Mech 2600 Fluid Mechanics

8/3/2019 Mech 2600 Fluid Mechanics

1/19

MECH 2600 FLUID MECHANICS

LABORATORY ASSSIGNMENT:

EXPERIMENT ONE: FORCE ON A SUBMERGED BODY

Aim:

The main aim of this experiment was to find the magnitude of the force F subjected to the vertical

surface of the float and also to calculate the point on the vertical surface at which the force acts in

relation to the liquids surface.

Apparatus:

Method:

For this experiment, the apparatus was set up in the manner shown above with a float and

counterweight suspended on a pivot. The tank was filled up until the water line reached the top ofthe vertical surface of the float. At this point, due to force of the water on the vertical wall of the

float, the system tilted around the pivot and the counterweight dropped. The next step was to take

some measurements of the apparatus. This included the area, A, of the vertical surface of the float

(bd), the depth of the centre of area (y) (d/2), the distance of the bar to which the float was

connected to the waters surface (q) and the distance from the pivot to the position where further

weight will be added. After this, weights were added to the end of the bar on the same side of the

pivot as the float in increments of 0.05kg until the float and counterweight began to even out and

balance on the pivot when smaller weights were then added. When the float and counterweight

were balanced in equilibrium over the pivot, no more weights were added. A record of the added

weight was then taken.

Results:


2/19

(a) Calculate the magnitude of the force F acting on the vertical surface.b = 0.075m d = 0.1m q = 0.1m

Thus:

A = 7.5 x 10-3

m2

y = 50 x 10-3

m = 1000kgm

-3

g = 9.81N

( ) ( )

()(b) Using the apparatus provided, experimentally determine y.

( )

(

)

(Ignore minus sign for distances)

(c) Calculate the theoretical value of y and compare to (b).

As shown above, the theoretical value of y is 0.0667m whereas the experimentally determined

value of y is 0.0832m.The percentage error for this is 19.8%.

Analysis:

Firstly, I was asked to find the force acting on the vertical surface which caused the float to

move upwards in rotation to the pivot. This happened because the forces from the water on the

curved surface of the float act directly in line with the pivot and thus has no affect on the movement

of the float. Therefore only the force on the vertical surface is the only influential force and needed

to be found.

To find y the equation, ( ), was used by the method of pivots and moments.The values of mg are the weights added to the floats bar; the x value is simply the distance from the

pivot to the weights; F was calculated in part (a) and q is the distance from the pivot to the waters

surface. By rearranging the equation to make y the subject, we can work out its experimental value.

The possible reasons for the percentage difference between the experimentally determined

value for y and the theoretical value include parallax error, which concerns the user taking

readings of the water level, and specifically the level of meniscus in terms of eye level, accurately

and precisely; and also random error, which is the inaccurate yet unavoidable error caused by the


3/19

apparatus available. For example, the rulers were only accurate to 1mm which leaves a margin of +-

0.5mm for error when measuring distances.

Conclusion:

I think that the aim set out for this experiment was achieved and the further calculations madetherefore proved that bearing in mind the percentage differences although they were small, the

measurements were accurate enough to prove that the theory is relevant.

EXPERIMENT TWO: CONSERVATION OF MASS AND ENERGY

Aim:


4/19

The aim of this experiment was to use strategically positioned manometers along a tapered venturi

shaped nozzle which opens up again to calculate the velocity and mass flow rate of the liquid at

certain points in the tube.

Apparatus:

The apparatus consists of a wide tube containing water that runs into a tapered venturi nozzle in asmaller diameter tube then back to a wide tube again. At certain intervals along this tube are much

smaller vertical single limb manometer tubes that are required to measure the pressure at those

points. Alongside these manometer tubes are measuring lines marked by numbers to record height

of the water levels in them.

Theory:

Mass flow rate: Conservation of mass: 1 = 2

u1A1 =u2A2

Conservation of Energy (Bernoullis Equation): p1+u12 =p2 +

u22

Pitot Tube: Measures total pressure =p+ u2

Method:

From the apparatus shown above, water was pumped through the tubes at an unknown speed. The

cross sectional areas ofthe tubes were then recorded. Using these values and Bernoullis Equation,

velocities at points 1 and 2 were calculated. Then the mass flow rate at the throat was calculatedfirstly using the equation for the conservation of mass and then by measurement (for example time

for 10litres to flow through). Finally, the velocity at the throat of the venturi nozzle, u2, was

calculated using only the pitot tubes and the equation total pressure. Density = 1000kg/m3


5/19

Results:

a) Determine the velocities at positions 1 and 2 by using Bernoullis Equation.Bernoullis Equation: p1+

u12 =p2 +

u22

p1p2= u22 u12[p1p2=g h][h h1 h5]

Conservation of mass: u1A1 =u2A2

u1=u2A2 / A1

A1,2= (d/2)2

where, d1 = 25mm and d2 = 10mm

A1 = 4.9 x 10-4

m2

A2 = 7.85 x 10-5

m2

h = 0.132m

g h

u22

(u2A2 / A1)2

1000 x 9.81 x 0.132 =x 1000 x u22 x 1000 x(u2x 7.85x10-5 / 4.9x10-4)2

1294.92 = u22x (487.17)

u22= 2.658

u2= 1.63m/s

u1= (u2A2 / A1)

u1= 0.26m/s

b) Calculate the mass flow rate of the flow at the throat.Use equation for the Conservation of Mass: u2A2 =

1000 x 1.63 x (7.85x10-5) = = 0.128kgs-1

From this calculation one has to question the accuracy of Bernoullis Equation at this point. Any

errors in calculating the velocity previously will be carried forwards and possibly exponentially

increasing the errors further on. The assumptions made include that the flow is ideal, incompressible

and ideal.

c) Determine the mass flow rate, , by measurement (1000litres = 1m3).

h1 (mm) h5 (mm) Difference

240 110 130

250 110 140

245 120 125

395

h = 131.67mm


6/19

Flow rate: This is the time it took for 10 litres of water to flow through the apparatus.

Flow rate = 17.84s

This was measured using a small marked tube which filled up according to how

much water was being pumped through the apparatus. The time for the water to pass between 0

and 10 litres was measured and this came to 17.84s.

10 litres 1000 = 0.01m30.01m3 over 17.84 seconds = 5.61x10-4m3/s

= 0.561kg/sThis value for mass flow rate by measurement is very different to that of the theoretical result. By

percentage difference, it is a 338.3% difference. This is most likely as a result of systematic and

random error in reading the measurements accurately and the manual use of the stopwatch to time

it. Also, at some points when the plug was put in the tank to allow the over flow of water fill up the

measuring tube that told us how many litres were flowing, the water level would initially fluctuate

and jump up and down making it difficult to differentiate when it actually passed 0litres and the user

could start timing on the stopwatch. It was also very difficult to read off the monometer the exact

location of the meniscus as it moved up and down the tube quickly. Not only this but there were twomeniscus readings to take at the same time, meaning that two users were needed to read them and

this could result in inconsistent readings between them.

d) Determine the velocity at the throat, u2, using the pitot tube.p1p2= u22 dynamic pressure[p1p2=g h][h h8 h5]

h m

1000 x 9.81 x 0.168 = 500 x u22

u2 = u2 = 1.816ms-1

Having compared the two values calculated for the velocity u2, the percentage difference shows to

be 10.8% difference between them. Therefore, since these values are relatively close together

having been found by different means the application of the conservation of mass and the

conservation of energy and the use of the pitot tube it is safe to say that both techniques are

accurate and reliable.

e) Explain what happens when you move the pitot tube beyond point 2 towards the exit?Towards the exit of the tube it starts to open up and widen. This causes turbulence and eddies in the

water. An eddy is a swirling of fluid and a reverse current is created as the tube opens up. In terms of

h1 (mm) h5 (mm) Difference

290 120 170

285 120 165

290 120 170

505

h = 168.3mm


7/19

the conservation of energy, the average velocity stays the same, yet turbulence causes the waste

and dissipation of energy within the system. Pressure will also decrease.

Conclusion:

As it was set out to achieve, the velocity and mass flow rate of the liquid through the tapered venturi

tube was calculated and compared when different methods and techniques were adopted giving areliable value for the velocity of the liquid yet due to the huge difference in mass flow rate between

the different methods of calculation, it is difficult to state an accurate mass flow rate value from

these results.


8/19

EXPERIMENT 3: CONSERVATION OF MOMENTUM

Aim:

The aim is to vary the flow rate of water out of a known diameter jet nozzle onto targets of various

shapes and use the application of momentum equation to compare which shape target produces the

maximum force exerted on the plate.

Apparatus:

The experiment is set up in the manner shown below which a nozzle for shooting a jet of water at a

target plate flat or curved in our case connected to a spring weight pan for measuring the force

of the water.

Method:

This experiment consisted generally of measuring the impact of jets of water and their subsequent

deflection on targets of various shapes. A vertical jet of water was aimed at a target. The vertical

force exerted on the target by the water was measured by placing weights on the pan until the force

of the jet matched the downward weight. This was done by zeroing the weights pan to be in line

with the level gauge with no water jet and no weights. Then once weights were added, the force ofthe jet was increased until the pan lined up with the level gauge. To measure the flow rate of water

from the nozzle, the time for 5litres of water to flow out was recorded. Since the force on the plate

was countered by the weight on the pan, this equation was used:

Looking at Newtons 3

rdLaw, the force applied to the fluid is equal to the rate of change of

momentum in the vertical direction:

u2 - u1 =Q(u2u1)Where u1 and u2are the velocities before and after impact, is the water density (1000kgm-3) and Qis the volumetric flow rate.


9/19

y = 0.8694x

0

0.5

1

1.5

2

2.5

3

3.5

0 0.5 1 1.5 2 2.5 3 3.5

Force on Plate (g)

Q/A

Flat Plate - 90o

By the use of Newtons 2nd

Law, the equation for the force exerted on the plate is given by:

u1 - u2 =Q(u2u1)=Q(u1u1 cos)

=Q(1 cos)And Q = Au1

()Results:

a) For a particular target, plot the force on the plate on the y-axis and the quantity on thex-axis. Measure the gradient of this and compare it to the theoretical value (1-cos) for your

particular target at = 90 and = 180.

Force on plate F = mg

A = x r2

A = 5.0265x10-5

m2

Flat Plate - 90

Force mg (g) Flow rate (litres/s)

0.49 0.17 0.575

0.98 0.23 1.0521.47 0.29 1.673

1.96 0.34 2.2998

2.45 0.39 3.026

2.94 0.4 3.183


10/19

This first set of data corresponds to the flat plate where the water was turned 90o. The force on the

plate was equal to mass x gravity as previously stated and for each force, a flow rate was calculated

for the water jet in litres/s. Lastly, the quantity was found using the

equation as stated in thequestion where = 1000kg/m

3and the area being 5.0265x10

-5m

2. Using this data, the above graph

was plotted and the gradient was measured to give 0.8694. Comparison between this and the

theoretical as follows:

Calculated gradient value = 0.8694

Theoretical gradient = (1 cos)

For = 90o, = (1 cos90)

= 1

Compare: 1 0.8694 = 0.1306

This slight error difference will be due to systematic and random error that cannot be avoided.Common examples of systematic errors include faulty calibration of measuring instruments, poorly

maintained instruments or faulty reading of instruments by the user. This last type is commonly

known as parallax error which is the user reads the instrument at an ang le resulting in reading

data that is consistently high or consistently low. To correct this error, the measurement method or

technique must be refined. Random errors affect the precision of the measurements. Measurements

subject to random error differ from one another due to unpredictable variations in the readings and

measurement process. The precision of measurements can be increased if number of readings in

increased also. An example of this in this particular experiment is when starting and stopping the

stopwatch there needs to be some degree of a quick reaction and the human being can only react at

a certain speed leaving some margin for error. This will cause a variation in readings and some

margin for error.

Curved Plate:

Curved Plate - 180

Force mg (g) Flow rate (litres/s)

0.49 0.125 0.311

0.98 0.17 0.575

1.47 0.19 0.718

1.96 0.24 1.146

2.45 0.26 1.345

2.94 0.29 1.673


11/19

This second set of data relates to the curved plate through which the water was turned 180o. These

values were calculated in the same way as previously finding the flow rate, force on the plate and

quantity for the curved plate apparatus. As a result, the above graph was plotted and from this, the

gradient was found. The comparison of the two gradients is as follows:

Calculated gradient value = 1.7819

Theoretical gradient = (1 cos)

For = 180o, = (1 cos180)

= 2

Compare: 2 1.7819 = 0.2181

As you can see there is also some percentage error between the measured gradient and the

theoretical gradient again due to systematic and random error. This includes parallax error in

reading the water level in the thin volume meter tube and also the users reactions at starting and

stopping the stopwatch which could results in some readings showing higher than normal and some

showing lower than normal.

b)

Determine which angle of gives the maximum force exerted on the plate.

The plate of angle 180o

gives the maximum force exerted on the plate because the gradient

calculated from the measurements and the graph is higher than that of the plate of angle 90o. Since

the gradient is higher and the curve on the graph is steeper for the curved plate than the flat plate,

the force exerted is higher at a specific value of the quantity.

Conclusion:

As a result of the data collected for this experiment and the calculations made using it, I feel that the

aim set out was achieved and the target of certain shape that gave the maximum force exerted on

the plate was found.

y = 1.7819x

0

0.5

1

1.5

2

2.5

3

3.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Force on Plate (g)

Q/A

Curved plate


12/19

EXPERIMENT FOUR: DIMENSIONAL ANALYSIS

4a) In Drop Tubes

Aim:

The aim of this experiment was to measure the total drag on a set of spheres that varied in diameter

and density and to investigate how this drag varies when the spheres flow through fluids of different

viscosities.

Apparatus:

Glass tubes in excess of 1meter high containing water, glycerine and oil (XHVI 8.2). A selection of spheres of differing diameters and densities thus made from different

materials including steel, glass, nylon and ptfe (polytetrafluoroethylene).

Stopwatch Measuring callipers

Method:The very first task for this experiment was to measure the diameters of each sized sphere for each

material since although they appeared the same size there were slight variances in their diameters.

Having found the diameters, the volumes, areas and frontal areas were then calculated. The

practical part to the experiment consisted of dropping the spheres individually into a tube of fluid

water, glycerine or XHVI 8.2 then starting a stopwatch once it passes a marked point and stopping

the time when it passes the second mark which indicated a whole meter. The top marker was a

small distance beneath the surface of the liquid to allow the sphere to reach terminal velocity whilst

it travelled between the two markers. By measuring the time it took to travels 1 meter, the terminal

velocity could be calculated simply using speed = distance/time. After recording all the dropping

times, the theoretical part to the experiment began. This comprises of finding the drag force and the

Reynolds number for each case. Finally, the coefficient of drag had to be calculated using ourmeasurements of drag force, terminal velocity and frontal area.

When the spheres are falling at terminal velocity:

DF B

DF = drag force

B = buoyancy

mg = weight x gravity

mg

Coefficient of Drag: Reynolds number: Buoyancy:

=

A = Frontal area (r2) d = diameter V = volume of sphere

u = Terminal velocity = dynamic viscosity g = 9.81N


13/19

Densities

Steel = 7800kg/m3 Water = 1000kg/m3

Glass = 2530kg/m3 Glycerine = 1262kg/m

3

Nylon = 1140kg/m3

XHVI 8.2 = 829kg/m3

PTFE = 2160 g/m3

Air = 1.225kg/m3

Results:

Using the above equations details about the spheres were obtained and are stated below:

TYPES OF SPHERE DIAMETER

S (m)

AREAS

(m)

VOLUMES

(m)

FRONTAL

AREA (m)

MASS

(kg)

mg

STEEL Large 0.0127 5.07E-04 1.07E-06 1.27E-04 8.36E-03 8.20E-02Medium 0.0095 2.84E-04 4.49E-07 7.09E-05 3.50E-03 3.44E-02

Small 0.00634 1.26E-04 1.33E-07 3.16E-05 1.04E-03 1.02E-02

NYLON Large 0.0127 5.07E-04 1.07E-06 1.27E-04 1.22E-03 1.20E-02

Medium 0.00957 2.88E-04 4.59E-07 7.19E-05 5.23E-04 5.13E-03

Small 0.00634 1.26E-04 1.33E-07 3.16E-05 1.52E-04 1.49E-03

GLASS Large 0.01266 5.04E-04 1.06E-06 1.26E-04 2.69E-03 2.64E-02

Medium 0.0054 9.16E-05 4.55E-07 2.29E-05 1.15E-03 1.13E-02

Small 0.00633 1.26E-04 1.33E-07 3.15E-05 3.36E-04 3.30E-03

PTFE Large 0.01268 5.05E-04 1.07E-06 1.26E-04 2.30E-03 2.26E-02

Medium 0.00951 2.84E-04 4.50E-07 7.10E-05 9.72E-04 9.54E-03

Small 0.00625 1.23E-04 1.28E-07 3.07E-05 2.76E-04 2.71E-03

The times taken for the spheres to drop through the 1 meter marks on the tubes of different fluid

are recorded below:

(All the nylon balls floated on the surface of the glycerine and wouldnt sink)

TYPES OF SPHERE TIMES (s) TO TRAVEL 1M

WATER GLYCERIN HVI 8.2

STEEL Large 0.62 2.91 0.65

Medium 0.68 3.6 0.9

Small 0.82 8.9 1.5

NYLON Large 4.68 FLOATS 7.79

Medium 5.31 FLOATS 11.28

Small 7.36 FLOATS 19.34

GLASS Large 1.8 13.72 2.12

Medium 2.09 22.19 2.88

Small - - -

PTFE Large 1.5 19.12 2.59

Medium 2.2 30.31 3.63

Small 1.8 62.63 5.69


14/19

(There were insufficient supplies of the small glass spheres therefore no data was recorded)

Using the data on drop times previously, the terminal velocity can be found by simply dividing 1 by

the time taken. This is taken from the equation speed = distance/time and with the distance being

1meter over which the sphere was timed, this is the distance.

TYPES OF SPHERE TERMINAL VELOCITYSTEEL Large 1.61 0.344 1.54

Medium 1.47 0.278 1.11

Small 1.22 0.112 0.667

NYLON Large 0.214 0 0.128

Medium 0.188 0 0.0887

Small 0.136 0 0.0517

GLASS Large 0.556 0.0729 0.472

Medium 0.478 0.0451 0.347

Small 0 0 0

PTFE Large 0.667 0.0523 0.386

Medium 0.455 0.0329 0.275

Small 0.556 0.016 0.176

In converging all this data and further values of density of fluid and dynamic viscosities, values for

drag force, buoyancy, Reynolds number and coefficient of drag were calculated and plotted on a

graph of CD against Re.

TYPES OF SPHERE BOUYANCY DRAG FORCE

Water Glycerine HVI 8.2 Water Glycerine HVI 8.2

STEEL Large 1.05E-02 1.33E-02 8.72E-03 7.15E-02 6.88E-02 7.33E-02

Medium 4.40E-03 5.56E-03 3.65E-03 3.00E-02 2.88E-02 3.07E-02

Small 1.30E-03 1.65E-03 1.08E-03 8.87E-03 8.53E-03 9.10E-03

NYLON Large 1.05E-02 1.33E-02 8.72E-03 1.47E-03 -1.28E-03 3.27E-03

Medium 4.50E-03 5.68E-03 3.73E-03 6.30E-04 -5.49E-04 1.40E-03

Small 1.30E-03 1.65E-03 1.08E-03 1.83E-04 -1.59E-04 4.06E-04GLASS Large 1.04E-02 1.31E-02 8.64E-03 1.59E-02 1.32E-02 1.77E-02

Medium 4.46E-03 5.63E-03 3.70E-03 6.83E-03 5.66E-03 7.59E-03

Small 1.30E-03 1.65E-03 1.08E-03 2.00E-03 1.65E-03 2.22E-03

PTFE Large 1.05E-02 1.32E-02 8.68E-03 1.21E-02 9.40E-03 1.39E-02

Medium 4.41E-03 5.57E-03 3.66E-03 5.12E-03 3.96E-03 5.88E-03

Small 1.26E-03 1.58E-03 1.04E-03 1.46E-03 1.13E-03 1.67E-03


15/19

From those two sets of data in drag force and buoyancy, the coefficients of drag and Reynolds

number for each size sphere was calculated and shown below:

To find the values of Reynolds number, the dynamic viscosity needed to be known for each fluid as

shown in the equation:

Reynolds number: ,where is the dynamic viscosity.

Water = 9.55x10-4

Ns/m2

Glycerine = 1.2Ns/m2

XHVI 8.2 = 9.12x10-2Ns/m2

By bringing together the data above, a graph was plotted showing the Coefficient of Drag on the y-

axis and the Reynolds numbers on the x-axis. The data for water, glycerine and XHVI 8.2 were all put

on the same graph to show the trend between them as the size and density are varied giving varying

results for the coefficient of drag and Reynolds numbers. Graph shown on next page.

TYPES OF SPHERE COEFFICIENTS OF DRAG REYNOLDS NUMBER

Water Glycerine HVI 8.2 Water Glycerine HVI 8.2STEEL Large 4.34E-01 7.28E+00 5.90E-01 21449.08 4.590 177.60

Medium 3.91E-01 8.34E+00 8.47E-01 14628.89 2.775 95.95

Small 3.78E-01 3.39E+01 1.56E+00 8096.029 0.749 38.42

NYLON Large 5.09E-01 0 3.78E+00 2841.545 0 14.82

Medium 4.94E-01 0 5.98E+00 1887.183 0 7.71

Small 6.27E-01 0 1.16E+01 902.0032 0 2.98

GLASS Large 8.21E-01 3.13E+01 1.53E+00 7364.747 0.970 54.28

Medium 2.61E+00 1.93E+02 6.63E+00 2705.479 0.256 17.04

Small - - - - - -

PTFE Large 4.33E-01 4.31E+01 1.79E+00 8851.658 0.697 44.50

Medium 6.98E-01 8.13E+01 2.63E+00 4526.416 0.330 23.81

Small 3.08E-01 2.28E+02 4.26E+00 3635.835 0.105 9.98


16/19

Graph Analysis:

From the graph, it is apparent that the further away from the y-axis, the more turbulent the fluid

becomes. For example, that would explain why the plots for water are more scattered rather than in

a neat line because the spheres moved quickly through the water making it difficult to record an

accurate time. The closer the plots are to the y-axis mean that the fluid is more viscous andtherefore less turbulent. The Stokes law shows that CD = 24/Re only for very viscous fluids, such as

XHVI 8.2 which very nearly fits this relationship. For other fluids like water that has high Reynolds

numbers, this relationship does not apply.

4b) In the Wind Tunnel

Aim:

The aim is to measure the drag on a number of different spheres in a range of different flow

conditions in order to investigate the effects that pressure has on the total drag with increasingReynolds numbers and therefore decreasing and less significant viscous effects.

Apparatus:

The body under test is mounted in a small wind tunnel on the sting of a drag force balance. Four different sphere/objects were used in this investigation:

1. Large Sphere polystyrene of diameter 0.07m2. Small Sphere polystyrene of diameter 0.03m3. Ellipsoid (large egg) polystyrene of diameter 0.044m4. Small egg polystyrene of diameter 0.04m

Air was passed through the wind tunnel around the body seated on the sting and pressurevalues were taken at the opening of the wind tunnel and a second one right underneath thebody.

Method:

Firstly, the drag created on the sting alone was found for air speeds of maximum, minimum and one

between which was subtracted from each value of drag for the bodies later on. Then the bodies

were added one by one to the wind tunnel and placed on the sting and a drag value was calculated

for each. To find the air speed u, in the tunnel for each wind speed, the upstream pitot pressure

tapping (po= p + ) and the static pressure (p) in the working section below the

body were used.

Equations used: Coefficient of Drag:

Reynolds number:


17/19

Results:

Firstly, the drag on the sting in open, closed and half positions was found:

Drag on sting

open closed half

Mass(kg) 0.012 0.027 0.019

static

pressure(Pa) 150 440 340

total pressure(Pa) 50 70 80

specific drag

force(N) 0.11772 0.26487 0.18639

Using this, the following data was calculated for the different spheres at varied wind speeds:

Closed Door tunnel big ball

small

ball big egg small egg

static pressure(Pa) 470 450 440 430

total pressure(Pa) 90 95 90 80

mass measured(kg) 0.162 0.057 0.07 0.046

total drag force(N) 1.58922 0.55917 0.6867 0.45126

specific drag force(N) 1.32435 0.55917 0.42183 0.18639

upstream velocity of

air(m/s) 24.58543 24.583 24.58179 24.58057

drag coefficient 0.455734 0.769684 0.24483 0.346892

Reynolds number 1.64E+05 8.19E+04 1.26E+05 7.05E+04

open door tunnel big ballsmallball big egg small egg



mass measured(kg) 0.066 0.022 0.032 0.017




air(m/s) 12.76182 12.76182 12.76182 12.76245


Reynolds number 8.53E+04 4.27E+04 6.57E+04 3.67E+04


18/19

half open door big ball

small

ball big egg

small

egg



mass measured(kg) 0.116 0.039 0.064 0.034




air(m/s) 20.59305 20.59203 20.59406 20.59


Reynolds number 1.38E+05 1.24E-05 1.91E-05

1.07E-

05

To find the density of air, certain values such as kinematic and dynamic viscosities had to be foundand also the density of atmosphere and then RT which is R = 287Jkg

-1K

-1x T = 288K:

kinematic viscosity of air dynamic viscosity

density

(atm) RT

big ball 1.50E-05 1.84E-05 101662.992 82656

small

ball 1.50E-05 1.84E-05 101662.992 82656

big egg 1.50E-05 1.84E-05 101662.992 82656

small

egg 1.50E-05 1.84E-05 101662.992 82656

Therefore, the density of air was found using these equations:

(abs) = (atm) = (static)

(atm) = (merc) x g x h ((merc) = 13600kg/m3, h = 0.762m)

(air) = (abs) / RT

thus:

Closed Open Half

p(abs)

density of

air p(abs) density of air p(abs)

density

of air

101193 1.224267 101502.992 1.228017228 101353 1.226202

101213 1.224509 101502.992 1.228017228 101363 1.226323

101223 1.22463 101502.992 1.228017228 101343 1.226081

101233 1.224751 101492.992 1.227896245 101383 1.226565

The graphs are as follows:


19/19

Graph analysis:

My results show a strong correlation similar to that of the graph shown in the lab report notes. It

represents that for high Reynolds numbers, the coefficient of drag is very low meaning that the

viscosity of the fluid is very low and turbulence is very likely to occur. In this case of the wind tunnel

we were dealing with air flow which causes a lot of turbulence when passing non aerodynamic

surfaces or bodies.

Discussion:

Judging by my graph of all the Reynolds number vs all the coefficients of drag, they seem to compare

well with the graph provided in the notes showing and steady negative correlation with increasing

Reynolds numbers and then plateau slightly before another steep drop in the coefficient of drag

which in my graph is represented by the wind tunnel results that appear to be a random scatter of

plots. This will probably be due to inaccuracies with in the apparatus and in taking measurements

too.

The Navier-Stokes law does apply to my results but only for the low Reynolds values where they areclose to the y-axis and the viscosity is high.

Date post:	07-Apr-2018
Category:	Documents
Upload:	sammy-leakey
View:	236 times
Download:	0 times

Mech 2600 Fluid Mechanics

Documents