Question 5.9: The siphon in Fig. 5–21a is used to draw water from the larg......
The siphon in Fig. 5–21a is used to draw water from the large open tank. If the absolute vapor pressure for the water is p_v = 1.23 kPa, determine the shortest drop length L of the 50-mm-diameter tube that will cause cavitation in the tube. Draw the energy and hydraulic grade lines for the length of the tube.
Fluid Description. As in the previous example, we will assume water to be an ideal fluid, and its level in the tank to remain essentially fixed, so that we have steady flow. \gamma _w = 9810 N/m^3.
Bernoulli Equation. To obtain the velocity at C, we will apply the Bernoulli equation at points A and C. With the datum at C, we have
0 + 0 + (L – 0.2 m) = 0 + \frac{V_C^2}{2(9.81 m/s^2)} + 0
V_C = 4.429 \sqrt{(L – 0.2 m) } (1)
This result is valid provided the pressure at any point within the tube does not drop to or below the vapor pressure. If this were to happen, the water would boil (cavitate), causing a “hissing” noise and energy loss. This, of course, would invalidate the application of the Bernoulli equation.
Since the flow is assumed to be steady, and the tube has a constant diameter, then due to continuity, V^2/2g is constant throughout the tube, and so the hydraulic head (p/\gamma + z) must also be constant.
Using standard atmospheric pressure of 101.3 kPa, the gage vapor pressure for the water is 1.23 kPa – 101.3 kPa = -100.07 kPa. To find the shortest drop length L, we will assume this negative pressure develops at B, where z, measured from the datum, is a maximum.
Applying the Bernoulli equation at points B and C, realizing that V_B = V_C = V, we have
\frac{- 100.07 (10^3) N/m^2}{9810 N/m^3} + \frac{V^2}{2g} + (L + 0.3 m) = 0 + \frac{V^2}{2g} + 0
L + 0.3 m = 10.20 m
L = 9.90 m
From Eq. 1 the critical velocity is
V_C = 4.429\sqrt{(9.90 m – 0.2 m)} = 13.80 m/s
If L is equal to or greater than 9.90 m, cavitation will occur in the siphon at B, because the pressure at B will then be equal to or lower than -100.07 kPa.
Notice that we can also obtain these results by applying the Bernoulli equation, between A and B to obtain V_B, and between B and C to obtain L.
EGL and HGL. Throughout the tube, the velocity head is
The total head can be calculated from C. It is
H = \frac{H_C^2}{2g} + \frac{p_C}{\gamma} + z_C = 9.70 m + 0 + 0 = 9.70 m
Both the EGL and the HGL for segment DBC are shown in Fig. 5–21b. Here the HGL remains at zero. We can determine the pressure head within the tube at D by applying the Bernoulli equation between D and C.
\frac{p_D}{\gamma} + 9.70 m + (9.90 m – 0.2 m) = 0 + 9.70 m + 0
\frac{p_D}{\gamma} = – 9.70 m
Therefore, the pressure head decreases from –9.70 m at D to p/\gamma =[-100.07(10^3) N/m^2]/9810 N/m^3 = -10.2 m at B, while the elevation
head z increases from 9.70 m to 9.70 m + 0.5 m = 10.2 m, Fig. 5–21b.
After rounding the top of the pipe at B, the pressure head increases, while the elevation head decreases by a corresponding amount.
