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# Part 7: Pumps and Turbines – The Bernoulli Equation

Essentially a pump adds energy to a system and a turbine takes it away. Therefore typically in the Bernoulli Equation the pump pressure (Pp) is added to the left-hand side of the equation and the turbine pressure (Pt) is added to the right. So for a system containing a pump and a turbine the Bernoulli equation would look something like this: $P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 + P_p = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g + P_t$

(16)

If you know the power (Wout) that is delivered by a pump, then it is possible to calculate the pressure it represents (Pp).

From Equations (6) and (7): $W_{out} = F.S/t$

From Equation (1): $v = S/t$

So we can write:
look something like this: $W_{out} = F.v$

(17)

Now from Equation (4): $P = F/A$

so: $F = P.A$

Combining this with Equation (17): $W_{out} = P.A.v$

From Equation (8): $Q = A.v$

So we can write: $W_{out} = P.Q$

Thus the energy the pump provides as a pressure (Pp) is given by: $P_p = W_{out}/Q$

(18)

Typically pumps have an efficiency (γ)which is the ratio of the power out (Wout ) to the power in (Win ). It represents the losses in the pump due to friction and electrical efficiency. It is usually expressed as a percentage and will always be less than 100%. It should be applied in the equation as a fraction. We can write therefore:
look something like this: $W_{out} = W_{in}.\gamma$

(19)

Combining Equations (18) and (19) gives:
look something like this: $P_p = W_{in}.\gamma /Q$

(20)

This can be substituted into the Bernoulli Equation (16) and allows the determination of the pump power requirement or alternatively the flow rate in a system for a given pump power.

Analysis of turbines will not be dealt with in detail here but is very similar to that of pumps (but in reverse).