Problem 3.122 (Fox & McDonald, 4th edition)
A rectangular container of water undergoes constant acceleration down an incline as shown.
Determine the slope of the free surface using the coordinate system shown.
At steady state,
the water in the
tank will behave
like a rigid body
 no relative
motion between
any two points in
the water
 water can be
treated as
inviscid
y

g
x
a x = 10
ft/s2
=
30o
Water is
incompressible
 Euler’s equation can be used



steady-state Euler:  p   g   a
components:
 p
  gx  a x ;
x
a y  a z  0, a x  10 ft / sec 2 ,

 p
  g sin 30  10  ;
x
 p  p x, y   dp 
 p
 p
  g y  ay ; 
  g z   az
 y
z
g x  g sin 30,
g y   g cos 30,
 p
  g cos 30  0;
 y
 p
 p
dx 
dy
x
 y
p free surface = patm = constant
 dp free surface = 0
 0   10  g sin 30dx   g cos 30dy
- 43 -
 p
0
z
gz  0
 10  g sin 30
10  32.2 sin 30  0.2187
dy 



free
dx  surface
 g cos 30
32.2 cos 30
NOTE: How is the direction of the surface
determined?
30o
Surfaces of constant pressure, including the free

surface, most be normal to p which is in the direction
of the maximum variation of p.




Euler:  p   g   a
direction of free surface
Example: The tank shown
is 4m long, 3m high, and
3m wide, and it is closed
except for a small opening
at the right end. It contains
oil (SG=0.85) to a depth of
2m in a stationary
situation. If the tank is
uniformly accelerated to
the right at a rate of
9.81m/s2, what will be the
maximum pressure
intensity in the tank during
acceleration?
g

p

a
Direction
of free
surface
opening
1
m
2
m
a x = 9.81 m/s2
y
oil (S.G. =
x0.85)
4
m
At steady state, oil will behave as a rigid body  no relative motion between fluid particles
 fluid is then inviscid
Oil is incompressible
 Euler’s equation can be used.



Steady-state Euler’s equation: p   g   a
Components:
 p
  gx  a x ;
x
 p
 p
  g y  ay ; 
  g z   az
 y
z
- 44 -
g x  g z  0,
g y   g  9.81 m / s 2 , a y  a z  0
a x  9.81 m / s 2

 p
  9.81;
x
 p
  9.81  0;
 y
 p  p x, y   dp 
 p
0
z
 p
 p
dx 
dy
x
 y
 dp free surface = 0
p free surface = patm = constant
 0   9.81dx  9.81dy
  9.81
dy 

 1  tan     135

free
dx  surface
 9.81
h  2 tan 45  2m but h 1m  oil will touch
the top of tank
1
m
h
 = 135o
45
o
y
2
m
b
 tan 45 b  4  a 
4  1
x
a
(4 – a)
=
135
o
y
x
b
Area occupied by the air remains the same
since the amount of oil is the same.
4  a  4  a
 1 4 


2
4  a  2
4  a  8  2.828
a  4  2.828  1.172 m
dp    9.81dx   9.81dy
- 45 -
8
  0.85 H 0  0.85 1000  850 kg / m3
2
 dp  8509.81 dx  8509.81 dy
p  8509.81x  8509.81y  const
At x  4m,
y  3  4  a   3  2.828  0.172 m,
p  patm
 patm  8509.814  8509.810.172  const
const  p atm  34788
 p  8338.5x  8338.5 y  patm  34788
p = pmax at x = 0, y = 0
pmax abs = 136113 Pa = 136.1 kPa
pmax gauge = 34788 Pa = 34.79 kPa
Euler’s Equation of Motion applied to a rotating fluid
z
Steady-state
free surface
Fluid rotates as a rigid body at steady state
Motor



Steady-state Euler: p   g   a

 g  Vg ; Vg   g z

r
(a conservative force is derivable from a
potential function)



moto
r
 g    gz
components:
 p
z
 g
  a r    r 2 
r
r

 p   g z     r 2
r
(1)
(r-component)
- 46 -
 p
z
 p
 g
  az 
  a z  g  (2)
z
z
z
(z-component)
Consider the r-component further

 p   g z     r 2 
r


r

r 2 2 
 p   g z  
0
2 


r 2 2 
  constant (3)
  p   g z  
2 

Example (Problem 5 from previous page)
At steady state liquid in tank behaves a rigid body -- liquid is inviscid since no relative motion
between liquid particles
liquid incompressible
diamete
Br
Euler applicable from (3) and
recognizing that V = r 
h
.
liquid
.
A
r
 pA   g zA 
VA  20 ft / s,
rA  r 

2
V A2  p B   g z B 

2
VB2
p A  30 psf  0.2083 psi
diameter   2  0.5  15
. ft
2
- 47 -
Example (Problem #4 from previous page)

.
At steady state, water is in sold-body rotation -no relative motion between fluid particles
--water is inviscid
--water is incompressible
--Euler’s equation is applicable
A
2m
Consider a point C at the bottom of the tank on
the axis of rotation
 pB   g z B 
g
2g
VB2  pC   g z C 
g
2g
VC2
z B  z C  0, VC  0, VB  rB  0.510  5 m / s
10009.815
p B  pC 
 12,500 Pa  12.5 kPa
2  9.81
2
Between C and A  p A   a z  g z A  pC   a z  g 
a z  4m / s 2 , z A  2m
pC  p A  10004  9.812  27,620 Pa  27.62 kPa
 p B  p A    p B  pC    pC  p A   12.5  27.62 kPa
 pB  p A   40.1 kPa
V A  rA   
rB  r 
V A 20

 13.33 rad / s
rA 1.5
diameter   2  0.5  2.5 ft
2
VB  rB  2.513.33  33.33 ft / s
z A  0, z B  h  1 ft
 g  S.G.  H O g  0.862.4  49.92 lbf / ft 3
2
- 48 -
. .
B
C
 p B  p A   g z A  z B  
g
V
2g
2
B

 V A2  30  49.920  1 

49.92
33.332  202
2  32.2

pB  531 psf  3.69 psi
Now consider the z-component further
 p
z
  a z  g  

 p   a z  g z   0
z
 p   a z  g z  const
(4)
General control-volume formulation of the rate of change of an extensive property of a
system
Let Pe be a general extensive property and Pi the corresponding intensive property
Pe  mPi
 Pe

system
 P dm    P dV
i
mass
system
i
V
system
system
system
III
II
streamlin
es
I
control
volume
(CV)
Fig. 4.1, p.99, (Fox & McDonald 4th
edition)
t = t + t
t=t


  Pe 
system  t
cv  t
Pe
control
volum
e





  Pe  Pe 
  Pe  Pe  Pe 

system  t  t
III  t  t
I
III  t  t
 II
 CV
Pe
- 49 -
 Pe system
dPe
 lim 
t 0 
dt


 Pe
CV
 lim 
t 0 




 Pe

CV
 t  t
 t

 Pe system 
t 
t


t  t
  
  
 
 
 Pe  
 Pe  
I  t  t
III  t  t
t 



 lim
 lim 
 t 0   t  t 0   t 











 
 

PeCV   Pi V  dA   Pi V  dA

 t CV
Ain flow
Aoutflow


 Pi dV
 t CV

 
P

V
 i  dA
CS: control surface
CS
rate of net mass efflux through an elemental area dA
(of the control surface) in unit time
In-flow boundary
Outflow boundary

dA

dA

V

V
 
V  dA  V dAcos180  V dA
 
V  dA  V dAcos0  V dA
Conservation of Mass (continuity of flow)
Pe = m,
Pi = m/m = 1
dm 
 0 (conservation of mass)

dt  system
- 50 -
 
dPe 


P

d
V

P

V
 dA

I
I

dt  system  t CV
CS
Recall:
 


dV


V
  dA
 t CV
CS
0
--control-volume form of the continuity equation(conservation of mass)
For incompressible flow of a homogeneous fluid,  = const
 

 V     V  dA

t
CS
0
but  V  m  const 

V   0
t
(if V is fixed)
 
   V  dA  0 
CS

 
V  dA  0 volumetric flow rate (Q = VA or Q   VdA )
CS
A
note: flow may or not be steady
For steady flow which is not incompressible,



0
t

 V  dA  0
CS






 V  dA  0  mass flow into the CV via the CS since V  dA  0 at an inflow boundary


 V  dA  0  mass flow out of the CV via the CS since V  dA  0 at an outflow boundary
4.28 Fluid with a 1050 kg/m3 density is flowing steadily through the rectangular box shown.




Given A1 = 0.05 m2, A2 = 0.01 m2, A3 = 0.06 m2, V1  4i m / s and V2  8 j m / s , determine

the velocity V3 .
- 51 -

n2
A
y
2
x
C.
V.
A
1
A
3

n1
60o
Conservation of mass for the control volume:
 

 dV    V  dA  0

t V
A
Flow is steady 

 dV  0
 t V
Assume flow is uniform at all exits and inlets:
3
 
 
   V  dA    V  A  0
i 1
A
 
 
 
  V1  A1  V2  A2  V3  A3  0


 




1050 4i  0.05i   ( 8 j )  (0.01 j )  V3  A3  0


 
3
V3  A3  40.05  (8)(0.01)  0.28 m sec ( 0)
flow at ‘3’ is an outflow


V3 n3  A3 n3  0.28
 V3 
0.28
 4.67 m / sec
0.06





V3  V3 (cos 30 i  sin 30 j )  4.67(cos 30 i  sin 30 j )



V3  (4.04i  2.34 j ) m / sec
- 52 -

n3
4.33 Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the
uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
 r2 
u  umax 1  2 
 R 
and umax = 10 ft/sec.
C
V
U
R
r
x
2
1
L
Conservation of mass for the control volume:
Given: flow is steady 
 

 dV    V  dA  0

t V
A

 dV  0
 t V
Water is the working fluid  flow incompressible   = const
 
  
 
   V  dA   V1  A1   V2  dA2   0
A
A2


 
 V1  A1 

V
2

 dA2  0
A2


V1  Ui ,
2




 3
2
A1   R ( i )     ( i )  0196
. i ft 2
 12 




 r2  
V2  u i  101  2  i , dA2  2 r dr i
 R 
d
r
r
- 53 -

r2 
u  101  2  ft / sec
 R 
R
 r2 
U R 2   101  2 2 rdr  0
0
 R 
R
r 2
 R2 R2 
r4 
 U R  20  

20

 2  4 
2 
 2 4R 0


2
 R2 
20  
 4  5
 U
 R2
4.38
ft / sec
A section of pipe carrying water contains an expansion chamber with a free surface
whaose area is 2 m2 . The inlet and outlet pipes are both 1 m2 in arewa. At a given
instant, the velocity at section ‘1’ is 3 m/sec into the chamber. Water flows out at
section ‘2’ at 4 m3/sec. Both flows are uniform. Find the rate of change of free surface
level at the given instant. Indicate whether the level rises or falls.
y
3
x
V1
V2
2
1
Conservation of mass for the cortrol volume:
 

 dV    V  dA  0

tV
A
Given: water is the working fluid  flow incompressible   = const

 

  dV    V  dA  0
t V
A

 

 V     V  dA  0
t
A
V
 

V

   V  dA  0
t
t A
( = const) (V is fixed)
- 54 -


  V  dA  0
A
Assume flow is uniform at all exits and inlets:


3
 
 
 
 
 
   V  dA    V  A   V1  A1  V2  A2  V3  A3  0
i 1
A


V1  3i ,


A1  1( i )

 

V2  A2  4m3 / sec, A3  2 j


 3i  1( i )  4  2V3  0


 
A3  V3  3  4  1 m3 / sec ( 0)  ’3’ is an inflow boundary

1
V3   j m / sec  liquid level falls
2
4.41
A tank of 0.5 m3 volume contains compressed air. A valve is opened and air escapes
with a velocity of 300 m/sec through an opening of 130 mm2 area. Air temperature
passing through the opening is –15 C and the absolute pressure is 350 kPa. Find the rate
of change of density of the air in the tank at this moment.
C
V
V = 0.5 m3
Ve = 300 m/sec
e
 

 dV    V  dA  0

t V
A
Pr 
Ae = 130 mm2 = 130 x 10-6 m2
Conservation of mass for the control volume:
P
0.35
T
258

 0.093 ; Tr 

 194
.
Pcrit 3.76
Tcrit 133
 z  1  air can be treated as an ideal gas
- 55 -
 

 V     e Ve  dAe  0
t
Ae
V
 

V

    e Ve  dAe
t
t
Ae
(V is fixed)
e 
pe
350

 4.73 kg / m 3
RTe 0.287  258
 

1
1

    e Ve  dAe  
4.73  300  130  10 6 
t
V Ae
0.5

 0.369 kg/m3 sec
t
4.44
A cylindrical tank, of diameter D = 50 mm, drains though an opening, d = 5 mm, in the
bottom of the tank. The speed of the liquid leaving the tank is approximately

V  2 g y , where y is the height from the tank bottom to the free surface. If the tank
is initially filled with water to yo = 0.4 m, determine the water depth at t = 12 sec.
D
y
x
i
C.V
.
yo
e
D = 50 mm = 0.05 m
d = 5 mm = 0.005 m
y0 = 0.4 m
Ve = 2 g y
Vi  0 (since D >> d)
Conservation of mass for the control volume:
 


dV


V
A  dA  0
 t V
d
Assume flow is uniform at ‘e’
  e
 
   V  dA    V  A
A
i
- 56 -
Assume  = const since the working fluid is a liquid
V 




D 2 y ; Ve  2 g y (  j ) ;
4



Ae  d 2 (  j )
4
e
e
 
 


  dV    V  A  0    V     V  A  0
t V
t
i
i
V
e  

V

  V  A  0
t
t
i
( = const)

D2
4
 
 
dy
 
  (0) D 2  2 g y (  j )  d 2 (  j ) 
dt
4
 4

dy
d2
 2
dt
D

y
y0
2g y 
dy
1
y2
1

d
y 2 dy   2 g  
 D
2
2
d
  2 g   dt
 D
12

0
dt
y
2
 12 
 0.005  12
2 y    2  9.81
 t 0
 0.5 
  y0
1
 21
2  y  0.4 2   0.044312


1
2
y 
0.044312
2
1
  0.4 2  0.2658  0.6325  0.3667
y = 0.134m
4.49
Water flows steadily past a porous flat plate. Constant suction is applied along the
porous section. The velocity profile at section cd is
u
 y  y
 3   2 
U
   
1.5
- 57 -
Evaluate the mass flow rate across section bc.
CV
U = 3 m/s
b
c
u
  1.5 mm
y
a
x


V  0.2 j mm / sec
U = 3Lm/s
=2m
d
 

 dV   V dA  0

A
t V
Conservation of mass for the c.v.:
Width,
w = 1.5
m
(c.v. is abcd)
Given: working fluid is water  flow is incompressible    const
Flow is steady 

 dV  0
 t V
 
 0
   V  dA  m
A









 Vab  Aab  m bc    Vcd  dAcd   Vad  Aad  0


Vab  3i ,
0



 1.5 
Aab  
1.5 i   0.00225 i
 1000 


   y   y 1.5 

Vcd  3i 3   2  , dAcd  1.5dy i
       

0.2 
Vad 
j,
1000



Aad  215
.   j   3 j




  0.2
 m bc    3i  0.00225 i    
 1000
 6.75   10
3
1.5
1000
0
 0.6  10    
3
1.5


   y   y 1.5 

j  3 j    1000 3i 3   2   1.5 dy i 
0


       
  y   y 1.5 
4.53   2   dy
       
- 58 -
Let
y

 Y  dy   dY
 m bc  6.15  10 3   
4.5 0 3Y  2Y 1.5 dY
1
1
2 2. 5 
 1. 5   3 2
 6.15  10    4.5
Y 
 Y 
2.5
 1000   2
0
3
  1000kg / m
2
3
1
m bc  6.15  10 3 1000   10004.51.5  10 3  1 
2.5 
2
 6.15  6.751.5  0.8  1.425 kg / s
3
Velocity profile development in pipes
“Sufficiently far from the pipe entrance, the boundary layer developing on the pipe wall
reaches the pipe centerline and the flow becomes entirely viscous. The velocity profile shape
changes slightly after the inviscid core disappears. When the profile shape no longer changes
with increasing distance, x, the flow is fully developed. The distance downstream from the
entrance to the location at which fully developed flow begins is called the entrance length. The
actual shape of the fully developed velocity profile depends on whether the flow is laminar or
turbulent. In Fig. 8.1 the profile is shown qualitatively for a laminar flow.
“For laminar flow, the entrance length, L, is a function of Reynolds number,
V D
L
 0.06
(8.1)
D

where D is pipe diameter, V is average velocity,  is fluid density, and  is fluid viscosity.
Laminar flow in a pipe may be expected only for Reynolds numbers less than 2300. Thus the
entrance length for laminar pipe flow may be as long as
L  0.06 Re D  0.062300D  138D
- 59 -
r
u
x
Uo
D
Fully
developed
velocity profile
Entrance
length
Fig. 8.1 Flow in the entrance region of a pipe
For fully-developed laminar flow in a pipe:

r2 
u  umax 1  2 
R 

umax : velocity on the centreline, i.e., pipe axis
u
x
r
R
y
umax
Q   u dA  
A
R
0
R

r2 
r3 
u max 1  2 2 r dr  2 u max   r  2  dr
0
R 
 R 

R
r 2
 R2
r4 
R4   R2
Q  2 u max  

2

u


u max
max 
2 
2 
2
 2 4R  0
 2 4R 
But Q  A u ave   R 2 u ave
 u ave 
 R2
2
u max 
u
1
 max
2
2
R
- 60 -
d
r
r