# Pressure and Flow Rate in a Tank

• Gwozdzilla
In summary, the pressure in the tank is initially atmospheric, but it drives the flows through the pipes because of the pressure differential. The initial flow rates through the pipes are proportional to the Reynold's numbers for the flow. The shear stress is highest at the points where the pipes meet the tank and lowest at the center line.
Gwozdzilla

## Homework Statement

A tank of cross-sectional area A is initially filled with fluid of density ρ and viscosity μ to height hi. The pressure above the fluid in the tank is atmospheric, Patm. At the base of the tank there are two pipes, which are both open to teh atmosphere, such that fluid flows out of them. The pipe on the left has diameter D1, length L1, and flow rate Q1. The pipe on the right has diameter D2, length L2, and flow rate Q2.
A. Write an expression for the initial pressure that drives the flows through pipes 1 and 2.
B. What is the ratio of the initial flow rates Q1/Q2?
C. What is the ratio of the initial Reynold's numbers for the flow in the two pipes, Re1/Re2?
D. Where is the shear force lowest and highest in the pipes? (a. At the wall, b. At the center line, c. Where the pipe meets the tank, d. Where the fluid flows out of the pipe to the atmosphere)

## Homework Equations

0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
Q =(ΔPπR4)/(8μL) = Au
Re = ρuD/μ

## The Attempt at a Solution

A. Pinit = ρg(hi) + (Patm)

B. Q1 =(ΔPπ(D1/2)4)/(8μL1)

Q2 =(ΔPπ(D2/2)4)/(8μL2)

Q1/Q2 = (D1/D2)4(L2/L1)

C. When I got to this one, I realized that I don't know how to find the velocity of the fluid, since I had assumed it to initially be zero, so I probably did A and B wrong.
Maybe I'm supposed to get it from Bernoulli's equation:
0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2

u = √((-2/ρ)(ΔP + ρgΔh))
u = √((-2/ρ)(Patm + ρg(hi)))

But I don't know why this would be the case, and it also isn't taking into account he difference in area or length between the pipes. Could you please send me in the right direction? Am I missing an equation?

D. I think the shear stress would be highest where the pipes meet the tank and lowest at the center line, but I don't know why.

You forgot to write one seemingly trivial, but essential equation: the one for continuity of flow, Q = A * V.

Since the problem specified that the flow rate at point 1 was Q1, you can calculate the velocity at this point. (BTW, a velocity of zero suggests no flow is occurring.)

With regard to question C, in my judgement, they are expecting you to assume fully developed quasi-steady flow in the pipes, with the flow rate given by the Hagen-Poiseulle equation.

If flow rate Q = Area * velocity, Q = Au, then A can be redone...

A. 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
Pinit = Patm + ρg(hi) + (1/2)ρ(Δuavg)2
where uavg = (1/2)(u1 + u2)
u1 = Q1/A1 = Q1/(π(D1/2)2)
u2 = Q2/A2 = Q2/(π(D2/2)2)

Would this Pinit value be correct?

Then for part B, do I just let the ΔP1 = ρg(hi) + (1/2)ρ(Q1/A1)2
and ΔP2 = ρg(hi) + (1/2)ρ(Q2/A2)2, keeping the rest the same? And then use the velocities from part A to calculate the Reynold's numbers?

Gwozdzilla said:
If flow rate Q = Area * velocity, Q = Au, then A can be redone...

A. 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
Pinit = Patm + ρg(hi) + (1/2)ρ(Δuavg)2
where uavg = (1/2)(u1 + u2)
u1 = Q1/A1 = Q1/(π(D1/2)2)
u2 = Q2/A2 = Q2/(π(D2/2)2)

Would this Pinit value be correct?

Then for part B, do I just let the ΔP1 = ρg(hi) + (1/2)ρ(Q1/A1)2
and ΔP2 = ρg(hi) + (1/2)ρ(Q2/A2)2, keeping the rest the same? And then use the velocities from part A to calculate the Reynold's numbers?
Bernoulli's equation should not be applied to this problem, unless you include the frictional pressure drop in the equation.

Chet

How do I find pressure without Bernoulli's equation? I've never seen a frictional pressure drop term in Bernoulli's equation; what would that look like?

So the Bernoulli equation is now:
0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD (L/D)(1/2)ρ(uavg)2
Should the lengths and diameters be averaged too, and then plug in the velocities from my last post to calculate Pinit?

Gwozdzilla said:
So the Bernoulli equation is now:
0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD (L/D)(1/2)ρ(uavg)2
Should the lengths and diameters be averaged too, and then plug in the velocities from my last post to calculate Pinit?
No, they shouldn't. You figure out the friction losses in each section of pipe which has a constant diameter. The friction losses don't scale in a linear fashion as diameter changes, unfortunately.

I think incorporation of friction losses into this problem has exceeded the scope of what was intended. There's too much background knowledge which is being glossed over in the application of Darcy-Weisbach.

SteamKing said:
I think incorporation of friction losses into this problem has exceeded the scope of what was intended. There's too much background knowledge which is being glossed over in the application of Darcy-Weisbach.
I respectfully disagree. I think that once they mentioned viscosity in the problem statement, they expected the students to include friction losses. However, in my judgement, they probably intended for the students to neglect inertia (i.e., the kinetic energy term in Bernoulli) so that the pressure drop was determined solely by the frictional loss (i.e., Hagen-Poiseulle).

Chet

This problem came from a practice exam recommended by my professor that belongs to a course different from, but similar to, what he taught us, so I can't actually speak for the original intentions of this problem. Since I hadn't been taught about friction, I'm fine with keeping it out. I don't mind keeping it in either though, as long as I know that I'm approaching the problem correctly and using the formulas correctly.

It sounds like if I want to keep the friction term, I should add it once for each of the pipes:
0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD(L1/D1)(1/2)ρ(u1)2 + fD(L2/D2)(1/2)ρ(u2)2

If friction isn't included then is the below formula correct, or is there a different issue?
Gwozdzilla said:
If flow rate Q = Area * velocity, Q = Au, then A can be redone...

A. 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
Pinit = Patm + ρg(hi) + (1/2)ρ(Δuavg)2
where uavg = (1/2)(u1 + u2)
u1 = Q1/A1 = Q1/(π(D1/2)2)
u2 = Q2/A2 = Q2/(π(D2/2)2)

Would this Pinit value be correct?

Then for part B, do I just let the ΔP1 = ρg(hi) + (1/2)ρ(Q1/A1)2
and ΔP2 = ρg(hi) + (1/2)ρ(Q2/A2)2, keeping the rest the same? And then use the velocities from part A to calculate the Reynold's numbers?

## 1. What is the relationship between pressure and flow rate in a tank?

The pressure and flow rate in a tank are directly related. As the pressure in the tank increases, the flow rate also increases and vice versa. This is because the pressure difference between the inlet and outlet of the tank determines the force that drives the fluid out of the tank.

## 2. How does the size of the tank affect pressure and flow rate?

The size of the tank has a significant impact on pressure and flow rate. A larger tank can hold more volume of fluid, resulting in a higher flow rate and lower pressure. On the other hand, a smaller tank will have a lower flow rate and higher pressure due to the limited volume it can hold.

## 3. How can I calculate the pressure and flow rate in a tank?

The pressure and flow rate in a tank can be calculated using the Bernoulli's equation, which considers factors such as the fluid's density, velocity, and elevation. Other methods include using the ideal gas law or using a pressure gauge and flow meter to directly measure the values.

## 4. What factors can affect the pressure and flow rate in a tank?

There are several factors that can affect the pressure and flow rate in a tank, such as the fluid's viscosity, temperature, and composition. Additionally, the shape and size of the tank, as well as the presence of obstructions or valves, can also impact the pressure and flow rate.

## 5. How does the opening size of the tank's outlet affect pressure and flow rate?

The size of the tank's outlet has a significant influence on pressure and flow rate. A larger outlet will result in a higher flow rate and lower pressure, while a smaller outlet will have a lower flow rate and higher pressure. This is because the size of the outlet determines the cross-sectional area through which the fluid can flow out of the tank.

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