/*! 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 54 A boat has a 250-mm-diameter pro... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A boat has a 250-mm-diameter propeller that discharges \(0.6 \mathrm{~m}^{3} / \mathrm{s}\) of water as the boat travels at \(35 \mathrm{~km} / \mathrm{h}\) in still water. Determine the thrust developed by the propeller on the boat.

Short Answer

Expert verified
The thrust developed by the propeller on the boat is 1500 N.

Step by step solution

01

Calculate the Area of the Pipe

First, we need to calculate the cross-sectional area of the propeller. The area \(A\) of a circle is given by the formula \(A = \pi r^{2}\), where \(r\) is the radius of the circle. Since the diameter is given as 250 mm, the radius would be half of that, which is 125 mm. But this should be converted to meters for consistency of units, thus \(r = 125 / 1000 = 0.125 m\). Hence, \(A = \pi (0.125)^{2} = 0.0491 m^{2}\).
02

Determine the initial and final velocities

Given that the boat travels at a speed of 35 km/hr in still water, this is the initial velocity of water relative to the boat. We convert this to m/s, \(V1 = 35 * 1000 / 3600 = 9.72 m/s\). The final velocity of the pulled water is unknown, but we can denote it as \(V2\). From mass conservation, given the water discharge, we can say, \(A * V2 = 0.6 m^{3}/s\). From this \(V2 = 0.6 / A = 0.6 / 0.0491 = 12.22 m/s\).
03

Calculate the Thrust

The thrust \(T\) developed by the propeller on the boat can be found using the momentum change principle, which states that the thrust is given as the rate of change of momentum, or the mass flow rate times the change in velocity. The mass flow rate \(\dot{m}\) is equal to the density \(\rho\) of water times the volumetric flow rate \(Q\), which is equal to \(A * V2\). Therefore, \(\dot{m} = \rho * Q = 1000 kg/m^{3} * 0.6 {m^{3}/s} = 600 kg/s\). The change in velocity is \(V2 - V1 = 12.22 m/s - 9.72 m/s = 2.5 m/s\). Hence, the thrust is \(T = \dot{m} * (V2 - V1) = 600 kg/s * 2.5 m/s = 1500 N\).

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.

Thrust Calculation
Calculating the thrust in fluid dynamics involves understanding how forces are generated through the movement of fluids. Thrust can be defined as the force exerted by a fluid jet on a body, which in turn propels the body in the opposite direction. In the context of a boat propeller, thrust is the force that moves the boat forward.
To calculate thrust, we primarily use the principle of momentum change. This means the thrust is the difference in momentum of the water before and after it is acted upon by the propeller. The calculation follows the formula:
  • Thrust, \( T = \dot{m} \times \Delta V \)
Where \( \dot{m} \) is the mass flow rate of water, and \( \Delta V \) is the change in velocity of the water. This formula helps in quantifying the exerted force by the moving water mass as it leaves the propeller.
Momentum Principle
The momentum principle is crucial in understanding thrust calculation. It is based on Newton's second law, which states that the force acting on an object is equivalent to the change in its momentum. In fluid mechanics, particularly for moving fluids, the momentum principle helps us calculate the changes occurring as a fluid is accelerated by a propeller.
In this exercise, the principle guides how we determine the change in momentum, which directly relates to thrust. Since momentum, \( p \), is given by the product of mass and velocity, \( p = m \times v \), the change in momentum or net force is calculated by:
  • The change in velocity of the water, \( \Delta V = V2 - V1 \).
  • Then, it connects to the thrust by \( F = \dot{m} \times \Delta V \).
Using the momentum principle, we turn the abstract forces within a fluid, like water being propelled by a propeller, into quantifiable computations.
Mass Flow Rate
Mass Flow Rate is an important concept when calculating the thrust developed by a propeller. It represents the amount of mass passing through a section per unit time. In our exercise, we're dealing with water, which has a density denoted by \( \rho \), and a given volumetric flow rate \( Q \).
These elements combine into the mass flow rate expression:
  • \( \dot{m} = \rho \times Q \)
For water, typically, the density \( \rho \) is \( 1000 \, kg/m^3 \). Thus, knowing the volumetric flow helps determine the mass flow rate needed to find thrust. In essence, it measures how much mass is being moved and provides critical data for how much momentum is being transferred at a specific time.
Propeller Dynamics
Propeller Dynamics refers to how the propeller interacts with the water to produce useful work, such as propelling a boat forward. It is essential to account for the geometric and physical aspects of the propeller to understand how effectively it works.
The core aspects of propeller dynamics include:
  • The diameter of the propeller, which affects the volume of water moved.
  • The velocity increase given to the passing fluid, influencing the overall thrust.
Moreover, with knowledge of propeller diameter, such as the 250 mm diameter given, we can compute essential areas and utilize them in thrust calculation. By understanding and manipulating these dynamic components properly, engineers enhance efficiency and effectiveness in marine transport and other fluid dynamics challenges.

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 motor boat is powered by a fan, which develops a slipstream having a diameter of \(1.5 \mathrm{~m}\). If the fan ejects air with an average velocity of \(50 \mathrm{~m} / \mathrm{s}\) relative to the boat, and the boat is traveling with a constant velocity of \(10 \mathrm{~m} / \mathrm{s}\), determine the force exerted by the fan on the boat. Assume that the air has a constant density of \(\rho_{\alpha}=1.22 \mathrm{~kg} / \mathrm{m}^{3}\) and that the entering air at \(A\) is essentially at rest relative to the ground.

The apparatus or "jet pump" used in an industrial plant is constructed by placing the tube within the pipe. Determine the increase in pressure \(\left(p_{B}-p_{A}\right)\) that occurs between the back \(A\) and front \(B\) of the pipe if the velocity of the flow within the 200 -mm-diameter pipe is \(2 \mathrm{~m} / \mathrm{s}\), and the velocity of the flow through the 20 -mm- diameter tube is \(40 \mathrm{~m} / \mathrm{s}\). The fluid is ethyl alcohol having a density of \(\rho_{e a}=790 \mathrm{~kg} / \mathrm{m}^{3} .\) Assume the pressure at each cross section of the pipe is uniform.

The \(10-\mathrm{Mg}\) jet plane has a constant speed of \(860 \mathrm{~km} / \mathrm{h}\) when it is flying horizontally. Air enters the intake \(I\) at the rate of \(40 \mathrm{~m}^{3} / \mathrm{s}\). If the engine burns fuel at the rate of \(2.2 \mathrm{~kg} / \mathrm{s}\), and the gas (air and fuel) is exhausted relative to the plane with a speed of \(600 \mathrm{~m} / \mathrm{s}\), determine the resultant drag force exerted on the plane by air resistance. Assume that the air has a constant density of \(\rho_{a}=1.22 \mathrm{~kg} / \mathrm{m}^{3}\).

The jet is traveling at a constant velocity of \(400 \mathrm{~m} / \mathrm{s}\) in still air, while consuming fuel at the rate of \(1.8 \mathrm{~kg} / \mathrm{s}\) and ejecting it at \(1200 \mathrm{~m} / \mathrm{s}\) relative to the plane. If the engine consumes \(1 \mathrm{~kg}\) of fuel for every \(50 \mathrm{~kg}\) of air that passes through the engine, determine the thrust produced by the engine and the efficiency of the engine.

Flow from the water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of \(2 \mathrm{~m} / \mathrm{s}\). The discharge from the 50 -mm-diameter nozzle is \(0.04 \mathrm{~m}^{3} / \mathrm{s}\). One-fourth of the discharge flows down the incline, and three- fourths flows up the incline.

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.