Question

Water is to be pumped from a lake to a reservoir located on a bluff \(290 \mathrm{ft}\) above. According to the specifications, the piping is Schedule 40 steel pipe having a nominal diameter of 1 inch and the volumetric flow rate is \(10 \mathrm{gal} / \mathrm{min}\). The total length of pipe is \(580 \mathrm{ft}\). A centrifugal pump is specified. Estimate the electrical power required by the pump, in \(\mathrm{kW}\). Is a centrifugal pump a good choice for this application? What precautions should be taken to avoid cavitation?

Short answer

The estimated electrical power required is 26.4 kW. A centrifugal pump is a good choice for this application. To avoid cavitation, maintain proper suction conditions and monitor pump speed.

Step-by-step solution

Convert volumetric flow rate to cubic feet per second (cfs)

Convert the given volumetric flow rate from gallons per minute (gal/min) to cubic feet per second (cfs). Use the conversion factors: \(1 \text{ gal} = 0.1337 \text{ ft}^3\) and \(1 \text{ min} = 60 \text{ s}\). The flow rate is \(10 \text{ gal/min}\). Calculation: \[ Q = 10 \text{ gal/min} \times 0.1337 \text{ ft}^3/\text{gal} \times \frac{1}{60} \text{ min/s} = 0.0223 \text{ ft}^3/s \]

Calculate the head loss due to elevation (static head)

The static head is the vertical distance the water needs to be pumped. In this case, it's 290 feet (the elevation difference).\[ h_{static} = 290 \text{ ft} \]

Calculate the Reynolds number to determine the flow regime

Reynolds number is calculated to determine if the flow is laminar or turbulent.First, estimate the velocity ( v ) in the pipe:\[ v = \frac{Q}{A} = \frac{0.0223 \text{ ft}^3/s}{ \frac{\text{π} (0.5 \text{ in})^2}{144} \text{ ft}^2} = 3.65 \text{ ft/s} \]Where inner diameter is 1 inch and area is \( A = \frac{\text{π} D^2}{4}\).Then use the kinematic viscosity ( u ) of water (1.22 \times 10^{-5} \text{ ft}^2/\text{s}): \[ \text{Re} = \frac{v \times D}{u} = \frac{3.65 \text{ ft/s} \times 0.0833 \text{ ft}}{1.22 \times 10^{-5} \text{ ft}^2/\text{s}} = 2.49 \times 10^4 \]

Determine the friction factor

Since the flow is turbulent (Re > 4000), use the Darcy-Weisbach equation and Moody chart or Colebrook equation for turbulent flow in smooth pipes:Assume roughness factor ( \text{k} ): \[ f = 0.02 \]

Calculate the head loss due to friction

Using Darcy-Weisbach equation:\[ h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.02 \frac{580 \text{ ft}}{0.0833 \text{ ft}} \frac{(3.65 \text{ ft/s})^2}{2 \times 32.2 \text{ ft/s}^2 } = 35 \text{ ft} \]

Calculate the total head required

Total head required ( h_{total} ) is the sum of static head and head loss due to friction:\[ h_{total} = h_{static} + h_f = 290 \text{ ft} + 35 \text{ ft} = 325 \text{ ft} \]

Calculate the hydraulic power

Hydraulic power ( P_{hyd} ) required to lift water to 325 feet: \[ P_{hyd} = \rho g Q h_{total} \rho = 62.4 \text{ lb/ft}^3 \] \[ P_{hyd} = 62.4 \text{ lb/ft}^3 \times 32.2 \text{ ft/s}^2 \times 0.0223 \text{ ft}^3/s \times 325 \text{ ft} = 14,579.7 \text{ ft-lb/s} \]Convert to kilowatt ( \text{kW} ):\[ 1 \text{ kW} = 737.56 \text{ ft-lb/s } \] \[ P_{hyd} = \frac{14,579.7 \text{ ft-lb/s}}{737.56} = 19.8 \text{ kW} \]

Estimate the electrical power requirement

Determine the electrical power considering the pump efficiency ( \text{η} ), assume pump efficiency as 75% (0.75):\[ P_{elec} = \frac{P_{hyd}}{\text{η}} = \frac{19.78 \text{ kW}}{0.75} = 26.4 \text{ kW} \]

Evaluate if centrifugal pump is a good choice

Centrifugal pumps are generally good for large discharge and lower head requirements. For a head of 325 feet, centrifugal pumps should be adequate, especially if they are high head types.

Precautions to avoid cavitation

To avoid cavitation:1. Ensure the suction pipe is short and has minimal elevation difference.2. Keep the suction pressure above the vapor pressure of the fluid.3. Maintain proper pump speed.4. Install an air release valve if necessary.

Key concepts

Water Flow Rate

The water flow rate indicates how much water is being transferred over a certain period. It is often measured in gallons per minute (gal/min) or cubic feet per second (cfs). The exercise gives a flow rate of 10 gal/min. Converting this to cfs helps simplify further calculations. The conversion is performed by recognizing that 1 gallon equals 0.1337 cubic feet and 1 minute contains 60 seconds, resulting in an ultimate rate of: \( Q = 10 \text{ gal/min} \times 0.1337 \text{ ft}^3/\text{gal} \times \frac{1}{60} \text{ min/s} = 0.0223 \text{ ft}^3/s \)

Static Head

The static head represents the vertical height the water needs to be pumped from one elevation to another. It is a primary component in calculating the total head. In this exercise, the water needs to be pumped 290 feet above the starting point. This elevation difference directly affects the energy required to move the water. Expressed mathematically: \( h_{\text{static}} = 290 \text{ ft} \)

Reynolds Number

The Reynolds Number (Re) helps determine the type of flow within the pipe—whether it’s laminar or turbulent. Re is calculated using the velocity of the water, the pipe's diameter, and the water’s kinematic viscosity. First, we calculate the velocity ( v ) using: \( v = \frac{Q}{A} = \frac{0.0223 \text{ ft}^3/s}{\frac{\text{π} (0.5 \text{ in})^2}{144} \text{ ft}^2} = 3.65 \text{ ft/s} \) Here A is the cross-sectional area of the pipe. Then, the Reynolds Number is calculated as: \( \text{Re} = \frac{v \times D}{u} = \frac{3.65 \text{ ft/s} \times 0.0833 \text{ ft}}{1.22 \times 10^{-5} \text{ ft}^2/\text{s}} = 2.49 \times 10^4 \)

Friction Factor

The friction factor ( f ) is crucial for determining the head loss due to pipe friction. Since our flow is turbulent (Re > 4000), we use the Darcy-Weisbach equation, often simplifying with the Moody chart or Colebrook equation. For this exercise, we assume a roughness factor ( k ) and select: \( f = 0.02 \)

Darcy-Weisbach Equation

The Darcy-Weisbach equation estimates the head loss ( h_f ) due to friction within the pipe. The formula used is: \( h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.02 \frac{580 \text{ ft}}{0.0833 \text{ ft}} \frac{(3.65 \text{ ft/s})^2}{2 \times 32.2 \text{ ft/s}^2 } = 35 \text{ ft} \) Here, L is the pipe length, D is the diameter, v is the velocity, g is gravitational acceleration, and f is the friction factor.

Hydraulic Power

Hydraulic power ( P_{\text{hyd}} ) measures the energy needed by the water to overcome elevation and friction losses. It is computed using the water density ( ρ ), gravitational acceleration ( g ), water flow rate ( Q ), and the total head ( h_{\text{total}} ). \( P_{\text{hyd}} = ρ g Q h_{\text{total}} \) Plugging in the numbers: \( P_{\text{hyd}} = 62.4 \text{ lb/ft}^3 \times 32.2 \text{ ft/s}^2 \times 0.0223 \text{ ft}^3/s \times 325 \text{ ft} = 14579.7 \text{ ft-lb/s} \)

Pump Efficiency

Pump efficiency ( η ) is a measure of how effectively the pump converts electrical power into hydraulic power. A pump’s efficiency plays a vital role in determining the electrical power requirement. Using an assumed pump efficiency of 75% (0.75), the electrical power needed ( P_{\text{elec}} ) is given by: \( P_{\text{elec}} = \frac{P_{\text{hyd}}}{η} = \frac{19.78 \text{ kW}}{0.75} = 26.4 \text{ kW} \)

Cavitation Prevention

Cavitation is a common problem that occurs when the water pressure drops below its vapor pressure, leading to the formation of bubbles that can damage the pump. Preventing cavitation is crucial for maintaining pump longevity. Here are a few precautions:

• Ensure the suction pipe is as short and direct as possible.
• Keep the suction pressure above the water's vapor pressure.
• Maintain appropriate pump speeds.
• Install air release valves if necessary.

By following these precautions, you can help to avoid the damaging effects of cavitation.