Undergrad Stoke's equation -- particle size

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To solve Stokes' equation for particle diameter, consistent unit conversion is essential, particularly for viscosity, which should be in mPa·s. The equation can be rearranged to isolate the particle size (d) as d = sqrt((18µV)/(g(Ps - Pw))). Using dimensionless variables is recommended to simplify calculations and analyze flow profiles effectively. The key is to ensure all units are uniform to avoid errors in the final result. Properly applying these principles will lead to an accurate determination of particle size.
Kbaba
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I want to solve stokes eq. for particle diameter in um size. Do i have to change the units to make them uniform. please help in solving the eq. by plugging these numbers.
velocity= 50 mm/h
Particle density= 2650 kg/m^3
Medium density= 1000 kg/m^3
viscosity = 1.002 mPa
Gravity= 9.8 m/s^2
 
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As long as you work with units consistently, you cannot make anything wrong.
Some unit conversions can be necessary to get a nice unit for the result.
 
Kbaba said:
I want to solve stokes eq. for particle diameter in um size. Do i have to change the units to make them uniform. please help in solving the eq. by plugging these numbers.
velocity= 50 mm/h
Particle density= 2650 kg/m^3
Medium density= 1000 kg/m^3
viscosity = 1.002 mPa
Gravity= 9.8 m/s^2

Your units of viscosity are incorrect, could/should be mPa*s.
 
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Thanks for advise. Also, I want to solve the following equation for particle size (d) using stokes eq
V= g (Ps – Pw) d2/18 µ

for unknown variable (d) i rearranged the eq. as follow and took the square root of my answer. Is this the correct way to rearrange and solve this eq for particle size determination. Please, advise.

d= 18 µ V/ g (Ps – Pw)
 
Something went wrong with the formatting. I guess you mean this:$$V= \frac {g(P_s-P_w)d^2}{18 \mu}$$
Solved for d:$$d= \sqrt \frac{18 \mu V} {g(P_s-P_w)}$$
Sure.
 
I would recommend that you use dimensionless variables. It will greatly simplify your calculations if you are planning to find the flow profile.
 

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