Centrifugal Pump: Pressure Drop & Power Calculation

[chemengineer23]

Homework Statement

A centrifugal pump is used to pump water from one tank to another through 1000 equivalent ft of 4 in. schedule 40 steel pipe as shown in the figure. The second tank is located 25 ft above the first.

Show that the pressure drop (in psi) in the pipeline between 2 and 3 is given in terms of the flow rate Q (ft3/s) by:

p2 –p3 =10.83+10,265fQ2 where f is the friction factor.

What is the pump power required if the flow rate is 0.2 ft3/s? (Assume negligible losses and elevation change in the suction line.)
Point 1 is where the level of water is in the first tank
Point 2 is right after the pump
Point 3 is where the level of water is in the second tank

Homework Equations

MEB: V^2/2 + P1/rho +gz1 + Ws = V^2/2 +P2/rho + gz2 +hf
hf=( 2*f*L)/D *V^2/2

The Attempt at a Solution

Delta P/rho= g(z3-z2)-V2^2/2+hf
but you don't know delta z, velocity or the L in hf

 

[frzncactus]

The second tank is located 25 ft above the first.

This tells us z, which is the vertical distance between points 2 and 3, which is used to incorporate gravitational potential energy into the energy balance.

Volumetric flow rate is given as 0.2 ft3/s, which can be used to find velocity because v*A = Q, where v is velocity, A is cross-sectional area, and Q is volumetric flow. Velocity at point 3 is negligible for a large tank: assuming a large cross-sectional area of the tank, velocity is small for a given Q because v and A are inversely related.

L is the length of pipe, which you stated was 1000 ft. By the way, "equivalent feet" accounts for pipe bends, sudden expansions/contractions, etc. Wherever such sources of friction pop up, an appropriate length is added to L to compensate in the calculation. Thus, the actual physical length of the pipe may differ from the length you actually use for L in the formula.