Question 10.3: Suppose the pump of Fig. 10-11 is installed in a system as s......
Suppose the pump of Fig. 10-11 is installed in a system as shown in Fig. 10-12. The pump is operating at 3500 rpm with the 6 in. impeller and delivering 200 gpm. The suction line is standard 4 in. pipe that has an inside diameter of 4.026 in. Compute the NPSHA, and compare it with the NPSHR. The water temperature is 60 F.
From Fig. 10-11 the NPSHR is 10 ft of head. The available net positive suction head is computed from Eq. 10-21a; however, the form will be changed slightly through the application of Eq. 10-1c between the water surface and the pump inlet:
NPSHA = \frac{P_{s} g_{c}}{ρg} + \frac{\bar{V}^{2}_{s}}{2g} – \frac{P_{\text{v}} g_{c}}{ρg} (10-21a)
\frac{g_{c}}{g} \frac{P_{1}}{ρ_{1}} + \frac{\bar{V}^{2}_{1}}{2g} + z_{1} = \frac{g_{c}}{g} \frac{P_{2}}{ρ_{2}} + \frac{\bar{V}^{2}_{2}}{2g} + z_{2} + \frac{g_{c} \text{w}}{g} + l_{f} (10-1c)
\frac{P_{B} g_{c}}{ρg} = \frac{P_{s} g_{c}}{ρg} + \frac{\bar{V}^{2}_{s}}{2g} + z_{s} + l_{f}
or
\frac{P_{s} g_{c}}{ρg} + \frac{\bar{V}^{2}_{s}}{2g} = \frac{P_{B} g_{c}}{ρg} – z_{s} – l_{f}
Then Eq. 10-21a becomes
NPSHA = \frac{P_{B} g_{c}}{ρg} – z_{s} – l_{f} – \frac{P_{\text{v}} }{ρ} \frac{g_{c}}{g} (10-21b)
Assuming standard barometric pressure,
\frac{P_{B} g_{c}}{ρg} = \frac{29.92 (13.55)}{12} = 33.78 ft of water
\frac{P_{\text{v}} g_{c}}{ρg} = \frac{0.2562 (144)}{62.4} = 0.59 ft of water
where P_{v} is read from Table A-1a at 60 F. Then from Eq. 10-21b
NPSHA = 33.78 − 10 – 5 − 0.59 = 18.19 ft of water
which is almost twice as large as the NPSHR. However, if the water temperature is increased to 160 F and other factors remain constant, the NPSHA becomes
NPSHA = 33.78 – 10 – 5 – (\frac{4.74 × 144}{61}) = 7.6 ftand is less than the NSPHR of 10 ft. Cavitation will undoubtedly result.