Experiment 2

Lab 2
Fluid Dynamics


The purpose of this lab was to understand further application for Bernoulli's Equation, which shows conservation of pressure in Fluid Dynamics (moving fluids). The Bernoulli's Equation says:

 p1+ ρ1gh1 + 1/2 ρ1v1^2 = p2 + ρ2gh2 + 1/2 ρ2v2^2

This lab used the following materials:
1. A bucket (rather cylindrical) with a drilled small hole on the lower side
2. Tap water
3. Ruler/ measurement
4. Stopwatch

The objective/ goal of this lab was to successfully, as accurate as possible, measure the radius of the drilled hole given a limited information. The height of the water surface (to the hole) and the time for the water to fill up a certain volume (16 ounces). Based on those scarce information, the radius of the hole, using bernoulli's, could be calculated to a certain accuracy.

Based on the equation,  p1 = p2 (both atmospheric pressure). At the top, h was measured to be 0.067 m +/- 0.0005 and having a velocity of 0. At another location (the hole), height is 0, and has a certain velocity which can be measured using conservation of energy. 

1/2 v^2 = gh

v = (2gh)^1/2

Also recall that Rate flow is measure of Volume per second (m^3 / s), or equivalently, Area * velocity (which also give m^3 /s). The area is the area of the hole, which was the shape of (assumed perfect) circle,  (π r^2)

Rate flow:

R = V / t; whereas t and V were both known. 

V/t = A v,whereas v (velocity) can be found by above method, (2gh)^1/2, and area π r^2.

Substituting the values, such formula is obtained:

V / t = π r^2 (2gh)^1/2.

r then, is

r =  (V / ((2gh)^1/2 π t))^1/2

Procedures

A volume of 16 ounces was to be obtained per experiment. 16 ounces is equivalent to 473 mL; since the beaker showed the unit increments of 10 mLs, the value including error would be 473.0 +/- 5.0 mL. The height of the water was measured to be 0.067 +/- 0.0005 m. We only measured the height once, because we were assuming the height would be relatively constant since the water used would be recycled over into the bucket. 

The experiment was done 6 times, each with time as follows:

1. 17.55 +/- 0.005 s

2. 17.08 +/- 0.005 s

3. 17.54 +/- 0.005 s

4. 17.50 +/- 0.005 s

5. 17.46 +/- 0.005 s

6. 17.50 +/- 0.005 s

Measured directly, the hole had a radius of 0.325 +/- 0.0005 cm (0.00325 +/- 0.000005 m)

t Calculation:

Using the direct measurement of the hole, theoretical time could be found using

t = V / (A (2gh)^1/2);

which yields t of 12.4 s

Analysis and Error:

Using 

((∂F/ ∂x * uX)^2 + (∂F / ∂y * uY)^2...)^(1/2),

whilst

Time

The average time was found to be A = (t1 + t2... +t6)/ 6 = 17.44 s

The accumulated error would be:

((∂E / ∂E1) uE1 + (∂E / ∂E2) uE2... + (∂E / ∂E6) uE6)^1/2; whereas En = the nth Error.

The average time, including error then, is 17.44 +/- 0.002 s

The ideal calculated time is 12.40s. It did not agree with the experimented time, even when uncertainties were included (17.42 - 17.46).

Time, in this case, was found using t = V / (A (2gh)^1/2), which implies that it was affected by volume, hole area, and height between hole and bucket's water surface. 

Analysis:

Our group counted only the volume and height only once, and assuming that they stayed constant. Throughout the experiments, water clearly, although not substantially, was lost when flowing from bucket into the beaker. Those water loss accumulated, and eventually, by the 5th or last try, the volume and height (hence the total pressure) was no longer the same as the first experiment.

The volume should have been measured upon each experiment; at the end, the average of those six measurements should be obtained.

Radius/ diameter of hole

Measured radius: 0.325 +/- 0.0005 cm = 0.00325 +/-0.000005 m

Calculated/ experimented radius (using the method mentioned at the beginning): 0.274 cm = 0.00274 +/- ur

r =  (V / ((2gh)^1/2 π t))^1/2

Whereas 

V = 0.472 +/- 0.005 L

h = 0.0067 +/- 0.0005 m

t = 17.44 +/- 0.002 s

∂r/∂V = 1/2 (V / t π(2gh)^1/2) ^ -1/2 * 1 = 3.2

∂r/∂h = - 1/4 (V / (π t (2gh)^1/2))^(-1/2) V( πt (2gh)^1/2)^(-2) (2gh)^(-1/2) 2g = 184.8

∂r/∂t = - 1/2 (V /(π t (2gh)^1/2))^(-1/2) V((2gh)^1/2 π t)^(-2) (2gh)^1/2 t = 0.0065

r's total uncertainty is:

ur=((∂r/∂V) uV)^2 + ((∂r/∂h) uh)^2 + ((∂r/∂t) ut)^2)^1/2 

= 0.09378 m

Experimented radius therefore, is 0.00274 +/- 0.09378 m or 0.274 +/- 9.38 cm.

Or twice of that, the diameter, is 0.548 +/- 9.38 cm

The calculated/ experimented radius/ diameter is definitely within the range of uncertainty error. 

% difference = 15.69%

So, 

Given diameter (measured directly) = 0.00650 +/- 0.000005 m

Calculated/ experimented diameter = 0.00548 +/- 0.09378 m

Error is within the uncertainty.