Need help with a unsteady state mass balance problem

[jeff25111]

if anybody can help,
the problem is finding out the amount of time it takes to drain a 12oz soda can through the shotgun method. this method of draining involves opening the tab on top of the can and then punching a hole at the bottom to drain the soda out.the velocity out obviously changes as the height of the fluid inside the can changes. I've taken the point on the top surface of the fluid (having p=1atm) and the point on the bottom, at the draining hole.
im looking for dM/dt or dV/dt. can i assume the velocity on the surface=0 in comparison to the velocit of water at the bottom hole?
ive come up with dV/dt=A2*Vel. exit
ive used the mechanical energy balance to solve for ave vel exit=sqrt(4*grav*y)*Area(of exit hole)
pls tell me if I am approaching this problem correctly.would appreciate any input

 

[stewartcs]

Sounds like you're on track. I think the Bernoulli Equation would yield you a good approximation.

Find the velocity of the fluid exiting out of the hole and multiply that by the cross-sectional area of the hole to find the flow rate. Then it's just a matter of how much fluid is in the can divided by the flow rate.

 

[stewartcs]

ive come up with dV/dt=A2*Vel. exit
ive used the mechanical energy balance to solve for ave vel exit=sqrt(4*grav*y)*Area(of exit hole)

The exit velocity using Bernoulli would be v = sqrt(2*g*h),

where,

h = height of the fluid column
g = gravitational accel.

The flow rate would then be the area times the velocity.

 

[stewartcs]

You know, on second thought I'm not sure if you could use that equation since you have such a small hole in the top of the can.

 

[jeff25111]

i wouldn't know which points to choose if the pressure on top is unknown. i don't think i can assume that the pressure on the draining hole is equal to the atmospheric pressure either..just at a loss

 

[stewartcs]

The pressure on top is atmospheric pressure unless you are applying more. If you are just punching a hole in the top of the can, then it's 1 ATM like you already stated.

The pressure at the outlet, where the fluid is draining to, is 1 ATM also since it is flowing freely.