Energy exchange within a flowing fluid

I want to fit another use of the energy equation in at this point. In my experience people find it difficult to explain how some fundamental devices actually work. I recall someone asking a collection of graduates in engineering how the force is generated on a rocket in space. Everyone knew about momentum and could calculate the force but no one knew how the force was actually applied to the casing of the rocket. It is not an unreasonable question because the casings of booster rockets are required to have a minimum weight yet to be strong enough to withstand whatever forces are applied to them when they are in use. We can make a start at an explanation with the following example.

 

Figure 5-14 is familiar to every student of applied mathematics. It shows a tank fitted with a nozzle on one side and mounted on wheels so that the whole tank can move across a horizontal plane. The tank is filled with water and a jet of water flows from the nozzle. Given the rate of flow  from the nozzle and the area of cross-section of the jet , the student is asked to calculate the force exerted on the tank and perhaps, given the weight and size of the tank and the depth of water, the acceleration of the tank. It is primarily an exercise in the use of force equals rate of change of momentum.

 

It is easy to calculate the force, it equals  where  is the instantaneous mass flow, which is equal to , and  is the velocity of the jet where the student is expected to ignore friction and put equal to . Then, as  equals  it also equals  and the force equals  or

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However the student is seldom asked to explain how the force is actually exerted on the tank.

 

One might start by noting that if there were to be no hole in the side of the tank, the tank would be in equilibrium because, for each small force exerted on a given small area of the inside wall of the tank, there would be an equal and opposite force on an equal area on the other side. The presence of the hole destroys this balance. Now as the jet does not expand sideways or contract as it emerges from the nozzle it follows that the jet emerges at the same pressure as its surroundings, that is at atmospheric pressure. It might then be supposed that the unbalanced force is simply caused by the difference between the atmospheric pressure acting on the area  of the jet and the pressure  acting on an equal area on the opposite side of the tank. This would equal . But this force is only a half of the force calculated from momentum above. This must mean that the force has another component that we have yet to consider. Somewhere another area is subject to pressure that is lower than that on the corresponding area on the other side of the tank. This can only be the internal surface of the nozzle and the region surrounding the nozzle where the water is accelerating to form the jet. The conclusions we can draw are that the pressure of the water in the immediate vicinity of the nozzle falls progressively from its static value to that of the atmosphere and that there is an associated exchange of pressure energy for kinetic energy. Figure 4-14a shows more or less hemispherical surfaces of uniform pressure around the nozzle each surface being at a lower pressure as the water approaches the nozzle. In 4-14b the junction of the surfaces and the wall of the tank are shown and successive circles are at lower pressures.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We can now see that in the tank there has been an exchange of potential energy for pressure energy as the liquid descends through the tank and then, in the relatively small region near the nozzle, an exchange of potential energy and pressure energy for kinetic energy.

 

 and the flow through the nozzle =  where  is the diameter of the jet.

In figure 5-15a I have shown imaginary hemispheres centred on some point just inside the nozzle and, as I did in figure 3-1, supposed the flow to be radial. Using the fact that the surface area of a hemisphere is  the velocity through a hemisphere of radius  is given by :-

                                                                     

The sum of the potential energy, the pressure energy and the kinetic energy is the same for every point in the flow. As the kinetic energy increases as the water approaches the nozzle the sum of the potential energy and the pressure energy must fall. As the potential energy of the water is the same at any given level the increase in kinetic energy near the nozzle will reduce the pressure compared with an identical point on the other side of the tank. It follows that if we find the drop in pressure at radius  and then the force on an elemental ring of radial width  as shown in figure 5-15b we can integrate to find the reduction in force on the nozzle side of the tank.

 

We have . It then follows that the force on the elemental area is given by :-

                                             which reduces to:-

                                                                  

We now have to decide on some limit for the integral and the lower limit must be  the other limit is not so easy to decide so I will look at the outcome of the integral. It is :-

                                                                   

The lower limit will be  and if the upper limit is large the term becomes vanishingly small.

 

The result is that the reduction in force on the side of the tank round the nozzle is  and we have the other half of the force.

 

The booster rocket works this way but there we have a burning gas flowing through a nozzle that first converges and then diverges and is much more complicated. There are two forces, one on the top end of the casing and one on the exit “cone” of the nozzle

 

You may think that this is just a fuss about nothing but, if you do, explain the children’s toy rocket. In this toy a soft drinks bottle is half filled with water and a nozzle fitted to replace the cap. Then air is compressed into the top half using a cycle pump and the toy launched upwards. The air drives out the water and the bottle rises to a respectable height. The child gets wet and all is well. If the air in the bottle is at 3 bar try deciding on the best ratio for the volume of water to the volume of air.