Advance Fluid Mechanics

Recommended Books: 1. Daugherty, R. L. Franzini B. & Finnemore E. J., Fluid Mechanics, McGraw Hill Book Co. 2. Douglus, Fluid Mechanics, McGraw-Hill Inc. 3. Jack P., Fundamentals of Fluid Mechanics , McGraw-Hill Inc.4. Merle Potter, Mechanics of Fluid, CL-Engineering (2011)

Laminar Flow in Circular Pipe:

For laminar flow

𝝉= 𝜇𝑑𝑢

𝑑𝑦(1)

where,

u = velocity at a distance ‘y’ from the boundary

as,

𝑦 = 𝑟𝑜 - 𝑟

𝑑𝑦 = −𝑟

𝑟𝑜 = constant for a particular pipe

Now equation (1) becomes

𝝉= − 𝜇𝑑𝑢

𝑑𝑦

-ve sign indicates that ‘u’ decreases as ‘𝑟’ increases.

To determine the velocity profile for laminar flow in a circular pipe 𝝉= − 𝜇𝑑𝑢

𝑑𝑟is substituted into expression ℎ𝐿 = 𝝉

2𝐿

𝑟

Therefore,

ℎ𝐿 = −𝜇𝑑𝑢

𝑑𝑟

2𝐿

𝑟

𝑑𝑢 = −ℎ𝐿

2𝜇𝐿 𝑟𝑑𝑟 (2)

Integrating equation (2) and assuming that integration constant ‘c’ is equal to 𝑢𝑚𝑎𝑥

𝑑𝑢 = −

𝑟=0

𝑢=𝑢𝑚𝑎𝑥ℎ𝐿2𝜇𝐿

𝑟𝑑𝑟

𝑑𝑢 = − ℎ𝐿2𝜇𝐿

𝑟𝑑𝑟

𝑑𝑢 = −ℎ𝐿2𝜇𝐿

𝑟𝑑𝑟

𝑢 = −ℎ𝐿

2𝜇𝐿

𝑟2

2+ 𝑐

𝑢 = −ℎ𝐿

2𝜇𝐿

𝑟2

2+ 𝑢𝑚𝑎𝑥

𝑢 = 𝑢𝑚𝑎𝑥 −ℎ𝐿2𝜇𝐿

𝑟2

2(3)

𝑢 = 𝑢𝑚𝑎𝑥- k𝑟2 (4)

where,

𝑘 =ℎ𝐿

4𝜇𝐿

Substituting the boundary condition that u = 0 for 𝑟 = 𝑟𝑜 and noting that 𝑢𝑚𝑎𝑥= 𝑉𝑐 , centerline velocity

𝑢 = 𝑢𝑚𝑎𝑥− k𝑟2 (4)

0 = 𝑉𝑐− k𝑟𝑜2

𝑉𝑐= k𝑟𝑜2

𝑘 =𝑉𝑐

𝑟𝑜2 (5)

Substituting the value of ‘k’ in equation (4), we get

𝑢 = 𝑢𝑚𝑎𝑥−𝑉𝑐

𝑟𝑜2 𝑟2

since 𝑉𝑐 = 𝑢𝑚𝑎𝑥

Therefore,

𝑢 = 𝑉𝑐 −𝑉𝑐

𝑟𝑜2 𝑟2

𝑢 = 𝑉𝑐 (1−𝑟2

𝑟𝑜2) (6)

Comparing equations (3) and (6), we get

𝑢𝑚𝑎𝑥 −ℎ𝐿4𝜇𝐿

𝑟2=𝑉𝑐 (1−𝑟2

𝑟𝑜2)

𝑉𝑐 −ℎ𝐿

4𝜇𝐿𝑟2=𝑉𝑐 −

𝑉𝑐

𝑟𝑜2 𝑟2

ℎ𝐿4𝜇𝐿

𝑟2=𝑉𝑐

𝑟𝑜2 𝑟2

𝑉𝑐 =ℎ𝐿4𝜇𝐿

𝑟𝑜2

As 𝑟𝑜 =𝐷

2and 𝑟𝑜

2 =𝐷2

4

Therefore

𝑉𝑐 =ℎ𝐿𝐷2

16𝜇𝐿(7)

where,

𝑉𝑐 = centerline velocity

Since,

mean velocity 𝑉 =1

2𝑉𝑐 i.e; 𝑉𝑐 = 2V

Putting 𝑉𝑐 = 2V in equation (7), we get

2V =ℎ𝐿𝐷

2

16𝜇𝐿

V =ℎ𝐿𝐷2

32𝜇𝐿(8)

As we know that =𝜌g

V =ℎ𝐿𝜌g𝐷2

32𝜇𝐿

V =ℎ𝐿g𝐷2

32𝜇

𝜌𝐿

since,

𝜇

𝜌 = 𝜈 kinematic viscosity

V =ℎ𝐿g𝐷

2

32𝜈𝐿

ℎ𝐿 =32𝑉𝐿𝜈

g𝐷2(9)

Equation (9) is Hagen – Poiseuille law for Laminar Flow.

Recalling Darcy – Weisbach equation of head loss

ℎ𝐿 = 𝑓𝐿

𝐷

𝑉2

2g(10)

Comparing equations (9) and (10), we get

𝑓𝐿

𝐷

𝑉2

2g=32𝑉𝐿𝜈

g𝐷2

𝑓 = 64 𝜈𝑉𝐷

𝑓 = 64

𝑉𝐷𝜈

𝑓 = 64

𝑅𝑒(11)

We can determine pipe friction ‘𝑓’ if 𝑅𝑒 is less than 2000.

Entrance Conditions in Laminar Flow:

In the case of a pipe leading from a reservoir, if the entrance is rounded so as to avoid any initial disturbance of the emergingstream, all the particles will start to flow with the same velocity, except for a very thin film (layer) in contact with the wall.Particles next to the wall will have zero velocity, but the velocity gradient here is extremely steep and with this slight exception,velocity is uniform across the diameter as shown in figure.

As the fluid progresses along the pipe, the streamlines in the vicinity of the wall are slowed down by the friction emanating fromthe wall, but as Q (discharge) is constant for successive sections, the velocity in the center must be accelerated, until the finalvelocity profile is a parabola as shown in figure.

Theoretically an infinite distance is required for this but it has been established both by theory and by observation that themaximum velocity in the center of the pipe will reach 99% of its ultimate value in the distance 𝐿´ = 0.058𝑅𝑒𝐷.

Thus for critical value 𝑅𝑒 = 2000 , the distance 𝐿´ = 166 pipe diameters.

Unestablished Flow:

“ It is the region in the pipe where velocity profile is changing.”

i.e; in the entry region of length 𝐿´, the flow is unestablished.

Mathematically,𝐿´ = 0.058𝑅𝑒𝐷

Established Flow:

“ It is the region in the pipe where velocity profile does not change and it has attained a parabolic shape.”

Boundary Layer:

“ The outer zone which is in contact with the wall and increases in thickness as flow moves along the wall. It increases its thickness until the shear stress becomes maximum.”

Problem1: Oil (S = 0.85) with a kinematic viscosity of 6 × 10−4 𝑚2/s flows in a 15 cm pipe at a rate of 0.020 𝑚3/s. What is the head loss per 100 m length of pipe?

Solution: Given that

Discharge Q = 0.020 𝑚3/s

Pipe diameter D = 0.15 m.

Specific gravity of oil S = 0.85

kinematic viscosity 𝜈= 6 × 10−4 𝑚2/s

Head loss per 100 m = ?

Step#1:

Mean velocity v = 𝑄

𝐴

v = 𝑄

𝜋𝐷2 4

v =0.020

𝜋0.1524

v = 1.13m

s

Step#2:

Reynold’s number 𝑅𝑒 =𝑽𝑫

𝝊

𝑅𝑒 =𝟏.𝟏𝟑×𝟎.𝟏𝟓

6 × 10−4

𝑅𝑒 = 283

Step#3:

Since 𝑅𝑒 < 2000, the flow is laminar.

Step#4:

Head loss ℎ𝐿 =32𝑉𝐿𝜈

g𝐷2

ℎ𝐿 =32×1.13×100×of 6 × 10−4

9.81×0.152

ℎ𝐿 = 9.83m

Problem#2: An oil with a kinematic viscosity of 0.135 stokes flows through a pipe of diameter 15cm. Below what velocity will be the laminar flow?

Solution: Given that

Pipe diameter D = 0.15 m.

Specific gravity of oil S = 0.85

1 stoke 𝜈=1 c𝑚2= 1 × 10−4 𝑚2/s

Kinematic viscosity 𝜈= 0.135 × 10−4 𝑚2/s

Reynold’s number 𝑅𝑒 =𝑽𝑫

𝝊

The flow is laminar with the Reynold’s number less than 2000. i.e;

𝑅𝑒 =𝑽𝑫

𝝊˂ 𝟐𝟎𝟎𝟎

𝑽×𝟎. 𝟏𝟓

0.135 × 10−4˂ 𝟐𝟎𝟎𝟎

𝑉˂ 𝟎. 𝟏𝟖𝒎

𝒔

Therefore, for the velocity of flow below 0.18 m/s, the flow will be laminar.

Problem#3: An oil with a kinematic viscosity of 0.005𝑚2/s flow through a 7.5cm diameter pipe with a velocity of 3m/s. Is the flow is laminar or turbulent?

Solution: Given that

kinematic viscosity 𝝊= 0.005𝑚2/s

Pipe diameter D = 7.5cm = 0.075 m

Velocity of flow 𝑉 = 3 𝑚/𝑠

Nature of flow = ?

As 𝑅𝑒 =𝑽𝑫

𝝊

𝑅𝑒 =𝟑×𝟎.𝟎𝟕𝟓𝟎.𝟎𝟎𝟓

𝑅𝑒 = 𝟒𝟓𝟎 ˂ 𝟐𝟎𝟎𝟎 Therefore the flow is laminar.

Problem#4: An oil (s = 0.8, 𝝊 =1.8×10−5𝑚2/s) flow in a 10cm diameter pipe at 0.5L/s. Is the flow is laminar or turbulent?

Solution: Given that

kinematic viscosity 𝝊=1.8×10−5𝑚2/s

Pipe diameter D = 10cm = 0.1 m

Velocity of flow 𝑉 = 0.5L/s = 5 × 10−4𝑚3/s

Nature of flow = ?

v = 𝑄

𝐴

v =𝑄

𝜋𝐷2 4

v =5 × 10−4

𝜋0.124

v = 0.0637m

s

As 𝑅𝑒 =𝑽𝑫

𝝊

𝑅𝑒 =𝟎.𝟎𝟔𝟑𝟕×𝟎.𝟏1.8×10−5

𝑅𝑒 = 𝟑𝟓𝟒 ˂ 𝟐𝟎𝟎𝟎 Therefore the flow is laminar.

Problem#5: For the case of problem#4, find the centerline velocity, the velocity at r = 2cm, the friction factor, theshear stress at the pipe wall, and the head loss per meter pipe length.

Solution: Since we have come to know that the flow is laminar. Therefore,

𝑉𝑐 = 2V

𝑉𝑐 = 2×0.0638

𝑉𝑐 =0.1274 m/s

𝑢 = 𝑢𝑚𝑎𝑥− k𝑟2

𝑢𝑚𝑎𝑥 = 𝑉𝑐 =0.1274 m/s

𝐷 = .1𝑚

When 𝑟 =𝑟𝑜= 0.05m

u = 00 = 0.1274− k ×0.052

𝑘 =50.96

𝑚. 𝑠𝑒𝑠Now,

𝑢𝑎𝑡 2𝑐𝑚 = 𝑢𝑚𝑎𝑥− k𝑟2

𝑢𝑎𝑡 2𝑐𝑚 = 0.1274− 50.96×0.022

𝑢 2𝑐𝑚 = 0.107m/s

For laminar flow 𝑓 = 64

𝑅𝑒

𝑓 = 64354

𝑓=0.181

𝝉𝑜 =𝑓

4𝑉2

2g

𝝉𝑜 =𝑓

4𝜌𝑉2

2𝜌 = / g

𝝉𝑜 =0.181

4× 850×

0.06372

2

𝝉𝑜 =0.78N/𝑚2

ℎ𝐿𝐿= 𝑓

1

𝐷

𝑉2

2g

ℎ𝐿𝐿= 0.181×

1

0.1

0.06372

2×9.81

ℎ𝐿

𝐿=0.00374m/m

Problem: Prove that the centerline velocity is twice the average velocity when the laminar flow occurs in a circular pipe.

Proof: The velocity profile for the laminar flow in a circular pipe can be written as

𝑢 = 𝑉𝑐 (1−𝑟2

𝑟𝑜2) (1)

where,

u = average velocity

= centerline velocity

𝑟 = radius of the pipe at any point

𝑟𝑜= maximum radius of the pipe.

The flow rate in a circular pipe can be calculated as,

𝑄 = 𝑜𝑟𝑜 𝑢𝑑𝐴 (2)

Using equation (1) into (2), we get

𝑄 =

𝑜

𝑟𝑜

𝑉𝑐 (1−𝑟2

𝑟𝑜2)𝑑𝐴

𝑄 = 𝑜𝑟𝑜 𝑉𝑐 1−

𝑟2

𝑟𝑜2 2𝜋𝑟𝑑𝑟 𝑑𝐴= 2𝜋𝑟𝑑𝑟

𝑄 = 2𝜋𝑉𝑐

𝑜

𝑟𝑜

1−𝑟2

𝑟𝑜2 𝑟𝑑𝑟

𝑄 = 2𝜋𝑉𝑐

𝑜

𝑟𝑜

𝑟 −𝑟3

𝑟𝑜2 𝑑𝑟

𝑄 = 2𝜋𝑉𝑐𝑟2

2−𝑟4

4𝑟𝑜2𝑜

𝑟𝑜

𝑄 = 2𝜋𝑉𝑐𝑟𝑜2

2−𝑟𝑜4

4𝑟𝑜2

𝑄 = 2𝜋𝑉𝑐𝑟𝑜2

2−𝑟𝑜2

4

𝑄 = 2𝜋𝑉𝑐𝑟𝑜2

4

𝑄 = 𝜋𝑉𝑐𝑟𝑜2

2(3)

The average velocity can be calculated as

v = 𝑄𝐴

v =𝜋𝑉𝑐

𝑟𝑜2

2𝜋𝑟𝑜2

v =𝑉𝑐2

𝑉𝑐 =2V proved

Therefore the centerline velocity is twice the average velocity.

Problem: with laminar flow in a circular pipe, at what distance from the centerline does the average velocity occurs?

Proof: The velocity distribution in case of laminar flow in a circular pipe is

𝑢 = 𝑉𝑐 (1−𝑟2

𝑟𝑜2) (1)

where,

u = average velocity

𝑉𝑐 = centerline velocity

𝑟 = radius of the pipe at any point

𝑟𝑜= maximum radius of the pipe.

Since,

𝑉𝑐 = 𝑢𝑚𝑎𝑥 = 𝑉𝑎𝑣𝑔

Or

𝑉𝑎𝑣𝑔= V = 0.5𝑉𝑐 (2)

Put equation (2) into (1)

0.5𝑉𝑐 = 𝑉𝑐 (1−𝑟2

𝑟𝑜2)

0.5 = (1−𝑟2

𝑟𝑜2)

𝑟2

𝑟𝑜2 = 1−0.5

𝑟2

𝑟𝑜2 = 0.5

𝑟2

𝑟𝑜2 = 0.5

𝑟

𝑟0= 0.707

𝑟 = 0.707𝑟0

Therefore the distance from the centerline at which the average velocity occurs is

𝑟 = 0.707𝑟0