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A pump transports water from Tank 1 to Tank 2 through a constant-diameter piping system as...

A pump transports water from Tank 1 to Tank 2 through a constant-diameter piping system as shown below (not to scale). The fl

A pump transports water from Tank 1 to Tank 2 through a constant-diameter piping system as shown below (not to scale). The flow rate is controlled by two gate valves, the gate valve I controls the main pipeline, while the gate valve II controls the loop line from T-joint A to T-joint B. All pipes are galvanized steel pipe of diameter D = 4 in. It has a total length of Li2= 620 ft from tank 1 to tank 2. All fittings are flanged and the elbows are long radius. The elevation difference between the free surfaces of the two tanks is z2 -Z1 = 30 ft. For steady-state operation with the gate valve I fully open while the gate valve II fully-closed, determine the total flow rate Q in the pipe when the pump head is hpump = 120 ft. Also find out the power of the pump, W in hp. pump Note that you can write out both mass balance equation and energy equation of the head form for the control-volume enclosed by the free surfaces, (1) and (2), of the two tanks. You need to consider all minor head losses through fittings, valves, etc. (see the table given). To estimate the major head loss through the straight pipes, the Haaland equation is required to determine the friction factor f. An iteration process is required to find out the friction factor f and the flow rate Q simultaneously. You are required to set up the equations and carry out at least two iteration steps. B. Gate Valve II (closed) (1) Gate Pump Valve I (open) KL 1.3 Fittings and Valves 90° elbow (flanged & long radius) Basket Strainer T-joint (line flow) T-joint (branch flow) Exit Gate valve (fully open) 1.0
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Datas given d=4in = 0.3188ft ; A=0.0872 ft?; &= 0.00015 ft (steel) g=32-2ft/s? Properties of water is taken at 80F VelocityRelative roughness is given by, E 0.00015 = 0.00045 0.333 d To find friction factor, 1.11 1 -1.810g 10 &ld 6.9 + 3.7 Re 1.11Basket strainer, k=1.3 Total loss coefficient, K = 0.15+(7x0.2)+17+0.15+2.3 K = 21 Total head loss = Major loss + minor loss120 = 30 +(1861.86f +21)2·039Q? 90 =(1861.86f +21)2.039Q? f Q (GPM) Q (ft^3/s) A (ft^2) V (ft/s) Re 1/f60.5 h 100 0.22280093Flow rate is 0.908 ft/s (407.5GPM) Pump power is given by, P=pgQh P= 62.2x120x0.908 lbf.ft P=6801.28 S lbf.ft 1 = 0.0013558kW

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