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.
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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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