- #1
Bis2008
- 6
- 0
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
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