So,
Increased velocity causes lower pressure, or rather pressure is used to provide the extra push, which lowers the pressure and speeds up the flow.
This is caused by the gradient, which is caused by the thinning of the pipe.
So.. and this is the bit I'm struggling with..
Is the pressure now...
Hmm I don't think I'm explaining myself very well, sorry about that.
As the liquid goes through the pipe, the area reduces and causes a squeeze on the water.. this squeeze tries to increase the pressure in the water, but as the water is free to move it gathers more speed instead. And in this...
Sorry I was referring to the kinetic energy/pressure.
I'm just struggling to visualise it.
What would happen If the reduction in the pipe was a half, would the velocity be greater than double?
And this is where the pressure drop comes from?
Hi guys,
Could anyone explain Bernoulli's Principle to me so that it makes sense from an alternative point of view?
http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html
I can make sense of the maths but I'm trying to understand what actually happens so I can visualise it.
At first...
I have natural frequency given on the revision sheet as 3.51Hz
Are you hinting that this is wrong and 3.51 is the actual frequency?
Makes more sense to me, and it wouldn't surprise me if they were giving us wrong answers.
I think I've stumbled onto something.
Critical damping = 264.5Ns/m
But that's per second, so 264.5 x 0.57 = 150.765
leaving 113.735 actual damping?
I'm not sure if it makes sense but its the right number that I need?
i got k from the equation
Natural freq = 1/2pi x √k/m
(3.51 x2pi)2 x 6= k
=2915
"You need to understand this stuff rather than trying to memorize solutions"
I agree totally, but unfortunately we are given little more than equations to memorise.
I find myself having to spend hours online...
what's the simple relation connecting ω and 0.57 sec.?
Natural Frequency is square root of k/m = 22.04 (radians) divide by 2pi to get 3.51Hz
0.57 for 2 revs = 0.285 secs per revolution. 1/0.285 = 3.51Hz
What about the 5% and exp(-ct/2m)?
I can't for the life of me find a value for c...
Thanks, but is there a simpler way to explain what is going on?
It's all getting very complicated.
I'm going to have to dedicate my weekend to it I think.
I mentioned it in the second post.
I can work out Natural frequency,Spring stiffness and Critical damping.
And when given the one of the other 3, (damping ratio, actual frequency, actual damping coefficient), I can solve for the other 2.
I now need to differentiate to find x, once I have x...
OK this is what I have,
Natural frequency = 2/0.57 = 3.51Hz
Spring stiffness 3.51Hz = 1/2pi x square root of K/m (k works out to be about 2915)
Critical damping square root of 4mk = 264.5
I need the damping ratio to find actual damping and actual frequency, I don't think I can do it...
Hi guys,
Revision for an exam tomorrow, I hope someone can help?
I've been given the answers but I can't work out how to find the damping ratio without being given a velocity or distance for the mass to travel?A mass of 6kg is suspended on a spring and set oscillating. it is observed that...