# Season of Migration to the North and Orientalism

Topics: Regression analysis, Atmospheric thermodynamics, Linear regression Pages: 5 (834 words) Published: January 16, 2013
Date: 29/06/2009 .

Bernoulli’s Law

Name: Hibah Ismail . Section: 1 . Partner’s Name: Mohammad Kanso .

A) Drag Coefficient
What happens when you give the ball a gentle sideways push? Why?

When we give the ball a gentle sideways push, it returns to its initial position above the fan. The air particles around the fan have a velocity equal to zero since they are still air. Therefore, the velocity of the air particles above the fan is higher than the velocity of the air particles around the fan. Thus, according to Bernoulli’s Law:

ghf + Pf/ρ + Vf2/2 = ghs + Ps/ρ + Vs2/2
But hs = hf So : Pf/ρ + Vf/2 = Ps/ρ + Vs2/2

So as the velocity increases, the formula suggests that the pressure decreases. Since the velocity above the fan is larger than the velocity around it, the pressure around the fan is larger than the pressure above the fan. The net force F = ΔP.A that acts on the ball is from the high pressure area(surrounding) to the low pressure area(above the fan) which makes the ball return to its initial position.

Recorded air velocity = 5.4 +/- 0.5 m/s.
Air density = 1.3 kg/m3

Diameter= (5.040 +/- 0.005) ×10-2m
[pic]=[pic]

[pic]= [pic]=[pic]= 1.6×10-5 m2
Ball area= (7.980 +/- 0.016)[pic]m2
Ball mass = (3.30 +/- 0.05)[pic]kg
[pic]

Now F of air resistance equals the ball’s weight since it is just lifted above the fan (held in equilibrium).

W = Weight of the ball = [pic] = F
[pic]= 0.2140

[pic]= [pic]

= [pic]= 0.04

[pic](0.21+/-0.04)

B) Flow rate

|Test point |Cross-sectional |Area(m2) |1/A |Velocity(m/s) | | |Diameter (cm) | | | | |1 |4.5 |1.59[pic] |6 629 |14.2 | |2 |6 |2.83[pic] |3 353 |5.4 | |3 |8.5 |5.67[pic] |1 176 |4.3 | |4 |10 |7.85[pic] |1 127 |2.9 |

Plot v vs 1/A. From the slope determine the air flow rate in m3/s along with its rms error.

[pic]

Flow Rate and rms error:
Flow rate:[pic]
When plotting v = R(1/A) we get a linear relation with a slope = R. Using linear regression on the calculator we get:
Slope = R = 0.0217 [pic]
Y-intercept = -0.271 m/s
And the correlation coefficient r = 0.965

So Y=0.0217x - 0.271

[pic]

Error on slope [pic]with [pic]

and [pic]= y- yi = Axi + B – yi

[pic]= 5.38 m2/s2
[pic]5.68×105 m-4

[pic]=1.29 ×103 m-2

[pic][pic]= 4×5.68×105-(1.29 ×103)2
=607900

[pic]=> = 1.77[pic]10-5 m3/kg
So the error = 4.21[pic]10-3
= 0.004

R = (0.022 [pic]0.004) [pic]

C) Determining air density

|Test point |Pressure(Pa) |Velocity(m/s) |v2 | |1 |13 |4.6 |21.16 | |2 |27 |6.6 |43.56 | |3 |53 |9.2 |84.64 | |4 |57 |9.6 |92.16...