Question

QUESTIONS 300 250 200 Head, it 150 100 50 0 Q. cfs Using the pump curve shown, how many feet of head would the pump be able to move 1795 gallons per minute against? The pump curve shown can be plotted by evaluating the equation y 14-15x2) on the interval (1) then multiplying Xabies by 5 cs and y valies by 100 feel

Answer 1

solution, The initial curve equation b= 1 – x² on so basic outlaya can be expressed like this: given bra the interval 40 Co, 1) 4b n=1-x Now multiplying a values by 5 cfg and values by 300 ft the pump curve is plotted. 0 + Basic Outlay of Pump Curve: 300 250 Basically region Head It 2007 50f y valve a got divided into 300 divisions as well as a value Ott region got divided into o 1 2 3 4 5 5 divisions. 8 , caps + So now modified curre equestion combe represented like this: 091,5- 300 ( 1 - ()?)

448.83 We have to use this to evaluate care equation the head value" against 1795 gallons/min. We know; 448.83 gadsmin = Iefs on, 1 gel/min = 8 efs os, gal fomin = 0.002228 ofs So, at a= 1795 gal/min= 1795*0.002228 efs or, x= 3.99926 ~ 3.99g cofs E 300 (1 - (3.999)) 3 DE 108.096 It Therefore the pump would be able to move on 1795 goullonin against 108.096 ft. it 108.096 ft. (Answer.)