- #1
jeff25111
- 3
- 0
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
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