# Solutions to sheet 5

1  The new form of the equation is a power equation. It differs from the energy equation by having all the terms multiplied by the mass flow. This permits the inclusion of devices like pumps and turbines to the equation. The diagram shows the elements of such a system.

If the power equation is applied to sections 1 and 2 of this system we can write :-

+ the power  supplied between 1 and 2

=  + the power extracted between 1 and 2 + the power lost to                                                        internal friction between 1 and 2.

Clearly the new terms are the power input, the power output and the loss expressed not in energy per unit weight but in power lost to internal friction. The power input would come from a pump or compressor and the output would be through a motor or a turbine. The loss term will be the rate at which mechanical energy is absorbed into the flowing fluid as a result of internal friction arising from the flow in ducts and from the loss of un-recovered kinetic energy.

2 We have the power equation.

+ the power loss between 1 and 2.

.

Therefore  power loss 1 to 2.

power loss 1 to 2.

power loss 1 to 2.

Now the power loss is included in the difference between the useful power input to the oil, that is 4,875 watts, and the input power to the shaft.

So the input to the shaft is   watts.

3 Applying the energy equation to the turbine :-

+ the power extracted from the water between A and B + the loss of power to friction between A and B.

.

Therefore :-

power extracted + the loss =

= .

Of the power extracted 85% goes to the shaft and the rest to the loss into the fluid by friction. Therefore :-

output power =  kw.

4(i) Applying the energy equation to the pipe at the pressure gauge and the delivery point, that is, to A and B  we get :-

+ power to loss to friction.

Power to pipe loss = .

Therefore cancelling common term  gives :-

and

Therefore  and

4(ii)

Now the system can be considered without reference to the pressure gauge.

Applying the energy equation to D (a point in the free surface in the reservoir) and to B gives:-

power loss in pipes where P is the  power input from the pump to the water. Then:-

5    This question really contains a Venturi-meter question as well.

If  is the volume flow,

For a Venturi-meter :-  and

.

Therefore            and

That is :-

.

Applying the energy equation to the section at the gauge 1 and to the exit plane of the 70 mm pipe :-

+power to losses.

Therefore

.

Power to losses =

=

Cancelling

6 (i) .

(ii) Using the energy equation applied to the section just before the nozzle 1 and to the jet 2 we get :-

.

Therefore                        (Compare with 40m)

and  bar

(iii)  The friction loss in the pipe must be .

1 The diagram shows the tank and the gate valve. It also shows the instant at which the level in the tank is  above the outlet flange of the gate valve. I have shown point 1 in the free surface and 2 in the water leaving the valve.  The energy equation can be applied to 1 and to 2 to give :-

+ loss of energy in the vicinity of the valve.

The worst case for this is for the loss at entry to be  and at exit .

and then :-

.

The instantaneous value of the volume flow is  m3/s if  in metres.

In time  the outflow is  m3 and this will produce a drop in level  :-

.  Then   and Mathcad gives a time of 457 seconds

There is a question mark over the choice of loss at the valve. It is worth re-working  using other decisions to see just how much difference they make.

2 Read the coordinates from the graph to give this table.

## MPSetEqnAttrs('eq0073','',3,[[5,9,0,-1,-1],[7,11,0,-1,-1],[9,14,0,-1,-1],[8,13,1,-1,-1],[12,17,0,-1,-1],[13,21,1,-2,-2],[24,35,1,-3,-3]]); MPEquation(); h

0.5

0.00035

-0.301

-3.45

1

0.0005

0

-3.3

1.5

0.00061

0.176

-3.22

2

0.0007

0.301

-3.15

2.5

0.00079

0.4

-3.1

3

0.00087

0.477

-3.06

From graph :-

Slope  and k=

.

In time  the outflow will be  and the drop in level  =  Rearranging and integrating using Mathcad gives 1,800 seconds.

3 (i)   The volume of a cone is  where  is the base area and  is the height.  We are given that the radius is equal to the height then .

Therefore

(ii) We have the rate of flow . The outflow in a short time  will be given by:-

m3

This will produce a fall in level of  where A is the area of the free surface at depth . But  so  .

This can be rearranged for integration. .

(When this question was written programs were written to find the data to plot the graph. This was an opportunity to practice. It is not necessary now.) Mathcad gives:-