/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 A reciprocating pump has two dou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A reciprocating pump has two double-acting cylinders each \(200 \mathrm{~mm}\) bore and \(450 \mathrm{~mm}\) stroke, the cranks being at \(90^{\circ}\) to each other and rotating at \(2.09 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (20 rev/min). The delivery pipe is \(100 \mathrm{~mm}\) diameter, \(60 \mathrm{~m}\) long. There are no air vessels. Assuming simple harmonic motion for the pistons determine the maximum and mean water velocities in the delivery pipe and the inertia pressure in the delivery pipe near the cylinders at the instant of minimum water velocity in the pipe.

Short Answer

Expert verified
Mean Velocity: 7.52 m/s, Maximum Velocity: 11.81 m/s, Inertia Pressure: 118243 Pa

Step by step solution

01

- Calculate Pump Displacement Volume

The displacement volume of one double-acting cylinder per stroke is given by: \[ V = \frac{\text{Bore Area} \times \text{Stroke}}{2} \] Compute the bore area: \[ A = \frac{\frac{\text{bore}}{2}}^2 \times \text{π} = \frac{\frac{200 \text{ mm}}{2}}^2 \times π = 0.0314 \text{ m}^2 \] Now, compute the displacement volume: \[ V = \frac{0.0314 \text{ m}^2 \times 0.45 \text{ m}}{2} = 0.007065 \text{ m}^3 \]
02

- Compute Pump Flow Rate

The flow rate is computed as the total volume displaced by both cylinders over time. Since there are two double-acting cylinders, each stroke contributes: \[ \text{Flow Rate} = 2 \times 2 \times V \times \text{pump speed} \] Given the pump speed is 2.09 rad/s: \[ \text{Flow Rate} = 4 \times 0.007065 \text{ m}^3 \times 2.09 \text{ rad/s} = 0.0590 \text{ m}^3/s \]
03

- Calculate Mean Water Velocity in the Pipe

The mean water velocity is given by: \[ \text{Mean Velocity} = \frac{\text{Flow Rate}}{\text{Cross-sectional area of the pipe}} \] The cross-sectional area of the pipe is: \[ A_{\text{pipe}} = \frac{\frac{100 \text{ mm}}{2}}^2 \times \text{Ï€} = 0.00785 \text{ m}^2 \] Now, compute mean velocity: \[ \text{Mean Velocity} = \frac{0.0590 \text{ m}^3/s}{0.00785 \text{ m}^2} = 7.52 \text{ m/s} \]
04

- Determine Maximum Water Velocity

Assuming simple harmonic motion for the pistons, the maximum velocity is: \[ \text{Maximum Velocity} = \text{Mean Velocity} \times \frac{Ï€}{2} \] Substituting the mean velocity: \[ \text{Maximum Velocity} = 7.52 \text{ m/s} \times \frac{Ï€}{2} = 11.81 \text{ m/s} \]
05

- Calculate Inertia Pressure

Inertia pressure is given by: \[ \text{Inertia Pressure} = \text{Density} \times \text{Length} \times \text{angular velocity}^2 \times \text{Stroke} / 2 \] Assuming water density as 1000 kg/m\textsuperscript{3}: \[ \text{Inertia Pressure} = 1000 \text{ kg/m}^3 \times 60 \text{ m} \times (2.09 \text{ rad/s})^2 \times (0.45 \text{ m}/2) \] Calculation results in: \[ \text{Inertia Pressure} = 118243 \text{ Pa} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Pump Displacement Volume
In a reciprocating pump, the displacement volume is crucial for understanding how much fluid is moved per stroke. The displacement volume depends on the bore (diameter) and stroke (distance traveled) of the cylinder. For a double-acting cylinder, each stroke fills and discharges fluid from both sides, so we consider half the stroke in the calculation. The formula used is:
\[ V = \frac{\text{Bore Area} \times \text{Stroke}}{2} \]
To find the bore area, use:
\[ A = \frac{\left( \frac{\text{bore}}{2} \right)^2 \times \pi}{2} \]
Then multiply by the stroke to find the displacement volume. In our example, with a bore of 200 mm and a stroke of 450 mm, the displacement volume per stroke becomes 0.007065 m³ for one cylinder.
Flow Rate Calculation
The flow rate of the pump is the volume of fluid displaced over a certain period. With two double-acting cylinders, each contributing volume from both ends, the total flow rate can be calculated using:
\[ \text{Flow Rate} = 2 \times 2 \times V \times \text{pump speed} \]
Here, V is the displacement volume, and the pump speed in rad/s converts the revolutions per minute into a useful factor. For the given example with a displacement volume of 0.007065 m³ and a pump speed of 2.09 rad/s, the flow rate comes out to be 0.0590 m³/s. This tells us how much water is pumped through the system every second.
Mean Water Velocity
The mean water velocity in the delivery pipe helps us understand how fast the water is moving on average. This is calculated by dividing the flow rate by the cross-sectional area of the delivery pipe:
\[ \text{Mean Velocity} = \frac{\text{Flow Rate}}{\text{Cross-Sectional Area of the Pipe}} \]
First, find the pipe's cross-sectional area with:
\[ A_{\text{pipe}} = \frac{\left( \frac{\text{diameter}}{2} \right)^2 \times \pi}{2} \]
With a pipe diameter of 100 mm, we get an area of 0.00785 m². Now, using the calculated flow rate of 0.0590 m³/s, the mean velocity is 7.52 m/s. This mean velocity is a key factor in designing efficient piping systems.
Maximum Water Velocity
In conditions of simple harmonic motion, the pistons create a situation where water velocity periodically reaches a maximum value. This peak water velocity, which we call maximum water velocity, can be calculated using the mean velocity multiplied by a factor of \( \frac{\pi}{2} \). Here's the formula:
\[ \text{Maximum Velocity} = \text{Mean Velocity} \times \frac{\pi}{2} \]
Plugging in the mean velocity of 7.52 m/s, we get:
\[ \text{Maximum Velocity} = 7.52 \, m/s \times \frac{\pi}{2} = 11.81 \, m/s \]
This value represents the highest speed that water can reach inside the delivery pipe, which is essential for pressure-related calculations and for avoiding surges.
Inertia Pressure
Inertia pressure in the delivery pipe arises due to the accelerating and decelerating water masses. This pressure depends on factors like water density, pipe length, angular speed, and stroke length. The formula to calculate inertia pressure is:
\[ \text{Inertia Pressure} = \text{Density} \times \text{Length} \times \text{Angular Velocity}^2 \times \frac{\text{Stroke}}{2} \]
For our example, with water density assumed to be 1000 kg/m³, pipe length of 60 m, angular velocity of 2.09 rad/s, and stroke of 450 mm, we get:
\[ \text{Inertia Pressure} = 1000 \, kg/m^3 \times 60 \, m \times (2.09 \, rad/s)^2 \times \frac{0.45 \, m}{2} = 118243 \, Pa \]
Understanding inertia pressure is important for designing robust pumping systems that can handle the additional stresses due to accelerating fluids.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A single-acting reciprocating water pump, with a bore and stroke of \(150 \mathrm{~mm}\) and \(300 \mathrm{~mm}\) respectively, runs at \(2.51 \mathrm{rad} \cdot \mathrm{s}^{-1}(0.4 \mathrm{rev} / \mathrm{s}) .\) Suction and delivery pipes are each \(75 \mathrm{~mm}\) diameter. The former is \(7.5 \mathrm{~m}\) long and the suction lift is \(3 \mathrm{~m}\). There is no air vessel on the suction side. The delivery, pipe is \(300 \mathrm{~m}\) long, the outlet (at atmospheric pressure) being \(13.5 \mathrm{~m}\) above the level of the pump, and a large air vessel is connected to the delivery pipe at a point \(15 \mathrm{~m}\) from the pump. Calculate the absolute pressure head in the cylinder at beginning, middle and end of each stroke. Assume that the motion of the piston is simple harmonic, that losses at inlet and outlet of each pipe are negligible, that the slip is \(2 \%\), and that \(f\). for both pipes is constant at 0.01. (Atmospheric pressure \(10.33 \mathrm{~m}\) water head.)

During a laboratory test on a water pump appreciable cavitation began when the pressure plus velocity head at inlet was reduced to \(3.26 \mathrm{~m}\) while the total head change across the pump was \(36.5 \mathrm{~m}\) and the discharge was \(48 \mathrm{~L} \cdot \mathrm{s}^{-1} .\) Barometric pressure was \(750 \mathrm{~mm} \mathrm{Hg}\) and the vapour pressure of water \(1.8 \mathrm{kPa}\). What is the value of \(\sigma_{\mathrm{c}}\) ? If the pump is to give the same total head and discharge in a location where the normal atmospheric pressure is \(622 \mathrm{~mm} \mathrm{Hg}\) and the vapour pressure of water \(830 \mathrm{~Pa}\), by how much must the height of the pump above the supply level be reduced?

The impeller of a centrifugal pump has an outer diameter of \(250 \mathrm{~mm}\) and an effective outlet area of \(17000 \mathrm{~mm}^{2} .\) The outlet blade angle is \(32^{\circ} .\) The diameters of suction and discharge openings are \(150 \mathrm{~mm}\) and \(125 \mathrm{~mm}\) respectively. At \(152 \mathrm{rad} \cdot \mathrm{s}^{-1}(24.2 \mathrm{rev} / \mathrm{s})\) and discharge \(0.03 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the pressure heads at suction and discharge openings were respectively \(4.5 \mathrm{~m}\) below and \(13.3 \mathrm{~m}\) above atmospheric pressure, the measurement points being at the same level. The shaft power was \(7.76 \mathrm{~kW}\). Water enters the impeller without shock or whirl. Assuming that the true outlet whirl component is \(70 \%\) of the ideal, determine the overall efficiency and the manometric efficiency based on the true whirl component.

A large centrifugal pump is to have a specific speed of \(1.15 \mathrm{rad}\) \((0.183 \mathrm{rev})\) and is to discharge liquid at \(2 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against a total head of \(15 \mathrm{~m}\). The kinematic viscosity of the liquid may vary between 3 and 6 times that of water. Determine the range of speeds and test heads for a one-quarter scale model investigation of the full-size pump, the model using water.

The following duties are to be performed by rotodynamic pumps driven by electric synchronous motors, speed \(100 \pi / n \mathrm{rad} \cdot \mathrm{s}^{-1}(=50 / n \mathrm{rev} / \mathrm{s})\), where \(n\) is an integer: (a) \(14 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) of water against \(1.5 \mathrm{~m}\) head; \((b)\) oil (relative density \(0.80\) ) at \(11.3 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(70 \mathrm{kPa}\) pressure; \((c)\) water at \(5.25 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(5.5 \mathrm{MPa}\). Designs of pumps are available with specific speeds of \(0.20,0.60,1.20,2.83,4.0\) rad. Which design and speed should be used for each duty?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.