Energy Conservation (Bernoulli’s Equation) If one integrates Euler’s eqn. along a streamline,...
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Transcript of Energy Conservation (Bernoulli’s Equation) If one integrates Euler’s eqn. along a streamline,...
Energy Conservation (Bernoulli’s Equation)
If one integrates Euler’s eqn. along a streamline, between two points , &
Which gives us the Bernoulli’s Equation
Constant 22 2
222
1
211 gz
Vpgz
Vp
Flow work + kinetic energy + potential energy = constant
02
1
2
1
2
1 gdzVdV
dp
0 gdzVdVdp
Recall Euler’s equation:
Also recall that viscous forces were neglected, i.e. flow is invisicd
We get :
Bernoulli’s Equation (Continued)
p
A
x
t
W
AV
p
pAV
t
xA
p
t
xpA
t
W
1
,
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.
Due to the fluid pressure, the fluid element moves a distance x within time t. Hence, the work done per unit time W/t (flow power) is:
Flow work per unit mass
Flow work or Power
1/mass flow ratepv
p
Flow work is often also referred to as flow energy
t)unit weighper (energy g where,22 2
222
1
211
z
g
Vpz
g
Vp
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.
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
H
1
2
p1 = p2, V1=0
)/(5.69
)85.8()1.0(4
*1000
)/(85.84*8.9*2
2)(2
2
212
skg
AVm
sm
gHzzgV
Bernoulli’s Equation (Cont)
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.
h(t)
1
2
First, choose the control volume as enclosedby the dotted line. Specify h=h(t) as the waterlevel as a function of time.
0 20 40 60 80 1000
1
2
3
4
time (sec.)
wat
er h
eigh
t (m
)
4
2.5e-007
h( )t
1000 tsec 90.3 t
0.0443t- 20
4
h
Energy exchange (conservation) in a thermal system
1
211
2z
g
Vp
2
222
2z
g
Vp
Energy added, hA
(ex. pump, compressor)
Energy extracted, hE
(ex. turbine, windmill)
Energy lost, hL
(ex. friction, valve, expansion)
pump turbine
heat exchanger
condenser
hEhA
hL, friction lossthrough pipes hL
loss throughelbows
hL
loss throughvalves
Energy conservation(cont.)
2
222
1
211
22 z
g
Vphhhz
g
VpLEA
Example: Determine the efficiency of the pump if the power input of the motoris measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.
Mercury (m=844.9 lb/ft3)water (w=62.4 lb/ft3)
1 hp=550 lb-ft/s
No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2
6-in dia. pipe 4-in dia.pipe Given: Q=300 gal/min=0.667 ft3/s=AVV1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s
kinetic energy head gain
V V
gft2
212 2 2
2
7 54 3 33
2 32 20 71
( . ) ( . )
* .. ,
p z z p z z
p p z
lb ft
w o m w o w
m w
1 2
2 1
29 62 1 25 97813
( )
(844. .4)* . . /
pump
Z=15 in
1 2
zo
If energy is added, removed or lost via pumps turbines, friction, etc.then we use
Extended Bernoulli’s Equation
Looking at the pressure term:
Energy conservation (cont.)
Example (cont.)
Pressure head gain:
pump work
p pft
hp p V V
gft
w
Aw
2 1
2 1 22
12
97813
6215 67
216 38
.
.4. ( )
. ( )
Flow power delivered by pump
P =
Efficiency =P
P
w
input
Qh
ft lb s
hp ft lb s
P hp
A
( .4)( . )( . )
. ( / )
/
.
.
.. .
62 0 667 16 38
681 7
1 550
1 24
1 24
1 50 827 82 7%
Frictional losses in piping system
loss head frictional
22 equation, sBernoulli' Extended
21
2
222
1
211
L
LEA
hppp
zg
Vphhhz
g
Vp
P1
P2
Consider a laminar, fully developed circular pipe flow
p P+dp
w
Darcy’s Equation:
R: radius, D: diameterL: pipe lengthw: wall shear stress
[ ( )]( ) ( ) ,
,
p p dp R R dx
dpR
dx
p p ph
g
L
Df
L
D
V
g
w
w
Lw
FHIK FHIKFHGIKJ
2
1 22
2
2
4
2
Pressure force balances frictional force
integrate from 1 to 2
where f is defined as frictional factor characterizing
pressure loss due to pipe wall shear stress
w
f VFHIK
FHGIKJ4 2
2
24
2Vfw
g
V
D
Lf
D
L
gh w
L 2
4 2
When the pipe flow is laminar, it can be shown (not here) that
by recognizing that as Reynolds number
Therefore, frictional factor is a function of the Reynolds number
Similarly, for a turbulent flow, f = function of Reynolds number also
. Another parameter that influences the friction is the surface
roughness as relativeto the pipe diameter D
Such that D
Pipe frictional factor is a function of pipe Reynolds
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 D
fVD
VD
f
f F
f F
FH IK
FHIK
64
64
, Re ,
Re,
(Re)
.
Re, :
.
DFf
Re,
D
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.
Minor Losses: Additional components (valves, bends, tees, contractions, etc) inpipe flows also contribute to the total head loss of the system. Their contributionsare generally termed minor losses.
The head losses and pressure drops can be characterized by using the loss coefficient,KL, which is defined as
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 intoconsideration this effect as described in the next page.
Kh
V g
pp K V
L
L
LV
2 2
12
2
2 12/
,
so that
12 0
3 7
2 51
f fD FHG
IKJ. log
.
.
Re,
valid for nonlaminar range
Minor Loss through flow entrance
V2 V3
V1
(1/2)V22 (1/2)V3
2
KL(1/2)V32
pp
ghK
zzgK
VVppp
g
VKhz
g
Vphz
g
Vp
LL
LLL
1
2)(2(
11,0,
2,
22:Equation sBernoulli' Extended
313131
23
3
233
1
211
gzV
p
2
2
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) (Ch. 1 YAC)E = U + KE (kinetic energy) + PE(potential energy)
dW Work done by the system where W = Wext(external work) + Wflow(flow work)
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 ?
mechanical energy
Energy Conservation (cont.)
U = mu, u(internal energy per unit mass), KE = (1/2)mV2 and PE = mgzFlow work Wflow= m (p/)
It is common practice to combine the total energy with flow work.Thus:
The difference between energy in and out is due to heat transfer (into or out)and work done (by or on) the system.
Energy flow rate: m(u +V
2 plus Flow work rate m
p
Flow energy in Energy out =
2
)
( ) , ( )
FHGIKJ
gz
m up V
gz m up V
gzin in out out
2 2
2 2
p
Energy Conservation (cont.)
( )m up V
gzin in
2
2 ( )m u
p Vgzin out
2
2
Heat in, =dQ/dt
Work out dW/dtFrom mass conservation:
From the First law of Thermodynamics (Energy Conservation):
dQ
dt or
dQ
dt
where is defined as "enthaply"
( ) ( ) ,
( ) ( )
m m m
m up V
gz m up V
gzdW
dt
m hV
gz m hV
gzdW
dt
h up
in out
in out
in out
2 2
2 2
2 2
2 2
Q
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
system
“Enthalpy”
Conservation of Energy – ApplicationExample: Superheated water vapor enters a steam turbine at a mass flow rate1 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.
P=1.4 MpaT=350 CV=80 m/sz=10 m
P=0.5 Mpa100% saturated steamV=50 m/sz=5 m
10 kw
From superheated vapor tables: hin=3149.5 kJ/kg
From saturated steam tables: hout=2748.7 kJ/kg
dQ
dt
( ) ( )
( ) ( )[( . . )
( )
( . )( )]
. . .
. ( )
m hV
gz m hV
gzdW
dtdW
dt
kW
in out
2 2
2 2
2 2
10 1 3149 5 2748 7
80 50
2 1000
9 8 10 5
1000
10 400 8 1 95 0 049
392 8
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.
Q – total heat transfer (J)
– rate of total heat transfer (J/s, W)
q – heat transfer per unit mass (J/kg)
– Heat Flux, heat transfer per unit area (J/m2)
Q
q
Q, q … ?!%
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