/*! 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 63 An irrigation pump takes water f... [FREE SOLUTION] | 91Ó°ÊÓ

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An irrigation pump takes water from a river at \(10^{\circ} \mathrm{C}, 100 \mathrm{kPa}\) and pumps it up to an open canal at a 50 -m higher elevation. The pipe diameter into and out of the pump is \(0.1 \mathrm{~m}\), and the motor driving the pump is 5 hp. Neglect kinetic energies and friction and find the maximum possible mass flow rate.

Short Answer

Expert verified
The maximum possible mass flow rate is approximately 7.59 kg/s.

Step by step solution

01

Understand the Problem

We need to find the maximum mass flow rate when water is pumped from a lower elevation to a higher one (50 m higher) through a pipe. Given data include the initial conditions of temperature and pressure, pipe dimensions, and motor power. We will use the principles of energy conservation, disregarding kinetic and friction losses.
02

Convert Power Units

The power output of the motor is given as 5 horsepower (hp). We will first convert this to watts, as the SI unit system is used in fluid dynamics problems: \[ 1 ext{ hp} = 745.7 ext{ W} \]Thus, \[ 5 ext{ hp} = 5 \times 745.7 = 3728.5 ext{ W} \]
03

Use Bernoulli's Equation Simplified for Pumping

For a pump increasing the water to a height \(h\), the work done per unit mass is:\[ W = g h \]where \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity and \(h = 50 \text{ m}\) is the elevation height.
04

Calculate the Maximum Mass Flow Rate

The power available to lift the water (neglecting losses) can be given as the product of mass flow rate (\(\dot{m}\)) and the work done per unit mass:\[ Power = \dot{m} \times g h \]Rearrange for mass flow rate:\[ \dot{m} = \frac{Power}{g h} \]Substitute known values:\[ \dot{m} = \frac{3728.5}{9.81 \times 50} \approx 7.59 \text{ kg/s} \]
05

Verify Result

Check that all unit conversions and substitutions have been done correctly. The calculation shows that a mass flow rate of approximately \(7.59 \text{ kg/s}\) can be achieved with the given power and elevation.

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Key Concepts

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

Bernoulli's Equation
When dealing with fluid dynamics, Bernoulli's Equation is a fundamental concept. It's all about the relationship between pressure, velocity, and height in a flowing fluid. Picture it as the conservation of energy for fluids. Bernoulli's Equation tells us that the total energy along a streamline is constant.
For incompressible, non-viscous fluids without energy loss due to friction, Bernoulli's Equation is expressed as:\[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]where:
  • \( P \) is the pressure energy
  • \( \frac{1}{2} \rho v^2 \) is the kinetic energy per unit volume
  • \( \rho g h \) is the potential energy per unit volume
  • \( \rho \) is the fluid density
  • \( g \) is the acceleration due to gravity
In our specific problem of pumping water, we simplify Bernoulli's Equation. We focus primarily on the potential energy change, given the height difference. By neglecting kinetic energies and friction, only the pressure and height terms matter. This simplification is crucial in finding the mass flow rate efficiently.
Mass Flow Rate
In fluid mechanics, understanding mass flow rate is key. It's the mass of fluid passing through a section of the pipe per unit time. For practical purposes, we describe it using units like kilograms per second (\(\text{kg/s}\)). Mass flow rate is an essential factor in ensuring efficient system design and energy usage.
For this problem, the motor's power drives the water through elevation, impacting the mass flow rate. The equation we use is:\[ \dot{m} = \frac{Power}{g h} \]Here:
  • \( \dot{m} \) is the mass flow rate
  • \( Power \) is the energy input from the motor
  • \( g \) is gravity's acceleration again
  • \( h \) is the height difference
We calculated an approximate maximum flow rate using the available power and the elevation height. This flow rate indicates how much water can be moved with the given energy, neglecting any loss aspects.
Energy Conservation
In the realm of fluid dynamics, energy conservation is a principle that holds everything together. It means the total energy in a closed system remains constant, although energy can change forms.
When pumping the water against gravity, we witness this principle at play. The mechanical energy supplied by the motor (in watts) is transmitted to the water to do work, specifically lifting it vertically. This energy transformation is described in terms of work-energy principle:\[ Work = mgh \]The balance of energy translates directly into the movement of water through elevation. The motor's power facilitates lifting the fluid to the specified height, a clear demonstration of energy conservation.
  • We don't create energy from nothing; we transform it from electrical to mechanical forms.
  • This transformation helps us achieve the desired lifting effect in the system.
  • It's crucial to note how critical energy evaluations are in optimizing systems for performance.
Energy conservation aids in calculating the limits and capabilities of such systems to ensure operation within expected standards.

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