homework assignment (viscous flow)
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7/27/2019 Homework Assignment (Viscous Flow)
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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 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
mailto:eng.fm@fit.ac.cymailto:eng.fm@fit.ac.cy -
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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 chart
Substitute all above information in the equation for ,calculate and finally substitue
in equation for
L L
pump
h h
W
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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 thesystem 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= -41.5 10 and 1 .in=0.0254 m
Figure 4: Pump/pipeline configuration
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8. In the turbulent region the friction factor associated with pipe flow isapproximated by the formula:
10 0.9
0.5
5.74log
3.7 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 5.74because lim 0.
Relog
3.7
f
D
Liquid with specific gravity
is flowing in a
vertical pipe. If the diameter of thepipe is and the viscosity of
the fluid is
determine the direction of the flowand the mean velocity if the piperelative roughness is
310 kN/mg
15 cmD3 2
3 10 N m/s
/ 0.008D .The pressures shown are staticpressures. Hint: Assume a highReynolds number and verify.
2 2
1 1 2 21 2
2
2 2
where the losses are estimated using2
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)
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2 2 2
1 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 2
1 1 2 21 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= 1445003 10
mm m
uu u
g
uD
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9. Estimate the elevation required in the upper reservoir to produce a water discharge of minimum pressure in the pipeline and what is the pressure there?
10
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10. Water flows from a reservoir through a pipe 150mm diameter and 180m longto a point below the surface of the reservoir where it branches into twopipes, 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 other60m long discharging to atmosphere 24m below reservoir level. Assuming
that 0.032 calculate the discharge from each pipe, neglecting all losesother than friction.
f
60m
48m
180m 18m
24m
Figure 2: Reservoir pipeline configuration
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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 areneglected determine the flowrate in each pipe.
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12. Water is to be pumped from one large, open tank to a second large, opentank 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 onthe figure, and the friction factor for the pipe can be assumed constant
and equal to 0.02 . A certain centrifugal pump having the performancecharacteristics shown in the figure is suggested as a good pump for thisflow system. With this pump, what would be the flowrate between thetanks? Do you think this pump would be a good choice?
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