Worked examples on vortices etc.

 

Q1 An open-topped, cylindrical tank is mounted with its axis vertical and arrangements are made for it to be rotated at constant speed. The tank has a diameter of 400 mm and a height of 600 mm. It is filled to a depth of 500 mm with water. Calculate the speed at which the water just starts to spill over the top edge.

 

Q2 Find the speed at which the bottom of the tank in Q1 is exposed.

 

Q3 Suppose that the tank in Q1 was fitted with a lid with a small hole in the middle of it and the tank completely filled with oil of density 900 kg/m³. Calculate the forces that would be exerted on the top and bottom of the tank when:-

a) when the tank is stationary,

b) when the tank and the oil rotate at 100 rpm.

 

Q4 Calculate the total amount of mechanical energy imparted to the oil in Q3 during the process of raising its rotational speed to 100 rpm.

 

Q5 Water flows radially inwards between two plane, horizontal, circular plates. At a radius of 300 mm the pressure is 0.5 bar absolute and the radial speed is 3 m/s. Calculate the pressure at a radius of 100 mm.

 

Q6 Suppose that, in the system in Q5, the water formed a free vortex as well as the inward radial flow and the volume flow remained unchanged. If the radial and tangential velocities are equal at every point in the flow calculate the new velocities at 300 mm and 100 mm and, if the pressure at 300 mm is now1 bar absolute, calculate the pressure at the radius of 100 mm.

 

Q7 An open-topped tank of 700 mm diameter is set up with its axis vertical and filled with water to a depth of 800 mm. A stirrer made up of two crossed blades as shown in the diagram is set up coaxially with the tank and rotated at 60 rpm. This rotation of the stirrer will impart angular momentum to the water and ultimately the water will rotate steadily with the water inside the containing cylinder of the stirrer approximating to a forced vortex and the rest of the water rotating as a free vortex. The stirrer has a diameter of 75 mm.

 

The water that rotates as a free vortex will have a depression that may be too small to detect by the unaided eye. Ignoring the meniscus at the wall estimate the difference in level between the wall and the free surface at the radius of the stirrer. Could you draw this profile to scale?

 

 

Q8 We have seen that a fee spiral vortex can be formed in cylindrical tank by giving the water an initial rotation and then allowing the water to discharge through a central hole in the bottom of the tank.

 

Suppose that such a system was rotating with a fully developed, air entraining, vortex and the outflow was stopped. The central core cannot now be maintained and the flow pattern will change to produce a so-called compound vortex. Suppose that this is made up of a central forced vortex surrounded by a free vortex. If the radius at which the two vortices meet is  show that the depth of the central depression below the outer level of the surface is given by  where  is the angular velocity at radius .

 

 

 

 

Solutions

Q1  As no water is spilt the volume of water in the can is unchanged and therefore the volume of air is unchanged. All that has happened to the air is that its shape has changed from a cylinder to a paraboloid of revolution. We shall need to be able to relate the volume of the paraboloid to its height and diameter.

 

I have marked points 1 at the lowest point of the paraboloid and on the axis, 2 at the upper edge of the tank and 3 on the inside surface of the tank at the same level as 1. A plane surface is shown dotted and a free vortex will exist in this surface.

 

For that vortex .  In this case  and . So :-

                                                

.

We now have an expression for the height of the paraboloid of air and its diameter. We need an expression for the volume in order to find  .

 

Volume .

             .

We have . Therefore:-

Volume  or the volume equals one-half of the containing cylinder. This means that  = 0.2 m. Hence:-  from which  rpm.

 

Q2 In this case . Also  from which . Speed of revolution =  

 

Q3a Force on top is zero.

 Force on bottom

3b When the tank and the oil are rotating at 100 rpm there will a forced vortex distribution of pressure over both the inside of the lid and over the bottom and the pressure at the middle of the lid will be zero. If we find the force on the lid the force on the bottom will greater by 666 N.

 

This requires an integration of the force on an elemental ring of area forming part of the lid. If the elemental ring has a radius  its area is . For a forced vortex with zero pressure at its centre the pressure head at radius . Therefore:-

Pressure at radius .  radians/sec.

Pressure at radius .

From this the force on the elemental ring is  and the total force on the inside of the lid .

The force on the bottom is 790 N

 

Q4 During the starting phase an unknown amount of energy is imparted to the internal energy of the oil but, once start up is complete, the oil contains pressure energy and kinetic energy that it has acquired. Two integrations will be required to calculate these two amounts.

 

If the volume of oil is taken to be made up of many cylindrical shells the weight of a shell of radius  and thickness   where  is the length of the shell. The weight =  which is  Newton.

 

 

Kinetic energy/unit weight at radius  J/N.

But the pressure energy per unit weight, , also equals  =  J/N and the total energy imparted to the shell is  = .

 

So the total energy stored is

 

Note The question implies that the oil is made to rotate by the action of viscosity that starts at the solid surfaces of the tank and gradually reaches a steady state. Energy will be lost in this process. However the oil could be made to rotate by the fitting of radial blades inside the tank and this would make the start-up much shorter and waste less energy. But suppose that the tank were to be open topped as in Q1 and oil could be fed into the bottom. Then the oil could be continuously lifted by this tank and its blades and it would be the basis of a crude centrifugal pump. Its efficiency would be less than 50%.

 

Q5 This is really a convergence to be dealt with using the energy equation and continuity of flow. The total energy at 300 mm radius is given by :-

                    

 

At 100 mm the velocity will be  and the kinetic energy per unit weight will be . This is converging flow and if losses are ignored, the total head is unchanged and the pressure head is . The pressure will be .

 

 

At 300 mm the radial velocity is still 3 m/s and, as the tangential velocity equals the radial velocity, the absolute velocuty will be . So the total head at the 300 mm radius is

 

At 100 mm radius the radial velocity of the water will again be 9 m/s so the absolute velocity will be . Then the pressure head will be . This corresponds to 0.28 bar.

 

Q7 This complicated question is teling us the tangential velocity of the water at a radius of 37.5 mm. The tangential velocity is . (This is very small because 60 rpm is very low. Wind powered generators turn at about 12 rpm.)

 

As this is a free vortex the product of the radius and the tangential speed is constant. Therefore the tangential speed of the water near to the wall is .

The difference in level will be  or 2.7 mm. This is the difference in level over a distance of 312.5 mm!

 

Q8 If the diameter of the tank is large compared with the radius  the kinetic energy head at  will be given fairly accurately by . This means that the surface level at  will be lower that the level at the wall by . The further drop in level to the centre of the forced vortex will also be given by  so the total depression will be .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 The case of the flow between paralel circular plates is clearly controlled flow that an engineer might put to practical use. Most people will only ever see the uncontrolled swirl that occurs in eddies and in