How Does a Siphon Work with Different Liquid Heights and Pressures?

[Destrio]

Siphon with a liquid of density P and negligible viscosity. With the bottom of the tube in the contained at point A d + h1 from point B at the top of the siphon, and the bottom of the tube extracting the liquid at point C d + h1 + h2 from the top of the siphon and h2 away from point a. (All distances are vertical).
a) With what speed does the liquid emerge from tube at C?
b) What is the pressure at topmost point B?
c) What is the greatest height that a siphon may lift water?

I figure since there are no changes in area, the velocity must be constant at all point in the tube. But as the water level drops, will the pressure in the siphon change, causing the velocity to drop?

 

[Shooting Star]

What is the pressure difference that is making the siphon work?

You have not said what is d. Give a better description of the whole apparatus.

 

[Moetasim]

a) The speed at which the liquid emerges from tube C can be calculated using the Bernoulli's equation, which states that the total pressure at any point in a fluid system is equal to the sum of the static pressure, dynamic pressure, and potential energy per unit volume. In this case, the dynamic pressure can be neglected due to the negligible viscosity of the liquid. Therefore, the speed of the liquid at C can be calculated using the equation v = √(2gh2), where g is the acceleration due to gravity, h2 is the distance from point C to the top of the siphon, and v is the speed of the liquid.

b) The pressure at the topmost point B can be calculated using the equation P = ρgh1, where ρ is the density of the liquid, g is the acceleration due to gravity, and h1 is the distance from point B to the bottom of the siphon. This is the pressure exerted by the weight of the liquid in the siphon.

c) The maximum height that a siphon can lift water is limited by the atmospheric pressure. As the liquid rises in the siphon, the pressure at the topmost point B will decrease due to the weight of the liquid in the siphon. Once the pressure at point B becomes equal to the atmospheric pressure, the siphon will stop working. Therefore, the greatest height that a siphon can lift water is equal to the difference between the atmospheric pressure and the pressure at point B. This can be calculated using the equation hmax = (P0 - P)/ρg, where P0 is the atmospheric pressure, P is the pressure at point B, ρ is the density of the liquid, and g is the acceleration due to gravity.