Determine the available NPSH for the system shown in Fig. 13.38(a). The fluid reservoir is a closed tank with a pressure of -20 kPa above water at 70°C. The atmospheric pressure is 100.5 kPa. The water level in the tank is 2.5 m above the pump inlet. The pipe is a DN 40 Schedule 40 steel pipe wi

Question

Determine the available NPSH for the system shown in Fig. 13.38(a). The fluid reservoir is a closed tank with a pressure of -20 kPa above water at 70°C. The atmospheric pressure is 100.5 kPa. The water level in the tank is 2.5 m above the pump inlet. The pipe is a DN 40 Schedule 40 steel pipe with a total length of 12.0 m. The elbow is standard and the valve is a fully open globe valve. The flow rate is 95 L/min. Then calculate the maximum allowable [latex] NPSH_R [/latex] for the pump in this system.

Accepted Answer

Use Eq. (13–13). First, find [latex] h_{sp}[/latex]:

[latex] NPSH_A = h_{sp}\pm h_s - h_f - h_{vp}[/latex] (13–13)

Absolute pressure = Atmospheric pressure + Tank gage pressure

[latex] p_{abs} = 100.5 kPa - 20 kPa = 80.5 kPa [/latex]

But we know that

[latex] h_{sp} = p_{abs}/\gamma [/latex]

[latex] = \frac{80.5 imes 10^3 N/m^2}{9.59 imes 10^3 N/m^3} = 8.39m [/latex]

Now, based on the elevation of the tank, we have

[latex] h_s = +2.5 m [/latex]

To find the friction loss [latex] h_f [/latex], we must find the velocity, Reynolds number, and friction factor:

[latex] v = \frac {Q}{A} = \frac {95L/min}{1.314 imes 10^{-3}m^2} imes \frac {1.0m^3/s}{60000L/min} = 1.21m/s [/latex]

[latex] N_R = \frac {vD}{ u} = \frac {\left(1.21 ight)\left(0.0409 ight) }{4.11 imes 10^{-7}} = 1.20 imes 10^5[/latex] (turbulent)

[latex] \frac{D}{\epsilon} = \frac{0.0409m}{4.6 imes 10^{-5}m} =889 [/latex]

Thus, from Fig. 8.7, f = 0.0225. From Table 10.5, [latex] f_T = 0.020.[/latex] Now we have

[latex] h_f = f\left(L/D ight) \left(v^2/2g ight) +2f_T\left(30 ight) \left(v^2/2g ight) +f_T\left(340 ight) \left(v^2/2g ight) +1.0\left(v^2/2g ight) [/latex]

(pipe) (elbows) (valve) (entrance)

The velocity head is

[latex] \frac{v^2}{2g} = \frac{\left(1.21 m/s ight)^2 }{2\left(9.81 m/s^2 ight) } =0.0746m[/latex]

Then, the friction loss is

[latex] h_f = \left(0.0225 ight) \left(12/0.0409 ight) \left(0.0746 ight) +\left(0.020 ight) \left(60 ight) \left(0.0746 ight) +\left(0.020 ight) \left(340 ight) \left(0.0746 ight) +0.0746[/latex]

= (0.0746 m) [(0.0225) (12/0.0409) + (0.020) (60) + (0.020) (340) + 1.0]

= 1.16m

Finally, from Table 13.2 for the vapor pressure head of water, we get

[latex] h_{vp} = 3.25 [/latex]m at 70°C

Combining these terms gives

[latex] NPSH_A = 8.39 m + 2.5 m - 1.16 m - 3.25 m = 6.48 m[/latex]

We can compute the maximum allowable [latex] NPSH_R [/latex] for the pump from Eq. (13–12),

[latex] NPSH_A\gt 1.10 NPSH_R[/latex]

Rearranging, we get

[latex] NPSH_R \lt NPSH_A/1.10[/latex]

Then,

[latex] NPSH_R \lt 6.48 m/1.10 = 5.89 m[/latex]

The result indicates that any pump that requires 5.89 m or less for NPSH is acceptable.

TABLE 13.2 Vapor pressure and vapor pressure head of water

Vapor Pressure Head (ft)	Specific Weight ([latex]lb/ft^3[/latex])	Vapor Pressure (psia)	Temperature °F	Vapor Pressure Head (m)	Specific Weight [latex] (kN/m^3)[/latex]	Vapor Pressure kPa (abs)	Temperature °C
0.2043	62.24	0.08854	32	0.06226	9.806	0.6105	0
0.2807	62.43	0.1217	40	0.08894	9.807	0.8722	5
0.4109	62.41	0.1781	50	0.1253	9.804	1.228	10
0.5917	62.37	0.2563	60	0.2388	9.789	2.338	20
0.8393	62.3	0.3631	70	0.4345	9.765	4.243	30
1.173	62.22	0.5069	80	0.758	9.731	7.376	40
1.618	62.11	0.6979	90	1.272	9.69	12.33	50
2.205	62	0.9493	100	2.066	9.642	19.92	60
3.948	61.71	1.692	120	3.25	9.589	31.16	70
6.775	61.38	2.888	140	4.967	9.53	47.34	80
11.18	61	4.736	160	7.405	9.467	70.1	90
17.55	61.58	7.507	180	10.78	9.399	101.3	100
27.59	60.12	11.52	200
35.36	59.83	14.69	212

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

Nominal Pipe Size

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 13.3: Determine the available NPSH for the system shown in Fig. 13...

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