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

Question

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?

Accepted Answer

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,

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

[latex]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} }[/latex] (13.52)

where we found the following values at the pump inlet: p_in = 800 lb_f/ft², and [latex]\bar V_{in}[/latex] = 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 lb_f/ft³, and p_v= 0.2563 psia = 36.9 lb_f/ft². Applying Eq. 13.50, we have NPSHA = p_in/ρg + [latex]\bar V^2_{in}[/latex]/2g− p_v/ρg, with

[latex]\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},[/latex] [latex]\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}[/latex]

and

[latex]-\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}[/latex]

Thus the net positive suction head available in this case is

[latex]NPSHA=\frac{p_{in}}{\rho g}+\frac{\bar V^2_{in}}{2 g}-\frac{p_v}{\rho g}=[/latex] 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 H_new, 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 H_new, 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

[latex]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}[/latex]

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 H_new = 10 ft − 6.9 ft = 3.1 ft will allow the pump to operate properly.

Question 13.18: Find the NPSHA for the pump and inlet system described in Ex...

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