Engineering Manual o.. - HVAC.Amickracing

VALVE SELECTION AND SIZINGSTEAM VALVE SIZING EXAMPLESEXAMPLE 1:A two-way linear valve (V1) is needed to control highpressuresteam flow to a steam-to-water heatexchanger. An industrial-type valve is specified. Steampressure in the supply main is 650 kPa with nosuperheat, pressure in return is equal to atmosphericpressure, water flow is 5.2 L/s, and the watertemperature difference is 10 kelvins.Use the steam valve Kv formula to determine capacityindex for Valve V1 as follows:Kv =Where:Q= The quantity of steam required to passthrough the valve is found using theconverter valve formula:Q = L/s • ∆Tw • 6.462Where:L/s = 5.2 L/s water flow through exchanger∆Tw = 10 kelvins temperature difference6.462 = A scaling constantSubstituting this data in the formula:Q∆PWhere:P 1= 329.7 kilograms per hour= The pressure drop across a valve in amodulating application is:∆P = 80% x (P 1 – P 2 )P 2Q0.224 ∆P • P o= Upstream pressure in supply main is650 kPa.= Pressure in return is atmospheric pressure or101.325 kPA.The critical pressure drop is found using the followingformula:∆P critical∆P critical= 50% x kPa= 0.50 x 650 kPa= 325 kPaThe critical pressure drop (∆P critical ) of 325 kPa is used incalculating Kv, since it is less than the pressure drop (∆P)of 439 kPa. Always, use the smaller of the two calculatedvalues.Q = 329.7 kg/h∆P = ∆Pcritical = 325 kPaP o = P 1 – ∆P = 650 kPa – 325 kPa =325 kPa0.224 = A scaling constant.Substituting the quantity of steam and pressure drop in theKv formula shows that the valve should have a Kv of 4.5.Kv ==329.70.224 325 • 325329.70.224 x 325 = 4.53Select a linear valve providing close control with acapacity index of 4.5 and meeting the required pressureand temperature ratings.NOTE: For steam valves downstream from pressurereducing stations, the steam will be superheated inmost cases and must be considered.EXAMPLE 2:In Figure 19, a linear valve (V1) is needed for accurateflow control of a steam coil that requires 325 kilogramsper hour of steam. Upstream pressure in the supply mainis 150 kPa and pressure in the return is 90 kPa minimum.Substituting this data in the pressure drop formula:∆P = 0.80 x (650 – 101.3)= 0.80 x 548.7= 439 kPaENGINEERING MANUAL OF AUTOMATIC CONTROL443