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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 \(2.0 \times 10^{7}\) Pa in the hydraulic oil, and the piston has a radius of 0.050 \(\mathrm{m}\) . In a stroke lasting 25 \(\mathrm{s}\) 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 3.8 kW.

Step by step solution

01

Calculate the area of the piston

The area of a circle is calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius of the piston. Given \( r = 0.050 \, \mathrm{m} \), the area \( A \) becomes: \[ A = \pi \times (0.050)^2 \approx 0.00785 \, \mathrm{m}^2. \]
02

Find the force exerted by the piston

The force exerted by the piston is calculated by multiplying the pressure \( P \) by the area \( A \). Given \( P = 2.0 \times 10^7 \, \mathrm{Pa} \) and \( A = 0.00785 \, \mathrm{m}^2 \), \( F \) becomes: \[ F = P \times A = 2.0 \times 10^7 \times 0.00785 \approx 1.57 \times 10^5 \, \mathrm{N}. \]
03

Calculate the work done by the piston

The work \( W \) done is the product of force \( F \) and the distance \( d \) the piston moves. Given that the distance \( d = 0.60 \, \mathrm{m} \), \( W \) becomes: \[ W = F \times d = 1.57 \times 10^5 \times 0.60 \approx 9.42 \times 10^4 \, \mathrm{J}. \]
04

Compute the power required

Power \( P \) is defined as work done over time, \( P = \frac{W}{t} \). Given \( W = 9.42 \times 10^4 \, \mathrm{J} \) and \( t = 25 \, \mathrm{s} \), the power \( P \) is: \[ P = \frac{9.42 \times 10^4}{25} \approx 3.8 \times 10^3 \, \mathrm{W}. \]

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Key Concepts

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

Pressure Calculations
In hydraulic systems, understanding how pressure is calculated is essential. Pressure in a fluid is defined as the force exerted by the fluid per unit area on a surface. It's denoted by the letter 'P' and measured in Pascals (Pa). In the context of a hydraulic piston, we're often dealing with large forces because the system multiplies the force through this pressure mechanism.
To calculate pressure in our example, the piston experiences a pressure of \(2.0 \times 10^{7} \) Pa. This high pressure enables the hydraulic oil to execute tasks like splitting a log by translating pressure into force.
  • Formula: Pressure \( P = \frac{Force \,(F)}{Area \,(A)} \)
  • Units: 1 Pascal (Pa) = 1 Newton (N) per square meter \((m^2)\)
Piston Force
The piston force in a hydraulic system shows how pressure generates a large force over a relatively small area. This force is crucial for performing work.
In the example, the force \( F \) generated by the piston is derived by multiplying the pressure exerted by the hydraulic fluid \( P = 2.0 \times 10^{7} \) Pa by the area \( A = 0.00785 \, \mathrm{m}^2 \). This results in: \[ F = P \times A = 1.57 \times 10^5 \, \mathrm{N} \]. Key points to remember include:
  • Force is directly proportional to both the pressure applied and the piston area.
  • The unit of force is Newton (N).
  • Even a modest pressure applied over a small area can produce a significant force.
Work-Energy Principle
The work-energy principle explains how the work performed by the piston translates into movement. Work \( W \) is computed as the product of the force applied and the distance moved by the piston in the direction of the force.
In the exercise, the work done is calculated when the force of \( 1.57 \times 10^5 \, \mathrm{N} \) moves the piston through a distance of \( 0.60 \, \mathrm{m} \): \[ W = F \times d = 9.42 \times 10^4 \, \mathrm{J} \]. Understanding this helps in grasping:
  • Work is measured in Joules (J).
  • The direction of force and movement should be consistent.
  • This principle is foundational to explaining energy transformations.
Power Calculation
Power describes how quickly work is done or energy is transferred over time. In our log splitter example, calculating the power required involves determining how much work is done within a specific timeframe.
Using the formula \[ P = \frac{W}{t} \] where \( W = 9.42 \times 10^4 \, \mathrm{J} \) and \( t = 25 \, \mathrm{s} \) gives \[ P = \frac{9.42 \times 10^4}{25} \approx 3.8 \times 10^3 \, \mathrm{W} \]. Let鈥檚 simplify:
  • Power is represented in Watts (W) which are Joules per second.
  • Higher power implies more rapid energy usage.
  • Efficiency is key in power consumption.

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Most popular questions from this chapter

A \(58-\mathrm{kg}\) skier is going down a slope oriented \(35^{\circ}\) above the horizontal. The area of each ski in contact with the snow is 0.13 \(\mathrm{m}^{2}\) Determine the pressure that each ski exerts on the snow.

Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of \(8.30 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}\) . The weight of the input piston is negligible. The radii of the input piston and output plunger are \(7.70 \times 10^{-3} \mathrm{m}\) and \(0.125 \mathrm{m},\) respectively. What input force \(F\) is needed to support the 24500 \(\mathrm{-N}\) combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.30 m above that of the input piston?

A cylindrical storage tank has a radius of 1.22 m. When filled to a height of 3.71 m, it holds 14 300 kg of a liquid industrial solvent. What is the density of the solvent?

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(=9.50 \times 10^{-3} \mathrm{m} )\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(=1.90 \times 10^{-2} \mathrm{m}\) ), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is 9.00 \(\mathrm{N}\) . Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.
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