The high-side pressure is related to the ambient air temperature. When servicing a unit with the condenser inside an air-conditioned building, expect the head pressure to be considerably lower than a unit with its condenser on the roof in the hot summer sun. The warmer the surrounding air, the higher the head pressure, and the more difficult it becomes for the unit to cool.
As a general rule, to determine the expected high-side pressure in a unit with an air-cooled condenser, add 30°F to the current ambient (surrounding the condenser) temperature; check the chart on page 133 for the pressure reading at that added temperature for the type of refrigerant used in the unit. This figure indicates the expected head pressure in the unit. For water-cooled condensers, the head pressures should be calculated by adding 20°F to the exhaust-water temperature. (See the section on Water-Cooled Condensers.)
EXAMPLE:Suppose there is a commercial central air-conditioning unit using R-22 operating in an ambient temperature of 90°F. Assume also that this is a new unit with no malfunctioning parts. What head pressure should you expect to read on the high-pressure gauge? For the answer, add 30°F to the ambient temperature (90 + 30 = 120°F), find 120°F on the chart, and read across to the R-22 column. Hence, 259.9 lb/in2 will be the expected high-side pressure.
If the system is overcharged, if the condenser is dirty and/or linted, if there is a slow or inoperative condenser fan, or if there is a restriction in the system, the head pressure increases dramatically, causing a higher than normal head pressure. A temperature- pressure chart should always be carried with you on service calls. The amount and type of refrigerant used in the unit can be determined by the unit nameplate.
PROBLEM: Determine the head and the back (low side) pressures of an ice cream freezer with a cabinet temperature of -20°F using refrigerant 502. The unit operates in a room with an ambient temperature of 80°F.
SOLUTION: The evaporator temperature must be adjusted to 20°F below the desired temperature produced. Therefore, 20°F – 20°F = -40°F. Referring to the temperature-pressure chart, at -40°F, the corresponding pressure for units using refrigerant 502 is 4.3 psi, which should be the low-side pressure. To determine the high-side pressure, add 30°F to 80°F, which will be equal to 110°F. The corresponding pressure on the chart in the R-502 column at 110°F is 245.8 psi = the proper head pressure for this unit in this environment.
EXAMPLE 1: What evaporator temperature is required to maintain the cabinet temperature of a floral display case at 40°F?
ANSWER: 40°F – 20°F = 20°F.
EXAMPLE 2: Suppose service is needed on a delicatessen case with a cabinet temperature of 35°F using refrigerant R-12. The ambient temperature is 75°F. Determine the high- and low-side pressures.
ANSWER: The evaporator temperature should be regulated at 15°F. (35°F-20°F=15°F.) According to the chart, 17.7 psi back pressure is required for R-12 to create 15°F. To determine the high-side pressure: 75°F + 30°F = 105°F. On the chart, the corresponding pressure for 105°F in the refrigerant R-12 column is 126.6 psi.
EXAMPLE 3: Determine the head and back pressures of a frozen-yogurt machine, which uses R-502 with evaporator temperature of -10°F. The condenser is cooled by water, which is in tubes that run through the condenser coil. (Many commercial units have water-cooled condensers for improved heat exchange. See figure 87 and the pages concerning water-cooled condensers for more detail).
ANSWER: In the case of water-cooled condensers, measure the water temperature as it leaves the condenser through the return pipe by placing a thermometer on that pipe. Consider that the ambient temperature and add 20°F to that reading. Assume that this exhaust water temperature is 60°F. To determine the expected head pressure, 60°F + 20°F = 80°F. On the chart, 80°F creates 159.9 lbs/in2 of head pressure in the system when 502 refrigerant is used.
To determine the low-side pressure: the chart shows that under R-502, 22.8 psi produces -10°F.
