Pressure Drag On A Circular Cylinder Engineering Essay

Published:

The Experiment is to study the flow of air or fluid on a circular body and to understand the pressure distribution around the object which is cylinder in our case. There are laminar and turbulent flows around the object which depends on the value of Reynolds number, we will look how it affects our lab measurements. This is relevant to many engineering applications.

The drag force caused due to the pressure forces exerted by the fluid on a relatively moving object in the streamwise direction is called pressure drag. This is the most dominant component of drag force for blunt shapes like a circular cylinder.

The flow pattern past a circular cylinder at high Reynolds number is shown in Figure 1(a), (b) and (c) [1]. The fluid becomes stationary at the leading windward edge of the cylinder; the pressure of the fluid here is the stagnation pressure. The fluid particles flow around the surface of the cylinder on either side of the stagnation point with an increase in velocity in the streamwise direction. This causes a drop in pressure. Figure 1(b) shows the flow with a laminar boundary layer; here separation of the boundary layer occurs upstream of the maximum thickness, which translates to larger turbulent wake; this causes large pressure force in the streamwise direction. This condition occurs for Reynolds numbers typically less than 2.5Ã-105. On the other hand at significantly higher Reynolds number the boundary layer is turbulent and is much less susceptible to the adverse pressure gradient in the flow. Thus delaying the separation until further downstream from the point of maximum thickness as shown in Figure 1(c). The consequent wake is much smaller and the net pressure forces acting on the cylinder surface are much less as compared to the condition discussed earlier. Thus in low Reynolds number flow, forced transition to turbulent boundary layer is used in several design applications to reduce drag; the dimples on a golf ball are an example of which induce transition of the boundary layer due to the surface roughness.

Evaluation of Drag Force

dCYL

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Figure [2]: Calculation of pressure drag

Consider a cylinder of unit length as shown in Figure [2]. If P is the pressure exerted by the fluid on an infinitesimally small area ds which is located at an angle θ from the leeward point on the surface of a cylinder shown in Figure 2, then the force can be given by,

The streamwise component of this pressure force, which is the pressure drag can be given by,dDp=

The integral of streamwise drag force around the entire surface of the cylinder gives the total drag pressure on the cylinder per unit span d.

Now, it is conventional to work in terms of the non-dimensional drag coefficient,

and pressure coefficient,

Where' D' is the cylinder diameter. We therefore have,

The second integral is zero, giving,

Equipments:

Circular cylinder model with pressure tap on curved surface.

Pitot tube is used for free-stream flow velocity measurement.

Inclined U-tube manometer is used for surface pressure measurement

Procedure:

Place the cylinder model in the wind tunnel test section with the angle of pressure tap set to 180Ëš; Ensure that the model is properly secured to the test section walls.

Record the initial reading from the manometer.

Record the ambient temperature and pressure.

Start the wind tunnel fan. When up to speed, adjust the damper set the free-stream dynamic pressure 10 mmH2O (98 Pa).

Once the flow has attained steady-state condition, record dynamic pressure from manometer.

Record the surface pressure from manometer. Increment the angle of pressure tap by 30° and record the surface pressure from the manometer.

Repeat step 6 for mapping surface pressure up to 360° and continue up to 180°.

Repeat steps 5 through 7 for free-stream dynamic pressure of 10, 15, 20 and 25 mmH2O.

Result and Calculations:

Ambient Air

Pressure (Patm) = 757.562 mm Hg = 101000 Pa

Temperature (Tatm) = 22 ℃ = 295.15 K

Free-Stream Flow Velocity

Free-stream velocity of the flow is calculated from the dynamic pressure measured by the pitot tube (total pressure) and the static pressure tap.

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Set 1: Free-Stream Flow Velocity

Dynamic Pressure (PDYNAMIC) = 5 mmH2O = 49.0319 Pa

Velocity (U∞) = 9.069054 m/s

Set 2: Free-Stream Flow Velocity

Dynamic Pressure (PDYNAMIC) = 10 mmH2O = 98.0638 Pa

Velocity (U∞) = 12.825 m/s

Set 3: Free-Stream Flow Velocity

Dynamic Pressure (PDYNAMIC) = 15 mmH2O = 147.0957 Pa

Velocity (U∞) = 15.708 m/s

Set 4: Free-Stream Flow Velocity

Dynamic Pressure (PDYNAMIC) = 20 mmH2O = 196.1276 Pa

Velocity (U∞) = 18.13809 m/s

Set 5: Free-Stream Flow Velocity

Dynamic Pressure (PDYNAMIC) = 25 mmH2O = 245.1595 Pa

Velocity (U∞) = 20.279 m/s

Density of Air

ρair=Patm/Rair*Tatm =101000/287*295 = 1.1923 kg/m³

Reynolds Number

The Reynolds Number for each set can be calculated relative to the diameter (d) of the cylinder as,

Example for the set 1:

Lab measurements:

Measuring point

surface pressure profile

set 1

set 2

set 3

set 4

set 5

theta

P

P

P

P

P

degrees

Pa

Pa

Pa

Pa

Pa

180

0

0

0

0

0

210

-2

-9

-20

-20

-20

240

-75

-158

-274

-303

-400

270

-155

-295

-440

-580

-720

300

-140

-262

-405

-520

-650

330

-144

-266

-400

-535

-675

0

-141

-276

-405

-540

-676

30

-145

-270

-400

-530

-682

60

-144

-270

-399

-535

-665

90

-140

-265

-393

-525

-660

120

-158

-292

-430

-570

-720

150

-1

-210

-250

-380

-440

 

Set 1

Set 2

Set 3

Set 4

Set 5

Pdynamic (Pa)

49.0319

98.0638

147.0957

196.1276

245.1595

Pstatic (Pa)

 

-49.0319

-98.0678

-147.0957

-196.1676

-245.1895

U∞ (m/s)

 

9.069054

12.825

15.708

18.13809

20.279

Re

 

7555.21

10681.82

13083.05

15107.05

16890.19

Evaluation of Drag Force

Example for the set 1:

Set 1: 180

Dp= -0.0065 * 0 * Cos 180 * π/6 = 0

Dp= -0.0065 * 0 * Cos 180 * 0.523598775 =0 N

210:

Dp = -0.0065 * -2 * Cos 210 * 0.523598775 = -5.6625*10ˉ°Â³ N

240:

Dp=-0.0065 * -75 * cos240 *0.523598775 = -0.12269 N

270:

Dp =-0.0065 * -155 * cos270 * 0.523598775 = 0.4993 N

300:

Dp=-0.0065 * -140 * cos300 * 0.523598775 = -0.22903 N

330:

Dp =-0.00625 * -144 * cos330 * 0.523598775 = -0.4080275 N

0:

Dp=-0.00625 * -144 * cos0 * 0.523598775 = -0.4613 N

30:

Dp=-0.00625 * -145 * cos30 * 0.523598775 = -0.4108 N

60:

Dp=-0.00625 *-144 * cos60 * 0.523598775 = -0.235575 N

90:

Dp=-0.00625 * -140 *cos90 * 0.523598775 = -0.20528 N

120:

Dp=-0.00625 * -158 * cos120 * 0.523598775 = 0.25847 N

150:

Dp=-0.00625 * -1 * cos150 * 0.523598775 = 2.833125*10ˉ°Â³ N

Drag force final values

 

Dp Set1*Radi

Set2*Radi

Set3*Radi

Set4*Radi

Set5*radi

180

0

0

0

0

0

210

-0.00566831

0.025507393

0.056683095

0.056683095

0.0567

240

-0.122741101

0.258574585

0.448414154

0.495874047

0.6546

270

-7.04938E-05

0.000134166

0.00018192

0.000263783

0.0003

300

0.229006437

-0.42856919

-0.662482908

-0.850595339

-1.063

330

0.408052793

-0.753764187

-1.13347998

-1.516029473

-1.913

0

0.461407813

-0.90318125

-1.325320313

-1.76709375

-2.212

30

0.410930462

-0.765180861

-1.133601276

-1.50202169

-1.933

60

0.235625104

-0.441797069

-0.652877891

-0.875412711

-1.088

90

2.12239E-05

-4.01739E-05

-5.95786E-05

-7.95898E-05

-1E-04

120

-0.258491612

0.477718676

0.70348983

0.932533031

1.1779

150

-0.002833852

0.595108832

0.708462895

1.076863601

1.2469

Sum =

1.355238465

-1.935489079

-2.990590051

-3.949014996

-5.073

Coefficient of Drag Force

Example for the set 1:

0

Cd= 0.4613125/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.752671

30

Cd= 0.4108125/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.670276685

60

Cd=0.235575/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.384361308

90

Cd=0.20528/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.00003349323

120

Cd=0.25847/(1/2)*1.1923*9.069054*9.069054*0.0125=0.421716

150

Cd=2.83315*10ˉ°Â³/(1/2)*1.1923*9.069054*9.069054*0.0125=0.004622533

180

Cd=0/(1/2)*1.1923*9.069054*9.069054*0.0125=0

210

Cd=-5.6625*10ˉ°Â³/(1/2)*1.1923*9.069054*9.069054*0.0125=0.009238866

240

Cd=-0.12269/(1/2)*1.1923*9.069054*9.069054*0.0125=0.200179

270

Cd=0.4993/(1/2)*1.1923*9.069054*9.069054*0.0125=0.000114651

300

Cd=-0.22903/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.374037

330

Cd=-0.4080275/(1/2)*1.1923*9.069054*9.069054*0.0125=-0.66573

Coefficient of Drag Cd

(Ѳ) Theta

set 1

set2

set3

set4

set5

0

-0.752808044

-0.736790851

-0.720773659

-0.721

-0.721841472

0.523583

-0.670451928

-0.624213864

-0.61650752

-0.613

-0.630687193

1.047167

-0.384433181

-0.360406108

-0.355066758

-0.357

-0.355066758

1.57075

-3.46278E-05

-3.27728E-05

-3.24018E-05

-3E-05

-3.26491E-05

2.094333

0.42174094

0.389709983

0.382591992

0.3804

0.38437149

2.617917

0.004623559

0.4854737

0.385296587

0.4392

0.406873196

3.1415

0

0

0

0

0

3.665083

0.009248107

0.020808242

0.030827025

0.0231

0.018496215

4.188667

0.200257745

0.210938158

0.243869431

0.2023

0.213608261

4.71225

0.000115014

0.000109449

9.89367E-05

0.0001

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This Essay is

a Student's Work

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This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

Examples of our work

0.000106852

5.235833

-0.373634523

-0.349615161

-0.360290433

-0.347

-0.346946343

5.759417

-0.665756878

-0.614900449

-0.616441553

-0.618

-0.624147073

Sum=

-2.211133816

-1.578919675

-1.626428353

-1.611

-1.655265474

Coefficient of Pressure Evaluation

Cd

-2.211

-1.579

-1.626

-1.611

-1.655

Set

Set 1

Set 2

Set 3

Set 4

Set 5

Measuring Point

Coefficient of Pressure Profile

Set 1

Set 2

Set 3

Set 4

Set 5

θ

Cp

Cp

Cp

Cp

Cp

0

0.753

0.737

0.721

0.721

0.722

30

0.670

0.624

0.617

0.613

0.631

60

0.384

0.360

0.355

0.357

0.355

90

0.000

0.000

0.000

0.000

0.000

120

-0.422

-0.390

-0.383

-0.380

-0.384

150

-0.005

-0.485

-0.385

-0.439

-0.407

180

0.000

0.000

0.000

0.000

0.000

210

-0.009

-0.021

-0.031

-0.023

-0.018

240

0.000

-0.211

-0.244

-0.202

-0.214

270

0.374

0.000

0.000

0.000

0.000

300

0.666

0.350

0.360

0.347

0.347

330

0.753

0.615

0.616

0.618

0.624

Cp

3.164

1.579

1.626

1.611

1.655

Coefficient of Pressure equation:

Cp=P-P∞/0.5*ρair*U∞²

Example for set 1

Cp180 =0+1294/0.5*1.1923*(9.069054)²=26.666

Cp210=-2+1294/0.5*1.1923*(9.069054)²=26.6248

Cp240=-75+1294/0.5*1.1923*(9.069054)²=25.1204

Cp270=-155+1294/0.5*1.1923*(9.069054)²=23.47187

Cp300=-140+1294/0.5*1.1923*(9.069054)²=23.78098

Cp330=-144+1294/0.5*1.1923*(9.069054)²=23.69855

Cp0=-141+1294/0.5*1.1923*(9.069054)²=23.76038

Cp30=-145+1294/0.5*1.1923*(9.069054)²=23.67795

Cp60=-144+1294/0.5*1.1923*(9.069054)²=23.69855

Cp90=-140+1294/0.5*1.1923*(9.069054)²=23.78090

Cp120=-158+1294/0.5*1.1923*(9.069054)²=23.41005

Cp150=-1+1294/0.5*1.1923*(9.069054)²=26.64542

Variation of Coefficient of Pressure with θ

As velocity near the surface of the cylinder is a function of θ, the local static pressure (P) is also influenced by θ. Since this pressure distribution on the surface is used to determine the force acting on the cylinder, it is plotted against θ.

Graph 1; Pressure Co-efficient versus for "set 1"

Graph 2; Pressure Co-efficient versus for "set 2"

Graph 3; Pressure Co-efficient versus for "set 3"

Graph 4; Pressure Co-efficient versus for "set 4"

Graph 5; Pressure Co-efficient versus for "set 5"

Graph 6; Drag versus Reynolds Number

Discussion:

The measurement is taken from the cylinder with a small whole in the middle and the measurement are taken with reference to that point, the fluid becomes stationary at the leading windward edge of the cylinder; the pressure of the fluid here is the stagnation pressure at this point the total pressure is '0' zero, i.e. theta 180 is zero. The fluid particles flow around the surface of the cylinder on either side of the stagnation point with an increase in velocity in the streamwise direction. This causes a drop in pressure. The flow corresponds to the Reynolds number when the Reynolds number is <0.1 then the flow around the object is normal it shows the object is not moving with great velocity.

But when R=10³ to 2.5*10^5 then velocity at the stagnation point is 0, shown on the fig1(b) the flow with a laminar boundary layer. Here the separation of the boundary layer occurs upstream of the maximum thickness, which gives a larger turbulent wake or more drag force when compared to fig (a) or (c). This causes large pressure force in the streamwise direction if R is typically less than 2.5*10^5 and at the separation point the pressure decreases and the velocity increases since velocity and pressure are inversely proportional to each other laminar flow is smooth flow and no energy is given to the surrounding layers.

In the fig (c) the Reynolds number R>2.5 the boundary layer is turbulent and is much less susceptible to the adverse pressure gradient in the flow, thus delaying the separation until further downstream from the point of maximum thickness as shown. The wake or the drag force is less when compared to the flow on the fig (b) here the flow is turbulent, this has the property of mixing with adjacent layers.

From the lab measurements we came to know that R is more than 2.5*10^5 from 1-5 sets, we come to know that the flow is turbulent. We can notice that the measured pressure is decreasing from set 1 to set 5 it can be shown at measuring point at 270 deg, where the pressure for set 1 is -155 pa and for set 5 was -720pa. So from this we come to know the velocity increases since it's inversely proportional to the pressure which can be checked by the calculated equations for Free-stream flow velocity 'U∞' where the velocity for set1 is (9.0690m/s) and for set 5 is (20.279m/s).

From the Graph 6 shows the Drag against Reynolds number. At the beginning the drag was too less at Cd= -2.21 at R=7400 and it increases rapidly Cd=-1.59 vs R=10683 from there we can see slight fluctuations in the graph where Cd=-1.62 vs R=13083 point where the drag has a slight decrease when compared to the previous point and so on.

Conclusion:

The objectives of the experiment have been achieved by successfully obtaining the results and the pressure distribution has been analysed. There were some difficulties at some stage which have been solved by finding another helpful guiding book; there may be some errors in the log book due to calculation faults, computational and reading errors.