CE 319 F
Daene McKinney
Elementary Mechanics of Fluids
Bernoulli
Equation
Euler Equation
•
•
Fluid element accelerating in l
direction & acted on by pressure
and weight forces only (no friction)
Newton’s 2nd Law
 Fl  Mal
pA  ( p  p ) A  W sin   lAal
p  ( p  p )  l sin   lal
dp
dz
 g  al
dl
dl
a
d p
 (  z)  l
dl 
g

Ex (5.1)
•
•
Given: Steady flow. Liquid is decelerating
at a rate of 0.3g.
Find: Pressure gradient in flow direction in
terms of specific weight.

a
d p
(  z)  l
dl 
g
dp

dz
  al  
dl
g
dl

  ( 0.3g )   sin 30o
g
  (0.3  0.5)
dp
 0.2
dl
Flow l
30o
EX (5.3)
•
•
vertical
Given:  = 10 kN/m3, pB-pA=12 kPa.
Find: Direction of fluid acceleration.
A
d p
a
(  z)  z
dz 
g
1 dp dz
az   g(
 )
 dz dz
p  pB
az   g( A
 1)
1m

B

 12,000
 1)
10,000
a z  g (1.2  1)  0
az   g(
(acceleration is up)
HW (5.7)
• Ex (5.6) What
pressure is needed to
accelerate water in a
horizontal pipe at a
rate of 6 m/s2?
a
d p
(  z)  l
dl 
g
dp
dz
 
 a l
dl
dl
dp

 al  1000kg / m 3 * 6m / s 2
dl
dp
 6000 N / m 3
dl

Ex (5.10)
•
•
Given: Steady flow. Velocity varies linearly
with distance through the nozzle.
Find: Pressure gradient ½-way through the
nozzle
ax
d p
 (  z) 
dx 
g
dp
dV
  a x    (V
)
dx
dx
V1/2=(80+30)/2 ft/s = 55 ft/s
dV/dx = (80-30) ft/s /1 ft = 50 ft/s/ft
 ( 1.94 slugs / ft 3 ) * (55 ft / s ) * (50 ft / s / ft )
 5,355lbf / ft 2 / ft
HW (5.11)
Bernoulli Equation
d p
1
 (  z )  at
ds 
g
1 dV
 V
g ds
d  V 2 

ds  2 g 
d  p
V 2 
z
0


ds  
2g 
V2
z
 Constant

2g
p
• Consider steady flow
along streamline
• s is along streamline,
and t is tangent to
streamline
p

 z  Piezometric head
V2
Velocity( dynamic) head
2g
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
Ex (5.47)
Point 1
•
•
•
Given: Velocity in outlet pipe from reservoir
is 6 m/s and h = 15 m.
Find: Pressure at A.
Solution: Bernoulli equation
V12 p A
V A2
 z1 

 zA 

2g

2g
p1
0
pA
V A2
h

0

2g 
2g
0
pA
pA
V A2
18
  ( h  )  9810(15 
)
2g
9.81
 129.2 kPa
Point A
Example
Point 1
•
•
Given: D=30 in, d=1 in, h=4 ft
Find: VA
•
Solution: Bernoulli equation
V12 p A
V A2
 z1 

 zA 

2g

2g
p1
0 0
V A2
h
 0

2g 
2g
0
V A  2 gh
 16 ft / s
Point A
Example – Venturi Tube
•
•
•
Given: Water 20oC, V1=2 m/s, p1=50 kPa,
D=6 cm, d=3 cm
Find: p2 and p3
Solution: Continuity Eq.
V1 A1  V2 A2
A
D
V2  V1 1  V1  
A2
d 
•
2
V2 p
V2
 z1  1  2  z2  2

2g

2g
p1
 p1 

2

2
(V12  V22 )
[1  D / d 4 ]V12
1000
[1  6 / 34 ]22 Pa
2
p2  120 kPa
 150,000 
D
d
2
1
Bernoulli Eq.
p2  p1 
D
3
Nozzle: velocity
increases, pressure
decreases
Diffuser: velocity
decreases, pressure
increases
Similarly for 2  3, or 1  3
p3  150 kPa
Pressure drop is fully recovered, since we
assumed no frictional losses
Knowing the pressure drop 1  2 and
d/D, we can calculate the velocity and
flow rate
V2 
2( p1  p2 )
 [1  d / D 4 ]
Ex (5.48)
•
•
•
Given: Velocity in circular duct = 100
ft/s, air density = 0.075 lbm/ft3.
Find: Pressure change between
circular and square section.
Solution: Continuity equation
Vc Ac  Vs As

100( D 2 )  Vs D 2
4

Vs  100( )  78.54 ft / s
4
•
Air conditioning (~ 60 oF)
Bernoulli equation
Vc2 ps
Vs2
 zc 

 zs 

2g 
2g
pc
pc  p s 

2
(Vs2
 Vc2 )
0.075lbm / ft 3
pc  p s 
(78.542  1002 )
2 * 32.2 lbm / slug
 4.46 lbf / ft 2
Ex (5.49)
•
•
•
Given:  = 0.0644 lbm/ft3 V1= 100 ft/s,
and A2/A1=0.5, m=120 lbf/ft3
Find: h
Solution: Continuity equation
V1 A1  V2 A2
V2  V1
•
A1
 100 * 2  200 ft / s
A2
Bernoulli equation
V2 p
V2
 z1  1  2  z2  2

2g 
2g
p1
p1  p2 

2
(V22
 V12 )
0.0644lbm / ft 3
p1  p2 
( 2002  1002 )
2 * 32.2 lbm / slug
 30 lbf / ft 2
Heating (~ 170 oF)
•
Manometer equation
p1  p2  h( m   air )
(120  0.0644)lbm / ft 3
30  h
* 32.2 ft / s 2
32.2 lbm / slug
h  0.25 ft
HW (5.51)
Stagnation Tube
V12 p2
V22
 z1 

 z2 

2g

2g
p1
V12 p2


 2g 
2
V12  ( p2  p1 )
p1


2

( (l  d )  d )
V1  2 gl
Stagnation Tube in a Pipe
H
p

z
V2
2g
V2
2g
p

Pipe
2
Flow
1
z0
z
Pitot Tube
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
p1 V12 p2 V22



 2g  2g
V2  2 g[(
p1

 z1 )  (
V  2 g ( h1  h2 )
p1

 z1 )
Pitot Tube Application (p.170)
V
1
z1-z2
2
p1  ( z1  z2 ) k  l k  y Hg  (l  y ) k  p2
p1  p2  y ( Hg   k )  ( z1  z2 ) k
p1  p2
k
l
 z1  z2 
y ( Hg   k )
k
h1  h2  y ( Hg /  k  1)
V  2 gy ( Hg /  k  1)  24.3 ft / s
y
HW (5.69)
HW (5.75)
HW (5.84)
HW (5.93)