/*! 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 15 Show that the ideal head for a r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that the ideal head for a radial-flow pump can be determined from \(\Delta H=\left(U_{2} V_{2} \cos \alpha_{2}\right) / g,\) where \(V_{2}\) is the velocity of the water leaving the impeller blades. Water enters the impeller blades in the radial direction.

Short Answer

Expert verified
The ideal head \(\Delta H\) for a radial-flow pump is given by \(\Delta H = \frac{U_{2}V_{2}\cos\alpha_{2}}{g}\), where \(V_2\) is the velocity of the water leaving the impeller blades. This is a result of considering the energy, momentum, and angular momentum balances of the water as it flows through the impeller blades.

Step by step solution

01

Clarify the variables

Before tackling the problem, it's crucial to understand what each variable in the equation stands for. \(\Delta H\) is the ideal head, a measure of the energy transfer to the fluid, \(U_2\) is the speed at which the fluid leaves the pump, \(V_2\) is the velocity of the fluid as it exits the impeller blades, \(\alpha_2\) is the angle at which the fluid leaves the impeller, and \(g\) is the acceleration due to gravity.
02

Apply the given

The equation for this problem has already been given: \(\Delta H = \frac{U_{2}V_{2}\cos\alpha_{2}}{g}\). The exercise is to show that the ideal head can be determined from this equation.
03

Conceptual understanding

The equation is a result of considering energy, momentum and angular momentum balances in a flow through the impeller blades. The radial direction of ingress and the perpendicular egress of water drives this equation. The multiplication of the fluid exit velocity \(V_2\) and the speed at the outlet of the pump \(U_2\) represents the kinetic energy of the fluid. This kinetic energy is then divided by the gravitational acceleration to convert the energy to a 'head' (a height equivalent), and adjusted by the angle of exit \(\alpha_2\) to capture the fact that not all of the fluid's kinetic energy is used to increase its vertical height.

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.

Fluid Mechanics
At the heart of radial-flow pump operations is fluid mechanics, the study of fluids (liquids and gases) and the forces on them. Fluid mechanics encompasses multiple concepts including flow velocity, pressure, energy, and how these elements interact within a pump system.

The performance of a radial-flow pump depends on the behavior of the fluid within the device. As water enters the pump radially and is forced outward by the rotation of the impeller, it experiences changes in velocity and pressure. The concept of an 'ideal head' in this context refers to the potential energy per unit weight that the pump imparts to the water, which translates to the height that the water could be raised by the pump's operation in an ideal scenario.
Energy Transfer in Fluids
Energy transfer in fluids is essential to understanding pump mechanics. A pump serves to transfer mechanical energy from the motor to the fluid, moving it from one location to another or raising it to a higher elevation. The head created by a pump, denoted as \( \Delta H \), is effectively a measure of this energy transfer.

When a pump imparts energy to a fluid, it increases the fluid's kinetic energy, potential energy, or both. In the equation \( \Delta H = \frac{U_{2}V_{2}\cos\alpha_{2}}{g} \), the product \( U_{2}V_{2}\cos\alpha_{2} \) reflects the kinetic energy imparted to the fluid due to the combined effect of the blade speed and fluid velocity at the discharge point. The cosine term adjusts for the direction of fluid discharge as not all energy contributes to producing head, especially if the discharge is not perfectly vertical. Finally, dividing by \( g \) shifts kinetic energy into the potential energy domain, expressing the energy transfer as a height or head.
Impeller Blade Dynamics
The dynamics of impeller blades are critical to pump function. Impeller blades accelerate the fluid, and the design and arrangement of these blades impact the pump's effectiveness. In a radial-flow pump, the fluid enters the impeller radially and is acted upon by the blades, acquiring energy and being propelled outward.

The angle \( \alpha_{2} \) in the equation is of particular importance as it defines the direction of fluid outflow relative to the blades. The effectiveness of the impeller is determined by how well it can convert rotational energy into fluid motion. The fluid's velocity as it leaves the impeller \( V_{2} \) and the rotational speed of the blades \( U_{2} \) are therefore integral parameters. Energy is conserved as per the First Law of Thermodynamics, ensuring the input mechanical energy from impeller rotation is transferred to the fluid as kinetic and potential energy.

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 200-mm-diameter impeller of a radial-flow water pump rotates at \(150 \mathrm{rad} / \mathrm{s}\) and has a discharge of \(0.3 \mathrm{~m}^{3} / \mathrm{s}\). Determine the discharge for a similar pump that has an impeller diameter of \(100 \mathrm{~mm}\) and operates at \(80 \mathrm{rad} / \mathrm{s}\)

Water flows from a lake through a 300-m-long pipe having a diameter of \(300 \mathrm{~mm}\) and a friction factor of \(f=0.015 .\) The flow from the pipe passes through a 60 -mm-diameter nozzle and is used to drive the Pelton wheel, where the bucket deflection angles are \(160^{\circ} .\) Determine the power and torque produced when the wheel is turning under optimum conditions. Neglect minor losses.

The buckets of the Pelton wheel deflect the 100-mm-diameter water jet \(140^{\circ}\) as shown. If the velocity of the water from the nozzle is \(30 \mathrm{~m} / \mathrm{s}\), determine the power that is delivered to the shaft when the wheel is rotating at a constant angular velocity of \(2 \mathrm{rad} / \mathrm{s}\). How fast must the wheel be turning to maximize the power developed by the wheel?

Rater flows through the 400 -mm-diameter delivery pipe at \(2 \mathrm{~m} / \mathrm{s}\). Each of the four 50 -mm-diameter nozzles is aimed tangentially at the Pelton wheel, which has bucket deflection angles of \(150^{\circ} .\) Determine the torque and power developed by the wheel when it is rotating at \(10 \mathrm{rad} / \mathrm{s}\).

Water flowing at \(4 \mathrm{~m} / \mathrm{s}\) is directed from the stator onto the blades of the axial-flow turbine, where the mean radius of the blades is \(0.75 \mathrm{~m}\). If the blades are rotating at \(80 \mathrm{rad} / \mathrm{s}\) and the flow is \(7 \mathrm{~m}^{3} / \mathrm{s}\), determine the torque produced by the water.

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.