Question

Design a system to move 100 gpm of water from a river with surface water elevation of 201.35 ft msl to a tank with surface water elevation = 246.55 ft msl. The tank is pressurized; the air pressure in the tank is 60 psig. The total length of pipe is 125 ft with 6 standard elbows all made of Sch 40 steel pipe. Your design should specify the nominal diameter of pipe (use only commercially available sizes) and the power of the pump. If the pump is 65 % efficient, compute the cost of running the pump continuously for 6 months. (you may assume the system is in Corvallis).

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 located at water surface elevation of river and tank respectively

P1 / g + V1²/2g + Z1 + Hp = P2/g + V2²/2g + Z2 + H_f + H_m

Since, river is open to atmosphere, pressure is only atmospheric hence gauge pressure P1 = 0

Also, there is no velocity at surface so V1 = V2= 0

Elevation from datum Z1 = 201.35 m and Z2 = 246.55 m

H_f = head loss in system due to friction

H_m = Minor head loss

Hp = head added by pump

Hence, above equation reduces to,

0 + 0 + Z1 + Hp = P2/g + 0 + Z2 + H_f + H_m

=> Hp = (Z2 - Z1) + H_f + H_m  

=> Hp = 45.2 + H_f + H_m

To determine head loss due friction , determine Darcy friction factor , f

k = roughness of steel pipe  = 0.0018 in

Let  us assume Sch 40 pipe with diameter 6 in

then velocity = Q / A

Discharge (Q) = 100 gpm or 0.222 ft³/s

Area = (/4) D² = 0.785(0.5)² = 0.19625 ft²

=> V = 0.222 / 0.19625 = 1.13 ft/s

  Now, Reynold Number ,Re =  V D/

where, V = Flow velocity

= Kinematic viscosity of water at 20 deg C = 1.13 x 10^-6 m² /s

=> Re = (1.13) (0.5) / 1.13 x 10^-5

=> Re = 50000

Relative roughness , k/D = 0.0018/6 = 0.003

According to Moody Diagram, for Re = 50000 and k/D = 0.003, f = 0.0219

H_f = f L V² / 2g D

=> H_f = 0.0219 (125) (1.13)² /(2 x 32.2 x 0.5) = 0.108 ft

Minor head loss due to elbows ,H_m = 6 x K V²/2g

Loss coefficient for elbows , K = 0.3

=> H_m = 1.8(1.13)² / (2 x 32.2 ) = 0.036 ft

Putting values,

=> Hp = 45.2 + 0.108 + 0.036 = 45.35 ft

We know Pump power (P) = Q Hp /

=> P = 62.4 x 0.222 x 45.35 / 0.65

=> P = 966.5 lb ft/s or 1.76 HP   

Hence, provide diameter (D) = 6 in or 0.5 ft

Power of pump (P) = 1.76 HP