Question

2. From the textbook 5.99. (Statement: A siphon is used to draw water at 70°F from a large container as indicated in the figure. The inside diameter of the siphon line is 1 in. and the pipe centerline rises 3 ft above the essentially constant water level in the tank. Show that by varying the length of the siphon below the water level, h, the rate of flow through the siphon can be changed. Assuming frictionless flow, determine the maximum flowrate possible through the siphon. The limiting condition is the occurrence of cavitation in the siphon. Will the actual maximum flow be more or less than the frictionless value? Explain E DE

Answer 1

Solution :

The flowrate, Q, can be determined with To obtain A and B ve we apply the energy equation (Eg. 582) between points in the sketch above to obtain +974 + 4 shaft -loss (2) net in 13) i - 918,-?) - 1055 un ve • /z [9(h) -1058 ] With Eq. 3 we conclude that as h vories, so does ve and thus @ For no loss, the maximum flow will accur when the pressure at point c is just egnal to the vapor pressure of water at o'c. We apply the energy equation (Eg. 5.82) between points A and to get using absolute instead of gage pressures we obtain with Eg. 4 or V= /29 (2, -2) + Profe V- 202,4,)(-3 ft (0.3087) + (12304, 125, ) = 90485 (Con't)" ( 97 7 ) - Yên8;

Since Q = Ac" = PDV we have for the maximum flowrate through the siphon, O r u in." (0.3048 m (9.048 m) - 4.58x10 4(144 in.") # 57 5 With Egs. 3 and 4 we conclude that any loss would out to lower the value of V in the siphon and thus make the actual maximum flowrate with friction less than the maximum flowrate without friction

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