Transcript Energy Conservation (Bernoulli’s Equation) Integration of Euler’s equation Bernoulli’s equation  dp    VdV   gdz  0 V1 p2 V22   gz1    gz2  2  p1 Flow.

Energy Conservation (Bernoulli’s Equation)
Integration of Euler’s equation
Bernoulli’s equation

2
1
dp

2
2
  VdV   gdz  0
1
1
2
V1
p2 V22

 gz1 

 gz2
 2

2
p1
Flow work + kinetic energy + potential energy = constant
Dx
p
Under the action of the pressure, the fluid element
moves a distance Dx within time Dt
A The work done per unit time DW/Dt (flow power) is
P
DW pADx  p  Dx

   A
 AV  ,
Dt
Dt
   Dt

 1  DW 

 
  work done per unit mass flow rate
  AV  Dt 
p
Energy Conservation (cont.)
2
V1
p2 V22

 z1 

 z2 , where  g (energyper unit weight)
 2g
 2g
p1
It is valid for incompressible fluids, steady flow along a streamline,
no energy loss due to friction, no heat transfer.
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
Examples:
1
H
V2  2 g ( z1  z 2 )  2 gH
2
 2 * 9.8 * 4  8.85(m / s )
m  AV  1000*
 69.5 (kg / s )

4
(0.1) 2 (8.85)
Energy Equation(cont.)
Example: 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
2
water height (m)
4
4
3
h( t ) 2
2.5e-007
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,
dh
 0.0443 h ,
dt
1
dh
 0.0443dt, integrate
h
h(t) = H  0.0215t, h  0, tdrain  93 sec.
0
0
0
20
40
60
t
time (sec.)
80
100
100
Energy conservation (cont.)
Energy added, hA
(ex. pump, compressor)
Generalized energy concept:
p1
p2
V1

 z1
 2g
hL
loss through
valves
2
V2

 z2
 2g
2
Energy extracted, hE
(ex. turbine, windmill)
Energy loss, hL
(ex. friction, valve, expansion)
heat exchanger
hA
hE
turbine
pump
hL, friction loss
through pipes
condenser
hL
loss through
elbows
Energy conservation(cont.)
2
V1
p2 V22
ExtendedBernoulli's equation, 
 z1  hA  hE  hL 

 z2
 2g
 2g
p1
Examples: 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.
6-in dia. pipe
4-in dia.pipe
hE=hL=0, z1=z2
2
1
Q=300 gal/min=0.667 ft3/s=AV
pump
V1=Q/A1=3.33 ft/s
zo
V2=Q/A2=7.54 ft/s
Z=15 in
kinetic energy head gain
V22  V12 (7.54)2  (3.33)2

 0.71 ft,
2g
2 * 32.2
Mercury (m=844.9 lb/ft3)
water (w=62.4 lb/ft3)
1 hp=550 lb-ft/s
p1   w zo   m z  p2   w zo   w z
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
97813
.

 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
ExtendedBernoulli's equation, 
 z1  hA  hE  hL 

 z2
 2g
 2g
p1  p2 Dp

 hL  frictionalhead loss
P
p1

1

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
P+dp
Darcy’s Equation:
Pressure force balances frictional force
 dp 
2 w
dx, integrate from 1 to 2
R
Dp
p1  p2



4
 hL  w
g
F
V I
LL IV
F
 LI
  f F
 K
  f HK
G
J
H
D 2H
g2 g K
D
D
2 2
22
where f is defined as frictional factor characterizing
f

V
F
I
f

V


F
I
 w   G J
44 K
22 Kpressure loss due to pipe wall shear stress
H
H
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
F I
H K

D
.
(Re, :) Pipe frictional factor is a function of pipe Reynolds
Such that f fFF Re,
DD
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.
HD K
Energy Conservation (cont.)
Energy: E=U(internal thermal energy)+Emech (mechanical energy)
=U+KE(kinetic energy)+PE(potential energy)
Work: W=Wext(external work)+Wflow(flow work)
Heat: Q heat transfer via conduction, convection & radiation
dE=dQ-dW, dQ>0 net heat transfer in dE>0 energy increase and vice versa
dW>0, does positive work at the expense of decreasing energy, dE<0
U=mu, u(internal energy per unit mass), KE=(1/2)mV2, PE=mgz
Wflow=m(p/)
F
I
G
HJ
K
V2
p
P


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
Their difference is due to external heat transfer and work done on flow
Energy Conservation (cont.)
p
Heat in q=dQ/dt
2
V

min (u  
 gz)in
 2
V2
m in (u  
 gz )out
 2
p
Work out dW/dt
From mass conservation: m in  m out  m
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"

Energy Conservation(cont.)
Example: Superheated water vapor is entering the steam turbine with a mass
flow rate of 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
power output of the turbine.
From superheated vapor table:
P=1.4 Mpa
hin=3149.5 kJ/kg
T=350 C
V=80 m/s
2
2
dQ
V
V
dW
z=10
m



m
(
h


gz
)

m
(
h


gz
)

10 kw
dt
2
in
2
out
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 table: hout=2748.7 kJ/kg
dt