, this is a quality of 1, or 100% gas. To determine how much of the refrigerant is liquid, and how much is gas, you can construct a triangle like the one below. The total liquid available is the area under the triangle which is found by ½ X (1.23) X .77 = .296 percent, round up to 30%, is liquid at any given moment in the evaporator. The rest is saturated vapor, or gas.
The energy can only be removed by boiling liquid refrigerant so, if 30 percent of the content is liquid, only 30 percent of the 9 pounds per minute is in the evaporator, or, .3 X 9 = 2.7 pounds of liquid at any moment in the evaporator. The specific volume of the refrigerant liquid in the evaporator is .012738 cubic foot per pound so, the liquid occupies .012738 X 2.7 = .0343 physical cubic foot. We know 30% of that is liquid and the other 70% is gas occupying the remaining space in the evaporator. It’s safe to say that the amount of space the gas occupies must be minimized because the gas doesn’t remove any heat to speak of. The gas and liquid are at the same temperature and pressure so the relation (P1 X V1)/T1 = (P2 X V2)/T2, The P & T’s cancel out and you have V1 = V2, so the gas is approximately occupying .0343 cubic foot also. The amount of refrigerant gas would then be .0343 multiplied by the vapor density, 1.6421, or, .0343 X 1.6421 = .046 pounds of vapor in the evaporator. Total volume of the evaporator, .0686, would have the equivalent volume of 127 foot of 3/8 copper line. The total evaporator charge during operation is 2.75 pounds. Next determine how much refrigerant is in the suction line. Information about tube dimensions and applications of copper tubing: https://www.copper.org/applications/plumbing/cth/ ¾” line has .00252 cubic foot of contents per foot. The line is 50footlong, so the entire volume of the suction line is .00252 X 50 = .126 cubic foot. To know how much refrigerant is in the line, divide the volume of the line by the specific volume of the refrigerant at 55 degrees, .126/.6287 = .2 pounds in the suction line. So, the low side of the system has, 2.75 + .2 = 2.77 pounds of refrigerant. Moving to the highpressure side of the refrigerant circuit we address the small amount of discharge line the same way as the suction line except we use the specific volume at 256 degrees, which is .30017. Two foot of line multiplied by the internal volume, 2 X .00252 = .005, now divide the volume by the specific volume, .005/.30017 = .017 pounds in the discharge line. The latent process of the condenser must remove the same energy as the evaporator, the condenser also must make at least 2.7 pounds of liquid per cycle in its cycle to keep the balance of energy happy. The specific volume of the refrigerant liquid in the condenser is .014364 cubic foot per pound so, the liquid occupies .014364 X 2.7 = .0388 physical cubic foot. The amount of refrigerant gas would then be .0388 multiplied by the vapor density, 4.941, or, .0343 X 4.941 = .19 pounds of vapor in the latent area of the condenser. The total refrigerant held up in the condenser is then, .19 + 2.7 = 2.89 pounds. For the desuperheating region of the condenser we know the low side charge must be desuperheated as it continues to the condenser so, we need estimate how much line it will take. Total heat transfer is equal to the total area in square foot multiplied by heat transfer coefficient and the temperature difference. The formula can be rearranged to find the total area, and then the length of tubing. Btu/hour/ (Transfer Coefficient X Temperature difference) = Area in square foot. Our Btu/hour = 9 pounds X (209.43 180.20) X 60 =15,784.2 Divided by (220 Btu/hour for copper X a temperature difference of 35) (air side) = 2.4 square foot. 3/8 copper pipe has an internal area of .0821 foot per foot, so 2.4/.0821 = 29 foot of pipe The internal volume of 3/8 copper is .00054 cubic foot, so total area is, .00054 X 29 = .015 cubic foot The density of the discharge gas is 3.3314 pounds per cubic foot multiplied by the volume of .015, you get .05 pounds in the desuperheating region. The subcooling process, 112 – 109.3 = 2.7 Btu/pound to cool the condensed liquid below its saturation point. Find the Btu/hour, 9 pounds X 2.7 X 60 = 1,458 Find the area needed, 1,458/ (220 X 35) = .189 Find the length of pipe, .189/.0821 = 2.3 Find internal area, 2.3 X .00054 = .0012 The density is 69.618 pounds per cubic foot multiplied by .0012 you get .086 pounds of liquid being subcooled. All we need to complete the total is the refrigerant in the liquid line. 3/8” line has an internal volume of .00054 cubic foot per foot. Multiply by the 50 foot of line, 50 X .00054 = .027 cubic foot in the liquid line. The amount of refrigerant is found by multiplying the volume of the line by the specific density of the refrigerant at 110 degrees. The chart only give us specific volume, but no problem, the specific density is just the reciprocal of the volume, so 1/.014364 = 69.61 pounds per cubic foot of liquid at those conditions. So, 69.61 X .027 = 1.879 pounds in the liquid line. Tallyho: Evaporator and suction – 2.77 pounds. Discharge Line  .017 pounds. Desuperheating – .05 Condensing – 2.89 Subcooling – .086 Liquid line – 1.879 Total estimated charge – 7.7 pounds. This is just an estimate of the operating charge of a direct expansion air cooled system. The charge should be added at 80% of the estimated value then adjusted to meet the subcooling value of 10 degrees. I’ll start this discussion with a plot of the system design perimeters. I do this a lot because if you don’t know how the system is supposed to operate, you can’t know when something is broke! Our unit is an air cooled 134a cycle with 10 degrees of superheat and 9 degrees of subcooling. The saturation temperature of the evaporator is 45 degrees Fahrenheit, and the saturation temperature of the condenser is 120 degrees F. The system capacity is 36,000 Btu/hour, or three tons. Below is the cycle plotted and some of the system characteristics calculated from the information on the diagram. There a varying degrees of low system charge, very low system charge will freeze the coil up and have ice hanging off the suction line. This condition doesn’t require much diagnostic effort, if the filter is okay, and the blower is okay, then low charge is probably it, and then you find the leak. The low charge condition that’s tough to find is one that leads to capacity problems with no other apparent symptoms. This condition leads to very high energy costs and should be corrected as soon as its determined. Refrigerant mass flow multiplied by the heat of compression and divided by 42.43 is the compressor horse power, so, (9.8 X 35)/42.43 = 8 horse power. To find the compressor amp draw, we multiply horse power by 746 (watts per horse power) and divide by the (volts X Efficiency X power factor). Efficiency and power factor can be considered 1 for most circumstances, so, (8 X 746) / 220 volts = 27.4 amps we would expect to see on a singlephase unit compressor. Next, we plot the low charge conditions against the design conditions. The plot will look like low load or low ambient, but one thing to remember is that low load and low ambient have ample cooling capacity, where low charge everything is low (pressures, capacity, subcooling) but two characteristics, the heat in the compressor and out. The superheat with an expansion valve system will be elevated depending on the extent of undercharge since the valve will try to compensate for the high suction line temperature. On fixed metering devices, such as pistons and capillary tubes, the superheat will be notably high. I have plotted a cycle of the same unit during the same operating conditions but the unit is under charged. The subcooling dropped to 1 degree and the suction superheat bumped up to 13 degrees. The compressor superheat (discharge temperature minus the saturation temperature) also increased from 108 to 111 degrees.
The evaporator enthalpy dropped to 60 Btu/pound lowering the net refrigeration effect and the inlet volume to the compressor changed to .99 cubic foot per pound. The volume of flow, 8.7 cfm, calculated on the first diagram, is the compressors physical displacement because that was the design flow at the design inlet volume. If we divide the fixed volume of 8.7 by our new suction density of .99, the mass flow is reduced to 8.7/.99 = 8.78 pounds per minute of refrigerant flow. 11% lower. The new mass flow multiplied by the lower evaporator enthalpy, 8.78 X 60 Btu/pound = 527 Btu/minute. There is 200 Btu/min in a ton of cooling, so, 527/200 = 2.6 Tons of cooling 14% less cooling than the 3ton design. That’s enough to cause problems. The compressor is running hotter because the gas coming back to it is hotter. The pressure ratio dropped down some so there is a little less work going on. The heat of compression, 33 Btu/pound, multiplies by the new flow, 8.78 pounds per minute, divided by 42.43 is 6.8 horse power. Finding the new motor current, (6.8 X 746)/220 = 23 amps, about 5 amps less than the fully charged system. Hot gas bypass (also called discharge bypass) is a feature in a refrigeration system uses to satisfy the mechanical needs of the system during low load conditions. Low load conditions can lead to frosting or freezing the evaporator, refrigerant flood back to the compressor, system shutdown, oil return problems, and several other undesirable conditions. The hot gas bypass utilizes a line from the discharge of the compressor to deliver high temperature, high pressure refrigerant either to the evaporator distributor (preferred) or the suction line to the compressor (which may overheat the compressor). The components usually are a solenoid valve and a pressure/flow control. Larger systems will likely only have an automatic valve, but no solenoid valve. Below, is a system sketch of a basic discharge bypass line on a simple aircooled system. Item 1above, is the solenoid valve, the valve is to prevent the addition of discharge bypass gas when it isn’t wanted. The solenoid valve is usually controlled by a pressure sensing switch which activates if the suction pressure drops below a certain pressure, for airover evaporators this should be the saturation pressure that coincides with 36 degrees Fahrenheit. The valve could just as easily be controlled by the evaporator temperature. Item 2 above, is the pressure regulator/flow control. The valve is a springloaded regulator that monitors the line pressure downstream of the valve and opens, or closes, to maintain a preset pressure against the internal spring. Some have sensing lines, some don’t. They are set to the normal operating pressure of the suction line. Item 3 above, is the distributor. The distributor is installed after the expansion valve and “distributes” the refrigerant to the various loops of the evaporator for even cooling across the coil. Some distributors have a tap specifically for the hot gas, others will simply be a tee in the line prior the distributor itself. As mentioned before, but not shown on the sketch, the line can be tapped into the suction line somewhere before the compressor (not recommended). The location should be as far from the compressor as possible. Air cooled systems that are going to operate during low load and/or low ambient conditions should have a head pressure controller or fan cycling control capable of keeping the condenser pressure at a saturation temperature of 80 degrees Fahrenheit. The system will not function correctly without the control. Now that we have a description of the system we can add the details. The refrigerant is R134a designed to operate at a 120degree saturated condenser and a 45degree saturated evaporator. There is 9 degrees of subcooling and 10 degrees of superheat. The design load is 1 ton, or 12,000 Btu/hour. We know when the ambient temperature is below 70 F, the system will be expected to maintain half the design load, or 6000 Btu/hour. First, I will plot the two operating conditions without the hot gas. So, the blue cycle is the design cycle, and the pink is the cycle at low load. The low load cycle evaporator pressure is below freezing (41 psia/ 30 F). This will cause the evaporator to freeze up and slug the compressor with liquid refrigerant. Hot gas works by modifying two conditions in the evaporator, one, it raises the saturation pressure above freezing, and two, it displaces some liquid refrigerant that does the cooling. So we next need to determine how much hot gas we need to displace 6000 Btu/hour of cooling. The design cycle has a refrigeration effect of 61 Btu/pound of refrigerant. The 12,000 Btu/hour design divided 60 minutes, is 200 Btu/minute. Divide 200 by 61, and the mass flow of refrigerant per minute is 3.28 pound per minute. This is the design flow rate. The flow at low load depends on the volume change of the refrigerant at the compressor inlet. For the design cycle the volume is .89 cubic foot per pound, or .89 X 3.28 pounds, 2.93 cubic foot per minute. For the low load inlet conditions the volume is 1.19 cubic foot per pound, so the compressor design cfm, 2.93, divided by the low load specific volume, 1.19, gives 2.46 pounds per minute flow of refrigerant at low load. (Volumes can be found on standard P&H charts). The refrigeration effect at low load is 72 Btu/pound. Multiply by the flow, 2.46, and the system capacity is 177 Btu/minute. As stated earlier, the system needs to operate at half load, or 6000 Btu/hour = 100 Btu/minute. We need to apply 77 Btu/min (177 – 100) of hot gas bypass to reduce the amount of liquid available for refrigerating with discharge gas. What happens when we inject hot gas into the evaporator inlet is we displace liquid droplets that are capable of cooling, the result is a change to the “quality” of the gas in the evaporator, and in turn, a lowering of the refrigeration effect. The specific enthalpy of the gas at the discharge of the low load compressor is 76 Btu/lb. We need 77 for the evaporator, or 77/76 = 1.01 pound per minute of hot gas bypass. Another advantage to this approach is that hot gas diverted from the discharge isn’t fed into the condenser and expansion valve so less liquid refrigerant is available for the evaporator. So, we now have 1.01 pound per minute of hot gas and 1.45 pound per minute of liquid feed to the evaporator at an unknown quality. We can approximate the quality using a mixing equation to find the new quality; {(Mass1 X Quality1) + (Mass2 X Quality2)}/ (Mass1 + Mass2), {(1.45 X .15) + (1.01 X 1)}/ (1.45 + 1.01) = .499 is the new quality of the gas in the evaporator. This looks like the process plotted below. Changing the quality of the gas reduced the refrigeration effect to 41 Btu/pound. We multiply 41 by the new evaporator flow of 2.46 lb/minute we get 100.86 Btu/minute or 6052 Btu/hour, pretty much half the load.
The introduction of the gas at the prescribed rate will also raise the line pressure to the regulator set pressure of 46 psia (36 degrees F saturation temperature mentioned earlier) getting the coil above freezing. The discharge pressure and temperature will also increase proportionately while the suction superheat will remain controlled by the expansion valve. When discharge gas is injected in the suction line the behavior is a little different. The effect is to remove gas that could pass through the condenser and evaporator then raise the back pressure to raise the overall saturation temperature above freezing. If the hot gas injected into the suction line is before the sensing bulb, the expansion valve will try to compensate by adding liquid to the evaporator to lower the superheat which will lead to a roller coaster ride between the expansion valve and the hot gas regulator. If the injection point is down stream of the expansion bulb, the heat from the discharge gas will raise the temperature of the gas in the suction line to about 75 degrees which could be above the manufacturer inlet temperature specification for the compressor. Efficiency is always what you get out of a system divided by what you put in. If something is 50% efficient, then we needed two times the energy to get half the work in return. Mathematically speaking the formula for 50% efficiency is ½. Which is also to say, if we got 500 Btu/hour from our airconditioner and used 1000 Watts to get it, 500 / 1000 = ½ = 50% efficient. Thermal efficiency can be viewed in the same way. A coil operates at 45 degrees Fahrenheit, and the condenser operates at 100 degrees Fahrenheit. The efficiency becomes: What we got divided by what it took, minus were we started. Got that? What we got – 45, what it took – 100 minus where we started – 45. Mathematically, 45 / (10045) = .82 or 82%, that is thermal efficiency. On the 410a diagram below I drew a cycle of the above conditions with the compression line following the entropy line, or line of perfect compression, which put the temperature at 135F. I then marked the enthalpy line for the heat of compression at 185 and 197 for a specific heat of compression of 12 Btu/pound of gas. Since the thermal efficiency of this process is 82%, dividing 12 by .82 gives a new specific enthalpy of 14.6 Btu/pound. Using 185 (the original starting point) we add 14.6 to get 199.6, or about 200 Btu/lb Replotting the compression line below the new discharge temperature emerges. (Temperature lines are in 10 degree increments). The process can be used in the field using the gauge pressures as previously described. Actual discharge temperature will be little higher depending on the amount of friction in the compressor. If the temperature is much lower, determine if liquid may be entering the compressor
Adding more heat to a gas increases the temperature of the gas above its saturation temperature (the temperature in which it boiled into a gas). The temperature increase, or difference between the saturation temperature and the new temperature, is called superheat. On the 410a cycle below, the refrigerant has gone through the latent process (boiling) in the evaporator and had extra heat added to raise it 10 more degrees, most which, was added in the final stage of the evaporator. The added superheat was about 3.5 Btu/minute of added enthalpy, or about a 4% cost to the system over the cooling capacity.
The superheat adds heat to the compressor and raises the overall discharge temperature. So, it would seem less superheat would be better. Liquid refrigerant will destroy compressors. Superheat protects compressors from liquid refrigerant at a very small cost to overall performance. Note: heat in the compression process is cumulative, too much superheat will also be detrimental to the compressor causing overheating. Manufacturers publicize the maximum allowable inlet temperature for their specific model compressors, if that isn’t available, keep it below 30 degrees. Subcooling serves several purposes, two of which are to ensure a solid stream of liquid gets to the metering device, and increase the refrigeration effect. So, why is a solid column of liquid required? Performance of the metering device will be greatly reduced. All metering devices will only meet their rated capacity while being provided all liquid at a specified pressure drop. If a metering device gets liquid and flash gas the performance suffers greatly. The valve will hunt for set point, and the system pressures will behave erratically which could lead to other control problems. Only liquid droplets can evaporate and remove latent heat from the load. Flash gas is already boiled liquid so there is no latent energy potential to pick up the load in the evaporator. A metering device, in its simplest form, can be considered an orifice (plate with a hole in it). In the illustration below, the upper would be something like an orifice getting full liquid pushed against the plate and developing large droplets to feed the evaporator. The lower orifice is getting a mix of gas and liquid emitting a spray of gas and tiny droplets which will be inefficient at removing the heat energy in the evaporator. On the pressureenthalpy diagram, subcooling occurs in the final stage of the condenser then continues, to a lesser extent, in the liquid line until it reaches the metering device. The diagram shows what 10 degrees of liquid subcooling looks like on a R410a cooling cycle. The chart also shows nearly a 10 Btu/pound loss of capacity because the specific enthalpy difference is now less. 10 Btu/lb loss on a cycle that is 70 Btu/lb would be a little over 14%. Another point to consider is how much subcooling is enough? This can be found using the pressure enthalpy diagram and the piping information. Linesets, filter dryers, friction loss, and condenser pressure drop all add up to give a total pressure loss on the highside of the system. The diagram below has 5 degrees of subcooling marked against 10 psi incremental pressure drops. After 20 psi the metering device will start getting flash gas. 20 psi seems like a lot of pressure loss, but add it up, 80 equivalent foot of 3/8 liquid line (50 linear foot and three elbows) has a pressure loss of about 10 psi, filter dryer 3 psi, leaving 7 psi for the condenser, that’s cutting it close. A subcooling of 10 degrees allows for almost 40 psi pressure drop before failing at the valve. Most recommendations for split systems are rarely less than 9 degrees of subcooling.
The unit is a 400 ton, R134a, fourstage compressor, with a twostage flash gas economizer. To perform a heat balance and diagnose a condenser problem, the total flow of refrigerant through the condenser needs to be known. The full flow through the condenser is the sum of the evaporator and the compressor side loads from the economizer. The gas in the economizer is part liquid, part vapor, the vapor goes to the compressor, the liquid proceeds toward the evaporator. Multistage compressors have an ability to introduce gas in between impellor stages. This gas is taken from the economizer which lowers the pressure in the economizer stage, the result is an increase in refrigeration effect and a reduction of heat exchange surface area (which reduces first costs, maintenance costs, and is very desirable). Below is a basic arrangement of the twostage economizer and fourstage compressor system (no valves or controls are shown). The arrows mark the basic direction of flow and the numbered points mark the location of where the data points are associated with the points on the pressure enthalpy diagram (next figure). Point 1, is the compressor suction, point two, is the discharge of the first stage. Point 3, is the mixture of gas from the first stage and the low side economizer is, point 4. Point 5, is the discharge of the second stage, point 6 is the mixture of the second stage and the high side economizer is, point 7. Point 9, is the discharge of the compressor, point 10, the liquid leaving the condenser, point 11, is the high side liquid condition, and point 12, is the low side liquid condition. The system above is plotted on a pressureenthalpy diagram below: The first step is to determine the mass flow of refrigerant through the evaporator. The unit is a 400ton unit, to find the mass flow of refrigerant multiply 400 X 200 Btu/minute/per ton, and divide by the enthalpy difference between point 1, and point 12, or, 163 – 73 = 90 Btu/pound/minute, so (400 X 200)/90 = 888.8 pounds per minute.
889 pounds per minute are flowing through the evaporator (I rounded up), we can now determine the ratio of the pounds of flash gas to pounds of liquid in the low side from the enthalpy values at points 11, 12, and 4 as follows: (Point 11 – Point 12)/ (Point 4 – Point 11), or (83 – 73)/ (165 – 83) = .122 To find the flash gas flow multiply .122 by 889 to get, 108.5 pound per minute of flow for the flash gas from the low side economizer. Next step is to find the flow from the high side economizer. The flow is cumulative through the compressor, so, our new flow is 889 + 108.5 = 997.5 pound per minute to the third stage. We need to find the ratio of gas to liquid again using the same approach except this time using points, 10, 11, and 7. (Point 10 – Point 11)/ (Point 7 – Point 10) = (101 – 83) / (171 – 101) = .257 Using the flow of 997.5 multiplied by .257 we get, 256 pounds per minute flow from the high side economizer. So, to find the full flow to the condenser, 889 + 108.5 + 256 = 1253.5 pounds of refrigerant per minute through the condenser. Note: Point 3 and 6 are internal to the compressor and not likely obtainable in the field, since you have the two ratios needed now, the points can be found using the pressure enthalpy diagram as follows: Point 3 = (Point 2 enthalpy + (low side ratio X Point 4 enthalpy))/ (1 + (low side ratio)) Point 3 = (169 + (.122 X 165))/ (1 + .122) = 168.6 Point 6 = (Point 5 enthalpy + (high side ratio X Point 7 enthalpy))/ (1 + (high side ratio)) Point 6 = (176 + (.257 X 171))/ (1 + .257) = 174.9 Point 3 and point 6 are in reasonable agreement to the design information with error likely attributed to me reading the scaling on the chart. You are working on a 2ton aircooled condenser using 410a refrigerant. The system just almost cools the load but, there seems to be a capacity problem and low subcooling (difference between the actual liquid line temperature and the saturation temperature of the refrigerant), so, you suspect a condenser problem. The condenser is a little dirty, but the fan appears to be moving air okay. There is some fin damage, and the condensing fan looks like it has been changed recently, so you want to investigate if the condenser is moving enough air. This is a 2ton unit, so, it is cooling 400 Btu/min. The system is designed for 100degree Fahrenheit condensing temperature while operating a 45degree Fahrenheit evaporator. We need to know the mass flow of the refrigerant, per minute, moving through the system. We plotted the cycle diagram on the pressureenthalpy diagram and dropped lines from the points of interest straight down to the enthalpy scale. Below is what we ended up with: The specific enthalpy of the evaporator, or refrigerant effect, is the difference between 182 Btu/lbm and 108 Btu/lbm, or, 74 Btu/lbm. This is all the cooling the evaporator was designed for, so, if we are getting 400 Btu/min of cooling from the system, we must be moving, 400÷74=5.4 lbm of refrigerant. The specific enthalpy of the condenser is 192108=84 Btu/lbm. Multiplied by our mass flow, 84X5.4=453.6 Btu/min of total heatrejection for the condenser. We now have a basic model of how our system was intended to operate, from this, finding the condenser airflow is straight forward. First convert the condenser heat to Btu/hour by multiplying by 60 minutes, 435.6X60=27,216 Btu/hour heat rejection. Next, we need to know the temperature difference from the condenser discharge and the ambient air. In the field, we would just take this measurement, for discussion we will use the “apparent” rise from the pressure enthalpy chart. I marked the chart up to find the temperature difference of the refrigerant cycle in orange below. The chart is a little crowded at the top, but you can make out that the lines end on 120 and 90 degrees Fahrenheit, or a temperature difference of 30 degrees. We don’t know the exact temperatures of the air from the chart, but the difference will be proportional.
Next, we use the formula CFMX1.08X(T2T1) =Btu/hour. We want to know CFM (cubic foot per minute) for our condenser fan, so, we must rearrange the formula to solve for CFM, or, (Btu/hour)/ (1.08XTemperature difference) Solving: 27216/(1.08X30) =840 cubic foot per minute of condenser air minimum. With the flow known, 840cfm, the condenser air flow can be tested and determined whether it is sufficient enough to reach capacity. All refrigerants have a pressureenthalpy diagram. In short, the diagram shows property conditions of the refrigerant at various stages of state. The properties inside the “thumbprint” shape is “saturation” properties (mixtures of liquid and vapor). The area to the left of the thumbprint are properties of liquid refrigerant, and to the right, properties of vapor refrigerant. The left vertical axis is the index for pressure (psia, pounds per square foot atmospheric). The bottom axis is the index for enthalpy, or energy, the capacity of the refrigerant at that stage. Other lines, temperature (shown), volume (not shown), entropy (not shown), and quality (not shown), allow for the state of the refrigerant in a cycle to be plotted to find unknown values that can be used to calculate energy, cooling capacity, efficiency, horse power, and many more valuable bits. Using the pressure and temperature from a system the cycle below was plotted. I want to know how much horse power I should expect the system to be using. The two red lines in the diagram are drawn to be straight drops for the intersection of the cycle point, compressor inlet, and compressor discharge. The lines cross the enthalpy line at 177.5 and 188.5 (approximately). By subtracting 177.5 from 188.5, I arrive at a specific enthalpy of compression of 11 Btu per pound of refrigerant compressed per minute of operation. The formula to find our horse power is: 42.43 is a mechanical constant, the mass flow of refrigerant we will find next. Let’s say we are working on a 2ton airconditioner, and since 200 Btu per minute equals 1 ton of cooling, our total cooling per minute needed is 200 X 2 = 400 Btu per minute. Since each refrigerant, and set of conditions, have a specific cooling capacity called the “net refrigeration effect” we must find the specific cooling capacity of the system plotted above. To do this, we need to drop two more lines as below: The lines intersect the enthalpy scale at 107.5 and 175.5 (approximately). Once again, we take the difference of these two numbers, 175.5 – 107.5 = 68 Btu per pound.
68 Btu/lb is the “net refrigeration effect” for our system, or, the amount of cooling capacity per pound of refrigerant flowing. To find the mass flow of refrigerant, in pounds per minute, we divide 400 (Btu per minute of cooling for the total system) by 68 Btu/min (net refrigeration effect). 400 / 68 = 5.88 pounds per minute is being circulated to achieve the 2 tons of cooling in this system. Now that we have mass flow, 6.88 lb/min, enthalpy of compression, 11 Btu/lb, we can work the first equation: (6.88 X 11) / 42.43 = 1.78 horse power. Knowing that our system should use 1.78 horse power, we can now determine if the unit is overloaded, under loaded, drawing too high of current, etc. 
AuthorI worked for over thirty years in the HVACR industry. I have designed, installed, serviced, and trouble shot units of various types throughout the years. The posts here are information based on that experience, I hope you find them useful. If you have a different experience, please comment. Archives
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