Calculating Water Viscosity from Flow Rate

[JosephL]

Homework Statement

Find the viscosity of water given that;

flow rate ($$cm^3 s^{-1}$$) = 2.0
height from tube (m) = 0.1

l (m) = 0.2
r (mm) = 1
$$\rho$$ = 1000
g = 9.81

Using the equations;

x = $$\frac{Flow Rate}{Height}$$

x = $$\frac{\pi \rho gr^4}{8l\eta}$$

The Attempt at a Solution

Okay so first, and the most important bit I don't know if I'm doing right I convert the flow rate from $$cm^3 s^{-1}$$ to $$m^3 s^{-1}$$ by multiplying by $$10^{-6}$$

So flow rate is now $$2 * 10^{-6} m^3 s^{-1}$$ ***I don't know if that bit is correct.***

Then x = $$\frac{2 * 10^{-6}}{0.1} = 2 * 10^{-5}$$

and $$2 * 10^{-5} = \frac{\pi \rho gr^4}{8l\eta}$$

After re-arranging and substituting I get;

$$\eta = 9.6*10^{-4}$$------

If anyone could check my answer it would be greatly appreciated I haven't done physics for so long I've pretty much forgotten everything.

Thanks in advance

 

[joeyar]

I got the exact same answer you got, using your data.

 

[thomate1]

Hi there,

I can confirm that your approach and calculations are correct. You correctly converted the flow rate from cm^3 s^{-1} to m^3 s^{-1} and used the appropriate units for the other variables. Your final answer for the viscosity of water is also correct.

One suggestion I would make is to include the units in your final answer, which would be 9.6*10^{-4} Pa s. This will help to ensure that your answer is clear and complete.

Overall, great job on your calculations and good luck with your studies!