Water is pumped from one reservoir to a second reservoir The

Water is pumped from one reservoir to a second reservoir. The pump characteristic curve (for an operating speed of 1, 200 rpm) and system curve are shown. Data points for the pump characteristic curve are provided in the table. a. Determine the flow (gpm) and head (ft) delivered by the pump operating at 1, 200 rpm. b. The pump operating speed is changed to 1,500 rpm. Plot a new pump curve for this condition. c. Assuming the system does not change, determine the flow (gpm) and head (ft) delivered by the pump when it operates at 1,500 rpm. d. The pump impeller is trimmed to 95% of its original diameter and is operated at 1,500 rpm. Plot a new pump curve for this condition. e. Assuming the system does not change, determine the flow (gpm) and head (ft) delivered by the pump with the trimmed impeller operating at 1,500 rpm

Solution

a. You just need to find the Head and flow value at the intersection of system and pump curve which is according to the picture around 6000 gpm and 240 ft.

b. Q2 = Q1*N2/N1 and H2 = H1*(N2/N1)².....so for N2 = 1500rpm, N1 = 1200rpm just substitue the Q1 and H1 values on the table and get the new Q2 and H2 values for the new angular velocity.

c. Q2 = 6000*(1500/1200) = 7500gpm......H2 = 240*(1500/1200)² = 375ft

d. to develop the new pump curve with trimmed impeller we first find the values for a pump at 1200 and trimmed impeller so q₂ = q₁ (d₂ / d₁) and H₂ = H₁ (d₂ / d₁)² with d2/d1 = 0.95......after that with the new tables values we find with the procedure shown in b the final pump curve...

e. Q2 = 6000*(1500/1200) = 7500gpm......H2 = 240*(1500/1200)² = 375ft now at 95% trimmed impeller...

finalQ2 = 7500*(0.95) = 7125gpm......H2 = 375*(0.95)² = 338.44ft