A Container measuring 2ft x 2 ft x 4 ft is submerged length wise (4 foot vertically) in water so that the top of the container is 1 foot below the surface and the bottom of the container is 5 feet below the surface; The water is potable and 70 degrees F; The container is filled with air and has an 8 inch valve at the top and an 8 inch valve at the bottom; both valves are opened at the same time; what time elapses to fill the container completely with water;
[COLOR=red]I have used this calculation: AM I CORRECT
Velocity at depth at top of tank ie1 foot = v1= sqrt (2gh1) = sqrt (2x32.18x1) =8.023foot/sec
Velocity at depth at bottom of tank (1+4) i.e. 5 foot = v2= sqrt (2gh2)
= sqrt (2x32.18x5) =17.939 foot/sec
Now the crossection area of 8inch dia valve = ? (82)/4 = 50.2655 inch2 = 0.3491 foot2
Now, area x velocity xtime = volume
So, 0.3491x (8.023) xt = V1 so, V1= 2.801t
And in the same way V2= 0.3491 x 17.939xt, so V2= 6.2625t
Now V1 + V2 = 16
So, 2.801t +6.2625t = 16
So, t= 1.765 sec
I think you forgot the air in the container. How much water could get in the top for the length of time that the air is escaping?
edit: I think that may not be a trivial question. Burping and turbulence, etc.
edit: I think that may not be a trivial question. Burping and turbulence, etc.
I suspect you need to know all kinds of stuff that most people don't really want to know .. some bits of it are here ..
http://www.engineeringtoolbox.com/reynold-...ipes-d_574.html
http://www.engineeringtoolbox.com/reynold-...ipes-d_574.html
In order to get a realistic answer to this question, you must deal with some very messy details, as Sapo has pointed out. On the other hand, if you want to simplify it without being too far away from reality, you can look at this as a problem with flow through two orifices.
You can basically pretend that the flow follows Bernoulli's equation, and then include "orifice losses" to account for turbulence and friction with the orifice. There are standard formulas for flow through a small orifice into a larger diameter pipe, which you could use here, considering the box to be a pipe. The pure Bernoulli formula analysis will be rather far from the real behavior if you don't include these "discharge coefficients" of the orifices.
However, there are clearly two things wrong with your analysis, besides the orifice losses (which perhaps in your coursework you are told to ignore):
First, I do not think that much water will enter through the top orifice, since it will have an escaping air stream rushing up rather rapidly through it. Therefore, you will not be filling the box from both ends. A small amount of water will probably enter through the top as the air jet weakens when the tank is nearly full, but it would be very difficult (probably impossible) to calculate that amount of leakage from theory. It is likely to be small, though, so you could probably just ignore it to get an approximate solution.
Second, your calculation of the velocities an the orifices uses the water surface as a reference. This does NOT make sense in this case, because the tank is full of air, and the pressure in air is nearly independent of depth. therefore, the pressure inside the tank is the same at both the top and the bottom orifice. The pressure difference at the top orifice drives the air stream out of the tank and the pressure at the bottom drives the water in. The trouble is, you don't know the pressure in the tank after the valves are opened.
You can try to handle this by requiring that the volume flow rate of the exiting air must equal the volume flow rate of the entering water. This makes sense, since the air in the tank is being replaced with water. Then you could calculate the tank pressure by using Bernoulli's equation for the air stream at the top and for the water stream at the bottom. In each case, you would compare the pessure just outside the orifice with the pressure in the tank to get the P terms in the equation. The pressure outside the orifice is controlled by the water depth, so you know those pressures. Setting the volume flow rates equal lets you calculate the pressure in the tank, and then substituting this back into the Bernoulli equation for water lets you calculate the water stream velocity and therefore the rate of filling.
Two difficulties arise here. First, the rate of fill will not be constant, since it is driven by the pressure difference, and as the tank fills with water, the pressure at the top orifice does not stay equal to the pressure at the bottom orifice. Instead they differ by an amount that depends on the depth of the water in the tank. Therefore you have a differential equation to integrate in order to get the filling time.
Second, the air is a gas, of course, and so it is not incompressible like the water is. Therefore, you must base the air stream Bernoulli equation on the mass flow rate, not the volume flow rate. In other words, the density of the air depends on the pressure, which is one of the terms in Bernoulli's equation. Therefore, you must divide the whole equation by density on both sides, with the density on each side written as a function of the pressure, via the ideal gas law. (You can assume constant T because the air will easily be held to a constant temperature by thermal contact with the water.)
If you do all that, you get a rather complicated differential equation to solve. You should modify it by including the discharge coefficients at the two orifices (remember to use the density and viscosity of air at one orifice, and of water at the other). Now you can integrate the differential equation and find the time to fill the tank.
Sorry this is so complicated, but that's the nature of fluid flow. I hope that my help is not scarier than the original problem! Good luck with your work.
--Stuart Anderson
You can basically pretend that the flow follows Bernoulli's equation, and then include "orifice losses" to account for turbulence and friction with the orifice. There are standard formulas for flow through a small orifice into a larger diameter pipe, which you could use here, considering the box to be a pipe. The pure Bernoulli formula analysis will be rather far from the real behavior if you don't include these "discharge coefficients" of the orifices.
However, there are clearly two things wrong with your analysis, besides the orifice losses (which perhaps in your coursework you are told to ignore):
First, I do not think that much water will enter through the top orifice, since it will have an escaping air stream rushing up rather rapidly through it. Therefore, you will not be filling the box from both ends. A small amount of water will probably enter through the top as the air jet weakens when the tank is nearly full, but it would be very difficult (probably impossible) to calculate that amount of leakage from theory. It is likely to be small, though, so you could probably just ignore it to get an approximate solution.
Second, your calculation of the velocities an the orifices uses the water surface as a reference. This does NOT make sense in this case, because the tank is full of air, and the pressure in air is nearly independent of depth. therefore, the pressure inside the tank is the same at both the top and the bottom orifice. The pressure difference at the top orifice drives the air stream out of the tank and the pressure at the bottom drives the water in. The trouble is, you don't know the pressure in the tank after the valves are opened.
You can try to handle this by requiring that the volume flow rate of the exiting air must equal the volume flow rate of the entering water. This makes sense, since the air in the tank is being replaced with water. Then you could calculate the tank pressure by using Bernoulli's equation for the air stream at the top and for the water stream at the bottom. In each case, you would compare the pessure just outside the orifice with the pressure in the tank to get the P terms in the equation. The pressure outside the orifice is controlled by the water depth, so you know those pressures. Setting the volume flow rates equal lets you calculate the pressure in the tank, and then substituting this back into the Bernoulli equation for water lets you calculate the water stream velocity and therefore the rate of filling.
Two difficulties arise here. First, the rate of fill will not be constant, since it is driven by the pressure difference, and as the tank fills with water, the pressure at the top orifice does not stay equal to the pressure at the bottom orifice. Instead they differ by an amount that depends on the depth of the water in the tank. Therefore you have a differential equation to integrate in order to get the filling time.
Second, the air is a gas, of course, and so it is not incompressible like the water is. Therefore, you must base the air stream Bernoulli equation on the mass flow rate, not the volume flow rate. In other words, the density of the air depends on the pressure, which is one of the terms in Bernoulli's equation. Therefore, you must divide the whole equation by density on both sides, with the density on each side written as a function of the pressure, via the ideal gas law. (You can assume constant T because the air will easily be held to a constant temperature by thermal contact with the water.)
If you do all that, you get a rather complicated differential equation to solve. You should modify it by including the discharge coefficients at the two orifices (remember to use the density and viscosity of air at one orifice, and of water at the other). Now you can integrate the differential equation and find the time to fill the tank.
Sorry this is so complicated, but that's the nature of fluid flow. I hope that my help is not scarier than the original problem! Good luck with your work.
--Stuart Anderson
Stuart,
Awesome answer, as always. May I mention viscosity? It does get complicated.
Awesome answer, as always. May I mention viscosity? It does get complicated.
energydog,
>So, t= 1.765 sec
Without all of the math I can tell you there is no way 5 feet of water is going to give enough pressure to slam 120 gallons of water through an 8 inch whole in less than 2 seconds. you would need over 7200 psi to do that. that pressure would probably crush your box first.
>So, t= 1.765 sec
Without all of the math I can tell you there is no way 5 feet of water is going to give enough pressure to slam 120 gallons of water through an 8 inch whole in less than 2 seconds. you would need over 7200 psi to do that. that pressure would probably crush your box first.