Bernoulli equations and applications .pdf

Bernoulli applications

1. Aplicaçõesda
Eq. de Bernoulli

2. Ex. 5.42 MEDIÇÃO DA PRESSÃO:
MANÔMETRO DIFERENCIAL –CASO PRÁTICO
h
P
A
P
B
ρ
f
ρ
m
•Determine P
A
-P
B
em função da altura
h e das densidades dos fluidos.
Resp.:
P
A
-P
B
= (
ρ
m
-
ρ
f).g.h
Pergunta: se os fluidos forem água e
mercúrio, como fica a expressão?
Resp.:
P
A
-P
B
= (13600 -1000).g.h
•Pergunta: e se os fluidos forem ar e água?
Resp.:
P
A
-P
B
= (1000 –1.2)ghcomo
ρ
liq
>>
ρ
gas
→P
A
-P
B
≈
ρ
m
gh
•Um engano freqüente dos alunos:
aplicar a relação acima quando realizam medidas com dois
líquidos.

3. Thecavitationphenomenon
The cavitation
phenomenon is
characterized by the
formation and quick
growth of vapor bubbles in
the presence of a
depression, followed by a
violent implosion. Such an
implosion, often
supersonic, can generate a
spherical shock wave in
the liquid developing
pressures higher than the
solid material yield stress.

4. Erosão causada pela cavitação

6. FIXED CAVITATION IN A NOZZLE
incandescent light, strobe light,
exposure time 1/30 sec exposure time 5 µs

7. Veja solução em ‘exercícios resolvidos’

8. Components of Thermodynamic Cycles
•What can be the components of thermodynamic cycles?
•Turbines, valves, compressors, pumps, heat exchangers
(evaporators, condensers), mixers,

9. Vapor Vapor--cycle Power Plants cycle Power Plants
Refrigeration cycle Refrigeration cycle
Where, Exactly, Are They
Used For?

10. Equação da Energia: Regime Permanente
shaft
OUT
2
I
IN
2
I
W Q m
P
gz
2
V
u m
P
gz
2
V
u
& & & &− = ∑
















ρ
+ + + + ∑
















ρ
+ + + −






− =












ρ
+ + + −












ρ
+ + +
kg
Joules
w q
P
u gz
2
V P
u gz
2
V

shaft
IN h
2
I
OUT h
2
I
321 321
•Considere o V.C. com duas portas (uma entrada / uma
saída)
•Expressando em função do calor e trabalho específicos (dividindo por ),
m&

11. RepresentaçãoGenéricados Componentes
de CiclosTermodinâmicos
•Aplicação de um balanço de energia para dispositivos que operam
com fluxo de energia (entalpia), produzem trabalho e trocam
calor com um reservatório a T
0
mh
1
mh
2
Q
W
T
0

12. Turbine Turbine
•A turbine is device in which work is produced
by a gas passing over and through a set of
blades fixed to a shaft which is free to rotate.
in
m
•
out
m
•
CV
W
•

17. Output range up to 100 MW Live steam conditions: Temperature up to 540
o
C
Pressure up to 140 Bara
Exhaust steam conditions:
Back-pressure: 3-16 Bara/300
o
C
Condensing 0,03 - 0,25 Bar
Controlled extraction:
Pressure/Temperature 3-25
Bara/400
o
C
Turbina a vapor ATP 4 -ABB

18. Turbines Turbines
•We will draw turbines like this:
inlet
outlet
w
maybe q

19. TURBINES TURBINES
) ( ) (
1 2
2
1
2
2
1 2
z zg
2
h h wq− +
−
+ − = −
V V
Sometimes
neglected
eixo 1 2
w(hh)
=
−
Almost
always
neglected
Isentropic s
2
= s
1
Almost
always
neglected

20. T-s Diagram of an
Adiabatic Turbine
Isentropic s
2
= s
1
T
T
1
T
2a
T
2s

21. W=?
Inlet
P
1
= 5 MPa
T
1
=600
o
C
V
1
=30m/s
Outlet
P
2
= 7.5 kPa
x=0.95
V
2
=100m/s
) ( ) (
1 2
2
1
2
2
1 2
z zg
2
h h w q− +
−
+ − = −
V V
0
0
h
1
superaquecido = 3666 kJ/kg
h
2
saturado , 40.2
o
C
h
l= 168.8 kJ/kg & h
v
=2574.8 kJ/kg
→h
2
= 2454.5 kJ/kg
2
h h w
2
1
2
2
1 2
V V−
+ − = −) (
1212 kJ/kg
4.5 kJ/kg
1207 kJ/kg
→W = m.w = 603 MW !!!

22. Verdadeiro, processo politropico Verdadeiro, 1
a
lei
Falso, Falso, se rev. q=∫Tds Verdadeiro, S é propriedade
(Ps/Pe)

23. •Machines developed by engineers to make
life easier, decrease world anxiety, and
provide exciting engineering problems
from the industrial revolution for students.
•Analysis proceeds the same as for turbines,
although the signs may differ.
•Compressor-used to raise the pressure of
a compressible fluid
•Pump-used to raise pressure of an
incompressible fluid
•Fan-used to move large amounts of gas,
but usually has a small pressure increase
Compressors, pumps, and fans Compressors, pumps, and fans

24. Compressors, Compressors,
pumps, and fans pumps, and fans
Compressor
Side view End view
of pump of pump

27. Compressores de Deslocamento Positivo Compressores de Deslocamento Positivo

28. Identifique os fluxos para um Compressor
mh
e
mh
s
Q
W
T
0
•O compressor rejeita calor para o
ambiente.
•A temperatura T
0
éa temperatura
do ambiente
22
se
se se
qw(hh) g(z z)
2
−
−= − + + −
VV
neglected
neglected
es
w(hh)q =−+

29. Sample Problem Sample Problem
Air initially at 15 psiaand 60°F is compressed to 75 psiaand 400°F.
The power input to the air is 5 hp and a heat loss of 4 Btu/lbmoccurs
during the process. Determine the mass flow in lbm/min.
W
shaft
= 5 hp
75 psia
400
o
F
Q = 4 Btu/lb
Compressor
HYPOTHESIS
•Steady state steady flow (SSSF)
•Neglect potential energy changes
•Neglect kinetic energy changes
•Air is an ideal gas
es
w(hh)q =−+→
()
shaft
ps e
W
m 2.46lbm/min
qCTT
==
−−
&
&
15 psia& 60
o
F

30. es
w(hh)=−
(tab. A-2.1) h
e
= + 187 kJ/kg e
s
e
= 0.696 kJ/kgK
processoreversível, s
s
= s
e
=
0.696 kJ/kgK
(tab. A-2.2) P
s
= 1.6 MPa e s
s
=
0.696 kJ/kgKÆh
s
= 216 kJ/kg
(
)
kg kJ29 216 187 w/
−
=
− =
T
T
1
T
2a
T
2s

31. Cars & Intercoolers Cars & Intercoolers

32. Total Backpressure
5
Manifold Boost Pressure
4
Intercooler & Piping pressure drop
3
Turbo Compressor pressure increase
2
Intake System pressure drop
1
Muffler / Cat-back pressure drop
9
Main Cat / Midpipepressure drop
8
Pre-Cat / Downpipepressure drop
7
Turbo pressure drop
6
Compression with Compression with intercooling intercooling

33. Se vocêentenderesteproblemavocêcompreenderáa
razãodo intercooler Expressãoparacálculodo trabalhode eixocompressãoouexpansão 1
a
lei
proc. rev. &
relaçãotermd.
isolando∆h+q
eixo e s
w(hh)q=−+
qTds=
∫
dh vdP =−
∫
∫
(he hs) q vdP ⇒−+=−
∫
eixo
w vdP Eq. (5.56) =−
∫

34. Reversible work
relations for
steady-flowand
systems

35. Compressor e o DiagramaP-v
paraum processoreversível
P
v
v
i
v
f
Pv
1
= cte
Pvn
= cte
Pv
γ
= cte

36. Compressor e o DiagramaP-v
paraum processoreversível
P
v
Pv
1
= cte
Pv
γ
= cteA área amarela
representa o trabalho
necessário para um
processo isotérmico
eixo
wvdP=−
∫
A área achurada
representa o trabalho
extra necessário para
um processo
adiabático
Conclusão: menor trabalho éo processo isotérmico

37. eixo
wvdP=−
∫
Compressãocom Resfriamento
Intermediário
P
v
Pv
1
= c
te
Pv
γ
= c
te
1
o
estágio
Trabalho
‘poupado’
Conclusão: inter-resfriamento ‘poupa’trabalho eixo
Intercooler
2
o
estágio
Trabalhoeixo
Trabalho
isotérmico

38. Trabalhode EixoReversível, GásIdeal
Isotérmico: v = RT/P Æ
f
iso
i
P dP
wRT RTln
PP

=− =−

∫
Adiabático: Pv
γ
= c Æ
()
()
1
1
f
adiab i i i i
i
P dP
wPv Pv 1
P1P
γ−
γ
γ
γ


 γ


=
−=− − 

 γ−




∫
0
10
20
30
40
50
12345678910
Razão Pressão
(Wad-Wiso)/Wiso (%)

39. BUTTERFLY VALVE
SPHERE VALVE
GLOBE VALVE
PIPE CRACK
GATE VALVE

40. Throttling Devices (Valves) Throttling Devices (Valves)
Typical assumptions for throttling devices Typical assumptions for throttling devices 1. No work
2. Potential energy changes are zero
3. Kinetic energy changes are usually small
4. Heat transfer is usually small
5. Two port device

41. Look at energy equation: Look at energy equation:
) ( ) (
1 2
2
1
2
2
1 2
z zg
2
h h wq− +
−
+ − = −
V V
Apply assumptions from previous page:
000 0
We obtain:
0
1 2
=
−
)h h(
or
1 2
h h
=
Does the fluid temperature:
increase,
decrease, or
as it goes through an adiabatic valve?
remain constant

42. Look at implications: Look at implications:If the fluid is liquid/vapor:
During throttling process:
•The pressure drops,
•The temperature drops,
•Enthalpy is constant
s
T
h const.
h const.
P const.

43. Look at implications: Look at implications:
if fluid is an ideal gas
:
0
1 2
=
−
)T T(C
p
=
−
)h h(
1 2
C
p
is always a positive number, thus:
1 2
T T
=

44. s, kJ/(kgK)
T (K)
15MPa
611K (340
o
C)
373K (100
o
C)

45. John Connor (T1000) in Terminator 2
Fenômenode Fragilizaçãode Metais
porBaixasTemperaturas

46. Cold embrittlement
(
dureza e rigidezmas baixaresistência a tensão )
Low temperature embrittlementdoes affect most materials
more or less pronounced. It causes overloaded components to
fracture spontaneously rather than accommodating the stress by
plastic deformation. The picture shows a fractured fitting whose
material was not suitable for low temperatures.
Consequences of the Temperature Drop on Material Strenght

47. PP
vv
SS
TT
diminui Pressão
Entalpia
Temperatura
En. Interna
Entropia
constante
gás ideal h=h(T), portanto T fica constante
gás ideal u=u(T), portanto u fica constante
TT
ss
v
PP
1 1
> P > P
22
diminui, ∆s= Cpln(T
2
/T
1
)-Rln(P
2
/P
1
)

48. TEAMPLAY TEAMPLAY
Refrigerant 12 enters a valve as a saturated liquid at 0.9607 Mpa
and leaves at 0.1826 MPa. What is the quality and the temperature
of the refrigerant at the exit of the valve?
State (1) Liquid saturated, x=0 Psat= 0.9607 MPa Tsat= ? Hliq= ?
State (2)
Liq+vap x=?
Psat= 0.1826 MPa
Tsat= ?
Hliq= ?
40
o
C
75kJ/kg
-15
o
C
22 & 180 kJ/kg
0.33

49. Heat exchangers are used Heat exchangers are used
in a variety of industries in a variety of industries
•Automotive -radiator
•Refrigeration -evaporators/condensers
•Power production -boilers/condensers
•Power electronics -heat sinks
•Chemical/petroleum industry-mixing processes

50. Something a little closer to home.. Something a little closer to home..

51. Heat Exchangers Heat Exchangers

52. Condenser/evaporator Condenser/evaporator
for heat pump for heat pump

53. Heat Exchangers Heat Exchangers
•Now, we must deal with multiple inlets and outlets:
1
m&
2
m&
4
m&
3
m&
A 2 1
m m m& & &
=
=
B 3
m m m& & &
=
=
4
If we have steady
flow, then:

54. Conservation of energy can Conservation of energy can
be a little more complicated... be a little more complicated...
1
m&
2
m&
4
m&
3
m&
I’ve drawn the control
volume around the
whole heat exchanger.
Implications:
No heat
transfer from
the control
volume.
Fluid A
Fluid B

55. Heat Exchangers Heat Exchangers
•Now if we want the energy lost or gained
by either fluid
we must let that fluid be
the control volume, indicated by the red.
A
m m& &
=
1
2
m&

56. Heat Exchangers Heat Exchangers
()
()
A
A CV
z zg
V V
h h m W Q








− + − + − + −
2 1
2
2
2
1
2 1
2 2
& & &
()
()
0
2 2
4 3
2
4
2
3
4 3
=








− + − + − +
B
B
z zg
V V
h h m&
0
0, (sometimes
negligible)
0
0, (sometimes negligible)
0, (usually negligible)
0, (usually
negligible)

57. Heat Exchangers Heat Exchangers
•And we are left with
)h h(m)h h(m
B A3 4 2 1
−
=
−
& &
The energy change of fluid A is equal to the
negative of the energy change in fluid B.

58. (1)
(3)
(2)
m
v
= 0,1 kg/s
h
v
= 2776 kJ/kg
m
c
= ? kg/s
h
v
= 100 kJ/kg
m
v
+m
c
= ? kg/s
h
s
= 200 kJ/kg
123
11 22 33
mmm0
mh mh mh 0
−
−+=
−
−+=


Resp.:
m
c
= 2,57 kg/s

59. Nozzles and Diffusers Nozzles and Diffusers
•
Nozzle--a device which
accelerates a fluid as the
pressure is decreased.
This configuration is for sub-sonic flow.
V
1
, P
1
V
2
,
P
2
•
Diffuser--a device which
decelerates a fluid and
increases the pressure.
V
1
,
P
1
V
2
,
P
2

60. Nozzles
For supersonic flow, the shape
of the nozzle is reversed.

61. General shapes of General shapes of
nozzles and diffusers nozzles and diffusers
Supersonic flow
Subsonic flow
nozzle
diffuser
nozzle diffuser

62. conservation of energy conservation of energy
q = 0 (adiabatic)
w = 0 (these are not work producing devices;
neither is work done on them
)
) ( ) (
1 2
2
1
2
2
1 2
z zg
2
h h wq− +
−
+ − = −
V V
00
0
2
h h
2
2
2
1
1 2
V V−
= −) (

63. Sample Problem Sample Problem
An adiabatic diffuser is employed to reduce the velocity of a stream
of air from 250 m/s to 35 m/s. The inlet pressure is 100 kPaand the
inlet temperature is 300°C. Determine the required outlet area in
cm
2
if the mass flow rate is 7 kg/s and the final pressure is 167 kPa.
OUTLET
P
2
= 167 kPa
V
2
= 35 m/s
A
2
=?
Diffuser
INLET
T
1
= 300
°
C
P
1
= 100 kPa
V
1
= 250 m/s
= 7 kg/s
m&

64. Conservation of Mass: Steady State Regime Conservation of Mass: Steady State Regime
2
2 2
1
1 1A A
m
ν ν
V V
= =&
solve for A
2
2
2
2
m
A
V
ν
&
=
But we don’t know v
2
!
Remember ideal gas equation of state?
1
1
P
RT
=
1
ν
2
2
P
RT
=
2
ν
and
We know T
1
and P
1
, so v
1
is simple.
We know P
2
, but what about T
2
?
NEED ENERGY EQUATION!!!! NEED ENERGY EQUATION!!!!

65. Energy E
nergy Eqn
E
qn
If we assumed constant specific heats, we could get T
2
directly
()()
22
12
12 12
2
P
VV
hh CTT
−
−≡ −=
(
)
22
21 1 2
2602
P
TTVV C K =− − ⋅=
3
222
RT P 1.0352mkg
ν
==
The ideal gas law:
42 2
2
2
71.035
A 10 2070
35
m
cm
ν
⋅
=⋅=
=
&
V
And the area:

66. Turbojet Engine Basic
Components

67. T
ur
b
o
j
et
E
ng
i
ne
B
as
i
c
Components and T-s
Diagram for Ideal Turbojet
Process
q
in
=calor
combustão
h
6
h
1

68. Identifique os fluxos para um Motor a Jato
mh
e
mh
s
Q
W
T
0
•Háadição de calor a
pressão constante pela
queima do combustível.
•A temperatura T
0
éa
temperatura da câmara de
combustão
()










− − − +=
≅
321
0
2 2
2 2

e
I
s
I
V V hs he q w