Monday, September 5, 2011

Experiment 2 : Fluid dynamics


In this experiment the goal is to obtain the amount of time required for 200 ml of water to transfer from a large bucket to a 200 ml marked beaker underneath with two different method. First is to use the following setup and then theoretically.




A. 
Trials
1
2
3
4
5
6
Time to empty (tactual)
12.2 ± 0.1
12.1 ± 0.1
11.4 ± 0.1
11.6 ± 0.1
11.5 ± 0.1
12.1 ± 0.1
S = ((S(xi –xavg)⁄n-1) = 0.4
taverage = 11.9 ± 0.4

                                   Theoretically




dhole  = 0.5 ± 0.05cm = 0.005 ± 0.0005 m
h = 4.5 ± 0.05cm = 0.045 ± 0.0005 m
Areahole = ¼πd2 = 0.196 × 10-4 ± 0.0025 × 10-4 m
g = 9.81 m/s2
V = 2.1 × 102 ± 0.15 × 102 ml = 0.21 ± 0.015 L = 0.00021 ± 0.000015 m3
Beside all the errors in measuring different quantities, since the poured water made the top of the water in the 200 ml beaker turbulent every time we stop the flow the actual amount of water was higher than 200 ml, which is why we choose 210 ml with uncertainty of 15 ml.  This is the reason behind the large uncertainty for the final time.
ttheoretical= V⁄(A√2gh) = 11.4 ± 0.9 s

Error = 1 - ttheoretical taverage = 0.042 or 4.2%
Both values agree within their uncertainties. The huge uncertainty in the theoretical time is due to the big uncertainty in volume and also for the fact that the area is of the hole that let the water pass through is not exactly circle so the actual value should be smaller which makes the ttheoretical bigger and closer to experimental value.
If we rearrange the  ttheoretical= V⁄(A√2gh) to solve for d we get:
d = ((4V)/(πtactual2gh)) = 4.8 × 10-3 m
Error for d = 0.038 or 3.8 %

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