Water, ρ = 1.94 slugs/ft^3 and ν = 0.000011 ft^2/s, is pumped between two reservoirs at 0.2 ft^3/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig. E6.16. The roughness ratio is ε/d = 0.001. Compute the pump horsepower required.

Question

Water, ρ = 1.94 slugs/[latex]ft^3[/latex] and ν = 0.000011 [latex]ft^2[/latex]/s, is pumped between two reservoirs at 0.2 [latex]ft^3[/latex]/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig. E6.16. The roughness ratio is ε/d = 0.001. Compute the pump horsepower required.

Accepted Answer

Write the steady flow energy equation between sections 1 and 2, the two reservoir surfaces:

[latex]\frac{p_1}{ ho g} + \frac{V^2_1}{2g} + z_1 = \left(\frac{p_2}{ ho g} + \frac{V^2_2}{2g} + z_2 ight) + h_f + \sum{h_m} - h_p[/latex]

where [latex]h_p[/latex] is the head increase across the pump. But since [latex]p_1 = p_2[/latex] and [latex]V_1 = V_2 \approx 0[/latex], solve for the pump head:

[latex]h_p = z_2 - z_1 + h_f + \sum{h_m} = 120 ft - 20 ft + \frac{V^2}{2g}\left(\frac{fL}{d} + \sum{K} ight)[/latex] (1)

Now with the flow rate known, calculate

[latex]V = \frac{Q}{A} = \frac{0.2 ft^3/s}{\frac{1}{4} \pi (\frac{2}{12} ft)^2} = 9.17 ft/s[/latex]

Now list and sum the minor loss coefficients:

Loss	K
Sharp entrance (Fig. 6.21)	0.5
Open globe valve (2 in, Table 6.5)	6.9
12-in bend (Fig. 6.20)	0.25
Regular 90° elbow (Table 6.5)	0.95
Half-closed gate valve (from Fig. 6.18b)	2.7
Sharp exit (Fig. 6.21)	1.0
	Σ K = 12.3

Table 6.5	Nominal diameter, in
	Screwed				Flanged
	[latex]\frac{1}{2}[/latex]	1	2	4	1	2	4	8	20
Valves (fully open):
Globe	14	8.2	6.9	5.7	13	8.5	6.0	5.8	5.5
Gate	0.30	0.24	0.16	0.11	0.8	0.35	0.16	0.07	0.03
Swing check	5.1	2.9	2.1	2.0	2.0	2.0	2.0	2.0	2.0
Angle	9.0	4.7	2.0	1.0	4.5	2.4	2.0	2.0	2.0
Elbows:
45° regular	0.39	0.32	0.30	0.29
45° long radius					0.21	0.20	0.19	0.16	0.14
90° regular	2.0	1.5	0.95	0.64	0.50	0.39	0.30	0.26	0.21
90° long radius	1.0	0.72	0.41	0.23	0.40	0.30	0.19	0.15	0.10
180° regular	2.0	1.5	0.95	0.64	0.41	0.35	0.30	0.25	0.20
180° long radius					0.40	0.30	0.21	0.15	0.10
Tees:
Line flow	0.90	0.90	0.90	0.90	0.24	0.19	0.14	0.10	0.07
Branch flow	2.4	1.8	1.4	1.1	1.0	0.8	0.64	0.58	0.41

Calculate the Reynolds number and pipe friction factor:

[latex]Re_d = \frac{Vd}{ u} = \frac{9.17(\frac{2}{12})}{0.000011} = 139,000[/latex]

For ε/d = 0.001, from the Moody chart read f = 0.0216. Substitute into Eq. (1):

[latex]h_p = 100 ft + \frac{(9.17 ft/s)^2}{2(32.2 ft/s^2)} \left[\frac{0.0216(400)}{\frac{2}{12}} + 12.3 ight] = 100 ft + 84 ft = 184 ft[/latex] pump head

The pump must provide a power to the water of

[latex]P = ho gQh_p = [1.94(32.2) lbf/ft^3] (0.2 ft^3/s)(184 ft) = 2300 ft \cdot lbf/s[/latex]

The conversion factor is 1 hp = 550 ft · lbf/s. Therefore

P = [latex]\frac{2300}{550}[/latex] = 4.2 hp

Allowing for an efficiency of 70 to 80 percent, a pump is needed with an input of about 6 hp.

Question 6.16: Water, ρ = 1.94 slugs/ft^3 and ν = 0.000011 ft^2/s, is pumpe...

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