Transcript Slide 1

Example: Exercise 5.9.4 (Pump)
z
Flow
Pump
Q  70 Liters / s  70 m / s
3
Oil (S=0.82)
D 1  180 m m  0.18 m  r1  0.09 m
D 2  120 m m  0.12 m  r2  0.06 m
p 1  35 kN / m
p 2  120 kN / m
2
z1  0
2
z2  0
Want: Rate at which energy is delivered to oil by pump
V1 
Q
A1
3

0.07 m / s
 ( 0.09 m )
2
 2 .7508 m / s
V2 
Need to find hp associated with the pump:
Q
A2
3

0.07 m / s
 ( 0.06 m )
2
 6.18936 m / s
H 2  H1  h p  h p  H 2  H1
Example: Exercise 5.9.4 (Pump)
2
2
 p2
 p1
V2 
V1 
hp  H 2  H1  
 z2 
 z1 
  


2
g

2
g




2
2
2
2

120 kN / m
( 6 .18936 m / s )  
35 kN / m
( 2 .7508 m / s ) 
hp  
 0
 0
  

3
2
3
2
0
.
82
(
9
.
81
kN
/
m
)
2
(
9
.
81
m
/
s
)
0
.
82
(
9
.
81
kN
/
m
)
2 ( 9 .81 m / s ) 

 
h p  12 .1334 m
Rate of transfer of energy = power   Q h p
 ( 0.82  9 .81 kN / m ) ( 0.07 m / s )(12 .1334 m )
3
pow er  6.83
m  kN
s
 6.83 kW
3
Example: Exercise 5.9.4 (Pump)
• Pumps (and also turbines) are characterized by their efficiency.
• Say, in exercise 5.9.4 the pump is 90% efficient and we require
6.83 kW of output, then
input = 6.83 kW / 0.9 = 7.59 kW
• Pumps (and also turbines) are characterized by their efficiency.
Efficiency =  
pow er output
pow er input
General Energy Equation for Steady Flow of Any Fluid
First Law of Thermodynamics: For steady flow, external work done on any
system plus the thermal energy transferred
into or out of the system is equal to the change
of energy of the system
(I) Using the first law of thermodynamics, (II) taking into account non-uniform
velocity at a cross-section of flow region, and (III) assuming flow goes from
section 1 to section 2, we can derive the following:
2
2
 p1

 p2

V1
V2
 z1   1
 I1   hM  QH  
 z2   2
 I2 

2g
2g
1

2

General Energy Equation for Steady Flow of Any Fluid
2
2
 p1

 p2

V1
V2
 z1   1
 I1   hM  QH  
 z2   2
 I2 


2
g

2
g
 1

 2

 
1
AV
u dA
3 
3
V 
1
A
 u dA
•  is a correction factor accounting for non-uniform velocity in cross-section
• If velocity is uniform in cross-section, then   1
• This general equation also takes into account changes in density (via  )
energy changes due to machines (via h M ) and due to heat transfer to
or from outside the fluid (via Q H )
• It also accounts for the conversions of other forms of fluid energy into internal
heat ( I )
General Energy Equation for Steady Flow of Any Fluid
• On a unit weight basis, the change in internal energy is equal to the heat
added to or removed from the fluid plus the heat generated by fluid friction:
 I  ( I 2  I1 )  QH  h f

h f  ( I 2  I1 )  QH
• The head loss due to friction is equal to the internal heat gain minus any
heat added from external sources, per unit weight of fluid
• Energy loss due to friction gets converted to internal energy (proportional to
temperature)
Example: Exercise 5.3.5 (Friction Head Loss)
S of liquid in pipe = 0.85
A
p A  170 kN / m
h  10.5 m
B
V A  VB ;
z
2
HB  H A  hf
0 ; V A  VB
hf  H A  HB
hf 
( pA  pB )

(V A  V B )
2
 (z A  zB ) 
2
2g
2
Diameter at A = Diameter at B,
thus by continuity V A  V B  V
Want: Pipe friction head loss and direction of flow
Assume flow goes fom A to B:
p B  275 kN / m
Example: Exercise 5.3.5 (Friction Head Loss)
S of liquid in pipe = 0.85
A
p A  170 kN / m
h  10.5 m
B
hf 
z
( p A  pB )

 (z A  zB ) 
V A  VB ;
p B  275 kN / m
2
Diameter at A = Diameter at B,
thus by continuity V A  V B  V
(170  275) kN / m
0.85  9 .81 kN / m
2
2
3
Thus flow goes from B to A and h f  2 .09 m
 10.5 m   2 .09 m
Example: Exercise 5.3.5 (Friction Head Loss)
S of liquid in pipe = 0.85
A
p A  170 kN / m
h  10.5 m
B
hf 
V A  VB ;
z
( p A  pB )

 (z A  zB ) 
p B  275 kN / m
2
Diameter at A = Diameter at B,
thus by continuity V A  V B  V
(170  275) kN / m
0.85  9 .81 kN / m
2
2
3
 10.5 m   2 .09 m
Thus flow goes from B to A and h f  2 .09 m
Let p B  p A . If flow goes from B to A, h f  0 and
( p A  pB )

 (z A  zB )  0 
( pB  p A )

 (z A  zB )  ( pB  pB )   (z A  zB )
Role of pressure difference (pressure gradient)
A
B
z
h  (z A  zB )
pB  p A
Thus flow will go from B (high pressure) to A (low pressure), only if
( pB  pB )   (z A  zB )   h
Otherwise flow will from A (low pressure) to B (high pressure)
In general, the pressure force (resulting from a pressure difference) tends to
move a fluid from a high pressure region towards a low pressure region
For a flow to actually go from a high pressure region towards a low pressure
region, the pressure force must be higher than other forces that could be
trying to move fluid in opposite direction (e.g. gravitational force in exercise 5.3.5)
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Graphical interpretations of the energy along a pipeline may be obtained
through the EGL and HGL:
EGL 
HGL 
p

p


V
2
2g
z
z
EGL and HGL may be obtained via a pitot tube and a piezometer tube,
respectively
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
EGL 
p


V
2
2g
z
HGL 
p

h L  h f - head loss, say,
z
due to friction
EGL=HGL if V=0
EGL
HGL
hL
2
V2
piezometer
tube
2g
pitot tube
p 2 /
z2
z1
Datum ( z  0 )
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
EGL 
EGL and HGL
p


V
2
2g
z
HGL 
p

hL  h f
z
EGL
Large V2/2g because
smaller pipe here
HGL
Steeper EGL and HGL
because greater hL
per length of pipe
Head loss at
submerged discharge
p /
EGL and HGL
z
z 0
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
EGL and HGL
Positive
Positive
p
Negative

p

EGL
p /
HGL
z
V
2
2g
EGL 
z 0
HGL 
p


p

hL  h f
If HGL  z then
P

 0
and cavitation may be possible
V
2
2g
z
z
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Helpful hints when drawing HGL and EGL:
1. EGL = HGL + V2/2g,
EGL = HGL for V=0
2. If p=0, then HGL=z
3. A change in pipe diameter leads to a change in V (V2/2g) due to continuity
and thus a change in distance between HGL and EGL
4. A change in head loss (hL) leads to a change in slope of EGL and HGL
5. If HGL  z then
P

 0
and cavitation may be possible
Helpful hints when drawing HGL and EGL (cont.):
6. A sudden head loss due to a turbine leads to a sudden drop in EGL and HGL
7. A sudden head gain due to a pump leads to a sudden rise in EGL and HGL
8. A sudden head loss due to a submerged discharge leads to a sudden rise in EGL