Transcript Energy Conservation (Bernoulli’s Equation) dp Recall Euler’s equation:   VdV  gdz  0 Also recall that viscous forces were neglected, i.e.

Energy Conservation (Bernoulli’s Equation)
dp
Recall Euler’s equation:

 VdV  gdz  0
Also recall that viscous forces were neglected, i.e. flow is invisicd
If one integrates Euler’s eqn. along a streamline, between two points ,  &

We get :
2
1
dp

2
2
  VdV   gdz  0
1
1
Which gives us the Bernoulli’s Equation
2
V1
p2 V22

 gz1 

 gz2  Constant
 2

2
p1
Flow work + kinetic energy + potential energy = constant
Bernoulli’s Equation (Continued)
Flow Work (p/) :
It is the work required to move fluid across the control volume boundaries.
Consider a fluid element of cross-sectional area
A with pressure p acting on the control surface
as shown.
p
Dx
A
Due to the fluid pressure, the fluid element moves a distance Dx within
time Dt. Hence, the work done per unit time DW/Dt (flow power) is:
 p
DW pADx  p  Dx

   A
 AV  ,
Dt
Dt
   Dt

 1
 
  AV
p
 DW 


D
t


Flow work or Power
Flow work per unit mass
p

 pv
1/mass flow rate
Flow work is often also referred to as flow energy
Bernoulli’s Equation (Cont)
Very Important: Bernoulli’s equation is only valid for :
incompressible fluids, steady flow along a streamline, no energy loss due
to friction, no heat transfer.
2
V1
p2 V22

 z1 

 z2 , where   g (energy per unit weigh t)
 2g
 2g
p1
Application of Bernoulli’s equation - Example 1:
Determine the velocity and mass flow rate of efflux from the circular
hole (0.1 m dia.) at the bottom of the water tank (at this instant). The
tank is open to the atmosphere and H=4 m
p1 = p2, V1=0
1
V2  2 g ( z1  z 2 )  2 gH
H
 2 * 9.8 * 4  8.85 (m / s )
m  AV  1000 *
2
 69.5 (kg / s )

4
(0.1) 2 (8.85)
Bernoulli’s Eqn/Energy Conservation (cont.)
Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time
required to drain the tank to level 2.
1
First, choose the control volume as enclosed
by the dotted line. Specify h=h(t) as the water
level as a function of time.
h(t)
From Bernoulli' s equation, V = 2gh
dm
  Ahole V
dt
Ahole
dh
(0.1)2
since m  Atan k h,

V 2
2 gh
dt
Atan k
1
From mass conservation,
2
water height (m)
4
4
3
dh
 0.0443 h ,
dt
h( t ) 2
2.5e-007
dh
 0.0443dt, integrate
h
h(t) =0 H  0.0215t, h  0, tdrain  93 sec.
2 h  - 0.0443t
1
4
0
0
0
20
40
60
t
time (sec.)
80
100
100
t  90.3 sec
Energy exchange (conservation) in a thermal system
Energy added, hA
(ex. pump, compressor)
p2
p1


2
V2

 z2
 2g
2
V1
 z1
2g
Energy lost, hL
(ex. friction, valve, expansion)
Energy extracted, hE
(ex. turbine, windmill)
hL
loss through
valves
heat exchanger
hE
hA
turbine
pump
hL, friction loss
through pipes
condenser
hL
loss through
elbows
Energy conservation(cont.)
If energy is added, removed or lost via pumps turbines, friction, etc.then we use
2
V1
p2 V22
Extended Bernoulli’s Equation

 z1  hA  hE  hL 

 z2
 2g
 2g
p1
Example: Determine the efficiency of the pump if the power input of the motor
is measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.
No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2
4-in dia.pipe
6-in dia. pipe
2
1
pump
zo
Z=15 in
Given: Q=300 gal/min=0.667 ft3/s=AV
V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s
kinetic energy head gain
V22  V12 (7.54)2  (3.33)2

 0.71 ft,
2g
2 * 32.2
Looking at the pressure term:
p1   w zo   m z  p2   w zo   w z
Mercury (m=844.9 lb/ft3)
water (w=62.4 lb/ft3)
1 hp=550 lb-ft/s
p2  p1  ( m   w )z
 (844.9  62.4) * 1.25  97813
. lb / ft 2
Energy conservation (cont.)
Example (cont.)
Pressure head gain:
p2  p1
w
978.13

 15.67 ( ft )
62.4
pump work hA 
p2  p1
w
V22  V12

 16.38( ft )
2g
Flow power delivered by pump
P =  w QhA  (62.4)(0.667)(16.38)
 681.7( ft  lb / s)
1hp  550 ft  lb / s
P  1.24 hp
Efficiency  =
P
Pinput

1.24
 0.827  82.7%
1.5
Frictional losses in piping system
2
V1
p2 V22
Extended Bernoulli' s equation,

 z1  hA  hE  hL 

 z2
 2g
 2g
p1  p2 Dp

 hL  frictional head loss
p1

P1

P2
Consider a laminar, fully developed circular pipe flow
R: radius, D: diameter
L: pipe length
w: wall shear stress
[ p  ( p  dp)](R 2 )   w (2R)dx,
w
P+dp
p
Darcy’s Equation:

ww

Pressure force balances frictional force
 dp 
2 w
dx, integrate from 1 to 2
R
Dp
p1  p2



I
F I F IF
HK HKG
HJ
K
44w w L L  L  LV 2  V 
 hhLL 
  f  f   

 gg D D  D  D2 g 2 g 
2
f f  VV22  where f is defined as frictional factor characterizing


44 22  pressure loss due to pipe wall shear stress
I
FIF
G
HK
H J
K
When the pipe flow is laminar, it can be shown (not here) that
f 
64 
VD
, by recognizing that Re 
, as Reynolds number
VD

64
, frictional factor is a function of the Reynolds number
Re
Similarly, for a turbulent flow, f = function of Reynolds number also
Therefore, f 
f  F(Re). Another parameter that influences the friction is the surface
roughness as relativeto the pipe diameter

D
.
  I 
F
 Re, : Pipe frictional factor is a function of pipe Reynolds
Such that ff FF Re,
H D DK
number and the relative roughness of pipe.
This relation is sketched in the Moody diagram as shown in the following page.
The diagram shows f as a function of the Reynolds number (Re), with a series of
parametric curves related to the relative roughness
FI.
HDDK
Losses in Pipe Flows
Major Losses: due to friction, significant head loss is associated with the straight
portions of pipe flows. This loss can be calculated using the Moody chart or

Colebrook equation. 1  2.0 log D  2.51 , valid for nonlaminar range
F
G
H3.7
f
Re
IJ
fK
Minor Losses: Additional components (valves, bends, tees, contractions, etc) in
pipe flows also contribute to the total head loss of the system. Their contributions
are generally termed minor losses.
The head losses and pressure drops can be characterized by using the loss coefficient,
hL
Dp
KL, which is defined as
2
1
K 
L
2
V / 2g

1 V 2
2
, so that Dp  K V
L 2
One of the example of minor losses is the entrance flow loss. A typical flow pattern
for flow entering a sharp-edged entrance is shown in the following page. A vena
contracta region is formed at the inlet because the fluid can not turn a sharp corner.
Flow separation and associated viscous effects will tend to decrease the flow energy;
the phenomenon is fairly complicated. To simplify the analysis, a head loss and the
associated loss coefficient are used in the extended Bernoulli’s equation to take into
consideration this effect as described in the next page.
V
Minor Loss through flow entrance
1
V2
V
3
p
V 2
2
 gz
(1/2)V3
(1/2)V22
2
KL(1/2)V3
2
2
pp
p3 V32
V32
V1
Extended Bernoulli' s Equation : 
 z1  hL 

 z3 , hL  K L
 2g
 2g
2g
p1
p1  p3  p , V1  0, V3  1
1 KL
( 2 g ( z1  z3 ) 
2
gh
1 KL
Energy Conservation (cont.)
Let us now also account for energy transfer via Heat Transfer, e.g. in
a heat exchanger
The most general form of conservation of energy for a system can be
written as: dE = dQ-dW where
(Ch. 3, YAC)
dE  Change in Total Energy, E
and E = U(internal energy)+Em(mechanical energy)
E = U + KE (kinetic energy) + PE(potential energy)
dW  Work done by the system where
W = Wext(external work) + Wflow(flow work)
(Ch. 1 YAC)
mechanical
energy
dQ = Heat transfer into the system (via conduction, convection & radiation)
Convention: dQ > 0 net heat transfer into the system (Symbols Q,q..)
dW > 0, positive work done by the system
Q: What is Internal Energy ?
Energy Conservation (cont.)
U = mu, u(internal energy per unit mass),
KE = (1/2)mV2 and PE = mgz
Flow work Wflow= m (p/)
It is common practice to combine the total energy with flow work.
Thus:
F
I
G
J
H
 K

V2
 pp 


Energy flow rate: m(u +
 gz ) plus Flow work rate m  
  
2
V2
p V2
Flow energy in  m in (u  
 gz )in , Energy out = m out (u  
 gz )out
 2
 2
p
The difference between energy in and out is due to heat transfer (into or out)
and work done (by or on) the system.
Energy Conservation (cont.)
Hence, a system exchanges energy with the environment due to:
1) Flow in/out 2) Heat Transfer, Q and 3) Work, W
This energy exchange is governed by the First Law of Thermodynamics
p
2
V
m in (u  
 gz)in
 2
Heat in, Q =dQ/dt
V2
m in (u  
 gz )out
 2
p
system
From mass conservation: m in  m out  m
Work out dW/dt
From the First law of Thermodynamics (Energy Conservation):
dQ
p V2
p V2
dW
 m (u  
 gz )in  m (u  
 gz )out 
, or
dt
 2
 2
dt
dQ
V2
V2
dW


 m( h 
 gz )in  m(h 
 gz )out 
dt
2
2
dt
p
where h  u  is defined as "enthaply"

“Enthalpy”
Conservation of Energy – Application
Example: Superheated water vapor enters a steam turbine at a mass flow rate
1 kg/s and exhausting as saturated steam as shown. Heat loss from the turbine is
10 kW under the following operating condition. Determine the turbine power output.
10 kw
P=1.4 Mpa
T=350 C
V=80 m/s
z=10 m
From superheated vapor tables:
hin=3149.5 kJ/kg
dQ
V2
V2
dW
 m (h 
 gz )in  m (h 
 gz )out 
dt
2
2
dt
dW
 ( 10)  (1)[(3149.5  2748.7)
dt
80 2  50 2 (9.8)(10  5)


]
2(1000)
1000
 10  400.8  1.95  0.049
P=0.5 Mpa
100% saturated steam
 392.8( kW )
V=50 m/s
z=5 m
From saturated steam tables: hout=2748.7 kJ/kg
Q, q … ?!%
Q – total heat transfer (J)
Q – rate of total heat transfer (J/s, W)
q – heat transfer per unit mass (J/kg)
q – Heat Flux, heat transfer per unit area (J/m2)
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Internal Energy ?
•Internal energy, U (total) or u (per unit mass) is the sum of all
microscopic forms of energy.
•It can be viewed as the sum of the kinetic and potential energies of the
molecules
• Due to the vibrational, translational and rotational energies of the moelcules.
•Proportional to the temperature of the gas.
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