Any chance anybody can help me?
Wil the canl move to the left, right, or not move once the valve is opened and air is let inside
Make the following assumptions: Pressure outside can = P ; mass of can = m; Area of can ends(each) = A; area of valve hole = s.
Newton's second law :
m*a = P*s*e^(-bt) - C*v(t)
a = d^2x / dt
v = dx / dt
m*d^2x / dt = P*s*e^(-bt) - C*dx / dt
m*d^2x / dt + C*dx / dt = P*s*e^(-bt) >>> Differential equation
the solution for this differential eq is:
y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}
^^ according to this solution, which is velocity, the can will move to the right, but how can i prove that? we need to find something in common between all these relationships that prove it moves to the right...
Answer:
Other than with the equations you have, you could use the conservation of momentum. The center of mass of the can and the air that later fills it is slightly to the right of the center of mass of the can. That's how I would prove it.
With your equation, you can say that P is the pressure from outside on the left side area equal to the valve hole, which is directed to the right. Therefore, for the equations purpose it is a vector.
Okay, I went back to your original question, which "mr.perfesser" answered, and then evaluated your responses to the other questions you've asked on this subject. I came to the conclusion that the differential equation was solved incorrectly. However, after elaborate calculations I eventually came into agreement with your solution.
Now, to really answer your question:
***
You do not, as you expressed, want to prove that mb-c and e^(-ct/m)-e^(-bt) are both positive. You simply want to prove that they are OPPOSITE. This is relatively simple:
If e^... is to be positive, the first term must be greater than the second (since e^anything is always positive). Therefore, you need to prove that c/m is less than b (and therefore greater when multiplied by -t). Going from there:
c/m < b
c < mb (we know that m, c, and b are all positive)
0 < mb-c
So mb-c is positive also, and since the numerator and denominator are positive, the overall equation is positive.
However, if e^... is to be negative, the first term must be less than the second, and you need to prove that c/m is greater than b (and therefore less when multiplied by -t). Going from THERE:
c/m > b
c > mb
0 > mb-c
Therefore, the numerator and denominator are both negative, so the overall equation is still POSITIVE.
Since you know that the force acting on the can (from pressure) is directed to the right, you can prove that the velocity of the can is directed in the same direction.
Hope that helped. Have a good night.
The problem doesn't show where is the location of the hole where the valve is located. Let's assume it is on the top side. Due to the pressure in all the area is uniform so there will be no movement of the can until the instant the valve is being open. because all the forces due to the outside pressure will cancel. At the instant the can is open, there will be surge of pressure going into the can which sure depends upon the hole size. Let us assume again that the hole is exactly on the center of the top side. Upon opening the valve, the area in the top is smaller than the area at the bottom side, with the pressure in all direction are equal so there will be net upward force applied to the can. Therefore, I would say that the direction of the movement will depends upon which side the hole is located. Also, one suggestion in your equation why the initial velocity is not assume zero when in fact the can is at rest.
m*a = P*s*e^(-bt)
m*dv/dt = P*s*e^(-bt)
this shows that the velocity of the can is going upward right.
If the can is very small in diameter it will not move because the force directed to the left from the jet of inlet air hitting the opposite side of the inside of the can from the orifice will cancel out the force of the jet trying to move the can to the right.
If the can is very large in diameter so that the incoming jet of air cannot develop a force on the opposite inside surface of the can, then try something like this:
Try using v = (2gh)^0.5
h = velocity head of air entering can, in ft of air, based on pressure differential of air inside and outside of can.
v = velocity of the air in ft per sec.
g = acceleration due to gravity
Then P = wvA(v^2/2g) foot-lb./sec
P = power in ft-lb/sec developed by the air jetting into the can.
w = lb per cu. of pressurized air
A = area of inlet orifice in sq. ft.
v = velocity from above.
divide foot-lb/sec by W = weight of the can to get the velocity of the can?
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