Engineering Manual o.. - HVAC.Amickracing

CHILLER, BOILER, AND DISTRIBUTION SYSTEM CONTROL APPLICATIONSCONTROLVALVE(S)SUPPLY PIPINGV1V2V3PUMPHEATEXCHANGERSERIESPUMPSLOADCONTROLHEATING/VALVESCOOLINGSOURCE 123HEATINGORCOOLINGCOILSBALANCINGVALVESRETURN PIPINGHEAD24%LOADFULLLOADFLOW2 PUMPS1 PUMPC2410Fig. 67. System Operation for Series Pumps.ITEMHEATING OR COOLINGSOURCE ANDDISTRIBUTION PIPINGCOIL 1 LOOPCOIL 2 LOOPCOIL 3 LOOPTOTAL FLOW AND DROPFLOWINL/sDESIGN PRESSURE DROP IN kPaCOIL & PIPINGCONTROLVALVEBALANCINGVALVE2.5 69* _ _0.6 24 33 60.8 30* 33* 0*1.1 21 33 92.5 132*Dual Pump CurvesFor pumps in parallel (Fig. 66), assuming two identicalpumps, the curve is developed using the following formula:Where:flow 3flow 1p 1flow 3 = (flow 1 ) x 2 for any p 1= Total flow for both pumps= flow of one pump= Pressure in kPa for Pump 1 at flow 1 for anypoint on pump curveFor pumps in series (Fig. 67), assuming two identical pumps,the curve is developed using the following formula:* SUM OF SOURCE AND PIPING (69 kPa) AND COIL LOOP 2 (63 kPa) = 132 kPaC4603Fig. 68. Simplified Water Distribution System.In Figure 68 the flow and pressure considerations are:1. The flow through the heating or cooling source, the supplypiping, and the return piping (2.5 L/s) is the same as thesum of the flows through the three coil circuits:0.6 + 0.8 + 1.1 = 2.5 L/s.2. Design pressure drop includes the drop through theheating or cooling source, supply piping, return piping,and the highest of the three coil circuits:69 + (30 + 33) = 132 kPa.Where:p 1p 1p 3 = (p 1 ) x 2 for any flow 1= Total pressure in kPa for both pumps= Pressure in kPa for one pump at flow 1 (forany point on Pump 1 curve)NOTE: In this example, Coil 1 and 3 balancing valvesbalance each load loop at the 63 kPa designfor Loop 2. If the actual coil and control valvedrops were less than the design maximumvalues, the actual balancing valve effectswould be greater.DISTRIBUTION SYSTEM FUNDAMENTALSFigure 68 illustrates a closed system where static pressure(pressure within the system with pump off) does not need to beconsidered as long as all components are rated for the staticpressure encountered. The pump provides force to overcomethe pressure drop through the system and valves control theflow and pressure through the system. Figure 69 shows a graphof the system and pump curves for design load and reducedload conditions. The system curve indicates the pressure dropthrough the system (with the control valves full open) at variousflow rates. The pump curve shows the pump output pressure atvarious flow rates. Flow always follows the pump curve.In this example the pump must handle 2.5 L/s against a totalpressure of 132 kPa as shown in Figure 69. (This curve is takenfrom actual pump tests). The design drop across the valve is33 kPa with the valve fully open.If Figure 68 is a heating system, as the loads reduce valvesV1, V2, and V3 start to close. Hot water flow must be reducedto about 15 percent of full flow (0.375 L/s) to reduce heat outputto 50 percent. As flow through the coil is reduced the watertakes longer to pass through the coil and, therefore, gives upmore heat to the air.ENGINEERING MANUAL OF AUTOMATIC CONTROL345