Chapter 16 Worked examples.

 

Q1 Calculate the flow of oil of viscosity 0.08 kg/ms through a slit of width 50 mm and thickness 1 mm under a pressure difference of 2 bar. The length of the slit is 150mm. Treat this as a one-dimensional flow.

 

Text Box: Q2 The diagram shows a fitting in which oil of viscosity 0.08 kg/ms flows at a steady rate between two annular surfaces set 0.5 mm apart. If the pressure drop between inlet and outlet is 1.5 bar calculate the steady rate of flow. Suppose the flow to be laminar throughout.

 

 

 

 

Q3 For the fitting in Q2 show that ,if the oil discharges at atmospheric pressure, the pressure at radius  is given by  gauge

 

Q4  Calculate the force exerted by the oil on the lower plate in Q2.

 

Q5  The solution to Q4 is very tedious. Solve it using a mathematics package.

 

Text Box: Q6  The figure shows a cylindrical pot filled with oil. The pot contains a cylindrical steel block of diameter 100 mm and length 100 mm. The radial clearance between the pot and the block is 0.5 mm. Keeping in mind that there is a buoyant force on the block suppose that the block descends through the oil co-axially with the cylinder and calculate a value for the speed of  descent.

 

Take the density of the oil and of steel to be 7,860 kg/m3 and 850 kg/m3. Take the viscosity of the oil to be 0.08 kg/ms.

 

 

 

Q7 For the system shown in Q6 plot  graphs showing :-

 

(i) how the speed of descent varies with the clearance for values of clearance from 0.5 to 2 mm,

 

(ii) how the speed varies with an additional applied force acting downwards for a clearance of 0.075 mm. Use a range of force of 0 to 2000 N,

 

(iii) how the speed varies with diameter for a clearance of 0.5 mm and an added  force of 1000 N and,

 

(iv) how the speed varies with temperature over the range 10°C to 30°C if the diameter and the length are both 50 mm and a total force on the piston of 1,000 N. Take the relationship between temperature and kinematic viscosity over this range to be   where  is in °C and .

 

Q8 A thrust pad is 150 mm long and inclined so that the clearance at the leading edge is 0.01 mm and at the trailing edge is 0.003 mm. The collar that works with the pad moves at 1.5 m/s. If the viscosity of the lubricant that is fed continuously to the pad is 0.09 kg/ms :-

(i)   use a maths package to plot the graph of the pressure distribution in the lubricant and structure the programme so that the gaps at the leading and trailing edges can be changed at will,

(ii)  find the force exerted on the pad per metre width of the pad,

(iii) find the position of this force.

 

Q9 Once the programme in Q8 is available it is easy to find out the effect of changing the dimensions. The interesting changes are in the clearances. Try changing leading edge clearance whilst keeping the same trailing edge clearance. Then try vice versa.

 

 

Solutions

Q1 This computation serves only to get some idea of the numbers involved for these dimensions and for oil of this typical viscosity. The system has no practical use.

We have .           

Substituting 

This is 0.069 litres/s.

 

Text Box: Q2  It will be clear that the solution depends on the use of the expression  In this case the value of  varies with radius with  all constant. An integration is needed. The diagram shows an annular ring of radius  and width . Oil flows through the annular space associated with this ring and sustains a pressure drop of . Then we can write :- .

So the total pressure drop is  from which we can find a value for the flow.

. Then

 

Q3 We know the volume flow from Q2. We can use the expression  to find the pressure drop over the radial distance . Then, if  is the pressure at the inlet radius 0.025 m the pressure at radius  is given:-

                                                    from which :-

                                                  

In gauge pressure  and then :-

                          and 

                          .

 

Q4 The expression above can be further simplified to give :-

                     and

                     .

This pressure acts on an annular area  and exerts a force

                 .

From this we can find the force on the upper face by integration.

.

Now  and the first integral becomes :-

. The second integral becomes:-

 

The force on the annular area becomes

There is a force on the inner circular area given by .

                       The net force is 779+294=1073N

 

Q5   We have  and if we multiply by  we get the force on the elemental area. So the elemental force is given:-

              This can be integrated using Mathcad.

Add the force on the central area to give answer.[1]

 

Q6 The net downward force = volume of cylinder times the difference in the weight densities.

Using  

This block will fall very slowly indeed – it would take 5 minutes to fall though 100 mm.

 

Q7 This is an exercise in using   where  is the downward force that will be the sum of the net weight in oil and the added force.

 

Q8 This is an application of   for converging flat plates.

We need a value for .  Clearly, from the exaggerated diagram, Text Box: .

Text Box: By proportion  

(iv)
 
The programme and graph are given below. You have to keep a track of the units but I have substituted in an


[1] In my view there is no point in engineers trying to learn how to integrate. We have enough to learn. Let the package do it.