See the pumps section for the necessary background to solve this problem.

## The Problem

A pump is installed to fill an open water storage tank (elevation at the bottom of the tank is 200 ft asl) from a 70 degrees F reservoir (elevation at the piping outlet is 100 ft asl). The pump manufacturer has specified a minimum Net Positive Suction Head Required of 10 ft when pumping at 900 gpm. The six-inch Schedule 40 steel piping that connects the pump between the reservoir and the storage tank is 1000 ft long, contains six flanged 90 degree elbows, is connected to the reservoir at its bottom elevation, and discharges at the top of the upper strorage tank. If the depth of the water in the upper storage tank is 20 ft, what is the minimum depth of reservoir to prevent cavitation when the atmospheric pressure is 13 psi and the discharge of the pump is 900 gpm? Assume the Re of the flow is 100,000.

## The Solution

### Required Principals:

1. NPSHA
2. Friction Head Loss
3. Equivalent length of minor losses

### Governing Equations:

NPSHA = h_{atm} + h_{z(s)} - h_{f(s)} - h_{vp}

where:

• NPSHA = Net Positive Suction Head Available (ft)
• hatm = atmospheric head (ft)
• hz(s) = static suction head (ft)
• hf(s) = friction suction head (ft)
• hvp = vapor pressure head (ft)

h_{ft} = 144\frac{p_{psi}}{\gamma}

where:

• h = height of water above point (ft)
• p = pressure (psi)
• γ = specific weight of the fluid (lb/ft3)

\frac{4.72LQ^{1.85}}{C^{1.85}D^{4.87}}

where:

• hf' = friction head loss (ft)
• L = length of pipe (ft)
• Q = discharge (ft3/sec)
• C = Hazen-Williams Coefficient
• D = diameter of pipe (ft)

L_{eq} = \frac{kD}{f}

where:

• k = minor loss coefficient
• D = diameter of pipe (ft)
• f = friction factor

### Assumptions and Givens:

• Re = 100,000 (given)
• f = 0.018 (Moody Diagram using Re=10^5 and smooth pipe)
• D = 0.5 ft (given)
• k = 0.3 for flanged 90 degree elbows
• hvp = 0.84 ft for water at 70 degrees F (CERM Appendix 14.A)
• C = 140 for Schedule 40 welded steel pipe (CERM Appendix 17.A)
• Lpipe = 1000 ft (given)
• Q = 900 gpm = 2.005 ft3/sec (given and directly converted)

### Calculations:

h_{atm} = \frac{144*13}{62.4} = 30 ft

L_{eq/elbow} = \frac{0.3*0.5 ft}{0.018} = 33 ft_{/elbow} * 6 elbows = 50 ft

L_{total} = 1000 ft + 50 ft = 1050 ft

h_f = \text{ } \frac{4.72LQ^{1.85}}{C^{1.85}D^{4.87}} = \text{ } \frac{4.72*1050*2.005^{1.85}}{140^{1.85}*0.5^{4.87}} = 56.2 ft

h_z = 200 ft - 100 ft = X ft

NPSHA = 10ft = 30ft +h_{z(s)} - 56.2ft - 0.84ft

h_{z(s)} = 37.0 ft = \text{required depth of reservoir}

### By Table:

CERM Appendix 17.C gives friction head of 22.6 psi per 1000 ft of Schedule-40 steel pipe when discharging at 900 gpm. 22.6 psi = 52.2 ft of head, which is scaled to 55.1 ft (compared to the calculated 56.2 ft) for 1050 ft of pipe.

## References:

1. Michael R. Lindeburg, "Civil Engineering Reference Manual (11th Edition)", 2008

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