- #1
jayzedkay
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I posted this problem in another section, but it seems to have fizzled out and I’m trying to solve it, so resurrected it here hopefully to get somewhere.
I’m trying to calculate the impact force of a free falling cylinder of metal under water accounting for the drag. The cylinder is restricted to vertical falling, so cannot tumble down so to speak.
I started with the work energy principle and net work done to calculate the impact force, but now I am trying to include the drag from the water.
I got a couple of equations, one being stokes Fd = 6PIuvd and the other being Fd = 0.5CpAv^2.
Using the principle that the forces of buoyancy and drag are equal to the force of it falling; mg=Fb+Fd, I started to try and work things out.
Buoyancy force I used is the displaced fluid weight times the accel due to gravity. So here I used;
Fb = volume of cylinder X density of fluid X accel due to gravity = PIr^2hpg
For drag I used either Fd = 6PIuvd or Fd = 0.5CpAv^2 in each calculation.
I rewrote the equations with all the available parameters:
mg = PIr^2hpg + 6PIuvd
or
mg = PIr^2hg +0.5CpAv^2
I then rearranged for velocity, my objective is to isolate velocity inclusive of drag, then I can use that new calculated velocity in the original work energy principle and get the new impact force. So I get;
v = [ (m*g) – PI*r^2*h*p*g ] / 6*PI*u*d (eqn1).
Or
v = sqrt{ 2*[ (m*g) – PI*r^2*h*p*g ] / C*p*A } (eqn2).
I used the following for the parameters;
m = mass of metal cylinder, I used 10kg
g = acceleration due to gravity (9.8m/s^2)
PI = 3.1416
r = radius of cylinder, I used 7.5cm
h = height/length of cylinder, I used 30cm
p = density of fluid, I used 1000kg/m^3
u = viscosity of fluid, I used 0.00179 Pa.s
C = drag coefficient of cylinder, I used 0.82
d = diameter of cylinder, I used 15cm
A = cross sectional area of cylinder
Eqn1 doesn’t give me physically possible number as answer (>9k), eqn 2 could be, i.e. it’s less than 9.8, which I’d expect. Things fall slower under water than in air.
Am I going about it the right way, if not, where am I going wrong etc.
Thanks in advance for help/pointers etc.
I’m trying to calculate the impact force of a free falling cylinder of metal under water accounting for the drag. The cylinder is restricted to vertical falling, so cannot tumble down so to speak.
I started with the work energy principle and net work done to calculate the impact force, but now I am trying to include the drag from the water.
I got a couple of equations, one being stokes Fd = 6PIuvd and the other being Fd = 0.5CpAv^2.
Using the principle that the forces of buoyancy and drag are equal to the force of it falling; mg=Fb+Fd, I started to try and work things out.
Buoyancy force I used is the displaced fluid weight times the accel due to gravity. So here I used;
Fb = volume of cylinder X density of fluid X accel due to gravity = PIr^2hpg
For drag I used either Fd = 6PIuvd or Fd = 0.5CpAv^2 in each calculation.
I rewrote the equations with all the available parameters:
mg = PIr^2hpg + 6PIuvd
or
mg = PIr^2hg +0.5CpAv^2
I then rearranged for velocity, my objective is to isolate velocity inclusive of drag, then I can use that new calculated velocity in the original work energy principle and get the new impact force. So I get;
v = [ (m*g) – PI*r^2*h*p*g ] / 6*PI*u*d (eqn1).
Or
v = sqrt{ 2*[ (m*g) – PI*r^2*h*p*g ] / C*p*A } (eqn2).
I used the following for the parameters;
m = mass of metal cylinder, I used 10kg
g = acceleration due to gravity (9.8m/s^2)
PI = 3.1416
r = radius of cylinder, I used 7.5cm
h = height/length of cylinder, I used 30cm
p = density of fluid, I used 1000kg/m^3
u = viscosity of fluid, I used 0.00179 Pa.s
C = drag coefficient of cylinder, I used 0.82
d = diameter of cylinder, I used 15cm
A = cross sectional area of cylinder
Eqn1 doesn’t give me physically possible number as answer (>9k), eqn 2 could be, i.e. it’s less than 9.8, which I’d expect. Things fall slower under water than in air.
Am I going about it the right way, if not, where am I going wrong etc.
Thanks in advance for help/pointers etc.