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Chapter 3: Problem 42

Two hydraulic cylinders maintain a pressure of \(1200 \mathrm{kPa}\). One has a cross-sectional area of 0.01 \(\mathrm{m}^{2},\) the other \(0.03 \mathrm{~m}^{2}\). To deliver work of \(1 \mathrm{~kJ}\) to the piston, how large a displacement \(V\) and piston motion \(H\) are needed for each cylinder? Neglect \(P_{\mathrm{atm}}\)

Short Answer

Expert verified
Cylinder 1: 83.3 mm; Cylinder 2: 27.8 mm.

Step by step solution

01

Understand the Problem

We need to find the displacement (volume change) for each cylinder in order to deliver 1 kJ of work to the pistons. We're given the pressure and cross-sectional areas for each cylinder.
02

Recall the Work Formula

Work done by a hydraulic cylinder can be calculated using the formula, \[ \text{Work} = P \times \Delta V \] where \( P \) is the pressure in the cylinder and \( \Delta V \) is the displacement volume. We're given \( \text{Work} = 1 \text{ kJ} = 1000 \text{ J}\) and \( P = 1200 \text{ kPa} = 1200000 \text{ Pa}\).
03

Calculate Displacement Volume

Rearrange the formula to solve for \( \Delta V \): \[ \Delta V = \frac{\text{Work}}{P} = \frac{1000 \text{ J}}{1200000 \text{ Pa}} \] Simplify: \[ \Delta V = \frac{1000}{1200000} = \frac{1}{1200} \approx 0.000833 \text{ m}^3 \]
04

Determine Piston Motion

The displacement volume \( \Delta V \) is also equal to \( A \times H \), where \( H \) is the piston motion and \( A \) is the cross-sectional area of the cylinder. So, \[ H = \frac{\Delta V}{A} \] For Cylinder 1 with area 0.01 \( \text{m}^2 \): \[ H_1 = \frac{0.000833}{0.01} = 0.0833 \text{ m} = 83.3 \text{ mm} \] For Cylinder 2 with area 0.03 \( \text{m}^2 \): \[ H_2 = \frac{0.000833}{0.03} = 0.0278 \text{ m} = 27.8 \text{ mm} \]
05

Conclusion

Cylinder 1 requires a piston motion of 83.3 mm and Cylinder 2 requires a piston motion of 27.8 mm to deliver 1 kJ of work with a pressure of 1200 kPa.

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Key Concepts

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

Understanding Pressure in Hydraulic Cylinders
Pressure plays a vital role in the operation of hydraulic cylinders. In simple terms, pressure is the force exerted per unit area. For hydraulic systems, the pressure is measured in pascals (Pa) or kilopascals (kPa). One important formula associated with pressure is:
\[ P = \frac{F}{A} \]
Where:
  • \( P \) is the pressure
  • \( F \) is the force applied
  • \( A \) is the cross-sectional area
In the given exercise, we know that the cylinders maintain a constant pressure of 1200 kPa. This high pressure ensures that the pistons can perform significant work, necessary for moving loads or operating machinery.
Hydraulic systems are valuable because they can transfer large amounts of force through small changes in volume, leveraging pressure effectively.
Exploring Cross-Sectional Area
The cross-sectional area of a hydraulic cylinder is another crucial aspect to consider. It refers to the surface area of the piston through which the hydraulic fluid is pushing. It's presented in units of square meters (m虏) in the exercise.
The cross-sectional area determines how much force a cylinder can exert, due to the relation:
\[ F = P \times A \]
Where:
  • \( F \) is the force generated by the hydraulic cylinder
  • \( P \) is the hydraulic pressure
  • \( A \) is the cross-sectional area
In our example, the first cylinder has an area of 0.01 m虏, and the second one has 0.03 m虏. A larger area means a greater force can be applied, which is evident as the second cylinder needs less piston motion to perform the same amount of work.
Using the Work Formula
The work formula is a key element in calculations involving hydraulic systems. Work done by a cylinder is the product of pressure and volume change. The formula is:
\[ \text{Work} = P \times \Delta V \]
Where:
  • \( \text{Work} \) represents the energy exerted by the system
  • \( P \) is the pressure applied
  • \( \Delta V \) is the change in volume or displacement
In our exercise, to find out how much displacement is needed to perform 1 kJ of work, we rearrange the formula:
\[ \Delta V = \frac{\text{Work}}{P} \]
This gives a displacement volume of approximately 0.000833 m鲁 under 1200 kPa pressure. This volume corresponds to the variation needed to achieve the specified amount of work.
Understanding Piston Motion
Piston motion, which refers to the movement of the piston within the cylinder, is essential in converting hydraulic energy into mechanical energy. It's calculated by using the relationship between the displacement volume and the cross-sectional area:
\[ H = \frac{\Delta V}{A} \]
This tells us how far the piston needs to move to achieve a certain volume change at a given area.
  • For the first cylinder with 0.01 m虏 area, the piston moves 83.3 mm.
  • For the second cylinder with 0.03 m虏 area, the displacement is just 27.8 mm.
These movements are calculated by dividing the known displacement volume by the specific area of each cylinder.
Understanding piston motion is crucial for designing efficient hydraulic systems as it ensures the components move the required distances to perform their intended functions.

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Most popular questions from this chapter

Because a hot water supply must also heat some pipe mass as it is turned on, the water does not come out hot right away. Assume \(70^{\circ} \mathrm{C}\) liquid water at \(100 \mathrm{kPa}\) is cooled to \(45^{\circ} \mathrm{C}\) as it heats \(15 \mathrm{~kg}\) of copper pipe from 20 to \(45^{\circ} \mathrm{C}\). How much mass (kg) of water is needed?

A piston/cylinder assembly in a car contains \(0.2 \mathrm{~L}\) of air at \(90 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\), as shown in Fig. \(\mathrm{P} 3.163 .\) The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent \(n=1.25\) to a final volume six times smaller. Determine the final pressure and temperature, and the heat transfer for the process.

A \(1200-\mathrm{kg}\) car accelerates from zero to \(100 \mathrm{~km} / \mathrm{h}\) over a distance of \(400 \mathrm{~m}\). The road at the end of the \(400 \mathrm{~m}\) is at \(10 \mathrm{~m}\) higher elevation. What is the total increase in the car's kinetic and potential energy?

Air in a piston/cylinder assembly at \(200 \mathrm{kPa}\) and \(600 \mathrm{~K}\) is expanded in a constant-pressure process to twice the initial volume, state \(2,\) as shown in Fig. P3.182. The piston is then locked with a pin, and heat is transferred to a final temperature of \(600 \mathrm{~K}\). Find \(P, T,\) and \(h\) for states 2 and \(3,\) and find the work and heat transfer in both processes.

Consider a piston/cylinder setup with \(0.5 \mathrm{~kg}\) of \(\mathrm{R}-134 \mathrm{a}\) as saturated vapor at \(-10^{\circ} \mathrm{C}\). It is now compressed to a pressure of \(500 \mathrm{kPa}\) in a polytropic process with \(n=1.5 .\) Find the final volume and temperature and determine the work done during the process.
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