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Chapter 1: Problem 5

A hydraulic ram \(200 \mathrm{~mm}\) in diameter and \(1.2 \mathrm{~m}\) long moves wholly within a concentric cylinder \(200.2 \mathrm{~mm}\) in diameter, and the annular clearance is filled with oil of relative density \(0.85\) and kinematic viscosity \(400 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\). What is the viscous force resisting the motion when the ram moves at \(120 \mathrm{~mm} \cdot \mathrm{s}^{-1}\) ?

Short Answer

Expert verified
The viscous force resisting the motion is approximately 9688 N.

Step by step solution

01

Determine the annular gap

Calculate the annular gap, which is the difference between the cylinder diameter and the ram diameter. \[ \text{Gap} = \frac{200.2 \text{ mm} - 200 \text{ mm}}{2} = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} \]
02

Convert kinematic viscosity to dynamic viscosity

Calculate the dynamic viscosity, \( u \), using the kinematic viscosity and oil density. First, convert the relative density to absolute density (\( \rho \)). \[ \rho = 0.85 \times 1000 \text{ kg/m}^{3} = 850 \text{ kg/m}^{3} \] Now, calculate the dynamic viscosity with \( u \). \[ u = \text{kinematic viscosity} \times \rho = 400 \times 10^{-6} \text{ m}^{2}/\text{s} \times 850 \text{ kg/m}^{3} = 0.34 \text{ Pa} \text{ s} \]
03

Calculate relative velocity

The relative velocity (\( u \)) is given directly in the problem statement. \[ u = 120 \text{ mm/s} = 0.12 \text{ m/s} \]
04

Determine the shear area

First calculate the circumferential perimeter: \[ P = \text{diameter} \times \text{pi} = 200 \text{ mm} \times \text{pi} = 0.2 \text{ m} \times \text{pi} \] So, the shear area \( A \) is: \[ A = \text{pi} \times 0.2 \text{ m} \times 1.2 \text{ m} \]
05

Calculate the viscous shear force

The viscous shear force, \( F \), is given by: \[ F = \frac{u \times A \times u}{\text{gap}} = \frac{0.34 \text{ Pa} \text{ s} \times \text{pi} \times 0.2 \text{ m} \times 1.2 \text{ m} \times 0.12 \text{ m/s}}{0.1 \times 10^{-3} \text{ m}} \] This simplifies to \( 3084 \times \text{pi} \text{ N} \).

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Key Concepts

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

Annular Gap
The annular gap refers to the small space between the hydraulic ram and the concentric cylinder it moves within. This gap is essential in calculating the viscous force resisting the motion. In the provided exercise, the annular gap is found by taking the difference between the outer cylinder's diameter and the ram's diameter, then dividing by two. This results in a gap of 0.1 mm, or 0.1 × 10^{-3} meters. Accurately determining this gap is crucial for understanding the resistance encountered by the hydraulic ram due to viscous forces.
Kinematic Viscosity
Kinematic viscosity is a measure of a fluid's resistance to flow and shear under the influence of gravity. It's represented in units of mm²/s or m²/s. In the exercise, the oil's kinematic viscosity is given as 400 mm²/s. Kinematic viscosity is a key parameter because it affects how easily a fluid flows, influencing the shear stress and, consequently, the viscous force. High kinematic viscosity means the fluid flows more slowly, leading to higher resistance against the ram's movement.
Dynamic Viscosity
Dynamic viscosity describes a fluid's internal resistance to flow when an external force is applied. It is measured in Pa·s (Pascal-seconds). To find the dynamic viscosity in this example, we use the given kinematic viscosity and the oil's relative density to calculate the absolute density, which is 850 kg/m³. Multiplying the kinematic viscosity (converted to m²/s) with the absolute density gives a dynamic viscosity of 0.34 Pa·s. Understanding this concept is crucial as it directly impacts the calculation of the viscous force.
Relative Density
Relative density, or specific gravity, is the ratio of the density of a fluid to the density of a reference substance, typically water for liquids. In the given problem, the relative density of oil is 0.85. This means the oil is 85% as dense as water. Converting this to absolute density allows us to use it in further calculations such as finding dynamic viscosity. Knowing the relative and absolute densities helps to understand how heavy or light a fluid is compared to water, impacting the overall force calculations.
Shear Area
The shear area is the surface area over which shear force acts. In the example, it's calculated by determining the circumference of the ram (using the diameter and π) and multiplying it by the length of the ram. This gives us a shear area that interacts with the oil, contributing to the resistance against the ram's movement. Understanding the shear area helps in visualizing how a broader area would mean more fluid interaction and hence greater total resistance.
Viscous Shear Force
Viscous shear force is the resistance due to the fluid's viscosity acting against the motion of the hydraulic ram. Calculated using the formula \( F = \frac{{u \times A \times u}}{{\text{{gap}}}} \), where u is the dynamic viscosity, A is the shear area, and u is the relative velocity. In the problem, the force equates to 3084π N. This force results from the fluid's resistance, causing energy loss and impacting the efficiency of the hydraulic system. Understanding how to calculate and interpret this force is crucial for designing fluid systems that are energy efficient.

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Most popular questions from this chapter

A hydrogen-filled balloon is to expand to a sphere \(20 \mathrm{~m}\) diameter at a height of \(30 \mathrm{~km}\) where the absolute pressure is \(1100 \mathrm{~Pa}\) and the temperature \(-40^{\circ} \mathrm{C}\). If there is to be no stress in the fabric of the balloon what volume of hydrogen must be added at ground level where the absolute pressure is \(101.3 \mathrm{kPa}\) and the temperature \(15^{\circ} \mathrm{C}\) ?

Calculate the density of air when the absolute pressure and the temperature are respectively \(140 \mathrm{kPa}\) and \(50{ }^{\circ} \mathrm{C}\) and \(\mathrm{R}=\) \(287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

Calculate the Reynolds number for a fluid of density \(900 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and dynamic viscosity \(0.038 \mathrm{~Pa} \cdot \mathrm{s}\) flowing in a \(50 \mathrm{~mm}\) diameter pipe at the rate of \(2.5 \mathrm{~L} \cdot \mathrm{s}^{-1}\). Estimate the mean velocity above which laminar flow would be unlikely.

A liquid of kinematic viscosity \(370 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\) flows through an \(80 \mathrm{~mm}\) diameter pipe at \(0.01 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). What type of flow is to be expected?

A uniform film of oil \(0.13 \mathrm{~mm}\) thick separates two discs, each of \(200 \mathrm{~mm}\) diameter, mounted co-axially. Ignoring edge effects, calculate the torque necessary to rotate one disc relative to the other at a speed of \(44 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (7 rev/s) if the oil has a dynamic viscosity of \(0.14 \mathrm{~Pa} \cdot \mathrm{s}\).
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