Calculate the power supplied to the pump shown in Fig. 11.2 if its efficiency is 76 percent. Methyl alcohol at 25°C is flowing at the rate of 54.0 m^3/h. The suction line is a standard DN 100 Schedule 40 steel pipe, 15 m long. The total length of DN 50 Schedule 40 steel pipe in the discharge line

Question

Calculate the power supplied to the pump shown in Fig. 11.2 if its efficiency is 76 percent. Methyl alcohol at 25°C is flowing at the rate of 54.0 $m^3/h$ . The suction line is a standard DN 100 Schedule 40 steel pipe, 15 m long. The total length of DN 50 Schedule 40 steel pipe in the discharge line is 200 m. Assume that the entrance from reservoir 1 is through a square-edged inlet and that the elbows are standard. The valve is a fully open globe valve.

To begin the solution, write the energy equation for the system.

Accepted Answer

Using the surfaces of the reservoirs as the reference points, you should have

[latex] \frac{p_1}{\gamma} + z_1 + \frac{v^2_1}{2g} +h_A -h_L = \frac{p_2}{\gamma} +z_2 + \frac{v^2_2}{2g}[/latex]

Because [latex] p_1 = p_2 = 0 [/latex] and [latex] v_1 [/latex] and [latex] v_2 [/latex] are approximately zero, the equation can be simplified to

[latex] z_1 + h_A - h_L = z_2 [/latex]

Because the objective of the problem is to calculate the power supplied to the pump, solve now for the total head on the pump, [latex] h_A [/latex] .

The total head is

[latex] h_A = z_2 - z_1 +h_L[/latex]

There are six components to the total energy loss. List them and write the formula for evaluating each one.

Your list should include the following items. The subscript s indicates the suction line and the subscript d indicates the discharge line:

[latex] h_1 = K\left(v^2_s/2g ight) [/latex] (entrance loss)

[latex] h_2 = f_s\left(L/D ight) \left(v^2_s/2g ight) [/latex] (friction loss in suction line)

[latex] h_3 = f_{dT}\left(L_e/D ight) \left(v^2_d/2g ight) [/latex] (valve)

[latex] h_4 = f_{dT}\left(L_e/D ight) \left(v^2_d/2g ight) [/latex] (two 90° elbows)

[latex] h_5 = f_{d}\left(L/D ight) \left(v^2_d/2g ight) [/latex] (friction loss in discharge line)

[latex] h_6 = 1.0 \left(v^2_d/2g ight) [/latex] (exit loss)

Because the velocity head in the suction or discharge line is required for each energy loss, calculate these values now.

You should have [latex] v^2_s/2g = 0.17m [/latex] and [latex] v^2_d/2g =2.44m [/latex], found as follows:

[latex] Q = \frac{54.0m^3}{hr} imes \frac{1 hr}{3600s} = 0.015 m^3/s[/latex]

[latex] v_s = \frac{Q}{A_s} = \frac{0.015m^3}{s} imes \frac{1}{8.213 imes 10^{-3}m^2} =1.83m/s[/latex]

[latex] \frac{v^2_s}{2g} = \frac{\left(1.83 ight)^2 }{2\left(9.81 ight) } m=0.17m [/latex]

[latex] v_d = \frac{Q}{A_d} = \frac{0.015m^3}{s} imes \frac{1}{2.168 imes 10^{-3} m^2} = 6.92m/s[/latex]

[latex]\frac{v^2_d}{2g}= \frac{\left(6.92 ight)^2 }{2\left(9.81 ight) } m = 2.44m[/latex]

To determine the friction losses in the suction line and the discharge line and the minor losses in the discharge line, we need the Reynolds number, relative roughness, and friction factor for each pipe, and the friction factor in the zone of complete turbulence for the discharge line that contains a valve and pipe fittings. Find these values now.

For methyl alcohol at 25°C, [latex] ho = 789kg/m^3 [/latex] and [latex] \eta = 5.60 imes 10^{-4} pa·s[/latex] Then, in the suction line, we have

[latex] N_R = \frac{vD ho}{\eta} = \frac{\left(1.83 ight) \left(0.1023 ight) \left(789 ight) }{5.60 imes 10^{-4}} =2.64 imes 10^5[/latex]

Because the flow is turbulent, the value of [latex] f_s [/latex] must be evaluated from the Moody diagram, Fig. 8.7. For steel pipe, [latex] \epsilon = 4.6 imes 10^{-5} m[/latex]. Write

[latex] D/\epsilon = 0.1023/\left(4.6 imes 10^{-5} ight) = 2224 [/latex]

[latex] N_R = 2.64 imes 10^5 [/latex]

Then [latex] f_s = 0.018 [/latex] .

In the discharge line, we have

[latex] N_R = \frac{vD ho}{\eta} = \frac{\left(6.92 ight) \left(0.0525 ight) \left(789 ight) }{5.60 imes 10^{-4}} =5.12 imes 10^5[/latex]

This flow is also turbulent. Evaluating the friction factor [latex] f_d [/latex] gives

[latex] D/\epsilon = 0.0525/\left(4.6 imes 10^{-5} ight) = 1141[/latex]

[latex] N_R = 5.12 imes 10^5 [/latex]

[latex] f_d = 0.020 [/latex]

We can find from Table 10.5 that [latex] f_{dT} = 0.019[/latex] for the DN 50 discharge pipe in the zone of complete turbulence.

Returning now to the energy loss calculations, evaluate [latex] h_1 [/latex] , the entrance loss, in N·m/N or m.

The result is [latex] h_1 = 0.09 m.[/latex] For a square-edged inlet, K = 0.5 and

[latex] h_1 = 0.5 \left(v^2_s/2g ight) = \left(0.5 ight) \left(0.17m ight) =0.09m[/latex]

Now calculate [latex] h_2[/latex] , the friction loss in the suction line.

The result is [latex] h_2 = 0.45 m [/latex].

[latex] h_2 = f_s imes \frac{L}{D} imes \frac{v^2_s}{2g} = f_s \left(\frac{15}{0.1023} ight) \left(0.17 ight) m[/latex]

[latex] h_2 = \left(0.018 ight) \left(\frac{15}{0.1023} ight) \left(0.17 ight) m = 0.45m [/latex]

Now calculate [latex] h_3[/latex] , the energy loss in the valve in the discharge line.

From the data in Chapter 10, the equivalent-length ratio[latex] L_e /D [/latex] for a fully open globe valve is 340.

The friction factor is [latex] f_{dT} = 0.019 [/latex]. Then , we have

[latex] h_3 = f_{dT} imes \frac{L_e}{D} imes \frac{v^2_d}{2g} = \left(0.019 ight) \left(340 ight) \left(2.44 ight) m = 15.76m [/latex]

Now calculate [latex] h_4[/latex], the energy loss in the two 90° elbows.

For standard 90° elbows, [latex] L_e /D = 30 [/latex]. The value of[latex] f_{dT}[/latex] is 0.019, the same as that used in the preceding panel. Then, we have

[latex] h_4 =2 f_{dT} imes \frac{L_e}{D} imes \frac{v^2_d}{2g} = \left(2 ight) \left(0.019 ight) \left(30 ight) \left(2.44 ight) m = 2.78m [/latex]

Now calculate [latex] h_5[/latex], the friction loss in the discharge line.

The discharge-line friction loss is

[latex] h_5 = f_{d} imes \frac{L}{D} imes \frac{v^2_d}{2g} = \left(0.020 ight) \left(\frac{200}{0.0525} ight) \left(2.44 ight) m = 185.9 m [/latex]

Now calculate [latex] h_6[/latex], the exit loss.

The exit loss is

[latex] h_6 = 1.0 \left(v^2_d/2g ight) = 2.44m[/latex]

This concludes the calculation of the individual energy losses. The total loss [latex] h_L [/latex] an now be determined.

[latex] h_L = h_1 +h_2+h_3+h_4+h_5+h_6[/latex]

[latex] h_L = \left(0.09+0.45+15.76+2.78+185.9+2.44 ight) m[/latex]

[latex] h_L = 207.4m[/latex]

From the energy equation the expression for the total head on the pump is

[latex] h_A = z_2 - z_1 +h_L [/latex]

Then, we have

[latex] h_A = 10 m + 207.4 m = 217.4 m[/latex]

Now calculate the power supplied to the pump,[latex] P_A[/latex].

[latex] Power = \frac{h_A \gamma Q}{e_M} = \frac{\left(217.4m ight)\left(7.74 imes 10^3 N/m^3 ight) \left(0.015 m^3/s ight) }{0.76} [/latex]

[latex] P_A = 33.2 imes 10^3 N·m/s = 33.2kW[/latex]

This concludes the programmed example problem.

TABLE 10.5 Friction factor in zone of complete turbulence for new, clean, commercial Schedule 40 steel pipe

U.S. (in)	Metric (mm)	Friction factor, [latex] f_T [/latex]
½	DN 15	0.026
¾	DN 20	0.024
1	DN 25	0.022
1¼	DN 32	0.021
1½	DN 40	0.02
2	DN 50	0.019
2½	DN 65	0.018
3, 3½	DN 80, DN 90	0.017
4	DN 100	0.016
5.6	DN 125, DN 150	0.015
8	DN 200	0.014
10–14	DN 250 to DN 350	0.013
16–22	DN 400 to DN 550	0.012
24–36	DN 600 to DN 900	0.011

Question 11.1: Calculate the power supplied to the pump shown in Fig. 11.2 ...

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