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Chapter 13

Q. 13.18

Find the NPSHA for the pump and inlet system described in Example 13.9.This system is shown in Figure 13.44. If the pump manufacturer specifies a value of NPHSR = 20 ft, will the pump operate satisfactorily? If not, can you suggest a design modification that does not change the design flowrate or pipe dimensions?

13.44

Step-by-Step

Verified Solution

We are asked to evaluate a specified flow system to determine whether the suggested pump is satisfactory. If our calculations show that the present design is not satisfactory, we are to propose an improved design. Figure 13.44 serves as the appropriate sketch for this flow system. We will use Eq. 13.50 to evaluate NPSHA rather than Eq. 13.52 because the major and minor losses were evaluated in Example13.9,

NPSHA=\frac{p_{in}}{\rho g}+\frac{\bar V^2_{in}}{2 g}-\frac{p_v}{\rho g}               (13.50)

 

NPSHA=\left(\frac{p_{1}}{\rho g}+\frac{\bar V^2_{1}}{2 g}\right) -\frac{p_v}{\rho g}-H-\sum{\frac{h_L}{g}}-\sum{\frac{h_M}{g} }                (13.52)

where we found the following values at the pump inlet: pin = 800 lbf/ft2, and \bar V_{in} = 7.43 ft/s. For water at 60°F, Appendix A gives ρ = 1.938 slugs/ft3, so ρg = (1.938 slugs/ft3)(32.2 ft/s2) = 62.4 lbf/ft3, and pv = 0.2563 psia = 36.9 lbf/ft2. Applying Eq. 13.50, we have NPSHA = pin/ρg + \bar V^2_{in}/2g− pv/ρg, with

\frac{p_{in}}{\rho g}=\frac{800\ lb_{\mathrm{f}}/\mathrm{ft}^2 }{62.4\ lb_{\mathrm{f}}/\mathrm{ft}^3 }=12.8\ \mathrm{ft},        \frac{\bar V^2_{in}}{2 g}=\frac{(7.43\ \mathrm{ft}/s)^2 }{2(32.2\ \mathrm{ft}/s^2) }=0.86\ \mathrm{ft}

and

-\frac{p_v}{\rho g}=\frac{36.9\ lb_{\mathrm{f}}/\mathrm{ft}^2 }{62.4\ lb_{\mathrm{f}}/\mathrm{ft}^3 } =-0.59\ \mathrm{ft}

Thus the net positive suction head available in this case is

NPSHA=\frac{p_{in}}{\rho g}+\frac{\bar V^2_{in}}{2 g}-\frac{p_v}{\rho g}= 12.8 ft + 0.86 ft − 0.59 ft = 13.1 ft

Since we are told NPHSR = 20 ft, the pump will not operate satisfactorily. The design must be changed to increase NPSHA to at least 20 ft.

In many cases, as here, the desired flowrate is fixed, but changes can be made in the elevation of the pump, and in other design parameters that effect the major and minor losses. If we are to keep both the flowrate and the pipe dimensions the same, we must decrease the elevation of the pump above the free surface of the reservoir to a new value Hnew, as shown in Figure 13.44, while leaving the piping and inlet configuration unchanged. This will increase the pressure at the pump inlet and raise the NPSHA. To find the new conditions at the pump inlet with a decrease in elevation to a value Hnew, we need to repeat the head loss analysis of Example 13.9. However, in this case we do not actually need the conditions themselves but can make use of Eq. 13.52 to write

NPSHA=\frac{p_{A}}{\rho g}-H-\frac{p_v}{\rho g}-f\frac{L}{D}\frac{\bar V^2_{in}}{2 g}-K_{in}\frac{\bar V^2_{in}}{2 g}

This result shows that with the frictional losses the same, the available suction head increases in direct proportion to the decrease in H. Thus to increase NPSHA from 13.1 ft to 20 ft, we must decrease H by 6.9 ft. Thus we conclude that redesigning the system with Hnew = 10 ft − 6.9 ft = 3.1 ft will allow the pump to operate properly.