/*! 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 96 A log splitter uses a pump with ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 96

A log splitter uses a pump with hydraulic oil to push a piston, which is attached to a chisel. The pump can generate a pressure of \(2.0 \times 10^{7} \mathrm{Pa}\) in the hydraulic oil, and the piston has a radius of \(0.050 \mathrm{m}\). In a stroke lasting \(25 \mathrm{s},\) the piston moves \(0.60 \mathrm{m} .\) What is the power needed to operate the log splitter's pump?

Short Answer

Expert verified
The power needed is approximately 3.77 kW.

Step by step solution

01

Calculate the Area of the Piston

The area of the piston, which is circular, can be calculated using the formula: \( A = \pi r^2 \). Here, the radius \( r \) is given as \( 0.050 \text{ m} \).\[A = \pi (0.050)^2 \approx 7.85 \times 10^{-3} \text{ m}^2\]
02

Determine the Force Exerted by the Piston

The force exerted by the piston is derived from the pressure and the area of the piston using the formula: \( F = P \times A \). The pressure \( P \) is given as \( 2.0 \times 10^7 \text{ Pa} \).\[F = 2.0 \times 10^7 \times 7.85 \times 10^{-3} \approx 1.57 \times 10^5 \text{ N}\]
03

Find the Work Done by the Piston

Work done is calculated using the formula: \( W = F \times d \), where \( d \), the distance moved by the piston, is \( 0.60 \text{ m} \).\[W = 1.57 \times 10^5 \times 0.60 \approx 9.42 \times 10^4 \text{ J}\]
04

Calculate the Power

Power is the rate at which work is done, given by the formula: \( P = \frac{W}{t} \), where \( t \) is the time duration, \( 25 \text{ s} \).\[P = \frac{9.42 \times 10^4}{25} \approx 3.77 \times 10^3 \text{ W}\]
05

Determine the Power in Kilowatts

Finally, convert the power from watts to kilowatts by dividing by 1000.\[P = \frac{3.77 \times 10^3}{1000} \approx 3.77 \text{ kW}\]

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.

Pressure Calculation
In hydraulic systems, pressure is a crucial concept that signifies how much force is applied over a specific area. The log splitter in the example utilizes hydraulic oil to generate pressure. To find out how much pressure the system is enduring, you need to understand it's measured in pascals (Pa), which is a unit of force per square meter.
  • Pressure is denoted as \( P \) and calculated using the formula \( P = \frac{F}{A} \), where \( F \) is force and \( A \) is area.
  • In our exercise, the pressure given is \( 2.0 \times 10^7 \text{ Pa} \) for the oil, indicating a high level of force being exerted across the piston's surface area.
By understanding this pressure, you can move on to calculating other important properties, such as the force exertion by the piston. Remember, the higher the pressure, the more force is exerted on the piston.
Force Exertion
The force exerted by a piston in a hydraulic system is a result of the pressure multiplied by the area of the piston's face. This equation, \( F = P \times A \), helps determine the force in Newtons (N).
  • The area \( A \) of the piston's face is essential and can be found with the formula for the area of a circle: \( A = \pi r^2 \).
  • For a piston with a radius of \( 0.050 \text{ m} \), the area becomes \( 7.85 \times 10^{-3} \text{ m}^2 \). Substituting these values, the force \( F \) is \( 1.57 \times 10^5 \text{ N} \).
This immense force is what enables the piston to split the logs efficiently. Understanding this concept ensures you can apply pressure and area calculations to elliptical or complex shapes in more advanced problems.
Work Done by Piston
Work in physics refers to the energy transferred when a force is applied over a distance. In a hydraulic system like a log splitter, the work done by the piston is critical in ensuring tasks are completed efficiently. It is calculated using \( W = F \times d \), where \( W \) is work in joules (J), \( F \) is the force in newtons (N), and \( d \) is distance in meters (m).
  • In our exercise, the piston moves \( 0.60 \text{ m} \), pushing the chisel to split the log.
  • The force exerted was previously calculated as \( 1.57 \times 10^5 \text{ N} \), leading to a work done value of approximately \( 9.42 \times 10^4 \text{ J} \).
This calculation shows just how much energy is used to achieve the piston movement, a key parameter in determining the system's efficiency.
Power Conversion to Kilowatts
Power represents the rate at which work is done or energy is transferred over time. In hydraulic systems, knowing the power requirements is crucial for efficient operation. It's calculated as \( P = \frac{W}{t} \), where \( P \) is power in watts (W), \( W \) is work in joules (J), and \( t \) is time in seconds (s).
  • Our calculation shows the power needed during a 25-second operation is \( 3.77 \times 10^3 \text{ W} \).
  • To convert this to a more convenient unit, divide by 1000 to express it in kilowatts (kW), resulting in approximately \( 3.77 \text{ kW} \).
Power conversion to kilowatts provides a clearer understanding of the energy usage, establishing a useful baseline for designing or comparing hydraulic devices.

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

Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses \(A\) and \(B\) have the same length, but hose \(B\) has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's \(\operatorname{law}\left[Q=\pi R^{4}\left(P_{2}-P_{1}\right) /(8 \eta L)\right]\) applies to each. In this law, \(P_{2}\) is the pressure upstream, \(P_{1}\) is the pressure downstream, and \(Q\) is the volume flow rate. The ratio of the radius of hose \(\mathrm{B}\) to the radius of hose \(\mathrm{A}\) is \(R_{\mathrm{B}} / R_{\mathrm{A}}=1.50 .\) Find the ratio of the speed of the water in hose \(B\) to the speed in hose \(A\).

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of the aneurysm, the normal cross-sectional area \(A_{1}\) of the aorta increases to a value of \(A_{2}=1.7 A_{1} .\) The speed of the blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}\right)\) through a normal portion of the aorta is \(v_{1}=0.40 \mathrm{m} / \mathrm{s} .\) Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure \(P_{2}\) in the enlarged region exceeds the pressure \(P_{1}\) in the normal region.

An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is \(15.2 \mathrm{N}\). When completely submerged in water, its apparent weight is \(13.7 \mathrm{N}\). What is the volume of the object?

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{Pa}\) and a speed of \(2.1 \mathrm{m} / \mathrm{s}\). However, on the second floor, which is \(4.0 \mathrm{m}\) higher, the speed of the water is \(3.7 \mathrm{m} / \mathrm{s}\). The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

A blood transfusion is being set up in an emergency room for an accident victim. Blood has a density of \(1060 \mathrm{kg} / \mathrm{m}^{3}\) and a viscosity of \(4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s} .\) The needle being used has a length of \(3.0 \mathrm{cm}\) and an inner radius of \(0.25 \mathrm{mm} .\) The doctor wishes to use a volume flow rate through the needle of \(4.5 \times 10^{-8} \mathrm{m}^{3} / \mathrm{s} .\) What is the distance \(h\) above the victim's arm where the level of the blood in the transfusion bottle should be located? As an approximation, assume that the level of the blood in the transfusion bottle and the point where the needle enters the vein in the arm have the same pressure of one atmosphere. (In reality, the pressure in the vein is slightly above atmospheric pressure.)
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Engineering Physics

Read Explanation

Electricity

Read Explanation

Space Physics

Read Explanation

Classical Mechanics

Read Explanation

Dynamics

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.