Mechanical energy added to a fluid - A pump

 An important component of the general mechanical energy equation

$\dfrac{p_1}{\gamma}+z_1+h_A-h_R-h_L+\dfrac{v_1^2}{2g}=\dfrac{p_2}{\gamma}+z_2+\dfrac{v_2^2}{2g}$        Eq. (01)

is $h_A$ since it represents the mechanical energy added to the fluid so that it can continue its trip through the pipe. At this point, two cases are visualized,

• the energy $h_A$ is one the system requires in order to the pipe system to do its work, and
• the energy $h_A$ is brought from the technical features of a real life pump, for example, so that the conditions of the pipe system are changed in turn.

Of course, in most technical problems both cases presented above are part of the solution in an iterative procedure.

What is the relationsip between $h_A$ and a pump

Since the pump takes electrical energy and transforms it into mechanical energy to be later transferred into the fluid, these  two parameters should be related. This is done through the effeciency of the motor,

$\mathbf{e_M}=\dfrac{Power\, output\, from \, the\, motor}{Power\, delivered\, by\, fluid}=\dfrac{P_O}{P_R}$    Eq. (02)

where $P_O$ is the mechanical power the motor so that the impeller may turn at certain speed with certain force and $P_R$ is the power the fluid actually receives. Since part of the power $P_O$ is wasted as heat or lost due to wearing of mechanical parts, $P_O>P_R$ and as a consequence $\mathbf{e_M}<1$. If $\mathbf{e_M}=1$ you would have an impossible thermodynamical machine. $P_O$ is a parameter measured and supplied, in the name plate, by pump manufacturers so that this data is easy to get.

Notice that Eq. (02) is the same if instead of a pump the case were that of a turbine (actioned by a mechanical energy of the fluid). We would be talking about $h_R$ instead of $h_A$ too.

For a centrifugal pump the power transferred into the fluid and the energy added $h_A$ are related as follows,

$P_R=h_A\gamma Q$        Eq. (03)

where $Q$ is the volumetric flow rate and $\gamma$ a property of the fluid. Also, from Eq. (03) it is obvious that $P_R$ would be very hard to measured. Therefore, it is more common to speak of the efficiency of a pump which we know is smaller than 1 but with present technological advances could be in the range of 0.8 - 0.9 for new equipments. Then, the usual case would be that $P_R$ is unknown, so that,

$P_R=\mathbf{e_M}P_O$        Eq. (04)

and consequently, Eq. (01) becomes,

$\mathbf{e_M}P_O=h_A\gamma Q$        Eq (05)

From Eq. (05), $h_A$ is isolated to be,

$h_A=\dfrac{\mathbf{e_M}P_O}{\gamma Q}$        Eq. (06)

In this way $h_A$ can only be estimated if the efficiency of the pump $\mathbf{e_M}$ and power of the motor of the pump $P_O$ are known.