You are to choose the pipe diameter and pump for the system shown in Figure 13.39. The ﬂowrate is 1 ft³/s and the pipe is Schedule 40 commercial steel. The total length including an equivalent length allowance for minor losses is 1800 ft. Choose the pump based on the pump curves shown in

Question

You are to choose the pipe diameter and pump for the system shown in Figure 13.39. The ﬂowrate is 1 ft³/s and the pipe is Schedule 40 commercial steel. The total length including an equivalent length allowance for minor losses is 1800 ft. Choose the pump based on the pump curves shown in Figure 13.40 and assume that the water is at 50°F.

Accepted Answer

We are asked to use a given set of pump curves to select an appropriate pump for a well deﬁned single path system. Figure 13.39 serves as an appropriate sketch for this problem. There are many approaches one might take to optimize even a simple piping design problem such as this one. We will ﬁnd an equation for the system curve as a function of the pipe diameter. Then we will use the information provided in the manufacturer’s data to choose the most efficient pump by sketching the system curves over the pump curves. Therefore, in this example we will be minimizing operating costs only.

We begin by applying Eq. 13.38 from the reservoir surface to the free exit discharge:

[latex]\left(\frac{p_1}{\rho}+\frac{1}{2} \alpha _1\bar V^2_1+gz_1\right)-\left(\frac{p_2}{\rho}+\frac{1}{2} \alpha _2\bar V^2_2+gz_2\right)=h_T-H_{pump}+H_{turbine}[/latex]

Using gage pressures, and noting that p₁ = 0 psig, [latex]\bar V^2_1[/latex] ≈ 0, z₁ = 0, and p₂ = 0 psig, we have

[latex]H_{pump}=\frac{1}{2}\bar {V_2}^2+gz_2+h_T[/latex] (A)

where we have assumed α = 1 for turbulent ﬂow. We see that the pump must generate an amount of head [latex]H_{pump}=\frac{1}{2}\bar V^2_2+gz_2+h_T=H_{system}[/latex]. The system head consists of the increase of kinetic energy of the ﬂuid, the elevation potential and the total frictional head loss. The latter is the sum of the major and minor head losses and given by

[latex]h_T=f\frac{L}{D}\frac{\bar V^2_2}{2}+f\frac{L_E}{D}\frac{\bar V^2_2}{2}=f\left(\frac{L+L_E}{D}\right) \frac{\bar V^2_2}{2}[/latex]

We now want to superimpose the system head curve for each diameter choice. To do so we must transform (A )to units of head in feet and ﬂowrate in gallons per minute, with the understanding that H_system = H_pump. By using Q = [latex]\bar V_2[/latex]A, h_T = f[(L + L_E)/D] × ( [latex]\bar V^2_2[/latex] /2), and A = πD²/4, we obtain

[latex]H_{system}=\frac{Q^2}{2A^2} +gz_2+f\left(\frac{L+L_E}{D}\right)\frac{Q^2}{2A^2}=\left[1+f\left(\frac{L+L_E}{D}\right)\right]\frac{8Q^2}{\pi^2 D^4}+gz_2 [/latex]

[latex]\frac{H_{system}}{g} =\left[1+f\left(\frac{L+L_E}{D}\right)\right]\frac{8Q^2}{g\pi^2 D^4}+z_2 [/latex] (B)

Inserting numerical values into (B) yields

[latex]\frac{H_{system}}{g} =\left(1+f\frac{1800\ \mathrm{ft}}{D} \right)\left(\frac{0.0252(s^2/\mathrm{ft})Q^2}{D^4} \right)\left(\frac{2.228×10^{-3} \mathrm{ft}^3/s }{gal/min } \right)^2+10\ \mathrm{ft}[/latex]

[latex]\frac{H_{system}}{g} =\left(1+f\frac{1800\ \mathrm{ft}}{D} \right)\left(\frac{1.25×10^{−7} [\mathrm{ft}^5/(gal/min)^2]Q^2}{D^4} \right)^2+10\ \mathrm{ft}[/latex] (C)

TABLE 13.7 System Head (ft) as a Function of Pipe Size and Volume Flowrate

		Volume Flowrate, Q (gal/min)
Pipe Diameter:
Nominal D (Actual D)	0	100	200	300	400	450	500	600	700	600
4 in. (0.3355 ft)	10	18.9	45.7	90.4	152.9	*	*	*	*	*
5 in. (0.4206 ft)	10	12.8	21.0	34.8	54.1	65.8	78.9	109.2	*	*
6 in. (0.5054 ft)	10	11.1	13.2	19.6	27.0	31.5	36.5	48.2	62.0	77.9

*This combination of ﬂowrate and pipe diameter is not possible.

where D is in feet and Q is gallons per minute. The friction factor can be found by using the Moody chart or the Chen equation. You will ﬁnd that this system is fully turbulent such that f is independent of ﬂowrate for each diameter. The friction factors are 0.0163, 0.0155, and 0.0139 for the 4, 5, and 6 in. pipes, respectively. Table 13.7 gives system head in feet calculated by using (C) for three choices of pipe diameter and at several ﬂowrates. The entries for the design ﬂow rate of 1 ft³/s = 450 gal/min are given in boldface. These data are superimposed on the pump curves of Figure 13.40.

First of all we can see that this particular pump could not deliver the design ﬂowrate through a 4 in. pipe. A 6 in. pipe could be used, but the pump would be oversized. Thus the obvious choice for the design constraints of this problem is 5 in. diameter pipe. The design point on the 5 in. system curve is not directly on a pump curve. The system curve does intersect the 12.50 in. diameter impeller pump, providing 460 gal/min at 68 ft of head. At this operating point the pump would have an efficiency of 78% and would require a 10 hp motor. To obtain 450 gal/min, an additional 2.2 ft of head loss could be achieved through the use of a control valve. However, we also expect the head loss to increase in the system over time. Another method for moving the operating point to the design point is to change the pump curve by adjusting its rpm. In the case study of Section 3.3.3 we found that the dimensionless group Q/ωD³, the ﬂow coefficient, applies to pumps and fans. The pump curves in Figure 13.40 are for ω = 1160 rpm. By using the ﬂow coefficient we ﬁnd that the 12.50 in. diameter impeller operating at 1135 rpm would provide 450 gal/min as desired. However, the head coefficient, H/ω²D², must also be checked. Taking the head coefficient we ﬁnd that at 1135 rpm the pump would supply 65 ft of head, which is not enough to achieve the desired ﬂowrate in the system. Thus it should be understood that it is very difficult to make the operating point and design point coincide without control valves.

Before leaving this problem, consider the pressure of a ﬂuid particle moving through this system as indicated in Figure 13.41. At the surface of the reservoir, station 1, the pressure is atmospheric at 0 psig. The pressure increases due to hydrostatic head at the pipe inlet, station A, where it then is drawn into the pipe by the pump. The pressure decreases owing to frictional head losses in the pipe and reduced static head from A to B. From station B to the pump inlet, station C, the reduction in pressure is due to friction alone. Across the pump there is the pressure rise imparted by the pump. From the pump exit, station D, to the pipe exit, station 2, the pressure decreases to atmospheric pressure due to frictional losses. However, the ﬂuid at station 2 has greater mechanical energy than at station 1 because it is at a greater elevation and has greater kinetic energy.

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