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Water at \(20^{\circ} \mathrm{C}\) is to be pumped out of a reservoir at a rate of \(20 \mathrm{~L} / \mathrm{s}\) through a vertical 150 -mm-diameter ductile iron pipeline with an estimated roughness height of \(0.3 \mathrm{~mm}\). Appurtenances installed in the intake pipe are expected to contribute to a total local head loss coefficient of 12 . The pump being considered for installation has a required net positive suction head of \(5.5 \mathrm{~m}\). (a) What is the maximum elevation of the pump relative to the water surface in the reservoir? (b) If the total head loss coefficient can be reduced to 1.2 , how much higher can the pump be placed? Assume standard atmospheric conditions.

Short Answer

Expert verified
(a) Maximum pump elevation is approximately -3.25 m, (b) approximately 3.78 m higher.

Step by step solution

01

Identify Parameters for Calculation

We first gather all the given information:1. - Diameter of pipe, \(D = 150\) mm = 0.15 m - Roughness height, \(k_s = 0.3\) mm = 0.0003 m - Water temperature, \(T = 20^{\circ}C\) - Flow rate, \(Q = 20\) L/s = 0.02 m³/s - Local head loss coefficient, \(K= 12\) (part a) and \( K = 1.2 \) (part b) - Required net positive suction head (NPSH), \(NPSH_r = 5.5\) m2. Standard atmospheric pressure, \(P_{atm} = 101325\) Pa.3. Density of water at 20°C, \(\rho = 998\) kg/m³4. Dynamic viscosity of water at 20°C, \(\mu = 1.002 \times 10^{-3}\) Ns/m².
02

Calculate the Velocity of Water in the Pipe

To find the velocity \(v\) of water in the pipe, we use the formula: \[ v = \frac{Q}{A} \]where \(A\) is the cross-sectional area of the pipe.\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.15)^2}{4} = 0.01767 \text{ m}^2 \]Substitute the values:\[ v = \frac{0.02}{0.01767} \approx 1.131 \text{ m/s} \]
03

Determine the Reynolds Number

Reynolds number \(Re\) is given by:\[ Re = \frac{\rho v D}{\mu} \]Substitute the known values:\[ Re = \frac{998 \times 1.131 \times 0.15}{1.002 \times 10^{-3}} \approx 169290 \]This indicates that the flow is turbulent since \(Re > 4000\).
04

Calculate the Darcy-Weisbach Friction Factor

For turbulent flow, we use the Colebrook equation:\[ \frac{1}{\sqrt{f}} = -2.0 \log_{10} \left( \frac{k_s/D}{3.7} + \frac{2.51}{Re \sqrt{f}} \right) \]An iterative approach or a friction factor chart/table provides \(f \approx 0.018\) (a rough estimate for this flow).
05

Compute the Head Loss

Total head loss \( h_f \) is computed using the Darcy-Weisbach equation and local head loss:\[ h_f = f \frac{L}{D} v^2/(2g) + K \frac{v^2}{2g} \]Assuming a long enough pipe (just to acknowledge friction loss), replace \(L\) when rough estimates are required for the problem\[ h_f \approx f \frac{100}{0.15} \frac{1.131^2}{2 \times 9.81} + 12 \frac{1.131^2}{2 \times 9.81} \approx 0.23+7.8 \approx 8.03 \text{ m} \]
06

Compute Maximum Pump Elevation (Part a)

Complete the energy balance with the head term for maximum pump elevation:\[ h_s = \frac{P_{atm}}{\rho g} - \frac{v^2}{2g} - NPSH_r - h_f \]\[ h_s = \frac{101325}{998 \times 9.81} - \frac{1.131^2}{2 \times 9.81} - 5.5 - 8.03 \approx 10.35 - 0.065 - 5.5 - 8.03 \approx -3.245 \] mSince the pump can't be lower than the reservoir, realistically the height may need reinterpretation if negative, check assumptions or parameter changes.
07

Adjust Head Loss Coefficient for Part b

Applying \( K = 1.2\):\[ h_f \approx 0.23+0.78 \approx 1.01 \text{ m} \]
08

Compute New Maximum Pump Elevation (Part b)

Substitute into energy balance:\[ h_s = \frac{101325}{998 \times 9.81} - \frac{1.131^2}{2 \times 9.81} - 5.5 - 1.01 \approx 10.35 - 0.065 - 5.5 - 1.01 \approx 3.775 \] mThis means the pump can be raised higher by about \(3.78 \text{ m}\) than in part a.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pump Elevation
When considering where to place a pump in a fluid system, pump elevation is a critical factor. It's essentially the vertical height at which a pump can be positioned above the water surface while maintaining efficient operation. Elevation is crucial because it determines how much suction the pump can provide, impacting its ability to move fluids effectively.
In the context of the exercise, determining the maximum elevation the pump can be installed without compromising the net positive suction head (NPSH) is vital. This ensures efficient pump function without causing cavitation, which can damage the pump. Factors influencing pump elevation include:
  • The atmospheric pressure, which affects the amount of fluid a pump can lift.
  • Local and frictional head losses that reduce the pump's ability to elevate fluids.
  • The required NPSH, which is a measure of the pressure necessary to avoid cavitation.
Calculating the pump elevation involves understanding how these factors interact and maintaining a balance to ensure a successful fluid movement.
Darcy-Weisbach Equation
The Darcy-Weisbach Equation is a key tool in fluid mechanics used to calculate head loss due to friction and local disturbances in pipes. This equation is fundamental for engineers when designing or analyzing pipelines.
It states that the head loss (\( h_f \)) can be calculated using:\[ h_f = f \frac{L}{D} \frac{v^2}{2g}\]Where:
  • \( f \)
  • : Darcy-Weisbach friction factor, which accounts for the pipe's roughness and flow conditions.
  • \( L \)
  • : Length of the pipe.
  • \( D \)
  • : Diameter of the pipe.
  • \( v \)
  • : Velocity of the fluid.
  • \( g \)
  • : Acceleration due to gravity.
In addition to friction losses, local losses are usually accounted for with a head loss coefficient (\( K \)) which models losses due to bends, valves, and other fittings. The complete formula becomes:\[ h_f = f \frac{L}{D} \frac{v^2}{2g} + K \frac{v^2}{2g}\]Understanding this equation helps engineers gauge energy losses in a system, informing decisions on everything from pump specifications to the positioning of pipe fittings.
Head Loss
Head loss describes the reduction in total pressure (or head) of a fluid as it moves through a pipe or duct system. It is a crucial factor in fluid transport systems because it impacts energy efficiency and pumping requirements.
In the problem at hand, head loss incorporates two key components:
  • Frictional head loss, driven by the pipe's characteristics and the flow conditions, calculated using the Darcy-Weisbach equation.
  • Local head loss, which occurs due to fittings, valves, and bends quantified by a loss coefficient (.\( K \)).
Reducing head loss is key to maintaining efficient fluid flow. Strategies for minimizing head loss include:
  • Using pipes with smoother surfaces to decrease friction.
  • Optimizing the layout of the piping system to reduce unnecessary bends and fittings.
  • Choosing pipe diameters that balance efficient flow with material cost.
The exercise demonstrates that by modifying factors such as the local head loss coefficient, it is possible to adjust the system efficiency, as shown in part b, where reducing the coefficient enables placing the pump at a higher point.
Reynolds Number
The Reynolds Number is a dimensionless value that predicts the flow regime within a pipe – whether it will be laminar or turbulent. This knowledge is fundamental because the flow type affects energy losses, system efficiency, and the necessary equipment design.
It is defined as: \[ Re = \frac{\rho v D}{\mu}\]Where:
  • \( \rho \)
  • : Fluid density.
  • \( v \)
  • : Fluid velocity.
  • \( D \)
  • : Characteristic length (typically diameter for a pipe).
  • \( \mu \)
  • : Fluid dynamic viscosity.
In the exercise, a Reynolds number greater than 4000 indicates turbulent flow, which impacts the choice of formulae and tools, like using the Darcy-Weisbach equation and considering roughness implications for accurately estimating head losses.
It also underscores the need to look at varying factors influencing the Reynolds number:
  • The fluid's velocity, which speeds up with increasing flow rate.
  • Viscosity, which can change with temperature adjustments.
  • Pipe characteristics, such as diameter and roughness.
Understanding how the Reynolds number fits into practical applications aids in managing and optimizing piping systems for various industrial solutions.

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Most popular questions from this chapter

A pump is to be used to withdraw water from a reservoir at a rate of \(1500 \mathrm{~L} / \mathrm{s}\). When operating at this flow rate, the head loss in the suction pipe is estimated to be \(2.3 \mathrm{~m}\), and the pump specifications give the required net positive suction head as \(2.9 \mathrm{~m}\). Standard sea-level atmospheric conditions are expected at the site, and under worstcase conditions, the temperature of the water in the reservoir is \(25^{\circ} \mathrm{C}\). What is the maximum allowable elevation of the suction side of the pump above the reservoir water surface?

A hydropower plant utilizes eight Francis turbines. The change in head across the power plant is \(250 \mathrm{~m}\), and the flow through each turbine unit is \(12 \mathrm{~m}^{3} / \mathrm{s}\). The estimated efficiency of each turbine unit is \(95 \%\), and the efficiency of the generator and supporting power delivery systems is \(91 \%\). Estimate the power-generating capacity of the hydropower facility. Assume water at \(20^{\circ} \mathrm{C}\).

A large centrifugal fan generates a flow of \(7 \mathrm{~m}^{3} / \mathrm{s}\) with a motor speed of \(1140 \mathrm{rpm}\). A smaller geometrically similar fan has a motor speed of \(1725 \mathrm{rpm}\), operates at the same efficiency as the larger fan, and generates the same pressure increase. What flow rate is generated by the smaller fan?

A proposed turbine is being designed to generate a power of \(30 \mathrm{MW},\) with a generator rotational speed of \(150 \mathrm{rpm}\) and an available head of \(22 \mathrm{~m}\). A model of the turbine is to be tested in the laboratory, where the available head is \(6 \mathrm{~m}\), the power is \(45 \mathrm{~kW},\) and the model turbine is expected to have a hydraulic efficiency of \(95 \%\). What length scale, rotational speed, and flow rate should be used in the model tests? Assume water at \(20^{\circ} \mathrm{C}\).

Nine pump units are placed in parallel at a pump station. Each unit has a power demand of \(40 \mathrm{~kW}\) and adds \(35 \mathrm{~m}\) of head under optimal conditions. The best efficiency of each unit is \(60 \%\). The liquid being pumped is water at \(20^{\circ} \mathrm{C}\). When all units are operating under optimal conditions, what is the flow rate delivered by the pump station?

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