Exercise: Determining flow rate with shaft work

In this example, we are interested in determining the flow rate of the system when work shaft is present. We will solve one scenario in where friction is absent in the process.

Figure 1 showcases the setup of the system. In the water pumping system below, the tanks presented each have a diameter of 50' and the diameter of the pipes is 2". If z1 equals 40' and z2 equals 100', what is the flow rate of water if the shaft work is 2,850 lbm ft²/s³ ?

Figure 1. Diagram for Determining Flow Rate with Work Shaft

Solution

First we setup the mechanical energy balance of the system:

$\frac{P_2-P_1}{\rho }+\frac{1}{2}\left(\frac{v_2^2}{{\alpha }_2}-\frac{v_1^2}{{\alpha }_1}\right)+g\left(z_2-z_1\right)=\frac{{\dot{W}}_{shaft}}{\dot{m}}-\frac{{\dot{E}}_{viscous}}{\dot{m}}$

(1)

Then proceed to establish the parameters and assumptions according to the given information in the problem statement.

The system is isobaric as both tanks are open to atmospheric pressure, meaning that $P_2-P_1=0$.

We assume that the system has turbulent flow, ${\alpha }_i=1$.

To setup the mass balance. We assume steady state $m_{in}=m_{out}$, where $m_i={\rho }_i*V_i*A_i$, we obtain the following equation as the full mass balance ${\rho }_1*A_1*V_1={\rho }_2*A_2*V_2$.

In the system there is only one fluid present, which is water, meaning that the density (ρ) of the fluid coming in and out is constant.

The diameters of the tanks are equal to each other in this case, which indicates that the area of the tanks are the same, this leaves us with the following mass balance $V_2-V_1=0$.

With the given z values the total static head can be obtained:

$z_2-z_1=totalstatichead$

$z_2-z_1=100^{\prime }-40^{\prime }=60^{\prime }$

Friction is absent from the system so, ${\dot{E}}_{viscous}=0$.

Utilizing the obtained equations in conjuct with the mechanical energy balance equation (1) we obtain the following equation:

$\dot{m}*g*(z_2-z_1)={\dot{W}}_{shaft}$

Rearrange to solve for flow rate $\dot{m}={\dot{W}}_{shaft}/[g*(z_2-z_1)]$

The solution obtained is flow rate equal to 1.47 lbm/s