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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 \(25 \mathrm{s}\), 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 required is \( 3.77 \times 10^{3} \ \mathrm{W} \).

Step by step solution

01

Calculate the Area of the Piston

The area of the piston can be calculated using the formula for the area of a circle, \( A = \pi r^2 \), where \( r \) is the radius of the piston. Given that the radius \( r = 0.050 \ \mathrm{m} \), the area is:\[ A = \pi (0.050)^2 = 7.85 \times 10^{-3} \ \mathrm{m}^2 \]
02

Calculate the Force Exerted by the Piston

The force exerted by the piston is given by the formula \( F = P \, A \), where \( P = 2.0 \times 10^{7} \ \mathrm{Pa} \) is the pressure exerted by the pump. Using the area calculated in Step 1, the force is:\[ F = 2.0 \times 10^{7} \times 7.85 \times 10^{-3} = 1.57 \times 10^{5} \ \mathrm{N} \]
03

Calculate the Work Done by the Piston

The work done by the piston during its movement is the product of the force exerted and the distance moved. The distance moved by the piston is \( d = 0.60 \ \mathrm{m} \). Thus, the work done is:\[ W = F \, d = 1.57 \times 10^{5} \times 0.60 = 9.42 \times 10^{4} \ \mathrm{J} \]
04

Calculate the Power Required

Power is the rate at which work is done, defined as \( P = \frac{W}{t} \), where \( W \) is the work done (from Step 3) and \( t = 25 \ \mathrm{s} \) is the time taken. Therefore, the power required is:\[ P = \frac{9.42 \times 10^{4}}{25} = 3.77 \times 10^{3} \ \mathrm{W} \]
05

Conclusion

The power needed to operate the log splitter's pump is \( 3.77 \times 10^{3} \ \mathrm{W} \).

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Key Concepts

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

Pressure
Pressure in hydraulics is a fundamental concept that describes how force is applied over a specific area. It is measured in pascals (Pa), which is the unit of pressure in the International System of Units (SI). In our example of a hydraulic log splitter, the pump generates a pressure of \(2.0 \times 10^{7}\) pascals. This pressure ensures that the hydraulic fluid within the system exerts force onto the piston.
  • Pressure is defined mathematically as \( P = \frac{F}{A} \), where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area.
  • Knowing the pressure and area, we can calculate the force exerted, helping us determine the work and power of the system.
  • High pressure means the hydraulic system can exert a large force over small areas, which is crucial for cutting or splitting materials.
Force
In the context of our hydraulic log splitter, force plays a pivotal role in how the system operates. The force exerted by the piston is the product of the pressure and the piston鈥檚 area. Using the equation \( F = P \times A \), we can determine the force required for splitting the log.
  • In the example, the pressure \( P \) is given as \(2.0 \times 10^{7} \) Pa and the area \( A \) of the piston's face was previously calculated as \( 7.85 \times 10^{-3} \ \, \mathrm{m}^2 \).
  • Inserting these values into the formula, we find the force exerted by the hydraulic system is \(1.57 \times 10^{5} \ \, \mathrm{N} \).
  • This force is what pushes the chisel into the log, enabling it to split effectively.
Power
Power is a measure of how quickly work is done or energy is transferred. In our log splitter example, we are interested in the power necessary for the pump to operate. Power can be calculated using the equation \( P = \frac{W}{t} \), where \( W \) is work done and \( t \) is time.
  • In this situation, work was calculated as \( 9.42 \times 10^{4} \ \, \mathrm{J} \) and the time given for the work to occur was \( 25 \, \mathrm{s} \).
  • Using these values, we find the power required is \( 3.77 \times 10^3 \, \mathrm{W} \) or 3.77 kW.
  • This power measurement indicates how much energy per second is needed to keep the hydraulic pump functioning efficiently.
Work Energy Theorem
The Work-Energy Theorem is a crucial concept in understanding how energy is transformed in physics. It states that the work done on an object is equal to the change in its kinetic energy. In the case of the hydraulic log splitter, the piston does work on the log through a distance of \(0.60 \, \mathrm{meters} \).
  • The work done by the piston is calculated using the formula \( W = F \times d \), where \( F \) is force and \( d \) is the distance moved.
  • Here, the force was \( 1.57 \times 10^5 \, \mathrm{N} \) and the distance \( 0.60 \, \mathrm{m} \), leading to a work done of \( 9.42 \times 10^4 \, \mathrm{J} \).
  • This work is what allows the energy to be transferred from the hydraulic fluid to the kinetic energy needed to split the log, demonstrating the power of hydraulic systems in converting energy forms efficiently.

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Most popular questions from this chapter

As the initially empty urinary bladder fills with urine and expands, its internal pressure increases by \(3300\) \(\mathrm{Pa}\), which triggers the micturition reflex (the feeling of the need to urinate). The drawing shows a horizontal, square section of the bladder wall with an edge length of \(0.010 \mathrm{m}\). Because the bladder is stretched, four tension forces of equal magnitude \(T\) act on the square section, one at each edge, and each force is directed at an angle \(\theta\) below the horizontal. What is the magnitude \(T\) of the tension force acting on one edge of the section when the internal bladder pressure is \(3300\) \(\mathrm{Pa}\) and each of the four tension forces is directed \(5.0^{\circ}\) below the horizontal?

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an eardrum. Eardrums can rupture when the pressure difference is as little as \(35\) \(\mathrm{kPa}\). What is the depth at which this pressure difference could occur? The density of seawater is \(1025\) \(\mathrm{kg} /\mathrm{m}^{3}\).

A barber's chair with a person in it weighs \(2100\) \(\mathrm{N}\). The output plunger of a hydraulic system begins to lift the chair when the barber's foot applies a force of \(55\) \(\mathrm{N}\) to the input piston. Neglect any height difference between the plunger and the piston. What is the ratio of the radius of the plunger to the radius of the piston?

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance \(h,\) as the drawing shows. Using \(1.013 \times 10^{5}\) \(\mathrm{Pa}\) for the atmospheric pressure and \(1200\) \(\mathrm{kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is (a) \(0.15\) \(\mathrm{m}\) and (b) \(0.10\) \(\mathrm{m}\).

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about \(2000: \mathrm{Re}=2 \bar{v} \rho R / \eta\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{m}\right)\).
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