Question

fluid mechanics lesson (calculating popa power)

Calculate the power of the pump required to pump the water in hp (1 BG = 550 lbf.ft / sec)? Density = 62.4 lbm / ft³ and Viscosity = 1 cp. Water can send to a tank on a flow rate of 30 ft³ / min. data: g = 32.2 ft / sec², gc = 32,174 lbm.ft / lbf.sec², T = 20 degrees Celsius, Pump efficiency = 0.80.

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 loacted at water surface elevation of both tanks respectively,  

P1/ + /2g + Z1 + Hp = P2/ + /2g + Z2 + Hf +Hm

Since, both reservoirs are open to atmosphere, pressure is only atmospheric, hence gauge pressure P1 = P2 = 0

Velocity at surface is negligible so V1 = V2 = 0

Elevation, Z1 = 0 ft and Z2 = 100 ft

Hf is head loss due to friction

Hm is minor head loss due to bends and valves

Putting values,

0 + 0 + Hp = 0 + 100 + Hf + Hm

=> Hp = 100 + Hf + Hm.....................................(1)

Now,

Hf = f L /(2 g D)

where, f = friction factor

L = Pipe length = 10 + 400 + 140 + 50+50 = 650 ft

V = Flow velocity = Q / A

D = Pipe diameter = 5 in or 0.416 ft

Flow rate (Q) = 30 ft3/min or 0.50 ft3/s

Pipe area (A) = (/4) = 0.1358 ft2

=> V = 0.50 / 0.1358 = 3.68 ft/s

To determine friction factor calculate Reynold number (Re),

Re = V D /

where, = kinematic viscosity of water at 20 C = 1.2 x ft2/s

=> Re = 3.68 x 0.416 / (1.2 x )

=> Re = 127573

Roughness of smooth pipe = 0

=> Relative roughness (e/D) = 0

According to Moody diagram, for Re = 127573 and e/D =0 , friction factor (f) = 0.017

=> Hf = 0.017 x 650 x / (2 x 32.2 x 0.416) = 5.58 ft

Now,

Minor loss (Hm) = K / 2 g

Loss coeffcieint for entry = 0.5

Loss coefficient for exit = 1

Loss coeffcieint for 90 deg bend = 0.3

=> K = 0.5 + 3(0.3) + 1 = 2.4

=> Hm = 2.4 () / (2 x 32.2)

=> Hm = 0.50 ft

Putting values in equation 1,

Hp = 100 + 5.58 + 0.50 = 106.08 ft

We know,

Pump power = Q Hp /

where, = Unit weight of water = 62.4 lbm/ft3

Q = 30 ft3/min or 0.50 ft3/sec

Hp = Pump head = 106.08 ft

   = Efficiency = 0.80

Putting values,

=> Pump power = 62.4 x 0.50 x 106.08/ 0.80

=> Pump power = 4137.12 lb ft/s

1 hp = 550 lb ft/s

=> Pump power = 4137.12 / 550 = 7.52 hp