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The application of Appropriate Technology

Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10.

A Cylinder of Fluid

Figure 10

This body contains kinetic energy (energy of movement). If we imagine bringing this body to rest then this kinetic energy will be turned into another form.

What is the force that is required to bring this body to rest ?

From Equation (3) (Where a is the deceleration of the body as it is brought to rest):

F = M.a

And from Equation (2):

a=\left(v_2-v_1\right )/t

(In this case the final velocity (v2) of the fluid is zero, so the change in velocity is –v, negative, therefore a deceleration) So we can write (The body is brought to rest in time t):

Now from Equation (6):

E = F.S

So the kinetic energy E k is given by combining the last two equations :

E_k = M.v.S/t

However, the velocity is changing from v to 0 over the distance S, so the average value of velocity must be taken to calculate the overall energy. This is given by:

v_{av} = (v-0)/2 = v/2

So the equation is rewritten as:

Now from Equation (1):

v = S/t

So we can rewrite the energy equation as :

E_k = \frac{1}{2}.M.v^2

Now the mass of the fluid is given by Equation (5):

M = \rho.V

In this case, the Volume of the cylinder is given by:

V = A.S

So the Mass can be expressed as:

M = \rho.A.S

This is substituted into the kinetic energy equation to give:

E _k = \frac{1}{2}.\rho.A.S.v^2


In the Bernoulli equation, the energies are represented as pressures. From Equation (4) we get:

P = F/A

And from Equation (6):

E = F.S

Combining these two equations we get:

P = E/(A.S)

So if we divide the kinetic energy Equation (23) by (A.S) we get it in terms of pressure:

P=\frac{1}{2}.\rho .A.S.v^2/ (A.S)

The A.S terms cancel out to give the kinetic energy in terms of a pressure (Pk):



This equation represents the kinetic energy of a fluid in terms of a pressure.