Formal Lab 1g11r

ANSWER SHEET

EXPERIMENT: Drag Measurement on Cylindrical Bodies

DATE OF EXPERIMENT: 10 APRIL 2019

DATE OF SUBMISSION: 17 APRIL 2019 LECTURER’S NAME: DR SHAMSUL BAHARI BIN AZRAAI

GROUP :

NAME MUHAMAD SYAZWAN BIN ABDUL KADIR

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MATRIC NUMBER B041510140

ANSWER SHEET 1. OBJECTIVES To determine the drag coefficient of a circular cylinder using direct weighing and pressure distribution methods.

2. INTRODUCTIONS Fluid dynamic drag is the force resisting the motion of a body through a fluid, or, equivalently, the force of a moving fluid on a stationary body. It is necessary to know the drag force in order to calculate for the thrust required for an airplane to fly, the power required to propel a ship at the desired speed, or the necessary to keep a building upright on a windy day. In this experiment we will determine the drag on a circular cylinder using three different methods, any of which may be used for drag measurement on other bodies under similar flow conditions.

3. THEORY The resistance of a body as it moves through a fluid is of great technical importance in hydrodynamics and aerodynamics. In this experiment we place a circular cylinder in an air stream and measure its resistance, or drag by two methods; direct weighing method and pressure distribution method.

The curve had shown in Figure 1represents a section of an oval cylinder. Motion of the cylinder through stationary fluid produces actions on its surface, which give rise to a resultant force.

At any chosen point A of the surface of the cylinder, the effect of the fluid may conveniently be resolved into two components, pressure, p, normal to the surface and shear stress along the surface.

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ANSWER SHEET L p s

τδs

δs

δD



L P

τ

M

A A U

C

d

D

C P

Figure 1 A section of an oval cylinder

Let U denote the uniform speed of the motion and  the density of the fluid, then the ½U2, is

dynamic pressure in the undisturbed stream,

½U2 = Po - po

(1)

Where Po is the total pressure and po is the static pressure in the oncoming stream. This pressure is a useful quantity by which the gauge pressure p and shear stress  may be dimensionalised, and the following dimensionless term are defined Pressure coefficient, =

p − po ; ½ U 2

Skin friction coefficient = c f =

3

 1 U 2 2

(2)

(3)

ANSWER SHEET The combined effect of pressure and shear stress gives rise to resultant force on the cylinder. This resultant may conveniently be resolved into the following components acting at any chosen origin C of the section as shown in Figure 1. a. A component in the direction of U, called the drag force, of intensity D per unit length of cylinder. b. A component normal to the direction of U, called the lift force, of intensity L per unit length of cylinder c. A moment about the origin C, called the pitching moment, of intensity M per unit length of cylinder

These components may be expressed by definition of drag coefficients as follows:-

Drag coefficient, CD = Lift coefficient, C L =

D ½ U 2 d

(4)

L

(5)

1 U 2 d 2

Pitching moment coefficient, CM =

M 1 U 2 d 2 2

(6)

In which d denotes a suitable dimension which characterizes the size of the cylinder. In Figure 1 this is shown as the width measured across the cylinder, normal to U.

We may see how pressure and skin friction coefficients are related to lift and drag coefficients. Theoretical calculation shows that the drag of a cylinder may be found by measuring p and τ over the surface. For the case of circular cylinder, the effect of skin friction is very small compared to pressure drag and therefore may be neglected. This assumption allows us to calculate CD from the measured pressure distribution over the cylinder surface. In this “Pressure distribution method”, CD can be calculated as 2

CD =

 c Cos  d

(7)

p

0

Alternatively, by plotting cos θ against θ, CD may be obtained from the area beneath the curve. The area A beneath the mean curve is :4

ANSWER SHEET 

A =  c p Cos  d

(8)

0

which from equation (7), we recognize as the drag coefficient. This integration can be evaluated in various ways such as by using Simpson’s or the trapezium rule etc. (Hint: Refer to your Numerical Method notes on how to calculate area under a curve) In “direct weighing method”, the drag force is written as Dl, that is the product of the drag per unit length and the length l of the cylinder. Dl is measured in units of gram-force (gmf). By substituting Dl in equation 4, the experimental CD by direct weighing method can now be calculated as

Drag coefficient, CD =

Dl 1 U 2 dl 2

(9)

Where Dl is the experimental measured drag force in gmf, d is the diameter of cylinder and l is the length of cylinder

Assuming that the fluid is incompressible and non-viscous, the following theoretical formula can be applied:

p = pa − p0 = p=

1 1 U 2 − u 2 2 2

(10)

)

(11)

(

1 U 2 1 − 4 sin 2  2

=

p 1 U 2

2

= 1 − 4 sin 2 

(12)

Equation 10 to equation 12 is the theoretical result for an incompressible, inviscid fluid, and forms the basis of comparison with experimental results.

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ANSWER SHEET 4. PROCEDURE AND APPARATUS 4.1 APPARATUS The AF10 Airflow Bench and AF12 Drag Measurement Apparatus is shown in Figure 2.

Figure 2 Diagram of apparatus Given data: Diameter of cylinder, d

= 12.5 mm

Length of cylinder, l

= 48 mm

1 mmH2O

= 0.0981 mbar

1 gmf

= 981 x 10-3 N

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ANSWER SHEET 4.2 Procedure A) Direct Weighing Method

1. Weighing arm fixed to the balance position by adjusting the adjustable counter balance. 2. Adjustable counter balance is clamped when the weighing arm at balance position and equipment is switched on. 3. Wind speed is opened and brought up to the maximum. 4. Weights are added to the scale pan. The wind speed is trimmed so that the weighing arm at balance positions. The balance position is a position when the weighing arm did not spring back when being push downward. Measure it as a drag force. 5. Value of ΔHo (pressure head caused by total pressure Ps) and Δho (pressure head caused by static pressure Po) is recorded in Table 1. 6. Step 4 until 6 is repeated by decreasing the weight in amount of 5 gm for each experiment for at least seven times.

B) Pressure Distribution Method

1. The circular cylinder is fixed with the protractor mounted in place, and equipment is switched on. 2. The wind speed is opened at some convenient value near the maximum. 3. The protractor is set up to 0°. The value of ΔHo (pressure head caused by total pressure Po), Δho (pressure head caused by static pressure Po) and Δh (pressure head caused by surface pressure p) is recorded in Table 2. The value of ΔHo and Δho are constant. 4. Step 3 is repeated by setting the protractor to another value. It is recommended that 20° intervals be used until it complete to 180°.

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ANSWER SHEET EXPERIMENTAL DATA Table 1 Drag Force Measured by Direct Weighing Drag Force (gmf) 32

 Ho (mm) 62

 ho (mm) 16

27

52

12

22

42

8

17

32

6

12

22

2

7

12

0

2

2

-2

Table 2 Pressure Distribution around a Cylinder

(Deg.) 00

h (mm) 60

200

48

400

2

600

-46

800

-54

1000

-48

1200

-48

1400

-48

1600

-50

1800

-50



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ANSWER SHEET Note: 1. The pressure should be relatively symmetrical about the line  = 0o, so assume the values of pressure are similar for reverse direction ( = -20o until -180 o). 2. The initial value of pressure head are shown in below: 𝐻0,𝑖=90𝑚𝑚 ℎ0,𝑖=88𝑚𝑚 ℎ𝑖=88𝑚𝑚

3. Δ𝐻0 is constant: 62 mm, so pressure head caused by total pressure

𝑃0=𝜌𝑔Δ𝐻0 𝑃0=1000×9.81×0.062 𝑃0=608.22𝑁/m^2 4.

Δℎ0 is constant: 16 mm, so pressure head caused by static pressure 𝑝0=156960𝑁𝑚2⁄. 𝑝0=𝜌𝑔Δℎ0 𝑝0=1000×9.81×0.016 𝑝0=156.96𝑁/𝑚^2

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ANSWER SHEET EXPERIMENTAL RESULT Table 3 Drag Force Measured by Direct Weighing Drag Force (gmf) 32

 Ho (mm) 62

Po (N/m2) 637.65

 ho (mm) 16

po (N/m2) 156.96

Po – po = ½U2 (N/m2) 480.69

27

52

510.12

12

117.72

392.40

22

42

412.02

8

78.48

333.54

17

32

313.92

6

58.86

255.06

12

22

215.82

2

19.62

196.20

7

12

137.34

0

0

137.34

2

2

19.62

-2

-19.62

39.24

Sample Calculation: 𝑃0=𝜌𝑔Δ𝐻0 𝑃0=1000×9.81×0.062 𝑃0=637.65𝑁/𝑚^2 𝑝0=𝜌𝑔Δℎ0 𝑝0=1000×9.81×156.96 𝑝0=480.69𝑁/𝑚^2 𝑃0−𝑝0=637.65 – 156.96 = 480.69Pa

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ANSWER SHEET Table 4 Pressure Distribution around a Cylinder  (Deg.) 00 200 400 600 800 1000 1200 1400 1600 1800

h (mm)

p (N/m2)

60.00 48.00 2.00 -46.00 -54.00 -48.00 -48.00 -48.00 -50.00 -50.00

588.60 470.88 19.62 -451.26 -529.74 -470.88 -470.88 -470.88 -490.50 -490.50

p - po (N/m2)

p − po 1 U 2 2 (N/m2) 0.96 0.70 -0.30 -1.35 -1.52 -1.39 -1.39 -1.39 -1.44 -1.44

=

431.64 313.92 -137.34 -608.22 -686.84 -627.84 -627.84 -627.84 -647.46 -647.46

Sample calculation : 𝑃0=608.22𝑁𝑚2⁄ 𝑝0=156.96𝑁𝑚2⁄ ½⁄𝜌𝑈^2=𝑃0−𝑝0 ½⁄𝜌𝑈2=608.22−156.96=451.26𝑁𝑚2⁄ 𝑝=𝜌𝑔Δℎ 𝑝=1000×9.81×0.06 𝑝=588.6𝑁/𝑚^2 𝑝−𝑝0=588.6−156.96=431.64𝑃𝑎 𝐶𝑝=(𝑝−𝑝0)/(½⁄𝜌𝑈^2)=(431.64)/(451.26)=0.96 𝐶𝑝cos𝜃=0.96cos(0)=0.96

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Cos  0.96 0.65 -0.23 -0.68 -0.26 0.24 0.70 1.07 1.35 1.35

ANSWER SHEET DISCUSSION

1. For the direct weighing method, plot the drag force against the dynamic pressure. Establish the slope and calculate the drag coefficient, 𝐶𝐷.

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ANSWER SHEET 2. For the pressure distribution method, plot the graphs of 𝑐𝑝 and 𝑐𝑝cos 𝜃 as functions of angle 𝜃. Determine the drag coefficient, 𝐶𝐷.

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ANSWER SHEET 3. Discuss the difference of the values obtained for the drag coefficient by those two methods. The direct weighing and pressure profiles give the values of coefficient of drags that show close agreement in the range of angle of incidence between 0˚ to 20˚. Begin with 20˚ to 180˚, the drag coefficient value of pressure distribution method starts deviating from the drag coefficient value of direct weighing method. The method which come out from drag coefficient, the theoretical value is pressure distribution method, while the drag coefficient found from the method that are experimental value is direct weighing method. Nevertheless, the most reliable method is direct weighing method because the pressure method ignored the existed skin friction.

4.

Discuss the differences of theoretical and experimental values of drag coefficient, 𝑪𝑫. The theoretical and experimental values of drag coefficient are totally a little bit different. The difference between both values are around 0.04++. Ignoring the skin friction, this experiment show that there is no a big different in getting the result. So, we can ignore the skin friction because its effect in the results is too small.

5. State the possible errors expected in this experiment and suggestions to reduce the errors. First, the possible error that can be found throughout the experiment is the parallex error which is when taking the data of Δ𝐻𝑜 and Δℎ𝑜, which are the observer might not accurate. The position of eyes must be parallel to the scale and take the reading from 2 to 3 times to get the average of the reading accurately. Then, the counter balance must be balance before starting the experiment because it will affect the experiment result. For example, it should on centre between up and down before starting the experiment.

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ANSWER SHEET CONCLUSION

This experiment are conducted to determine the drag coefficient of circular cylinder using direct weighing and pressure distribution methods. From this experiment we have calculated the valued of the drag coefficient from both methods which are not a big different. The value of drag coefficient for direct weighing method is the experimental value while the value of drag coefficient for pressure distribution method is the theoretical value. This show us that there are errors occurred during this experiment conducted, such as parallax errors and the unbalanced counter balance. These kinds of errors can lead to inaccuracy when taking the reading which eventually affecting the value of the drag force when calculated. As a conclusion, the student able to analyzed using pressure distribution method based on the experiment had been done. Students also already know about the drag force phenomena. Finally, the objective is achieved regarding the analyze the pressure distribution method and the hypothesis is accepted whis is angular rotation become bigger, the flow of pressure coefficient getting slower.

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