# Coefficient of viscosity equation/answer issue

So I feel like my current answers are wrong but I am unsure, i feel like some could be a little high at -223 but I dont have the best experience in the field as this is my first time working with viscosity measurements.

I have editted this so the equation I should have been using is near the bottom of the post

Given values, r is 1.59mm, 2.38mm and 3.175mm, g im using as 9.81, Ps im using as 7.850 for the density of steel as theyre steel balls, P1 ranges from 920, 1000, 1020 and 1430 as the liquids I am using are olive oil, water, washing liquid and syrup, finally u is the average velocity so, 24.46 for water, 54.16 for olive oil, 17.5 for washing liquid and 2.505 for syrup ms.

Just to note I think I did the equation wrong and meant to write 2 / 9 x 1.59^2 x 9.81 (7.850 - 1000) / 24.46, which even with that equation I get similar to my answer at -223.54.

Any help would be appreciated as I feel im missing something

• This is one use for a spreadsheet... Aug 15, 2020 at 4:17
• @Solar Mike it was 4am when I wrote that out, while I have a train to catch in the morning so was pretty pressed for time but ive handed in the work after doing some extra research Aug 15, 2020 at 4:25

This is another example of the importance of using consistent units, for example, use SI only, Also, immediately on getting negative values for viscosity the OP should know that something is dead wrong. He shows the density of steel as 7.850 (no units), steel is about eight times the density of water which is 1000$$kg/m^3$$ so $$\rho_s = 7.850 g/cm^3 = 7850 kg/m^3$$. I did the calculation for water:
$$\mu = \frac 2 9 \frac{.00159^2 m^2 9.81 \frac m {s^2} (7850-1000) \frac {kg}{m^3}} {24.46 \frac m s}$$
$$\mu = .001543 \frac {kg}{s\cdot m}= .001543 \frac{N \cdot s}{m^2}$$
Looked up viscosity of water, found $$.0010518 \frac{N \cdot s}{m^2}$$ , (from https://www.engineeringtoolbox.com/water-dynamic-kinematic-viscosity-d_596.html)