/*! 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 Water is pumped at the rate of \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 15

Water is pumped at the rate of \(1.2\) cubic metre per minute from a large tank on the ground, up to a point 8 metre above the level of the water in the tank. It emerges as a horizontal jet from a pipe of cross-section \(5 \times 10^{-3}\) square metre. If the efficiency of the apparatus is \(60 \%\), find the energy supplied to the pump per second.

Short Answer

Expert verified
2613.33 Watts.

Step by step solution

01

- Calculate the potential energy per minute

First, calculate the potential energy imparted to the water when it is raised 8 meters. Potential energy (PE) is given by PE = mgh where m = mass of the water g = acceleration due to gravity h = height. In one minute, the volume of water pumped is 1.2 cubic meters, which equates to 1200 kg of water (since the density of water is 1000 kg/m鲁). Thus, m = 1200 kg g = 9.8 m/s虏 h = 8 m. Substituting these values in, PE = 1200 kg 脳 9.8 m/s虏 脳 8 m = 94,080 Joules.
02

- Convert Potential Energy per minute to Potential Energy per second

Since we need the power, convert the potential energy calculated per minute to potential energy per second. There are 60 seconds in a minute. Thus, Potential Energy per second (Power, P) = 94,080 J/ 60 s = 1568 J/s or 1568 Watts.
03

- Calculate the actual power delivered to the water considering the efficiency

Given that the efficiency of the pump is 60%, the actual power delivered to the water is Actual Power = Efficiency 脳 Power = 0.60 脳 1568 W = 940.8 W.
04

- Calculate the energy supplied to the pump per second

To find the energy supplied to the pump, since 60% efficiency means only 60% of the supplied energy is converted to useful work, the energy supplied per second ( Input Energy) can be calculated as Input Energy = Actual Power / Efficiency = 1568 W / 0.6 = 1568 W 梅 0.60 = 2613.33 W.

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 91影视!

Key Concepts

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

potential energy
Potential energy is the stored energy of an object due to its position. In fluid mechanics, when water is pumped from a lower level to a higher level, it gains potential energy due to the elevation change. The potential energy (PE) can be calculated using the formula: PE = mgh, where m = mass of the fluid, g = acceleration due to gravity (9.8 m/s虏), h = height. For example, in this exercise, pumping 1200 kg of water (since water's density is 1000 kg/m鲁) to a height of 8 meters results in potential energy of 94,080 Joules.
efficiency
Efficiency is a measure of how well a system converts input energy into useful output energy. It's expressed as a percentage and for pumping systems, it tells us how much of the supplied energy actually does useful work. In the exercise, the pump's efficiency is 60%. This means only 60% of the energy supplied to the pump is used to move the water, while the rest is lost due to various inefficiencies like heat and friction. To calculate the actual power delivered to the water, multiply the efficiency by the power determined from potential energy per second. For our example, 60% efficiency applied to 1568 Watts gives us 940.8 Watts of actual power.
power conversion
Power is the rate at which energy is transferred or converted. When calculating power in this context, we initially determine the potential energy per minute and then convert it to per second to find the power in Watts. For instance, the potential energy per minute of 94,080 Joules equates to 1568 Joules per second. Considering efficiency, we calculate the delivered and input power, where input power considers the system's efficiency. From our example, the pump supplies 1568 Watts, but effectively uses 2613.33 Watts of energy considering the 60% efficiency.
fluid mechanics
Fluid mechanics is the study of how fluids (liquids and gases) behave under various forces. Key concepts include fluid dynamics, pressure, and energy transformations. In this exercise, understanding how water is pumped, the role of potential energy, and how these relate to mechanical systems are vital. The pumping system creates a flow, transferring water from one location to another, overcoming gravitational forces which leads to an increase in potential energy, thus fluid mechanics principles are applied to determine necessary power and efficiency.
pumping systems
Pumping systems are mechanical setups used to move fluids from one place to another. They are crucial in various applications, from household water supply to industrial processes. These systems convert mechanical energy into fluid motion, governed by concepts like potential energy and efficiency. The efficiency of pumping systems is crucial because it determines how much energy is effectively used for moving the fluid. In our exercise, a 60% efficient pump transfers the required potential energy to the water while considering the real power demands on the system. Understanding these dynamics ensures proper design and operation of pumping systems for optimal performance.

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 body of mass \(2.5\) kilogramme is attached to the end \(B\) of a light elastic string AB of natural length 2 metre and modulus \(5 g\) newton. The mass is suspended vertically in equitibrium by the string whose other end \(A\) is attached to a fixed point. (i) Find the depth below A of B when the body is in equilibrium. (ii) Find the distance through which the body must be pulled down vertically from its equilibrium position so that it will just reach A after release.

A ring of mass \(m\) can slide freely on a smooth wire in the shape of a circle of diameter \(2 a\), which is fixed in a vertical plane. The ring is fastened to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(m g\). The other end of the string is attached to the lowest point of the wire. The ring is held at the highest point of the wire and is slightly disturbed from rest. Find the velocity of the ring: (a) when it is level with the centre of the circular wire, (b) when the string first becomes slack, (c) when the string makes an acute angle \(\theta\) with the upward vertical.

A particle of mass \(2 m\) is attached to one end of an elastic string of modulus \(m g\) whose other end is fixed to a point \(P .\) The particle is dropped from P. It will first come to rest: (a) when the tension in the string = \(2 m g\), (b) when the kinetic energy is zero, (c) below the equilibrium position, (d) when the length of the string has doubled.

A particle P of mass \(2 \mathrm{~kg}\) is attached to two strings PA and PB. PA is an. elastic string of natural length \(0.5 \mathrm{~m}\) and modulus of elasticity \(9.8 \mathrm{~N}\), and PB is an inelastic string. \(A\) and \(B\) are fixed points in a horizontal line. If P rests in equilibrium with PA making \(30^{\circ}\) with \(\mathrm{AB}\) and PA perpendicular to PB find the lengths of \(\mathrm{PA}\) and \(\mathrm{PB}\) and the tension in the inelastic string.

A particle falls freely from rest through a distance \(d\). Its speed is then: (a) \(\sqrt{g d}\) (b) \(-\sqrt{2 g u}\) (c) \(-\sqrt{\frac{g d}{2}}\) (d) \(\sqrt{2 g d}\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation
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.