Carrier 50HJQ004-016 Product Data

Also see for APOLLO CONTROL 50HJQ004-016: Installation, operating and troubleshooting instructions

23SELECTION PROCEDURE (WITH 50TFQ005 EXAMPLE)A. Determine cooling and heating loads at design conditions.Given:Required Cooling Capacity (TC) 38,000 BtuhSensible Heat Capacity (SHC) 24,000 BtuhRequired Heating Capacity 35,000 BtuhOutdoor Entering-Air Temperature db 95_FOutdoor-Air Winter Design Temperature 0°FIndoor-Air Winter Design Temperature 70_FIndoor-Air Temperature80_F edb (entering air, dry bulb)67_F ewb (entering air, wet bulb)Indoor-Air Quantity 1600 cfmExternal Static Pressure 0.45 in. wgElectrical Characteristics (V-Ph-Hz) 230-3-60B. Select unit based on required cooling capacity.Enter Cooling Capacities table at outdoor enteringtemperature of 95_F, indoor air entering at 1600 cfm and67_F ewb. The 50TFQ005 unit will provide a total coolingcapacity of 49,900 Btuh and a sensible heat capacity of33,900 Btuh.For indoor-air temperature other than 80_F edb, calculatesensible heat capacity correction, as required, using theformula found in Note 3 following the cooling capacitiestables.NOTE: Unit ratings are gross capacities and do not include theeffect of indoor-fan motor heat. To calculate net capacities, seeStep E.C. Select electric heat.Enter the Instantaneous and Integrated Heating Ratings tableat 1600 cfm. At 70_F return indoor air and 0°F air enteringoutdoor coil, the integrated heating capacity is 18,000 Btuh.(Select integrated heating capacity value since deductions foroutdoor-coil frost and defrosting have already been made. Nocorrection is required.)The required heating capacity is 35,000 Btuh. Therefore,17,000 Btuh (35,000 – 18,000) additional electric heat isrequired.Determine additional electric heat capacity in kW.17,000 Btuh = 5.0 kW of heat required.3413 Btuh/kWEnter the Electric Heating Capacities table for 50TFQ005 at208/230, 3 phase. The 6.5-kW heater at 240 v most closelysatisfies the heating required. To calculate kW at 230 v, usethe Multiplication Factors table.6.5 kW x .92 = 5.98 kW6.5 kW x .92 x 3413 = 20,410 BtuhTotal unit heating capacity is 38,410 Btuh (18,000 + 20,410).D. Determine fan speed and power requirements at designconditions.Before entering Fan Performance tables, calculate the totalstatic pressure required based on unit components. From thegiven and the Pressure Drop tables, find:External static pressure .45 in. wgEconoMi$er IV .07 in. wgElectric heat .09 in. wgTotal static pressure .61 in. wgEnter the Fan Performance table for 50TFQ005 verticaldischarge. At 1600 cfm and 230-v high speed, the standardmotor will deliver 0.76 in. wg static pressure, 723 watts, and0.64 brake horsepower (bhp). This will adequately handle jobrequirements.E. Determine net capacities.Capacities are gross and do not include the effect ofindoor-fan motor (IFM) heat.Determine net cooling capacity as follows:Net capacity = Total capacity – IFM heat= 49,900 Btuh – (723 Watts x3.413Btuh/Watts)= 49,900 Btuh – 2468 Btuh= 47,432Net sensible capacity = 33,900 Btuh – 2468 Btuh= 31,432 BtuhIntegrated heating capacity is maximum (instantaneous)capacity less the effect of frost on the outdoor coil and theheat required to defrost it. Therefore, net capacity is equal to38,410 Btuh, the total heating capacity determined in Step C.50TFQ