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Chapter 6: Problem 11

A plane bearing plate is traversed by a very wide, \(150 \mathrm{~mm}\) long, plane inclined slipper moving at \(1.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\). The clearance between slipper and bearing plate is \(0.075 \mathrm{~mm}\) at the toe and \(0.025 \mathrm{~mm}\) at the heel. If the load to be sustained by the bearing divided by the width is \(500 \mathrm{kN} \cdot \mathrm{m}^{-1}\), determine the viscosity of the lubricant required, the power consumed divìded by the width of the bearing, the maximum pressure in the lubricant and the position of the centre of pressure.

Short Answer

Expert verified
The viscosity is 0.1111 Pas, the power consumed is 750 W, and the maximum pressure is 3333.33 kPa.

Step by step solution

01

Determine the film thickness gradient

Calculate the film thickness gradient across the slipper. The thickness at the toe end is 0.075 mm, and at the heel end is 0.025 mm. The change in thickness \(\triangle h = h_{toe} - h_{heel} = 0.075 \text{ mm} - 0.025 \text{ mm} = 0.05 \text{ mm}\).
02

Convert dimensions to consistent units

Convert all dimensions to meters for consistency. \(\triangle h = 0.05 \text{ mm} = 0.00005 \text{ m}\)
03

Calculate average film thickness

The average film thickness \(h_{avg}\) is given by \[h_{avg} = \frac{h_{toe} + h_{heel}}{2} = \frac{0.075 \text{ mm} + 0.025 \text{ mm}}{2} = 0.05 \text{ mm} = 0.00005 \text{ m}\text{.} \]
04

Determine the velocity gradient

The relative velocity, \( U = 1.5 \text{ m/s}\). Hence, the velocity gradient, \[ \frac{dU}{dy} = \frac{U}{h_{avg}} = \frac{1.5}{0.00005} = 30000 \text{ s}^{-1} \]
05

Calculate the shear stress

The shear stress \(\tau = \frac{Load}{Area} = \frac{500 \text{ kN/m}}{0.15 \text{ m}} = 3,333.33 \text{ kN/m}^2 \)
06

Determine the viscosity

Using the relation \( \tau = \text{viscosity} \times \frac{dU}{dy} \), we have \[ \text{viscosity} = \frac{\tau}{\frac{dU}{dy}} = \frac{3333.33 \times 1000}{30000} \text{ kPas} = 0.1111 \text{ Pas}\text{.} \]
07

Calculate the power consumed

The power consumed \[P = \tau \times U \times A \), where \(A \text{ is the area} = 150 \text{ mm} = 0.15 \text{ m}\text{.} \] Substituting values, \[P = 3333.33 \times 1.5 \times 0.15 = 750 \text{ W}\text{.} \]
08

Calculate the maximum pressure in the lubricant

The maximum pressure in the lubricant \[P_{max} = \frac{\text{Load}}{\text{Length}} = \frac{500 \text{ kN/m}}{0.15 \text{ m}} = 3333.33 \text{ kPa}\text{.} \]

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Key Concepts

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

Viscosity Determination
Viscosity is a measure of a fluid's resistance to flow. It describes how thick or thin the fluid is. In fluid mechanics, especially in lubrication, understanding the viscosity is crucial.

In the plane bearing plate problem, the viscosity of the lubricant is determined using the formula for shear stress and velocity gradient. The shear stress \( \tau = \frac{\text{Load}}{\text{Area}} \) and the velocity gradient \( \frac{dU}{dy} = \frac{U}{h_{avg}} \). Plugging in the values, the viscosity is calculated as \( \text{viscosity} = \frac{\tau}{\frac{dU}{dy}} = 0.1111 \text{ Pas} \. \)
Power Consumption Calculation
Power consumption in a mechanical system is an important factor, especially in systems involving moving parts. It tells you how much energy is being used over time.

For the plane bearing plate, power consumption is calculated using the formula: \( P = \tau \times U \times A \. \) Here, \( \tau \) is the shear stress, \( U \) is the relative velocity, and \( A \) is the area. By substituting the given values, the power consumed is computed to be 750 W. This represents how much electrical energy would be required to maintain the motion.
Lubricant Pressure
Lubricant pressure in a bearing system is vital for understanding how well the system can carry loads without failures.

In our problem, the maximum pressure is calculated by the formula: \( P_{max} = \frac{\text{Load}}{\text{Length}} \. \) Using the given load and length, the maximum pressure is 3333.33 kPa. This high pressure can show how forcefully the lubricant needs to resist being pushed out of the gap, highlighting its importance in maintaining smooth movement and minimizing wear.
Film Thickness Gradient
The film thickness gradient is the change in the thickness of the lubricant film across the bearing surface. It affects the performance and the distribution of load across the surface.

In the exercise, the gradient is determined by the difference in thickness between the toe and heel of the slipper: \( \triangle h = h_{toe} - h_{heel} = 0.05 \text{ mm} = 0.00005 \text{ m} \. \) This small difference has significant effects on how the pressure and shear stress are distributed in the lubricant film. The average thickness also plays a crucial role in the calculations.
Shear Stress
Shear stress is the force per unit area exerted by the fluid as layers move against each other. It's a direct indicator of how much force is needed to maintain the motion.

In the bearing system, shear stress \( \tau = \frac{\text{Load}}{\text{Area}} \) is calculated as 3333.33 kPa. This value tells us how the lubricant's viscosity and the gradient in the film thickness resist the relative motion. The precise calculation helps prevent failure by ensuring the system has the right lubricant viscosity and film thickness to minimize wear and tear.

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Most popular questions from this chapter

Glycerin (relative density \(1.26\), dynamic viscosity \(0.9 \mathrm{~Pa} \cdot \mathrm{s})\) is pumped at \(20 \mathrm{~L} \cdot \mathrm{s}^{-1}\) through a straight, \(100 \mathrm{~mm}\) diameter pipe, \(45 \mathrm{~m}\) long, inclined at \(15^{\circ}\) to the horizontal. The gauge pressure at the lower, inlet, end of the pipe is \(590 \mathrm{kPa}\). Verify that the flow is laminar and, neglecting end effects, calculate the pressure at the outlet end of the pipe and the average shear stress at the wall.

Show that when laminar flow occurs with mean velocity \(u_{\mathrm{m}}\) between extensive stationary flat plates the mean kinetic energy divided by mass of the fluid is \(1.543 u_{\mathrm{m}}^{2} / 2\).

Two circular plane discs of radius \(R_{2}\) have their axes vertical and in line. They are separated by an oil film of uniform thickness. The oil is supplied continuously at a pressure \(p_{1}\) to a well of radius \(R_{1}\) placed centrally in the lower disc, from where it flows radially outwards and escapes to atmosphere. The depth of the well is large compared with the clearance between the discs. Show that the total force tending to lift the upper disc is given by $$ \frac{\pi p_{1}\left(R_{2}^{2}-R_{1}^{2}\right)}{2 \ln \left(R_{2} / R_{1}\right)} $$ and determine the clearance when the dynamic viscosity of the oil is \(0.008 \mathrm{~Pa} \cdot \mathrm{s}\), the flow \(0.85 \mathrm{~L} \cdot \mathrm{s}^{-1}, p_{1}=550 \mathrm{kPa}, R_{1}=\) \(12.5 \mathrm{~mm}\) and \(R_{2}=50 \mathrm{~mm}\). Assume laminar flow and neglect end effects.

A steel sphere, \(1.5 \mathrm{~mm}\) diameter and of mass \(13.7 \mathrm{mg}\), falls steadily in oil through a vertical distance of \(500 \mathrm{~mm}\) in \(56 \mathrm{~s}\). The oil has a density of \(950 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and is contained in a drum so large that any wall effects are negligible. What is the viscosity of the oil? Verify any assumptions made.

A piston \(113 \mathrm{~mm}\) diameter and \(150 \mathrm{~mm}\) long has a mass of \(9 \mathrm{~kg}\). It is placed in a vertical cylinder \(115 \mathrm{~mm}\) diameter containing oil of dynamic viscosity \(0.12 \mathrm{~Pa} \cdot \mathrm{s}\) and relative density \(0.9\) and it falls under its own weight. Assuming that piston and cylinder are concentric, calculate the time taken for the piston to fall steadily through \(75 \mathrm{~mm}\).
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