For fully-developed turbulent flow in a pipe:
1
rn

u  umax 1  

R
where n varies from 6 to 10 depending on the flow Reynolds number and umax is the velocity on
the pipe axis.
d
r
r
u
1
Q   u dA  
A
Let s  1 
R
0
1
R
r n
r n

u max 1   2 r dr  2 u max  1   r dr
0
 R
 R
r
dr
 ds   ; r  R1  s
R
R
1
1
1
1 
0
0
 Q  2 u max  s n R1  s  R ds   2 R 2 u max    s n  s n ds 
1
1


0
0


 1
1
 n 1
2 n 1
 1  1


 s n 1

n
s 
ns n
ns n 
  2 R 2 u max 
 2 R 2 u max 


 n  1 2n  1 
  1   1 

1
1


1





 1
  n   n 

1

 1
1 
 2 R 2 u max 0  n 


 n  1 2n  1 

Q  2 R 2 umax n
Q   R 2 u ave 
2n  1  n  1  2 R 2 umax n 2
n  12n  1 n  12n  1
2 R 2 u max n 2
n  12n  1
- 61 -
uave
2n 2

u
n  12n  1 max
n = 7 in most cases (1/7th power-law velocity distribution in fully-developed turbulent pipe
flow)
249
u
with n = 7  uave 
815 max
uave 
98
49
u max 
u max
120
60
Flow along curved streamlines

n

V
s=s
t=t
d
ds
streamline
s=s+
ds
t = t + dt
s
R
Radius
of
curvatur
e
Vn
ds
s

 V
V
ds
s
d
Centre of
curvature
C
Fluid is accelerated both along the streamline (in the s-direction) and normal to the streamlines
(in the n-direction)
 
Velocity field: V  V  s, t 



V
DV V
 

V
Accleration: a  a s 
(along streamline)
Dt
dt
s
Acceleration normal to the streamline:
an 

V 
1  Vn  1 
ds  V 
ds sin d 

dt   s  dt 
s 

d is small  sin d dds =Rd; ds is small
- 62 -
an 
1
dt

V   1 
V  ds  V ds
1 V
 s d   V 
 s  

ds 2
V 
 s   dt 
 s  R  R dt Rdt  s

(second-order term)
ds
V
dt
6.34
 an 
V2
R
Water flows steadily up a vertical 0.1 m diameter pipe and out the nozzle, which is
0.05m in diameter, discharging to atmospheric pressure. The stream velocity at the
nozzle exit must be 20m/sec. Calculate the gage pressure required at section (1),
assuming frictionless flow.
2
Given: Flow is steady
V2
Frictionless flow can be assumed
Working fluid is water Flow is
incompressible
4m
From the foregoing, Bernoulli’s
equation can be applied along a
streamline
1
Flo
w
Central streamline is chosen
p1
2
2
V
p V
 1  g z1  2  2  g z 2

2

2
 p1  p 2 


1
 V2 2  V1 2   g  z 2  z1 
2
  1000 kg / m3 , z 2  4m, z1  0,
 p1g 

p2 g  0

1000 2
2
V2  V1  10009.814  0
2
Continuity: V1 A1  V2 A2  V1 
A2
 0.052
420
V2 
A1
4  0.12
- 63 -
Datu
m
V1  5 m / s
p1g
 20 2  5 2 
  10009.814  226.7 kPa
 1000
2


6.36
Water may be considered to flow without friction through the siphon. The water flow
rate is 0.03m3/sec, its temperature is 20C, and the pipe diameter is 75 mm. Compute the
maximum allowable height, h, so that the pressure at point A is above the vapor pressure
of the water.
A
Given: Flow without friction
frictionless flow
h
z
Flow of water
incompressible flow
1
D = 75
mm
Assume: Flow is steady
 Bernoulli’s equation can
be applied along a streamline
from (1) to (2)
2
p1
flo
w
2
V
p
V
 1  g z1  A  A  g z A

2

2
z A  h,
z1  0, V1  0 (since tank >> delivery pipe)
p1  patm  101.3 kPa,
p A  pvapourH2O  2.34 kPa,
T  20C
p A  2.34 kPa
(see handout)
h 
101.3  2.34  10 3
1000  9.81
2

VA
2  9.81
Continuity: Q  V A A A  V A 
Q
0.03

 6.79m 3 / sec
2
A A  0.075
4
- 64 -
h 
101.3  2.34  6.792
9.81
2  9.81
 7.74 m
hmax  7.73m
6.53
The flow system of parallel disks shown contains water. As a first approximation,
friction may be neglected. Determine the volume flow rate and the pressure at point
(C). (R = 300mm and rc = 150mm.)
Given: Flow of water
incompressible flow
1
Friction may be neglected 
frictionless flow
Assume: Container much
larger than space between the
disks  V1  0  flow is
steady
 Bernoulli’s equation can be
applied along a streamline from
(1) to (2)
2
p1
H=1
m
C
z
r
1
2
rC
H = 1.5
mm
R
2
V
p V
 1  g z1  2  2  g z 2

2

2
p1  p2  patm ,
z 2  0,
z1  H  1 m
2

V2
 g1  V2  29.811  4.43m / sec
2
Volume flow rate: Q  A2V2  2 RhV2  2 0.30.00154.43  0.0125m 3 / sec
Continuity: Ac Vc  A2V2  Q  Vc 
Q
Q
0.0125


 8.86m / sec
Ac 2 rc h 2 015
. 0.0015
Bernoulli’s equation applied along a streamline from (C) to (2)
- 65 -
2
pc
2
V
p
V
 c  g zc  2  2  g z2

2

2
z 2  z c  0;
pc  p 2 
p2  patm  101.3 kPa

 V 2 2  Vc 2
2
  101.3  10004.43
 8.86 2
2  1000
2

pc  71.9 kPa
abs
6.45
An Indianapolis racing car travels at a maximum speed of 350 km/hr. A pressure gage
attached to the airfoil reads -50mm of water (gage). Estimate the air speed relative to
the car at that location.
Compressibility effects in air become significant when the local speed is more than 30% of the
speed of sound at the prevailing temperature c  k RT : c  340m / s .

350km / hr 

350
m / sec  97.2m / sec  102m / sec
3.6
 Flow is incompressible
Let flow observer be located on the wing  Flow appears steady to the observer
Flow outside the boundary layer can be treated as inviscid, i.e., frictionless
Bernoulli’s equation can be applied along a streamline
1
V1
p1
p1

2

2
p2
2
V1
p V
 g z1  2  2  g z 2
2

2
z1  z 2  p2  p1
V

1
2
 V2
2
2
 V
  V 2 
1   2  
2   V1  


2
1
- 66 -
V2
But p 2  p1   H 2O g h
2
2
V1   V2  
1       H 2O g h

2   V1  


2  H 2O g  h
V 
1   2  
 V1 2
 V1 
2
 2  H 2O g  h 
V2

 1 
2

V1

V
1


V1 
350
 97.2m / sec
3.6
V2  97.2 1 
4.58
210009.81 0.05
1.23  97.2
2
 101 m / sec (relative to the wing)
A jet of water issuing from a stationary nozzle at 15m/sec (Aj =0.05m2) strikes a turning
vane mounted on a cart as shown. The vane turns the jet through angle =50.
Determine the value of M required to hold the cart stationary.
Ry
e
i
=
50o
CV
T
y
x
T
M
Mg
Assumptions
Flow is steady
Flow at (i) and (e) is uniform
No drag forces at the vane
- 67 -
Momentum equation for the C.V.:



Fs  Fb 

 

V dV   V V  dA
t
V
A
(Steady flow)
 x-component of the momentum equation:
e
 
Fs x  FB x   u V  A   Q V xi  V xe 
i
T  10000.051515  15 cos 50  1000015
. 15 cos 50
2
T  4020 N  Mg
M 
4.78
since cart stays stationary
T 4020

 409.8 kg
g 9.81
A turning vane, which deflects the water through 60, is attached to the cart under the
conditions of Problem 4.57. Determine the tension in the wire holding the cart
stationary and the force of the vane on the cart.
CV
a
ea
CV
eb
b
x
=
60o
Fx
y
ib
b
Fx
a
Fy
b
Assumptions
No drag forces on the vane
Flow is uniform at (i) and (e)
Given
Flow is steady, level in tank is held constant
Flow of water   = constant
- 68 -
Momentum equation for the C.V.



Fs  FB 


 

V dV   V V  dA
t
V
A
(Steady flow)
 x-component of the momentum equation for C.V.a
ea
 
Fs x a  FB x a   uV  A   Q V xia  V xea 
ia


Fx a  1000 600  10 6 1010 cos 60  30 N

Fxa = Tension in wire holding cart = 30N i
x-component of the momentum equation for C.V.b:
eb
 
Fxb  FBxb   uV  A   Q V xib  V xeb   1000 600  10 6 10 10  10 cos 60  30 N


ib
Force ( Px ) of vane on cart in the x-direction = -(Force of cart on vane=Fxb)
 Px   Fxb  30 N
y-component of the momentum equation for C.V.b:
eb
 
Fyb  FByb   v V  A   Q  V yib  V yeb


ib


Fyb  Mg  1000 600 10 6 100  10 sin 60  51.96 N  Fyb  Mg  51.96N
Force (Py) of vane on cart in the y-direction = -(Force of cart on vane=Fyb)
 Py   Fyb  51.96 N


P  30i  51.96 j N  60 N
4.80 A conical spray head is shown. The fluid is water and the exit stream is uniform.
Evaluate (a) the thickness of the spray sheet at 400mm radius and (b) the axial force exerted by
the spray head on the supply pipe.
- 69 -
=
30o
CV
Rx
e
Assumptions
Flow is uniform at (i) and
(e)
Flow is steady
D = 300
mm
pig
Q = 0.03
m3/s
x
i
Given
Flow of water =const
p1 = 150 kPa
y (abs)
V = 10 m/s
=
30o
(a) Conservation of mass
for steady flow:
Q  Vi Ai  Ve Ae  (10)2 Rt  0.03
Vi Ai  0.03
Vi 
t
0.03
2
 0.3
4
 0.424 m / s
0.03
 1194
.
 10 3 m  1194
. mm
10
2

0
.
4
   
(b) Momentum equation for the C.V.



Fs  FB 

 

V  dV   V V  dA
t
V
A
(Steady flow)
 x-component of the momentum equation for C.V.:
e
 
R x  pig Ai  FBx   u  V  A   Q V xi  V xe 
i
Rx  1000150  1013
. 

4
0.3  10000.030.424  10 sin 30  3605 N
Axial force (Ps) of spray head on supply pipe = -(axial force of supply pipe on spray head = Rx)
Ps   Rx   3605 N   3605N
- 70 -
4.106 Experimental measurements are made in a low-speed air jet to determine the drag force
on a circular cylinder. Velocity measurements at two sections, where the pressure is
uniform and equal, give the results shown. Evaluate the drag force on the cylinder, per
unit width.
U
2
U
U = 50 m/sec
 = 1.2 kg/m3
D = 30 mm
a = 2.2D
  y
u  U sin
 ; 0 ya
 2a 
u = U; y > a
CV
a
D
u
1
3
x
4
Rx
y
Assumptions
Flow is steady
Flow is incompressible since flow is low
Conservation of mass for steady flow
1  m
2  m
3  m
4  0
m
 2  2  uwdy  m
4  0
  U w2a  m
a
0
2  m
 4   U w2a  2  uwdy  2  U  uwdy
m
a
a
0
0
Momentum equation for the c.v.:
  
 
 
Fs  FB 
V

dV

V
A  V  dA
 t V
(Steady flow)
x-component of the momentum equation:
2  m
 4    U w2a(U )  2  u 2 wdy
Rx  FBx  U m
a
0


2  m
 4   2   u 2  U 2 wdy
Rx  U m
a
0
- 71 -




Rx  2  U 2  Uu wdy  2  u 2  U 2 wdy  2  u u  U wdy
a
0
a
0
a
0
Drag force on cylinder (FD) = -(force of cylinder on the c.v. = Rx)
FD = -Rx
FD  2  u U  uwdy  2  u U  uwdy
a
a
0
0
  y
u
sin

U
 2a 
a u 
FD
u
 2 U 2 
 1   dy
0
w
U  U
a
  y 
  y
FD
 2  U 2  sin
1  sin

 dy

0
w
 2a  
 2a  
y
2a
   d 

2a
dy

FD
2a
 2  U 2  2 sin  1  sin   d
0
w


2a 
 sin 2  2
(2.2  0.3)  

2
 2 U 2

cos




2
(
1
.
2
)(
50
)
(
2
)
  (1)



 
2
4 0

 4

FD
 54.1 N m
w
4.134 A jet of water is directed against a vane, which could be a blade in a turbine or in any
other piece of hydraulic machinery. The water leaves the stationary 50 mm diameter
nozzle with a speed of 20 m/sec and enters the vane tangent to the surface at A. The
inside surface of the vane at B makes angle  = 150o with the x direction. Compute the
force that must be applied to maintain the vane speed constant at U = 5 m/sec.
=
150o
V
CV
e
U=5
m/s
i
Rx
y
Ry
x
- 72 -
Assumptions
Flow is steady
Flow is uniform at ‘i’ and ‘e’
No drag forces at the vane
Momentum equation for the c.v.:
  
 
 
Fs  FB 
V

dV

V
A  V  dA
 t V
(Steady flow)
x-component of the momentum equation:
e
 
Rx  FB x   . u V  A   Q V xi  V xe 
i
Rx   AV  U V  U   V  U  cos 
Velocity is referenced to the C.V.
2
    50 
Rx  1000   
 20  5  1  cos150
 4   1000 
2
Rx  824.4 N
y-component of the momentum equation:
e
 
R y  Mg   v V  A   Q  V yi  V ye


i
2
    50 
Ry  Mg  1000   
 20  5 0  sin 150
 4   1000 
2
Ry  ( Mg  220.9) N



R  ( 824.4i  220.9 j ) N  8535
. N
165
- 73 -
4.144 Consider a single vane, with turning angle , moving horizontally at constant speed, U,
under the influence of an impinging jet as in Problem 4.135. The absolute speed of the
jet is V. Obtain general expressions for the resultant force and power that the vane could
produce. Show that the power is maximized when U = V/3.
CV
e

i
V
Rx
U
y
x
Assumptions:
Flow is steady
Flow is uniform at (i) and (e)
No drag forces at the vane
Momentum equation for the c.v.:
  
 
 
Fs  FB 
V

dV

V
A  V  dA
 t V
(Steady flow)
x-component of the momentum equation:
e
 
R x  FB x   u V  A   Q V xi  V xe 
i
Rx   AV  U V  U   V  U  cos 
Rx   AV  U  cos  1
2
Resultant force exerted by vane (Kx) = -(force of cart on vane = Rx)
K x   Rx   AV  U  1  cos 
2
- 74 -
Power produced by vane: Wout  K xU   AV  U  U 1  cos 
2
dW out
0
dU
For maximum power:


 A1  cos  V 2  4VU  3U 2  0
V  U V  3U   0  U  V
or U 
V
3
FLOW OVER A VANE (Divided Flow)
V1
n
A1
CV
V0
V0sin
Fn

V0
A0
Assumptions
t
No drag force at the vane
Flow is steady
Flow is incompressible
Flow is uniform at the inlet and
exits
Momentum equation for the c.v.
vane

V2



Fs  FB 

A2
V
(steady flow)
Momentum equation in the t-direction:
e
 
F

 st Vt  V  A  0
i
 0  Q0  V0 cos    V1Q1  V2 Q2
V0  V1  V2
(no drag!)
Q0 cos   Q1  Q2
(1)
Conservation of mass for steady flow:
 Q0  Q1  Q2   0  Q0  Q1  Q2
 

V dV   V V  dA
t
V0cos
(2)
Solve (1) & (2) simultaneously
- 75 -
A
 Q1 
Q0
Q
1  cos ; Q2  0 1  cos 
2
2
Line of action of the normal force
V1
n
b1
CV
V0
O
b0

V2
 M  M
0
b V
 h
0L
2
nf
1
2
1
:Fn n f   A1V1
 b2 V2
2
2
t
2
Fn
nf
b2
b1
2 b
  A2V2 2
2
2

2 Fn
n-component of the momentum equation:
e
 
2
Fs n   Vn  V  A    Q0  V0 sin    Q0V0 sin    A0V0 sin 
i
 nf 

 hV0 2 b1 2  b2 2
2  hb0V0 sin 
2
  b
1
2
 b2
2

2b0 sin 
Momentum flux correction factor
Consider flow through a duct of constant area, A
flow is incompressible
- 76 -
h = width

M
actual   M ideal
  correction factor
c
2
  u 2 dA    Vavg
A
A
2
1  u 
  
 dA
A A  Vave 
Vave
4.146 Water, in a 100mm diameter jet with speed of 30 m/sec to the right, is deflected by a
cone that moves to the left at 15 m/sec. Determine (a) the thickness of the jet sheet at a
radius of 200mm and (b) the external horizontal force needed to move the cone.
CV
e
i
Vj
y
=
60o
x
V
cone
Rx
Assumptions
Flow is steady
Flow is uniform at (i) and (e)
No drag force at the cone
Given
Flow of water   = const
Conservation of mass for steady flow:
e
 
V  A  0   V A  V A   A
i
i
i
e
e
e

Vi
Ai
Ve
Ae  2 R t; Vi  Ve  V  VC  30  15  45 m / sec
- 77 -
C
Ae  Ai  2 R t 
 D2
D 2 0.1

 6.25  10 3 m  6.25 mm
8R 80.2
2
t 
4
Momentum equation for the c.v.:



Fs  FB 
  
V V  dA

V dV  
 t V
(steady flow)
A
x-component of the momentum equation:
e
 
FS x  FB x   u V  A   Q  V xi  V x e


i


R x   V j  Vc A  V j  Vc   V j  Vc cos 60  100045
2

4
0.12 cos 60  1
 7952 N
Rx = 7952 N
4.153 A steady jet of water is used to propel a small cart along a horizontal track as shown.
Total resistance to motion of the cart assembly is given by FD=kU2, where k = 0.92 N sec2/m2.
Evaluate the acceleration of the cart at the instant when its speed is U=10m/sec.
y
V
e
=
30o
CV
x
D = 25.0
mm
V=
30.0 m/s
i
M = 15.0
kg
Assumptions
No drag forces at the vane
Flow is uniform at (i) and (e)
Given
Flow is steady
- 78 -
U = 10.0
m/s
FD
Momentum equation for the c.v.



Fs  FB   a 0 dm 

 

V dV   V V  dA
t
m
V
A
(flow is steady)
x-component of the momentum equation:
e
 
FSx  FBx  a 0 M   u V  A
i
 FD  15a 0   V  U 

4
0.0252  V  U   V  U sin 30    V  U 2  0.0252 1.5
4
0.9210  100030  10
2
a0 
2

4
.
0.025 2 15
15
 135
. m / sec 2
Example: Neglecting losses, determine the force needed to hold the Y shown in place.
Assume the Y to be in a horizontal plane.
30 cm
Assumptions
2
F2
15 cm
F3
30o
300
L/s
Ry
200
L/s
20
Given
3
o
Rx
Water as working fluid
  = const
y
Conservation of mass
CV
500
L/s
Flow is steady
Flow is uniform at 1,2 & 3
x

A




 V  dA  0   V  dA  0
A
since  = const
p1 = 60 kPa
F1
Flow is assumed uniform
1
 
 
  V  dA  V  A  0
A
45 cm
- 79 -

VA  Q   Q  0  Q1  Q2  Q3  0
Q1  500 L / s  0.5 m 3 / s; Q1  0 since (1) is an inlet
Q2  300 L / s  0.3 m 3 / s; Q2  0.2 m 3 / s
Q1
0 .5

 3.14m / s
A1  (0.45) 2
4
Q
0 .3
V2  2 
 4.24 m / s
A2  (0.3) 2
4
V1 
V3 
Q3
0. 2

 11.32 m / s
A3  (0.15) 2
4
Bernoulli is used along a central streamline from (1) to (2) and from (1) to (3) to obtain the
pressures p2 & p3
Note: flow is frictionless along the central streamlines since the rate of shear (du/dy) is zero
there. Flow has been assumed steady and water is the working fluid which is incompressible
2
2
V
V
 p1  g z1   1  p 2  g z 2   2
2
2
z1  z2  z3  0 (Y is in the horizontal plane)
p 2  p1 


 V2

10 3
3.142  4.242  55,941 Pa  55.9 kPa
2
V
2
2
 V3  60  10 3 
10 3
3.142  11.322  858.6 Pa  0.86 kPa
2
2
3
Similarly,
p3  p1 

2
1
  60 10

V
2


1
F1  p1 A1  60,000 

4
F2  p 2 A2  55,941 
F3  p3 A3  858.6 

2


4
4
0.452  9542.6 N
0.32
 3954.2 N
0.152  15.2 N
- 80 -
x-component of the momentum equation for a c.v.
e
 
F

V

V
 x  x  dA   Q1V1x  Q2V2 x  Q3V3 x 
i
(V1x = 0)
F2 x  F3 x  Rx  Q2V2 x  Q3V3 x 
3954.2 cos 30152
. cos 20 Rx  10000.3 4.24 cos 30  0.21132
. cos 20
Rx  2384.3 N
y-component of the momentum equation for a c.v.
F
y
e
 
  V y  V  dA   Q1V1 y  Q2V2 y  Q3V3 y


i

F1 y  F2 y  F3 y  R y   Q1V1 y  Q2V2 y  Q3V3 y


F1  F2 sin 30 F3 sin 20 R y   Q1V1 y  Q2V2 y  Q3V3 y

R y  9542.6  3954.2 sin 3015.2 sin 2010 3 0.5314
.   10 3 0.34.24 sin 30 
 
0.2 10 3 1132
. sin 20
R y  7720 N
Force needed to hold Y in place:



R  2384i  7720 j  N or R  8080 N
Problem 4.196 (Fox & McDonald, 4th edition)
- 81 -
y
O
x
CVI for linear momentum equation
CVII for conservation of mass
CV
Given
I
Ac/Ae = 10, Vc(t) = V0-kt
M0 is the initial mass
Water is the working fluid   = const
c
CVII
Assumption
VC = V0 kt
Flow is uniform at (e) & (c)
y
x
e
Conservation of mass for the CVI
 

 dV    V  dA  0

A
t V
e
 
  V  A
(since flow is assumed uniform at (e) & (c) )
c


e  
e  


  dV   V  A  0   V    V  A  0
t V
t
c
c
e  
e
V

V
  V  A  0   Q  0
t
t
c
c
 AcVc  AeVe  0  Ve  Vc
Ac
 10V0  k t 
Ae
Conservation of mass for the CVII




d
V


V

d
A
A x y  0
 t V
- 82 -
(a)
dM
dM
 Ve Ae  0 
  Ve Ae
dt
dt
dM
   V0  k t  Ac   m
dt

M
M0
t

t2 
dM    Ac  V0  kt dt  M  M 0    Ac V0 t  k

0
2


t2 
M  M 0   Ac V0 t  k  (b)
2

Momentum equation for the c.v.II
Consider the y-component
FSy  FB y   a 0 y  dV 
V
 Mg  a 0 y M 

 MV y   Ve y m
t
Ve y  Ve , m  
dM
dt
  Mg  a 0 y M  M
a0 y 



V

d
V

V

V

d
A
y
A y x y
 t V
dV y
dt
 Vy
dM

 Ve m
dt

 Mg M dV y m


V y  Ve
M
M dt
M
V y  V0  k t 

dV y
dt

 k

a0 y   g  k 


m
Ve  V y  k 
M
 V0  k t  Ac V y  1 
Ve 

V y 

g
1 2

M 0   V0 t  k t  Ac


2
- 83 -
(c)