# Draining tank washout test

1. Jan 28, 2008

### Bis2008

Building an experimental washout tank for testing automotive adhesives.

Need to calculate the maximum water velocity of the following set-up in the 3" pipe:

Vertical 3" pipe approximately 3 meters high with a "T" located towards the bottom with a guillotine type valve attached to a reducer. I have reducers that range from 2.5" to 1.5 inch diameter. The pipe is filled with water and the guillotine valve is opened allowing all the water to drain from the 3" pipe through the teed reducer.

I need to simulate fluid velocities at a max of 2.5m/s in the LARGER 3" pipe. I can use forms of Bernoullis equation to calculate maximum outlet velocity speed and time to drain the tank. The outlet speed can be multiplied by the area difference to get the 3" pipe speed. However, how do I take into account the time it takes for the water in the 3" pipe to accelerate to max velocity? My graphs show a maximum velocity at time zero and I know the 3" pipe velocity should never be more than what is allowed by acceleration due to gravity (9.81m/s)

Here are the equations I have used so far:

t=((A(2/g)^0.5)/Ao)(h)^0.5

Vo=(2gh/(1-(Ao^2/A^2)))

V=(Ao/A)

Where h = height delta from top of 3" pipe and reducer outlet
Vo = outlet velocity
V = 3" pipe velocity (downward)
A = area of 3" pipe
Ao = area of reducer outlet

2. Jan 29, 2008

### FredGarvin

My texts list a couple of correlations for calculating entrance region of a pipe:

For Laminar flow: $$\frac{L_e}{D}=.06 Re$$

For Turbulent Flow: $$\frac{L_e}{D}=4.4(Re)^{\frac{1}{6}}$$

After that, all they mention is that the calculation of velocity profiles in entrance regions is very difficult. I guess you need to ask yourself just how much detail do you really need in your model?

3. Jan 29, 2008

### Bis2008

I am not too concerned with the type of flow, only the max velocity of the falling water column. What I am struggling with is the acceleration due to gravity. The water column should start at an intial velocity of zero, accelerate to a maximum than deccellerate due to the delta in height. Due to this initial acceleration needed to reach max, the maximum velocity I calculated would be only for a system that would have constant height deltas so it could reach steady state.

4. Jan 29, 2008

### Q_Goest

I understand that you have a system with a known pipe geometry, an inlet and outlet, and some known pressure somewhere. If that's the case, you can determine flow through this system of pipe using the attached manual. Note that the Bernoulli equation is modified using the expression for permenant pressure drop (HsubL) per the Darcy Weisbach equation (shown in equation 16 of the attached).

#### Attached Files:

• ###### Pipe-Flo Pro.pdf
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5. Jan 29, 2008

### FredGarvin

I totally misread what you were looking for. For some reason I thought you were only looking at the entrance region. My bad.

6. Jan 29, 2008

### Bis2008

I am only worried about max fluid velocity, I have data from automotive plants that suggest specific fluid velocities that I am trying to match. We are looking at creating a similar situations where beads of adhesives are placed on steel panels that are put in the water column (in order to simulate plant conditions). I have someone who may obtain some flow mters for me to confirm my max speeds but I wanted to calculate velocity so I at least design the apparatus appropriately and can apply theory to measured data.

7. Feb 17, 2008

### gmax137

I don't think the other posters understood your question. You are unlikely to solve a transient problem analytically. What is normally done is to break your system up into a series of nodes and flowpaths, then write the equations of mass, energy, and momentum for each, then solve these simultaneously. then increment in time by a small step and re-solve the equations. This is how to find the transient that occurs as you open the valve with the liquid initially at rest. One poster suggested pipe-flo code. It may have changed since I last used it (many years ago) but then it was a steady-state code which will not answer the question you are asking. You could use a code like RELAP, or for a simple system like yours, write a small code yourself. Good luck, it is not an easy problem but it is certainly possible to do it.

8. Feb 17, 2008

### Q_Goest

Hi gmax,
Welcome to the board. I'd agree the methods pointed to in Pipe-flo talk only about steady state. And I agree that once you open the valve you have a transient that lasts for some finite time until steady state is obtained. It's a good point. However, assuming the verticle pipe is already filled with water when you open the valve, the maximum velocity will be obtained at steady state.

At least for this case, the transient isn't too important.

Last edited: Feb 18, 2008
9. Feb 18, 2008

### gmax137

Thanks - I kind of stumbled upon this forum only recently.

I thought the problem was, with the small size of the source tank, the upper surface elevation was dropping (significantly) as the flow accelerated from rest to the steady-state value. And due to the lower static head, the maximum velocity would be lower than the steady-state prediction based on the initial level.

Maybe I read too much into the problem.

Thanks Again
Gregg

10. Feb 18, 2008

### Bis2008

Q Goest and gmax 137,

You are exactly right, my problem is the small source size of the tank. I thought this would be a simple problem using Bernoulli's equation but I keep coming up with acceleration values for the source tank that are greater than acceleration due to gravity. Which in turn tells me even though I have acceleration due to gravity(g) in my equation I still am not taking everything into account.

I have not done anything like this before so any help would be greatly appreciated. I am assuming when you say throw in time increments I would set up a worksheet and vary flow height to calculate time to steady state?

11. Feb 18, 2008

### Q_Goest

Hi Bis,
Here's what I understand you have:
- Tank 3 meters above reference point. Open to atmosphere.
- 3" Sched 40 pipe extends vertically downward from the bottom of the tank to the reference point.
- T at reference point. One side plugged. Other side open to valve.
- Don't know what kind of valve that is, I'll assume you mean a gate valve with a Cv of aprox. 700.
- Outlet of valve has a reducer which is intended to restrict flow so that velocity is no greater than 2.5 m/s
- Outlet of reducer goes to atmosphere.

Is this much correct?

To finish any kind of analysis, the height of the water in the tank is required. If you can provide details like this, I'll do the calculation for you. A picture would also help.

Last edited: Feb 18, 2008
12. Feb 18, 2008

### stewartcs

What equation(s) are you using, and what acceleration values are you coming up with? Show what you've got so far and I'm sure the problem will become clear very quickly to us.

As Q pointed out, the max velocity will occur when the system has reached a steady state. I'm not sure why you are so concerned with the transient respone since it will be over with very quick.

CS

13. Feb 19, 2008

### Bis2008

t=((A(2/g)^0.5)/Ao)(h)^0.5

This is used to calculate time to drain the tank. The h value is actually the square root of initial height - square root of final height, since final height is zero it drops out. I graph the results and take the slope of the graph at any point to get the velocity.

Maybe that is the problem? I try calculating "t" using the initial height and final heights that start at the initial height and drop to zero.

The reason the transient response is needed is because the max speed is being used to correlate to max speeds found in automotive wash tanks prior to the painting process. The tank drains completely in just a couple seconds and the time it takes to get up to full velocity is likely measured in tenths of seconds. This could represent max speed differences as great as 20%.

14. Feb 19, 2008

### stewartcs

Well in a somewhat simplified approach, the time it takes to drain the tank would be represented by a differential equation. The rate of change of the volume with respect to time in the tank $$\frac{dV}{dt}$$ is equal to the flow rate out of the tank ($$A_0 \cdot \sqrt{2gh}$$).

Since the tank is open to the atomosphere the velocity would be a special case of the Bernoulli equation (sqrt term in RHS of the equation below), hence...

$$\frac{dV}{dt} = A_0 \cdot \sqrt{2gh}$$

where,

V is the volume
$$A_0$$ is the area of the outlet
g is grav. accel.
h is height

Now just integrate it and use the boundary conditions that you know, i.e. at t=0 you have a full tank (V=V) and that at t=t, you have an empty tank (V=0).

Then just solve for t.

I'm sure there is a "book" formula out there for this type of problem so you might try some standard handbooks.

Hope that helps.

CS

15. Feb 19, 2008

### Q_Goest

Hi bis,
I don't see the Darcy Weisbach equation (or other flow restriction) in your simplified model, without which there is no restriction to flow and you can't solve for flow.

Also, if this really is a transient issue, it would help if you gave ths size of the tank and other system geometry.

16. Feb 19, 2008

### Bis2008

Attached drawing

The set-up is simple, there is no tank on the top. I am actually draining a 3" PVC pipe that is 2.3 meters higher than the outlet. The pipe has a T where the outlet is reduced to 1.5", 2" or 2.5 inch reducers (Simple bolt on plates). When I release the valve the water quickly drains out. It seems very simple but I get an instantaneous velocity that is faster than 9.81m/s (accel due to gravity)allows. What I am missing is the time it takes for the water column to accelerate versus the constant deceleration due to the falling head pressure.

#### Attached Files:

• ###### washout test.doc
File size:
25 KB
Views:
106
17. Feb 19, 2008

### Q_Goest

Hi bis,
Yes, this is a transient. I was under the impression you had this "washout tank" you were trying to drain. What gmax said might work:
I would also suggest neglecting pipe restrictions as a first shot. I doubt frictional flow would matter much, and being this will be a very quick event, you're not likely to develop any kind of steady state flow anyway. The entire restriction will be the outlet orifice that you install.

18. Apr 22, 2011

### glassfish

If you are still interested in RELAP5 and learning a bit more about it, there's a new RELAP5 User Workshop being help in Columbia, Maryland June 7-9, 2011. Here's the link: http://www.islinc.com/training/