Question

Problem Statement 1 The pump in the system shown below is used to transport water from reservoir A to and discharge it to the open atmosphere at point B. The elevation difference between the two reservoirs is Az, and the piping has a 2000 ft length of 16-in diameter. The flow is fully turbulent with kinetic energy correction factor = 1.1 and the Darcy-Weisbach friction factor f = 0.0002. The minor losses in the system are negligible and can be ignored. Suppose the pump vendor has supplied the characteristics of the pump in a form of tabulated data (shown in the table below), answer the following: (a) For the given dimensions of the system, generate the general system curve equation. Remember that the pipe diameter is constant throughout (useful head as function of elevation and volume flow rate). [8 %] (b) Plot two system curves, one with|4z = 0 ft and Az = 1100 ft (hp Vs volume flow rate). In the same figure, also plot the pump supply head and efficiency curves Vs volume flow rate using the data provided by the vendor (use a secondary y-axis for efficiency). In this plot, the x-axis will represent the volume flow rate, the left side of the y-axis will represent the supply curve and the two system curves and the right hand side of the y- axis (secondary y-axis) will represent the efficiency data (total of 4 curves). [30 %] (c) From part 'c', determine the operating points (volume flow rate & head intersections) and the efficiencies corresponding to the operating points of the two different Azs. (8 %] (d) For the two operating points in part d', determine the power inputs required by the %] Note: With your solution attachments, please include the excel sheet containing all the data and plots in addition to the scanned documents of the hand calculations. pump. [10 B Volume Flow Rate (ft3/s) 53 Open to the atmosphere Az Pump A 81 109 133 Pump Supply Efficiency Head (ft) (%) 3080 68 2980 74 2880 79 2780 82.7 2680 85 2580 86.3 2480 87 2380 86.4 2280 85.3 2180 83.4 Water 156 183 198 215 235 249 264 278 2080 1980 1880 1780 1680 81 78 74 70 291 302 312 65.5

Answer 1

a) Data's given d=16 in =1.333ft A=1.396ft? g=32•2ft/s? L = 2000 ft Properties of water is taken at 80F Velocity of the flow is given by, Q = AV V = 1.396 V = 0.7162Q ft/s The friction factor is 0.0002 Total head loss = Major loss + minor loss hp =h,+h, fLV2 hz +0 2gd

h = hgthm hz fLV2 +0 2gd 0.002 2000 )(11.4599)? h = 1.333 2x32.2 h = 0.023992 + By applying Bernoulli's theorem for sump and tank surface, PV +ZA+ hp P V2 +Z2 +h; Pg 2 g Pg 2g P1 = Patm = P.; V =V2 = V h = 0.023992 Z2 - Z = 45 ho = A=+0.0239Q? b)

eff (%) 68 74 79 82.7 85 86.3 system curve Q (ft^3/s) supply head (ft) system curve (Z=0) (Z=1100ft) 53 3080 67.1351 1167.1351 81 2980 156.8079 1256.8079 109 2880 283.9559 1383.9559 133 2780 422.7671 1522.7671 156 2680 581.6304 1681.6304 183 2580 800.3871 1900.3871 198 2480 936.9756 2036.9756 215 2380 1104.7775 2204.7775 235 2280 1319.8775 2419.8775 249 2180 1481.8239 2581.8239 264 2080 1665.7344 2765.7344 278 1980 1847.0876 2947.0876 291 1880 2023.8759 3123.8759 302 1780 2179.7756 3279.7756 312 1680 2326.5216 3426.5216 87 86.4 85.3 83.4 81 78 74 70 65.5

System curve 4000 100 90 3500 80 3000 70 2500 60 H (ft) 2000 50 Efficiency 40 1500 30 1000 20 500 10 0 0 0 50 100 150 200 250 300 350 Q -supply head (ft) -system curve (Z=0) -system curve (Z=1100ft) --eff (%)

c) Operating point for Z=0 is 285 ft/s and head is 1920 ft. Operating point for Z=1100 ft is 225 ft/s and head is 2320 ft. d) Operating point for Z=0 is 285 ft/s and head is 1920 ft. Pump power is given by, P=pgQh, P=yQh P=62-2x285x1920 lbf.ft P = 34035840 S lbf.ft = 0.0013558kW S P = 46145.79kW Operating point for Z=1100 ft is 225 ft3/s and head is 2320 ft. Pump power is given by, P=PgQhp P=YQhp P=62-2x225x 2320 lbf.ft P=32468400 S P = 44020kW