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Civil Engineering
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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
π= πππ’
ππ¦(1)
where,
u = velocity at a distance βyβ from the boundary
as,
π¦ = ππ - π
ππ¦ = βπ
ππ = constant for a particular pipe
Now equation (1) becomes
π= β πππ’
ππ¦
-ve sign indicates that βuβ decreases as βπβ increases.
To determine the velocity profile for laminar flow in a circular pipe π= β πππ’
ππis substituted into expression βπΏ = π
2πΏ
π
Therefore,
βπΏ = βπππ’
ππ
2πΏ
π
ππ’ = ββπΏ
2ππΏ πππ (2)
Integrating equation (2) and assuming that integration constant βcβ is equal to π’πππ₯
ππ’ = β
π=0
π’=π’πππ₯βπΏ2ππΏ
πππ
ππ’ = β βπΏ2ππΏ
πππ
ππ’ = ββπΏ2ππΏ
πππ
π’ = ββπΏ
2ππΏ
π2
2+ π
π’ = ββπΏ
2ππΏ
π2
2+ π’πππ₯
π’ = π’πππ₯ ββπΏ2ππΏ
π2
2(3)
π’ = π’πππ₯- kπ2 (4)
where,
π =βπΏ
4ππΏ
Substituting the boundary condition that u = 0 for π = ππ and noting that π’πππ₯= ππ , centerline velocity
π’ = π’πππ₯β kπ2 (4)
0 = ππβ kππ2
ππ= kππ2
π =ππ
ππ2 (5)
Substituting the value of βkβ in equation (4), we get
π’ = π’πππ₯βππ
ππ2 π2
since ππ = π’πππ₯
Therefore,
π’ = ππ βππ
ππ2 π2
π’ = ππ (1βπ2
ππ2) (6)
Comparing equations (3) and (6), we get
π’πππ₯ ββπΏ4ππΏ
π2=ππ (1βπ2
ππ2)
ππ ββπΏ
4ππΏπ2=ππ β
ππ
ππ2 π2
βπΏ4ππΏ
π2=ππ
ππ2 π2
ππ =βπΏ4ππΏ
ππ2
As ππ =π·
2and ππ
2 =π·2
4
Therefore
ππ =βπΏπ·2
16ππΏ(7)
where,
ππ = centerline velocity
Since,
mean velocity π =1
2ππ i.e; ππ = 2V
Putting ππ = 2V in equation (7), we get
2V =βπΏπ·
2
16ππΏ
V =βπΏπ·2
32ππΏ(8)
As we know that =πg
V =βπΏπgπ·2
32ππΏ
V =βπΏgπ·2
32π
ππΏ
since,
π
π = π kinematic viscosity
V =βπΏgπ·
2
32ππΏ
βπΏ =32ππΏπ
gπ·2(9)
Equation (9) is Hagen β Poiseuille law for Laminar Flow.
Recalling Darcy β Weisbach equation of head loss
βπΏ = ππΏ
π·
π2
2g(10)
Comparing equations (9) and (10), we get
ππΏ
π·
π2
2g=32ππΏπ
gπ·2
π = 64 πππ·
π = 64
ππ·π
π = 64
π π(11)
We can determine pipe friction βπβ if π π 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 πΏΒ΄ = 0.058π ππ·.
Thus for critical value π π = 2000 , the distance πΏΒ΄ = 166 pipe diameters.
Unestablished Flow:
β It is the region in the pipe where velocity profile is changing.β
i.e; in the entry region of length πΏΒ΄, the flow is unestablished.
Mathematically,πΏΒ΄ = 0.058π ππ·
Established Flow:
β It is the region in the pipe where velocity profile does not change and it has attained a parabolic shape.β
Boundary Layer:
β 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.β
Problem1: Oil (S = 0.85) with a kinematic viscosity of 6 Γ 10β4 π2/s flows in a 15 cm pipe at a rate of 0.020 π3/s. What is the head loss per 100 m length of pipe?
Solution: Given that
Discharge Q = 0.020 π3/s
Pipe diameter D = 0.15 m.
Specific gravity of oil S = 0.85
kinematic viscosity π= 6 Γ 10β4 π2/s
Head loss per 100 m = ?
Step#1:
Mean velocity v = π
π΄
v = π
ππ·2 4
v =0.020
π0.1524
v = 1.13m
s
Step#2:
Reynoldβs number π π =π½π«
π
π π =π.ππΓπ.ππ
6 Γ 10β4
π π = 283
Step#3:
Since π π < 2000, the flow is laminar.
Step#4:
Head loss βπΏ =32ππΏπ
gπ·2
βπΏ =32Γ1.13Γ100Γof 6 Γ 10β4
9.81Γ0.152
βπΏ = 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 π=1 cπ2= 1 Γ 10β4 π2/s
Kinematic viscosity π= 0.135 Γ 10β4 π2/s
Reynoldβs number π π =π½π«
π
The flow is laminar with the Reynoldβs number less than 2000. i.e;
π π =π½π«
πΛ ππππ
π½Γπ. ππ
0.135 Γ 10β4Λ ππππ
πΛ π. πππ
π
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 0.005π2/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 π= 0.005π2/s
Pipe diameter D = 7.5cm = 0.075 m
Velocity of flow π = 3 π/π
Nature of flow = ?
As π π =π½π«
π
π π =πΓπ.ππππ.πππ
π π = πππ Λ ππππ Therefore the flow is laminar.
Problem#4: An oil (s = 0.8, π =1.8Γ10β5π2/s) flow in a 10cm diameter pipe at 0.5L/s. Is the flow is laminar or turbulent?
Solution: Given that
kinematic viscosity π=1.8Γ10β5π2/s
Pipe diameter D = 10cm = 0.1 m
Velocity of flow π = 0.5L/s = 5 Γ 10β4π3/s
Nature of flow = ?
v = π
π΄
v =π
ππ·2 4
v =5 Γ 10β4
π0.124
v = 0.0637m
s
As π π =π½π«
π
π π =π.ππππΓπ.π1.8Γ10β5
π π = πππ Λ ππππ 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,
ππ = 2V
ππ = 2Γ0.0638
ππ =0.1274 m/s
π’ = π’πππ₯β kπ2
π’πππ₯ = ππ =0.1274 m/s
π· = .1π
When π =ππ= 0.05m
u = 00 = 0.1274β k Γ0.052
π =50.96
π. π ππ Now,
π’ππ‘ 2ππ = π’πππ₯β kπ2
π’ππ‘ 2ππ = 0.1274β 50.96Γ0.022
π’ 2ππ = 0.107m/s
For laminar flow π = 64
π π
π = 64354
π=0.181
ππ =π
4π2
2g
ππ =π
4ππ2
2π = / g
ππ =0.181
4Γ 850Γ
0.06372
2
ππ =0.78N/π2
βπΏπΏ= π
1
π·
π2
2g
βπΏπΏ= 0.181Γ
1
0.1
0.06372
2Γ9.81
βπΏ
πΏ=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
π’ = ππ (1βπ2
ππ2) (1)
where,
u = average velocity
= centerline velocity
π = radius of the pipe at any point
ππ= maximum radius of the pipe.
The flow rate in a circular pipe can be calculated as,
π = πππ π’ππ΄ (2)
Using equation (1) into (2), we get
π =
π
ππ
ππ (1βπ2
ππ2)ππ΄
π = πππ ππ 1β
π2
ππ2 2ππππ ππ΄= 2ππππ
π = 2πππ
π
ππ
1βπ2
ππ2 πππ
π = 2πππ
π
ππ
π βπ3
ππ2 ππ
π = 2ππππ2
2βπ4
4ππ2π
ππ
π = 2πππππ2
2βππ4
4ππ2
π = 2πππππ2
2βππ2
4
π = 2πππππ2
4
π = πππππ2
2(3)
The average velocity can be calculated as
v = ππ΄
v =πππ
ππ2
2πππ2
v =ππ2
ππ =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
π’ = ππ (1βπ2
ππ2) (1)
where,
u = average velocity
ππ = centerline velocity
π = radius of the pipe at any point
ππ= maximum radius of the pipe.
Since,
ππ = π’πππ₯ = πππ£π
Or
πππ£π= V = 0.5ππ (2)
Put equation (2) into (1)
0.5ππ = ππ (1βπ2
ππ2)
0.5 = (1βπ2
ππ2)
π2
ππ2 = 1β0.5
π2
ππ2 = 0.5
π2
ππ2 = 0.5
π
π0= 0.707
π = 0.707π0
Therefore the distance from the centerline at which the average velocity occurs is
π = 0.707π0