Homework Assignment (Viscous Flow)

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Mechanical Engineering Programme of Study

Instructor: Marios M. Fyrillas

Fluid Mechanics Email:[email protected]

SOLVED EXAMPLES ON VISCOUS FLOW

1. Consider steady, laminar flow between two fixed parallel plates due to apressure 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 thevolumetric flow rate and the mean velocity.

b. Obtain an expression for the dimensionless pressure loss as a functionof 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

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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 yy 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:

dd 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 d2 d

d3 d(mean velocity)

2 3 d

3 3 612 24

1 1 2 2 R

2 2

h

h

m

m m

m mm m

h p hy

x

h p

xh p

Q h pxu

A h x

u p up

h u h h u hu u h

b. Dimensionless pressure drop

1

eh

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2. Working in a similar fashion as for the case of a horizontal cylinder, obtainthe velocity profile of Poiseuilles law in an inclined pipe using the controlvolume 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 2

d sin

d d 2

d

sin sind cons

2 2 2

Evaluate the constant using the boundary conditions:

sin( ) 0 0 consta

4

m r

p g

u p grr

u r

r

p g p g ru r r

p gu r R R

tant

2

nt

sinconstant=

4

p gR

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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 with2.0 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?

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4. Consider steady, laminar flow in a circular pipe due to a pressure gradient.Using a control volume of length and radius robtain an expression forthe 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 besimplified to:

2p

r

.

c. Assuming laminar flow of a Newtonian fluid and applying an appropriateboundary condition obtain that the velocity profile is:

22 21

16

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 ( ) p

p r r p p rr

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

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2 2

1 1 3 3

1 3 3 3 1

2 2

1 33 3 1

2

3 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 2

1 3

3

1 13

3

22 2

1 1 1 1 1

3 3

22 2

1 1 1 1 1

2

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

1

22 1 31

2fficient 1

2

L L

L

h gh AK

V AV

g

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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. Thetotal length of 2 in steel pipe in the discharge line is 200 m . Assume thatthe 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 roughnessof the pipe is = 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

pump

L

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

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3 3 3

suction

suction

discharge

discharge

3

-4

2

2

Given:

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

3600

4 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 forp

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 V h 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 discharge

discharge -4

-3

discharge

discharge

26500010

0.045 100.00044

0.1016

0.019 from Moody chart

790 7.4 0.0508530000

5.6 10

0.045 100.000089

0.0508

0.014

D

f

V DRe

D

f

from Moody