Question

1. Using the flow rate of 250 gpm, select a plastic pipe diameter (2”, 4”, 6”, 8”, 10”…) & determine the approximate pump head based on the following:
   1. Minor pipe losses include 6-90 degree bends (standard) and one plug valve straightway. Document your source for your K-values.
   2. Document which calculation method you used to calculate the total pump head.
   3. To save energy, choose the largest pipe diameter that provides the recommended 2.0 fps velocity of flow.
   4. Assume nominal pipe diameters as the inside pipe diameter.
   5. The connection is 10,500 feet from the pump station.
   6. The connection point located at an elevation of 768.6 feet above MSL. The low water elevation in the station (at which point the pump will turn off) is 755.0 feet above MSL.

Recommended Pipe Diameter = __________________ inch

Total Pump Head = _______________________ ft

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 located at water surface elevation of upper and lower points respectively,

P1/ + V1^2 / 2 g + Z1 = P2/ + V2^2 / 2 g + Z2 + Hf + Hm

Since, both point 1 and 2 are open to atmosphere, pressure is only atmospheric hence gauge pressure P1= P2 =0

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

Elevation Z1 = 768.6 ft and Z2 = 755 ft

Hf is Friction head loss

Hm is minor head loss

Hence,above equation reduce to,

768.6 + Hp  = 755 + Hf + Hm

=> Hf + Hm = 13.60 ...................................(1)

Also, Hf = 8 f L / ( g )

where, f = friction factor

L = Pipe length

Q = Flow rate = 250 gpm or 0.55 cfs

D = Pipe diameter

Since, friction factor is not known,solution is iterative,so assume initial roughness as 0.02 to began iteration

=> Hf = 8(0.02)(10,500) / ( g )

=> Hf = 1.60 ./

Minor head loss, Hm = K / 2 g

where, K = Sum of Loss coefficent due to bends and valves

For standard 90 degree bend, loss coefficient = 0.30

Loss coefficient for plug valve = 0.29

=> K = 6(0.30) + 0.29 = 2.09

A = Pipe area = (/4)

=> = (/4) x (/4) = ( / 16)

=> Hm = 2.09 / 2 g ( / 16) = 5.058 / g

=> Hm = 0.0159 /

Putting values in equation 1,

  => Hf = (1.60./ ) + (0.0159 /) = 13.60

On solving above equation , we get D 0.65 ft

Iteration 1 :

Now use above value of diameter to calculate more improved value of friction factor using Reynold number,

Re =  V D /

where, = Kinematic viscosity of water = 0.000011 ft2/s

V = Velocity = 2 ft/s

=> Re = 2 x 0.65 / 0.000011

=> Re = 118182

Roughness of PVC pipe = 0.00006 ft

=> Relative rougness, e/D = 0.00006/0.65 = 0.00009

According top Moodyb diagram, for Re = 118182 and e/D = 0.0009, friction factor, f = 0.02

Since, value of friction facor comes out to be same as assumed value, diameter will remain same as above iteration so, D = 0.65 ft or 7.8 in 8 in

  

Ans b) To calculte pump head again apply Bernoulli equation between point 1 and 2,

P1/ + V1^2 / 2 g + Z1 + Hp = P2/ + V2^2 / 2 g + Z2 + Hf + Hm

=> 0 + 0 + 755 + Hp = 768.6 + Hf + Hm

=> Hp = 13.6 + Hf + Hm

From above part, Hf = (1.60./ ) = 1.60 / = 20.58 ft

Also, Hm = 0.0159 / = 0.0159 / = 0.122 ft

=> Hp = 13.60 + 20.58 + 0.122 = 34.30 ft