Mechanical Engineering Programme of Study

Instructor: Marios M. Fyrillas

Fluid Mechanics Email: eng.fm@fit.ac.cy

SOLVED EXAMPLES ON VISCOUS FLOW

1. Consider steady, laminar flow between two fixed parallel plates due to a pressure gradient. Using a control volume of unit depth, height 2y , and width x (centred at 0y ) obtain an expression for the velocity profile. a. By integrating the velocity profile obtain an expression for the

volumetric flow rate and the mean velocity. b. Obtain an expression for the dimensionless pressure loss as a function

of the Reynolds number.

x

Consider -momentum conservation

0 (steady-state so net momentum flux is zero)

The forces acting are:

i. right surface:

ii. left surfac

out in x

r r

x

M M F

p A

Conservation of Momentum of the control volume

e:

iii. top surface:

iv. bottom surface:

l l

t t

b b

p A

A

A

w

y

Balance of forces: - ( )

Because of symmetry:

Areas are given by: 1 2 , 1

2 ( - ) 2

d (Newtonian fluid)

d

dconstant (pressure increases lin

d

r r l l t t b b

t b

r l t b

l r

p A p A A A

A A y A A x

y p p x

u

y

p

x

2

early)

d d d( )

d d d

Integrate above expression

d d dd d d constant

d d 2 d

To find the constant use the boundary conditions, i.e.

at and - the velocity i

u p uy p p p x y

y x y

y p y p y pu y u y

x x x

y h y h

2 2

22 2 2

22

s zero (u 0)

d d( ) 0 constant constant=-

2 d 2 d

d d dso u= - 1

2 d 2 d 2 d

To find the volumetric flow rate:

d d 1 d 1 d

2 d

h p h pu y h

x x

y p h p h p y

x x x h

h p yQ u A u y

x h

2

3

3

2

2 22 2 2

d 4

2 d 3

2 d

3 d

2 dd3 d

(mean velocity) 2 3 d

3 3 612 24

1 1 2 2 R2 2

h

h

m

m m

m mm m

h p hy

x

h p

x

h pQ h px

uA h x

u p up

h u h h u hu u h

b. Dimensionless pressure drop

1

eh

2. Working in a similar fashion as for the case of a horizontal cylinder, obtain the velocity profile of Poiseuille’s law in an inclined pipe using the control volume suggested in the figure.

2 2

Consider -momentum conservation

0 (steady-state so net momentum flux is zero)

The force balance can be written as:

( ) 2 sin

out in x

x

M M F

p p r p r r mg

Conservation of Momentum of the control volume

2

2

2

0

sin 2d sin

d d 2d

sin sind cons

2 2 2

Evaluate the constant using the boundary conditions:

sin( ) 0 0 consta

4

m r

p gu p gr r

u rr

p g p g ru r r

p gu r R R

tant

2

nt

sinconstant=

4

p gR

22

22

0 0

2 2 4

( sin )1

4

( sin )d 2 d 1

2

( sin ) ( sin )

2 4 8

R R

p g R ru

R

p g R rQ u A u r r r r

R

p g R R p g R

d

3. An oil with a viscosity of 2 and density 3 flows in a

pipe of diameter 0.2 m . (a) What pressure drop 1 2

0.4 N s/m 900 kg/m D p p , is needed to

produce a flowrate of 5 3 if the pipe is horizontal with 2.0 10 m / s Q 1 0x

and 2 10 m ? (b) How steep a hill, x , must the pipe be on if the oil is to

flow through the pipe at the same rate as in part (a), but with 1 2p p ? (c) For

the conditions of part (b), if a1 200 kPp , what is the pressure at section

3 5 mx where x is measured along the pipe?

4. Consider steady, laminar flow in a circular pipe due to a pressure gradient. Using a control volume of length and radius r obtain an expression for the velocity profile. Follow the steps below: a. Consider the control volume below (Figure 1) and indicate the forces

exerted on the control volume. Give a physical explanation.

Control Volume

Figure 1: Laminar flow in a circular pipe.

a. Doing a force balance show that the momentum equation can be simplified to:

2p

r

.

c. Assuming laminar flow of a Newtonian fluid and applying an appropriate

boundary condition obtain that the velocity profile is: 22 2

116

p D ru

D

.

d. Integrate above expression to find the volumetric flow rate.

2 2

The forces acting on the control volume are the shear forces acting on the perimetric area 2 ,

and pressure forces acting on the fore and aft cross-sectional areas and ( ) , respectively.

B

r

p r p p r

2 2 2y doing a force balance 2 ( )

pp r r p p r

r

5. Determine the head loss for a sudden expansion. Consider the control volume shown on the figure below and use conservation of mass and conservation of momentum.

1 1 1 2 3 3

3 3

3 1 3 3 3 1

1 3 3 3 3 3

density is constant

( )

( ) ( )

Assume that

(

a a b b c c out in

out in out out in in

a b c

V A V A m

p A p A p A p A M M

M M m V m V m V V V A V V

p p p

p A p A V A

Mass Conservation

Momentum Conservation

3 1)V V

2 21 1 3 3

1 3 3 3 1

2 21 3

3 3 1

23 3 1

2 2

From momentum equation: ( )

Substitute above in energy equation

( )2 2

Solve above for

L

L

L

p V p Vh

g g g g

p p V V V

V VV V V gh

Vgh V V V

Energy Equation (Bernoulli's equation)

2 21 3

31 1

33

2 2 21 1 1 1 1

3 3

22 2

1 1 1 1 12

1 3 3 3 3 3

2 2Substitute

From mass conservation:

1

2 2

1 1 1 12 1 1

2 2 2 2

The loss coe

L

L

V

VV A

VA

V A V V Agh

A A

gh A A A A A

V A A A A A

2

122

1 31

2fficient 1

2

L LL

h gh AK

V AVg

6. Calculate the power supplied to the pump shown in Figure 3 if its efficiency is 76%. Methyl alcohol ( 3 -4790 kg/m , 5.6 10 Pa s ) is flowing at the rate

of 3 . The suction line is a standard 4 in54 m /hr steel pipe, 15 m long. The total length of 2 in steel pipe in the discharge line is 200 m . Assume that the entrance from reservoir 1 is through a squared-edged inlet and that the elbows are standard. The valve is a fully open globe valve. The roughness of the pipe is є= m . 0.045 m

Figure 3: Pump/pipeline configuration

2 21 1 1 2 2 2

1 2 1 2

Consider a streamline joining the points 1 and 2. Applying the energy equation we obtain

1 1.

2 2

= = . If we take as the datum the point 1 then 0 and 10 m

pumpL

atm

Wp u gz p u gz gh

Q

p p p z z

1 2

2

.

If we further assume that 0 and 0 the energy equation simpilfies to

.pump L

u u

W Q gz gh

3 3 3

suction

suction

discharge

discharge

3

-4

2

2

Given:

5454 m /hr m /s=0.015 m /s

36004 in=0.1016 m

15 m

2 in=0.0508 m

200 m

= 790 kg/m

5.6 10 Pa s

g= 9.81 m/s

10 m

The only uknown in the equation for p

Q

D

D

z

W

2 2 2 2 2

major losses major losses minor losses minor losses fully minor losses suction discharge pipe entrance open globe valve

0.5 10

is

+ + + 22 2 2 2 2

L L

ump

L L L L

K K

V V V V Vh f f K K K

D g D g g g g

2

of minor lossesthe 2 standard elbows pipe exit

0.3 1

2 2suction suction

+ 2

The loss coefficients can be obtained from a table, and the velocities from

= /( / 4) 4 0.015/ 3.14 / 0.1016 1.8

L L

L

K K

VK

g

V Q D

2 2discharge discharge

suction suctionsuction

5 m/s

= /( / 4) 4 0.015/ 3.14 / 0.0508 7.4 m/s

To find the major losses we need to find the Reynolds number and the relative roughness

790 1.85 0.1016

5.6

V Q D

V DRe

-4

-3

suction

suction

discharge dischargedischarge -4

-3

discharge

discharge

26500010

0.045 100.00044

0.1016

0.019 from Moody chart

790 7.4 0.0508 530000

5.6 10

0.045 100.000089

0.0508

0.014

D

f

V DRe

D

f

from Moody chart

Substitute all above information in the equation for ,calculate and finally substitue

in equation for L L

pump

h h

W

7. For the system shown in Figure 4, compute the power delivered by the pump to the water to pump 30.0031545 m /s of water at 15 Co to the tank. The air in the tank is at 276 kPa gauge pressure. Consider the friction loss in the 225-ft-long discharge pipe, but neglect other losses. Then, redesign the system by using a larger pipe size to reduce the energy loss and reduce the power required to no more than 3729 W . The roughness of the pipe is є= -41.5 10 and 1 . in=0.0254 m

Figure 4: Pump/pipeline configuration

8. In the turbulent region the friction factor associated with pipe flow is approximated by the formula:

10 0.9

0.55.74

log3.7 Re

f

D

Find an expression for the friction factor f for large number. Re

0.9Re

10

For large Reynolds number (Re) above expression simplifies to

0.5 5.74 because lim 0.

Relog3.7

f

D

Liquid with specific gravity

is flowing in a vertical pipe. If the diameter of the pipe is and the viscosity of the fluid is determine the direction of the flow and the mean velocity if the pipe relative roughness is

310 kN/mg

15 cmD 3 23 10 N m/s

/ 0.008D . The pressures shown are static pressures. Hint: Assume a high Reynolds number and verify.

2 21 1 2 2

1 2

2

2 2

where the losses are estimated using 2

and we have assumed that the flow is directed upwards.

L

mL

p V p Vz z h

g g g g

uh f

D g

Energy Equation (Bernoulli's equation)

2 2 21 2

2 2

Using mass conservation and assuming uniform flow

.

So Bernoulli's equation simplies to

200000 1100000 10

10000 2 10000 2

20 11 10 1

Hence, our original assumption was wrong a

m

m mL

L L

V V u

u uh

g g

h h

2 21 1 2 2

1 2

2 2

10 10

nd the

flow is directed downwards, i.e.

2 2

1

If we assume that the flow has a high Reynolds number

0.25 0.25then 0.0352

0.008log log

3.7 3.7

L

L

L

p V p Vz h z

g g g g

h

f

D

h

22

3

101 0.0352 0.12 1 2.89 m/s

0.15 2

1019 2.89 0.15Verify Reynolds number Re= 144500

3 10

mm m

uu u

g

uD

9. Estimate the elevation required in the upper reservoir to produce a water discharge of in the system. What is the minimum pressure in the pipeline and what is the pressure there?

10 cfs

10. Water flows from a reservoir through a pipe 150mm diameter and 180m long to a point below the surface of the reservoir where it branches into two pipes, each 100mm in diameter (see Figure 2). One of the pipes is 48m long discharging to atmosphere at a point below reservoir level and the other 60m long discharging to atmosphere 24m below reservoir level. Assuming that 0.032 calculate the discharge from each pipe, neglecting all loses other than friction.

f

60m

48m

180m 18m

24m

Figure 2: Reservoir pipeline configuration

11. The three water-filled tanks shown in the figure (Figure P8.102 in textbook) are connected by pipes as indicated in the figure. If minor losses are neglected determine the flowrate in each pipe.

12. Water is to be pumped from one large, open tank to a second large, open tank as shown in the figure. The pipe diameter throughout is 15 cm and the total length of the pipe between the pipe entrance and exit is 61 m . Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure, and the friction factor for the pipe can be assumed constant and equal to 0.02 . A certain centrifugal pump having the performance characteristics shown in the figure is suggested as a good pump for this flow system. With this pump, what would be the flowrate between the tanks? Do you think this pump would be a good choice?