Question 3.24: The pump in Fig. E3.24 delivers water (62.4 lbf/ft^3) at 1.5...

• System sketch: Figure E3.24 shows the proper selection for sections 1 and 2.

• Assumptions: Steady flow, negligible viscous work, large reservoir (V_1 ≈ 0).

• Approach: First find the velocity V_2 at the exit, then apply the steady flow energy equation.

• Solution steps: Use BG units, p_1 = 14.7(144) = 2117  lbf/ft^2  and  p_2 = 10(144) = 1440  lbf/ft^2.

Find V_2 from the known flow rate and the pipe diameter:

V_2 = \frac{Q}{A_2} = \frac{1.5  ft^3/s}{(\pi/4)(3/12  ft)^2} = 30.6 ft/s

The steady flow energy equation (3.75), with a pump (no turbine) plus z_1 ≈ 0 and V_1 ≈ 0, becomes as shown in Figure 3.24a or:

\left(\frac{p}{\rho g} + \frac{\alpha}{2g} V^2 + z\right)_{in} = \left(\frac{p}{\rho g} + \frac{\alpha}{2g} V^2 + z\right)_{out} + h_{turbine} – h_{pump} + h_{friction}                  (3.75)

• Comment: The pump must balance four different effects: the pressure change, the elevation change, the exit jet kinetic energy, and the friction losses.
• Final solution: For the given data, we can evaluate the required pump head:

h_p = \frac{(1440 – 2117)  lbf/ft^2}{62.4  lbf/ft^3} + 20 + (1.07 + 7.5) \frac{(30.6  ft/s)^2}{2(32.2  ft/s^2)} = -11 + 20 + 124 = 133 ft

With the pump head known, the delivered pump power is computed similar to the turbine in Example 3.23:

If the pump is 80 percent efficient, then we divide by the efficiency to find the input power required:

• Comment: The inclusion of the kinetic energy correction factor α in this case made a difference of about 1 percent in the result. The friction loss, not the exit jet, was the dominant parameter.