Question

Two very large open tanks A and F (Fig. P12.93) contain the same liquid. A horizontal pipe BCD, having a constriction at C and open to the air at D, leads out of the bottom of tank A, and a vertical pipe E opens into the constriction at C and dips into the liquid in tank F. Assume streamline flow and no viscosity. If the cross-sectional area at C is one-half the area at D and if D is a dis- tance h1 below the level of the liquid in A, to what height h2 will liquid rise in pipe E? Express your answer in terms of h1.

Answer 1

The height h2 to which liquid rises in pipe E can be found using the principles of continuity and Bernoulli's equation, considering the fluid velocity at the constriction and the difference in heights between the tank A and point D.

Step-by-step explanation:

To determine the height h2 to which liquid will rise in pipe E, we can apply the principle of continuity and Bernoulli's equation. Since the scenario assumes an inviscid, incompressible fluid and streamline flow, these principles hold. As per the continuity equation, the volumetric flow rate must be constant throughout a pipe system, which gives us A1V1 = A2V2, where A1 is the cross-sectional area at the wider section and A2 is at the constriction. This means the velocity of the fluid at the constriction (C) will be greater than at the wider portion (D).

Bernoulli's equation, which relates pressure, velocity, and height for a fluid in motion, can be employed to find h2. At point D, the fluid is open to the atmosphere, which means the pressure is atmospheric and can be set as our datum level for gravitational potential energy. Using Bernoulli's principle, the pressure at the height h2 in pipe E must be equal to the pressure in the constriction at C. By equating the two and solving for h2 in terms of h1, we find that h2 is directly related to the velocity of the fluid at points C and D and the height difference h1.