This problem involves water being siphoned from a tank through a tube and asks us to calculate two things:

(a) The volume flow rate of water from the tank. (b) The maximum height $H$ over which water can be siphoned without cavitation.

Step-by-Step Solution:

Part (a) - Volume Flow Rate

1. Bernoulli's Equation: To find the volume flow rate, we can apply Bernoulli’s equation between the surface of the water in the tank and the outlet of the siphon tube. Assuming the tank is large (and therefore the velocity of the water at the surface is very small and can be neglected), and given that viscous effects are negligible, the energy equation simplifies as:

   $P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$

   Here, the subscripts:

   • $1$ corresponds to the water surface,
   • $2$ corresponds to the outlet of the siphon tube.

   We assume atmospheric pressure $P_1 = P_2$, so they cancel out. The velocity at the surface $v_1$ is approximately zero. The height difference is $h_1 - h_2$, which equals the total height the water has fallen, including the distance from the tank bottom.

2. Simplified Bernoulli’s Equation: After simplification, we get:

   $0 = \frac{1}{2} \rho v_2^2 + \rho g h_2$

   Solving for $v_2$:

   $v_2 = \sqrt{2 g h_2}$

3. Flow Rate: The volumetric flow rate $Q$ is given by:

   $Q = A v_2$

   Where $A$ is the cross-sectional area of the tube, and $v_2$ is the velocity at the outlet.

   The area of the tube is:

   $A = \pi \left( \frac{D}{2} \right)^2 = \pi \left( \frac{2 \text{ in}}{2} \right)^2$

   Converting inches to feet:

   $A = \pi \left( \frac{1 \text{ in} \times 1 \text{ ft}/12 \text{ in}}{1} \right)^2 = \pi \left( \frac{1}{12} \right)^2 \text{ ft}^2$

   Plugging these into the formula gives the flow rate.

Part (b) - Maximum Height Without Cavitation

1. Cavitation Condition: Cavitation occurs when the pressure in the fluid drops below the vapor pressure of the water. Using Bernoulli's equation again, we need to determine the maximum height $H$ such that the pressure does not drop below 0.26 psia, which is the vapor pressure.

2. Pressure Calculation: The pressure at the highest point of the siphon is related to the atmospheric pressure minus the hydrostatic pressure due to the height $H$:

   $P = P_{\text{atm}} - \rho g H$

   The condition for no cavitation is:

   $P \geq P_{\text{vapor}}$

   Solving for $H$:

   $H = \frac{P_{\text{atm}} - P_{\text{vapor}}}{\rho g}$

   Where:

   • $P_{\text{atm}} = 14.7 \, \text{psia}$,
   • $P_{\text{vapor}} = 0.26 \, \text{psia}$,
   • $\rho$ is the density of water, and
   • $g$ is the acceleration due to gravity.

This will give the maximum height $H$.

Would you like further clarification on any part or more detailed calculations for the flow rate and maximum height?

Related Questions:

1. What effect would an increase in tube diameter have on the flow rate?
2. How does atmospheric pressure affect the siphoning process?
3. What assumptions are made when applying Bernoulli’s equation in this context?
4. How does cavitation impact the structural integrity of a siphon system?
5. What other methods can be used to prevent cavitation in fluid systems?

Tip:

When solving fluid flow problems, always ensure the assumptions like negligible viscous effects and steady flow are reasonable for the scenario in question.