the upstream velocity Vu to calculate the velocity pressure when calculating head loss through the reducer: (in-H20) = kd [Vu (ft/min) / 4,005]2 hf Source: United McGill, 1990. Example Find the power, Wf, required to push 1,000 cfm of air through 200 ft of 12-inch duct with four 5-gore elbows assuming that 1 and 2 are open to the atmosphere and at the same elevation. P2 = P1 because 1 and 2 are open the atmosphere. V2= V1 because the area of duct at 1 and 2 are the same. z2 = z1 because 1 and 2 are at the same elevation. Thus: = = = 0 hpres hvel helev = + + + + = + htotal hpres hvel helev hp hf hp hf From the monograph, for 1,000 cfm through 12-inch duct, ~ 0.2 in-H20/100 ft. Pduct Hence, the total head loss through the duct is: = 0.2 in-H20/100 ft x 200 ft = 0.4 in-H20 hp The velocity is: V = V / A = 1,000 ft3/min / (3.143 x 12 ft2 / 4) = 1,273 ft/min From the fittings chart, for a 12-inch 5-gore elbow, kf = 0.18. For four elbows, the total head loss is: (in-H20) = Nf x kf (V / 4,005)2 where V is in ft/min hf (in-H20) = 4 x 0.18 x (1,273/4,005)2 = 0.073 in-H20 hf The total head loss through ducts and fittings is: = + = 0.4 in-H20 + 0.073 in-H20 = 0.473 in-H20 htotal hp hf
the upstream velocity Vu to calculate the velocity pressure when calculating head loss through the reducer: hf (in-H20) = kd [Vu (ft/min) / 4,005]2 Source: United McGill, 1990. Example Find the power, Wf, required to push 1,000 cfm of air through 200 ft of 12-inch duct with four 5-gore elbows assuming that 1 and 2 are open to the atmosphere and at the same elevation. P2 = P1 because 1 and 2 are open the atmosphere. V2= V1 because the area of duct at 1 and 2 are the same. z2 = z1 because 1 and 2 are at the same elevation. Thus: hpres = hvel = helev = 0 htotal = hpres + hvel + helev + hp + hf = hp + hf From the monograph, for 1,000 cfm through 12-inch duct, Pduct ~ 0.2 in-H20/100 ft. Hence, the total head loss through the duct is: hp = 0.2 in-H20/100 ft x 200 ft = 0.4 in-H20 The velocity is: V = V / A = 1,000 ft3/min / (3.143 x 12 ft2 / 4) = 1,273 ft/min From the fittings chart, for a 12-inch 5-gore elbow, kf = 0.18. For four elbows, the total head loss is: hf (in-H20) = Nf x kf (V / 4,005)2 where V is in ft/min hf (in-H20) = 4 x 0.18 x (1,273/4,005)2 = 0.073 in-H20 The total head loss through ducts and fittings is: htotal = hp + hf = 0.4 in-H20 + 0.073 in-H20 = 0.473 in-H20
EnergyEfficientFluidFlow.docx