Question

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

Answer 1

Data's given d=4in = 0.3188ft ; A=0.0872 ft?; &= 0.00015 ft (steel) g=32-2ft/s? Properties of water is taken at 80F Velocity of the flow is given by, Q = AV 2 V А V= 0.0873 V = 11.459Q ft/s Reynolds number is given by, pVd Re= Vd Re= V 11.459Q x 0.333 Re= 0.925x10-5 Re=4129429

Relative 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 1.11 1 -1.810g 10 0.00045 3.7 + 6.9 4129429 ✓f 6.9 -1.8log10 4.511x10-5+ 234822.70 Note: Using moody diagram also f can be calculated. | Minor loss coefficient in pipe, Flanged T joint, k=0.15 7*90 elbow, k=0.2 Gate valve 3/4 closed, k=17 Gate valve fully open, k=0.15 Exit loss=1

Basket 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 loss h = h, +h V2 h, fLV2 + = 2gd 2g fL h; + ΣΚ d 2g fx620 (11.459Q)? h = +21 0.333 2x32.2 hz = (1861.86f +21)2.039Q + + By applying Bernoulli's theorem for sump and tank surface, +Z+hp P, 12 - + Z₂ + H₂ Pg 28 Pg 2g P1 = Patm = P2; V = V2 =V;h> = 120ft Z2 - 24 = 30ft

120 = 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 0.22280093 0.0872 2.555057 92003.86 7.056778 0.020081 5.909844 200 0.44560186 0.0872 5.110113 184007.7 7.349356 0.018514 22.45814 300 0.66840279 0.0872 7.66517 276011.6 7.477608 0.017884 49.46292 400 0.89120372 0.0872 10.22023 368015.4 7.550598 0.01754 86.89655 407.5 0.90791379 0.0872 10.41186 374915.7 7.554833 0.017521 90.12418 500 1.11400465 0.0872 12.77528 460019.3 7.597904 0.017323 134.75 600 1.33680558 0.0872 15.33034 552023.2 7.631111 0.017172 193.0195 610 1.359085673 0.0872 15.58584 561223.6 7.633896 0.01716 199.4192 800 1.78240744 0.0872 20.44045 736030.9 7.674707 0.016978 340.7994 900 2.00520837 0.0872 22.99551 828034.8 7.689797 0.016911 430.3082 1000 2.2280093 0.0872 25.55057 920038.6 7.702082 0.016857 530.2288 1100 2.45081023 0.0872 28.10562 1012042 7.712279 0.016813 640.5611 1200 2.67361116 0.0872 30.66068 1104046 7.720879 0.016775 761.3048 1300 2.89641209 0.0872 33.21573 1196050 7.728231 0.016743 892.4598 1400 3.11921302 0.0872 35.77079 1288054 7.734588 0.016716 1034.026 1500 3.34201395 0.0872 38.32585 1380058 7.740139 0.016692 1186.003 1600 3.56481488 0.0872 40.8809 1472062 7.74503 0.016671 1348.392

Flow 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 S P=9.22kW