Two tanks connected by a horizontal pipe

[sigma128]

from Applied Mathematics and Modeling for Chemical Engneers (Richard G. Rice)

Two vertical cylindrial tanks, each 10 m high,are installed side-by-side in a tank farm, their bottoms at the same level.The tanks are connected at their bottoms by a horizontal pipe 2 meters long,with pipe inside diameter 0.03 m. The first tank (1) is full of oil and the second tank (2) is empty.Moreover,tank 1 has a cross-sectional area twice that of tank 2.The first tank also has another outlet(to atmosphere) at the bottom, composed of a short horizontal pipe 2 m long, 0.03 m diameter.Both of the valves for the horizontal pipes are opened simultaneously.What is the maximum oil level in tank 2 ? Assume laminar flow in the horizontal pipes, and neglect kinetic, entrance-exit losses.

please show me solving.
PS. why i can not post new topic in Engineer Forum

 

[berkeman]

I believe that this problem is being worked in this other thread that you posted in. Do not post a question a second time if it is being worked, especially a problem that is in the Homework Help forums.

I'm locking this thread for now. Please continue the discussion in the other original thread in Homework Help.

https://www.physicsforums.com/showthread.php?p=1387331#post1387331

 

[tacman]

Thank you for your question. To find the maximum oil level in tank 2, we can use the continuity equation, which states that the mass flow rate into a system must equal the mass flow rate out of the system. In this case, we can set the mass flow rate into tank 2 equal to the mass flow rate out of tank 1 (since the tanks are connected by a horizontal pipe).

We can write the mass flow rate into tank 2 as:

m_dot = ρ1 * A1 * V1

Where ρ1 is the density of oil in tank 1, A1 is the cross-sectional area of tank 1, and V1 is the velocity of oil in tank 1.

Similarly, the mass flow rate out of tank 1 can be written as:

m_dot = ρ2 * A2 * V2

Where ρ2 is the density of oil in tank 2, A2 is the cross-sectional area of tank 2, and V2 is the velocity of oil in tank 2.

Since we are assuming laminar flow in the horizontal pipes, we can use the Hagen-Poiseuille equation to calculate the velocity of oil in the pipes:

V = (ΔP * π * r^4) / (8 * μ * L)

Where ΔP is the pressure difference between the two ends of the pipe, r is the radius of the pipe, μ is the viscosity of the oil, and L is the length of the pipe.

Since we are neglecting kinetic and entrance-exit losses, we can assume that the pressure at the bottom of tank 1 is equal to the pressure at the bottom of tank 2. Therefore, the pressure difference (ΔP) between the two ends of the horizontal pipe is equal to the pressure at the bottom of tank 1 minus the pressure at the outlet to atmosphere (since the outlet is at the same level as the bottom of tank 2).

ΔP = P1 - P2

Now, we can set the mass flow rate into tank 2 equal to the mass flow rate out of tank 1:

ρ1 * A1 * V1 = ρ2 * A2 * V2

Substituting the expressions for V1 and V2 from the Hagen-Poiseuille equation, we get:

ρ1 * A1 * [(ΔP * π * r1^4) /