Isn't there some sort of online calculator that can measure the viscosity of a liquid?
Nov 03, 2009 by Neer H | Posted in Other - Internet
Without all the complicated stuff. My science project this year is about viscosity and temperature and stuff like that. So if I put in the time it took for a marble to travel down a tube from one point to another (with a liquid in it), and the distance, and then maybe the diameter or mass of the marble, can't there be a calculator for that? I've Googled so many but they're all so complicated and want all this crap that I don't even know what it is. I just need a SIMPLE. ONE. Thanks.
Well yeah see, MINERVA, I'm not stupid and I do get straight A's but that's not to brag or anything. I mean the thing is all those calculators had all these blanks you had to fill in with these units I'm pretty sure a lot of people haven't heard of. So. Yeah.
what, stick the liquid in the usb port and wait for the result?
studying would do you good wouldn't it?
Minerva Ashford | Nov 03, 2009
Here it is: http://webbook.nist.gov/chemistry/fluid/
Here are 18.000 different calculators.
One page split into different category's
http://www.martindalecenter.com/Calculat ors.html
Cheers from DOWNUNDER
Gamerpc | Nov 03, 2009
Terminal velocity calculator; what am I doing wrong?
Jan 09, 2011 by Philip J | Posted in Physics
I've tried several Stokes law calculators, and they all say a .22 inch sphere of lead has a terminal velocity in air of 10.8296 km/s. That's about 1000 times too fast.
http://www.ajdesigner.com/phpstokeslaw/s tokes_law_terminal_velocity.php
My inputs are:
acceleration of gravity = 9.8 m/s²
particle diameter = .22 inch
density of particle = 11.34 g/cubic centimeter
density of medium = 1.2 kg/m³
viscosity of medium = .0000178 kg/m·s
The only parameter that I am not familiar with is viscosity, but I checked multiple sources.
What am I doing wrong?
After much searching, I found an equation for terminal velocity in non-laminar flow.
http://en.wikipedia.org/wiki/Drag_(physi cs)#Velocity_of_a_falling_object
vt = √(2mg/ρACd)
vt is terminal velocity.
For comparing two bullets of similar shape and density, g, ρ, A and Cd cancel.
So terminal velocity is proportional to √(d³/d²) = √d.
Terminal energy is therefore proportional to √d³.
Twice the diameter has 1.414 times the speed and 2.828 times the energy.
Quite a contrast from Stokes law, which yields 4 times the speed and 128 times the energy!
The main problem is that Stokes drag does not apply in this situation.
There are two causers of drag:
Viscous forces
Stagnation pressure forces
Depending on the regime of flow, one of these will greatly dominate the other.
Stokes drag is a laminar flow case calculation. That means it best works with high viscosity and slow flow. I've consulted with Engineering Equation Solver, and indeed that is a reasonable dynamic viscosity for air. At 20 C, EES gives me mu=0.00001825 kg/m-s for dynamic viscosity of air.
The way to figure out whether Stokes drag or Newton drag applies is to calculate Reynolds number
Re = rho*v*[]/mu
Replace the brackets with the characteristic geometry value. In the case of a sphere, that is the diameter. The value of mu is the dynamic viscosity (as opposed to nu, the kinematic viscosity, which is a mathematical shortcut of mu/rho).
OF COURSE THOUGH, your entire goal is to find the velocity, so it is difficult to know Reynolds number in advance. What you do is select either regime, get a "seed velocity", and see if that "seed velocity" gives a Reyonlds number for which your drag model is accurate.
This would yield for your Laminar assumption Re=3.979E+06.
For my turbulent assumption, the yield is Re=19933.
A typical cutoff for an external flow drag Reynolds number is 3500. Higher than that and it is turbulent, lower than that and it is laminar.
What you really need to do to get a result accurate with experimental results is to consider turbulent drag, whereby which the terminal velocity can be calculated as:
v_terminal = sqrt(2*m*g/(rho_fluid *Cd *A_frontal))
Which of course occurs because the turbulent drag force is modeled as:
D = 1/2*Cd*rho_fluid*A_frontal*v^2
For a sphere, of drag coefficient of Cd = 0.47...the formula reduces as:
v_terminal = 2.33*sqrt(m*g/rho_fluid)/d
Where:
m is mass
g is Earth's gravitational field
rho_fluid is the density of surrounding fluid
d is the diameter
This calculates for air at 1.2 kg/m^3 and the mass and diameter of sphere that you mentioned (in San Francisco standard g=9.8 N/kg):
v_terminal=54.26 m/s
gintable | Jan 09, 2011
I think gintable is on the mark
his terminal v is in the ballpark for a lead bullet (150 mph measured, 120 mph calculated)
suggestion:
use an equation designed to find the velocity for a projectile downrange, a ballistic equation
because the airflow at various speeds is not consistent, it could be a messy equation
but
if you took the slope of the velocity curve you would have acceleration
set that equal to g and solve for v
unless I'm totally mistaken, that should be equivalent to terminal velocity
ballistic equations require a ballistic coefficient for the projectile but there are lots of tables
especially for spheres as in lead shot
Old Science Guy | Jan 10, 2011
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