Transcript CVE 240 – Fluid Mechanics

CHAPTER 7
ENERGY PRINCIPLE
Dr . Ercan Kahya
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
General Energy Consideration
V12 P1
V22 P2
E T  α1
  z1  α 2

 z2  hP  hf  h t
2g γ
2g γ
Z: the position
P/g: the pressure head
V2/2g: the velocity head
hp: head supplied by pump
ht :head given up to a turbine
hf : head loss
• local losses (bends, expansions, valves)
• frictional losses (function of pipe type,
length)
a : Velocity coefficient and can be set to unity for regular & symmetrical cross-section like
pipe
Bernoulli vs. Energy
V12 P1
V22 P2
  z1 
  z2
2g γ
2g γ
Z is the position
P/g is the pressure head
V2/2g is the velocity head
Relates velocity and piezometric pressure along a
streamline, steady, incompressible, inviscid flow.
V12 P1
V22 P2
E T  α1
  z1  α 2

 z2  hP  hf  h t
2g γ
2g γ
Relates energy at two points for viscous, incompressible flow
in a pipe, with accounting for additional energy addition /
extraction
Energy Principle
• So far, mechanical forces on a fluid
– Pressure
– Gravity
– Shear Stress
• Considering Energy, we can solve:
– Power required to move fluids
– Effects of pipe friction
– Flow rates of fluids moving through pipes & orifices
– Effects of obstacles, bends, and valves on flow
First Law of Thermodynamics
E  Q  W
• E = energy of a system
• Q = heat transferred to a system in a given time t
• W = work done by the system on its surroundings during the same time
• Energy forms: Kinetic and Potential energy of a system as a whole
and energy associated with motion of the molecules
(atomic structure, chemical energy, electrical energy)
E = Ek + Ep + Eu
dE  
 Q W
dt
Involves sign convention:
+ heat transferred to the system
+ work done by the system
- heat transferred from the system
‐ work done on the system
First Law of Thermodynamics
Derivation of Energy Equation
Reynolds Transport Theorem applied to First Law of
Thermodynamics
E: extensive property of the system
e: intensive (energy per unit mass)
dE d
  edV   eV .dA
dt dt cv
cs


d
Q W   edV   eV .dA
dt cv
cs
e  ek  e p  u
V2
ek 
2
e p  gz
u  internalenergy
Flow Work
Work is classified as: (work) = (flow work) +(shaft work)
Flow Work: Work done by pressure forces as the system moves through
space
Force (F) = p A
Work = F l
= (pA) (V t)

W f   pV. A
cs
At section 2, work rate done on surrounding fluid is → V2 p2 A2
At section 1, work rate done by surrounding fluid is → - V1 p1 A1
Shaft Work (any work not associated with a pressure force!)
• Work done on flow by a pump
– increases the energy of the system, thus the work is negative
• Work done by flow on a turbine
– decreases the energy of the system, thus the work is positive
Reynolds Transport Theorem: Simplified form
If the flow crossing the control surface occurs through a number of inlet and
outlet ports, and the velocity v is uniformly distributed (constant) across each
port; then



2

2
Vo
Vi
Q W   mo (  gzo  ho )   mi (  gzi  hi )
2
2
.
Steady-Flow Energy Equation
Q = rate of heat transfer TO the system (input)
.
W = rate of work transfer FROM the system (output)
.
m = rate of mass flow
h = specific enthalpy (h = u + p/ρ)
Example 7.2:
If the pipe is 20cm and the rate of flow 0.06m3/s, what is the
pressure in the pipe at L=2000m? Assume hl=0.02(L/D)V2/2g
V12 P1
V22 P2
α1
  z1  hp  α 2
  z 2  h t  hl
2g γ
2g γ
This energy equation assumes steady flow & constant density
Power Equation
Let’s relate “head” to “power & efficiency”

Pump power:

Wp  gQh p  m ghp
Power delivered to turbine:


Wt  gQht  m ght
Both pump & turbine lose energy due to friction which is accounted for by
the “efficiency” defined as the ratio of power output to power input.

Poutput
Pinput
If mechanical efficiency of the turbine is ηt ,
the output power supplied by the turbine:


Ws  t Wt
Example 7.4: Power produced by a turbine
Discharge Q = 14.1 m3/s ; Elevation drop = 61 m
Total head loss = 1.5 m ; Efficiency = 87% Power = ?
Evaluations:
V1 = V2 = 0
V12 P1
V22 P2
α1
  z1  hp  α 2
  z 2  h t  hl z1-z2 = 61m
2g
γ
2g
γ
p 1 = p2 = 0
ht = (z1-z2) - hL
= 61 – 1.5 = 59.5 m
Power equation:
Pinput toturbine  gQht
= (9810 N/m3) (14.1 m3/s) (59.5m)
= 8.23 MW
Efficiency equation:
Poutput  from generator  Pinput toturbine
= 0.87(8.23 MW) = 7.16 MW
Application of the Energy, Momentum and
Continuity Principles in Combination
V12 P1
V22 P2
α1
  z1  α 2

 z2  hf
2g γ
2g γ


 Fs  mV2  mV1
Neglecting the force due to
shear stress
p1A2  p2 A2  γA2 LSinα  ρQ(V2  V1 )
Sudden expansion head loss
hf 
V1  V2 2
2g
Example 7.5: Force on a contraction in a pipe
Find horizontal force which is required to hold the contraction in place if
P1=250kPa ; Q=0.707m3/s & head loss through the contraction
V22
h f  0.1
2g
Assume α1= α2 = 0 (kinetic energy correction factor)
SOLUTIONS:
Q , p1, V1 and V2 : known


p1 A1  p2 A2  Fx  m2 V2  m1 V1
Fx and p2 : unknown
To obtain unknown p2:
V12 P1
V22 P2
α1
  z1  α 2

 z2  hf
2g γ
2g γ
Q  A1V1  A 2 V2
HYDRAULIC & ENERGY GRADE LINES
GRADE LINE INTERPRETATION - PUMP
GRADE LINE INTERPRETATION TURBINE
GRADE LINES - NOZZLE
GRADE LINES - PIPE DIAMETER CHANGE
GRADE LINES - SUB-ATMOSPHERIC PRESSURE
CLASS EXERCISE: Q7.32
Find the head loss btw the reservoir
surface and point C.
Assume that the head loss btw the reservoir surface and point B is three
quater of the total head loss.
CLASS EXERCISE: Q7.36
CLASS EXERCISE: Q7.60
CLASS EXERCISE: Q7.71