chapter 7. application of thermodynamics to flow processes
TRANSCRIPT
Chapter 7. Application of Thermodynamics
to Flow Processes
Introduction
Thermodynamics of flow
Based on mass, energy, and entropy balance on open system
Flow also fluid mechanics problem
Thermodynamics of flow
- Mass conservation, laws of thermodynamics
Fluid mechanics
- Momentum balance
This chapter
Duct flow – pipe, Nozzle, Throttle
Turbines (expanders)
Compressors, Pumps
7.1 Duct Flow of Compressible Fluids (1)
Adiabatic, steady state, one-dimensional flow of compressible fluid
No shaft work and no change in potential energy
Energy Balance Equation – 1st law
WQmzgu2
1H
dt
)mU(d
fs
2CV
0 0 0 0
0u2
1H 2
ududH
Changes in enthalpy directly go to
changes in velocity
Steady state
7.1 Duct Flow of Compressible Fluids (2)
Mass balance equation – Continuity Equation
0mdt
dmfs
cv 0)V
uA(d)m(d
0
0A
dA
u
du
V
dV
equation 2.27
(page 47)
7.1 Duct Flow of Compressible Fluids (3)
Thermodynamic Relations
Replace V in terms of S and P
VdPTdSdH
PPP
Ps
S
T
T
V
S
V
dSS
VdP
P
VdV
PT
V
V
1
(eqn 3.2)
T
C
T
S p
P
(eqn 6.17)
PP C
VT
S
V
Relation from physics
7.1 Duct Flow of Compressible Fluids (4)
S
22
V
PVc
Velocity of sound in a medium
is related with pressure
derivative w.r.t volume
with constant enthalpy (S)
dSS
VdP
P
VdV
Ps
dPc
VdS
C
T
V
dV2
P
7.1 Duct Flow of Compressible Fluids (5)
Variables : dH, du, dV, dA, dS, dP
Equations : four
ududH
0A
dA
u
du
V
dV
VdPTdSdH
dPc
VdS
C
T
V
dV2
P
dS, dA : treat as independent
Can develop equations of
other derivatives with dS and
dA
7.1 Duct Flow of Compressible Fluids (6)
0dAA
u
M1
1TdS
M1
MC
u
udu
0dAA
uTdS
C
u1VdP)M1(
2
22
2
p
2
2
p
22
M : Mach number = u/c
Derive yourself!
7.1 Duct Flow of Compressible Fluids (7)
0dx
dA
A
u
M1
1
dx
dST
M1
MC
u
dx
duu
0dx
dA
A
u
dx
dST
C
u1
dx
dPV)M1(
2
22
2
p
2
2
p
22
According to second law, (dS/dx) ≥ 0
Pipe Flow
Pipe Flow : constant cross sectional area (dA/dx=0)
dx
dS
M1
MC
u
Tdx
duu
dx
dS
)M1(
C
u1
V
T
dx
dP
2
2
p
2
2
p
2
Subsonic flow :
Implies :
0)M1( 2
0dx
dP
0dx
du
Pressure drops
in the direction of flow
Velocity increases
in the direction of flow
Pipe flow
The velocity does not increase indefinitely.
If the velocity exceeds the sonic value,
0dx
dP 0
dx
du
Supersonic flow
Shock wave and turbulence
Unstable flow
Sonic Boom?
What is happening around the aircraft?
Example 7.1
Consider the steady-state, adiabatic, irreversible flow of an
incompressible liquid in a horizontal pipe of constant cross-sectional area.
Show that:
(a) The velocity is constant.
(b) The temperature increases in the direction of flow.
(c) The pressure decreases in the direction of flow.
Example 7.1
(a) From the continuity equation,
121212
1
11
2
22 uuso,VV,AAV
Au
V
Au0)
V
uA(d
(b) For the entropy of incompressible fluid,
0T
TlnCSVdP
T
dTCdS
1
2PP
(c) From dH = -udu, dH = 0 since du = 0
)PP(V)TT(CHHHVdPT1dTCdH 1212P12P
0)TT(C)PP(V0H 12P12
GG
j j
jcvfs SS0S
T
Q
dt
)mS(d)mS(
Nozzles
Flow within a pipe or a duct (variable cross-sectional area)
Assume isentropic flow (dS = 0) reversible flow
dx
dA
M1
1
A
u
dx
du
dx
dA
M1
1
VA
u
dx
dP
2
2
2
Nozzles
Subsonic (M<1) Supersonic(M>1)
Converging Diverging Converging Diverging
dA/dx - + - +
dP/dx - + + -
du/dx + - - +
Converging Diverging
Converging Nozzle
For subsonic flow in converging nozzle
Pressure Velocity
Maximum obtainable velocity = speed of sound, c
Converging subsonic nozzle can be used to deliver constant flow
into a region of variable pressure
As P2 decreases, velocity increases and maximum value at sonic
velocity.
Ultimately, the pressure ratio P2/P1 reaches a critical value at
which the velocity at the nozzle exit is sonic!
Further decrease in P2 has no effect on velocity.
Converging / Diverging Nozzle
Speed of sound is attained at the throat of converging/diverging
nozzle only when the pressure at the throat is low enough that
critical value of P2/P1 is reached.
Value of critical pressure ratio
VdPTdSdH
ududH VdPudu
2
1
P
P
/)1(
1
2112
1
2
2P
P1
1
VP2VdP2uu
dS=0 Adiabatic
/1
11
11
P
VPV
VPconstPV
Value of critical pressure ratio
2
1
P
P
/)1(
1
2112
1
2
2P
P1
1
VP2VdP2uu
Critical value u = c
22
2
2
S
S
222
2
VPu
V
P
V
P
V
PVcu
1
1
2
22
2
2
1
1
2
P
P
VPu
0u
Example 7.2
A high-velocity nozzle is designed to operate with steam at 700 kPa and 300
oC. At the nozzle inlet, the velocity is 30 m/s. Calculate values of the ratio
A/A1 (A1 is the cross-sectional area of the nozzle inlet) for the sections where
the pressure is 600 kPa. Assume that the nozzle operates isentropically.
From the continuity equation,
21
21
1
2
1
11
2
22
uV
Vu
A
A
V
Au
V
Au
From dH = -udu,
)HH(2uu 12
2
1
2
2 Compare the unit of u2 and H
Example 7.2
The initial values for entropy, enthalpy, and specific volume can be found
from steam table.
S1 = 7.2997 kJ/kg∙K, H1 = 3,059.8×103 J/kg, V1 = 371.39 cm3/g
Then,
)10059,3H(230uandu
V
39.371
30
A
A 3
2
22
2
2
2
1
2
Since S2 = S1 = 7.2997 kJ/kg∙K (isentropic), other values, H2 and V2, can
be found from steam table (at 600 kPa and S1 = 7.2997 kJ/kg∙K)
From page 727,
S2 = 7.2997 kJ/kg∙K, H2 = 3,020.4×103 J/kg, V2 = 418.25 cm3/g
Then, A2/A1 and u2 can be calculated!
Example 7.3
Consider again the nozzle in Ex. 7.2, assuming now that steam behaves as
an ideal gas. Calculate the critical pressure ratio and the velocity at the
throat.
Let = 1.3 for steam,55.0
13.1
2
1
2
P
P 13.1
3.1
1
1
2
mol/J17.765,4K15.573Kmol/J314.8RTVP,gasidealFor
obtainedbeshouldVPP
P1
1
VP2VdP2uu
111
11
P
P
/)1(
1
2112
1
2
2
2
1
22
11 s/m322,296kg/J322,296mol/J17.765,4VP
s/m35.544u
322,296]55.01[13.1
511,2643.12900
P
P1
1
VP2uu
2
3.1/)13.1(
/)1(
1
2112
1
2
2
Throttling Process
Throttling Process : Orifice , Partly closed valve, porous plug, …
Primary result : pressure drop
For ideal gases, H = H(T) and no change in T
For real gases, slight change in T
12 HH,0H
Example 7.4
Propane gas at 20 bar and 400 K is throttled in a steady-state flow
process to 1 bar. Estimate the final temperature of the propane and
its entropy change. Properties of propane can be found from suitable
generalized correlations.
!solveroriterationwithTforsolvecanweorC
HHTT
0HH)TT(CHHdTCHHH
2
H
ig
P
R
2
R
112
R
1
R
212H
ig
P
R
1
R
2
T
T
ig
P12
2
1
R
1
R
2
1
2T
T
ig
P SSP
PlnR
T
dTCS
2
1
Example 7.5 Joule-Thompson Coefficient
Temperature change resulting from a throttling a real gas.
Joule-Thompson coefficient
Temperature increase/decrease as a result of throttling process
HP
T
Joule/Thomson Coefficient and other properties
HP
T
TPTPH P
H
T
H
P
H
H
T
P
T
1
TP P
H
C
1
J-T coeff. comes from the
pressure dependence of H
Joule/Thomson Coeff. from PVT relation
PP
2
P
2
T
PT
T
Z
PC
RT
T
Z
P
RT
P
H
T
VTV
P
H
With Cp and PVT
relation, any
property can be
predicted.
7.2 Turbines (Expanders)
Expansion of gas Production of Work
Internal Energy Kinetic Energy Shaft Work
Turbines (Expanders)
Heat effects are negligible, Inlet and outlet velocity changes are small
Normally T1, P1 and P2 are given
If turbine expands reversibly and adiabatically…
isentropic process (adiabatic process): S = 0
Maximum work
)HH(HW
)HH(mHmW
12s
12s
Ss )H()isentropic(W
Turbines (Expanders)
Turbine Efficiency
Turbine efficiency of properly designed turbine : 0.7 to 0.8
S
s
s
)H(
H
)isentropic(W
W
Turbines (Expanders)
Adiabatic expansion process in a turbine or expander
Example 7.6
A steam turbine with rated capacity of 56,400 kW operates with steam at
inlet conditions of 8,600 kPa and 500 oC, and discharges into a condenser
at a pressure of 10 kPa. Assuming a turbine efficiency of 0.75, determine
the state of the steam at discharge and the mass of flow of the steam.
At the initial condition of 8,600 kPa and 500 oC,
H1 = 3,391.6 kJ/kg, S1 = 6.6858 kJ/kg∙K (from steam table, page 752)
For isentropic process, kPa10Pat6858.6SS 212
8047.0x1511.8x6493.0)x1(6858.6
SxS)x1(SxSxS,Then
)liquidvapor(!phasetwoisit6493.0Sand1511.8S,kPa10Pat
v
2
v
2
v
2
v
2
v
2
l
2
v
2
v
2
v
2
l
2
l
22
l
2
v
22
Example 7.6
kg/kJ2.274,16.391,34.117,2HHH
kg/kJ4.117,28.584,28047.08.191)8047.01(H
kPa10Pat8.191Hand8.584,2H,tablesteamFrom
12S
2
2
l
2
v
2
Now, a turbine efficiency is = 0.75,
kg/kJ0.436,26.9556.391,3HHH
kg/kJ6.9552.274,175.0HH
12
S
The composition of steam is
s/kg02.59m)6.391,30.436,2(m400,56W
),HH(mHmWFrom
Kkg/kJ6846.71511.89378.06493.0)9378.01(S
...and
9378.0x8.2584x8.191)x1(0.436,2
s
12s
2
v
2
v
2
v
2
Example 7.7
A stream of ethylene gas at 300 oC and 45 bar is expanded adiabatically
in a turbine to 2 bar. Calculate the isentropic work produced. Find the
properties of ethylene, assuming an ideal gas.
1
2
1
2
S
ig
P
1
2T
T
ig
P
12H
ig
P
T
T
ig
P12
P
PlnR
T
TlnC
P
PlnR
T
dTCS
)TT(CdTCHHH
2
1
2
1
Example 7.7
For isentropic process (S = 0),
)TT(CdTC)H()isentropic(W
,TcalculateweOnce
0P
PlnR
T
dTCSfromTforsolveorCgcalculatinbyTFind
045
2lnR
15.573
TlnC0
P
PlnR
T
TlnCS
12H
ig
P
T
T
ig
PSs
2
1
2T
T
ig
P2S
ig
P2
2
S
ig
P
1
2
1
2
S
ig
P
2
1
2
1
7.3 Compression Processes
Compression Devices : Rotating blades, Reciprocating pistons
Turbine operating in reverse
Compressors
Compressor efficiency : 0.7 to 0.8
H
)H(
W
)isentropic(W
S
s
s
Compressors
Adiabatic compression process
H
)H(
W
)isentropic(W
S
s
s
Isentropic Compression of Ideal Gas
For ideal gas,
Assuming S = 0 (isentropic process),
Remember that S is not zero for real compression process!
1
2
1
2
SP
1
2T
T
ig
PPP
PlnR
T
TlnC
P
PlnR
T
dTCS
P
dPR
T
dTCdS
2
1
)TT(CdTC
)isentropic(WH,thenP
PTT
1
'
2H
'T
T
ig
P
SS
C/R
1
21
'
2
P
2
1
S
'
P
Example 7.8
Saturated-vapor steam at 100 kPa (Tsat = 99.63 oC) is compressed
adiabatically to 300 kPa. If the compressor efficiency is 0.75, what is the
work required and what are the properties of the discharge stream?
For saturated steam at 100 kPa,
S1 = 7.3598 kJ/kg∙K and H1 = 2,675.4 kJ/kg
For isentropic compression to 300 kPa,
For steam at 300 kPa with this entropy,
Kkg/kJ3598.7SS 12
kg/kJ8.888,2H2
kg/kJ5.284HW,And
)723page,tablesteamfrom(Kkg/kJ5019.7SandC1.246T
kg/kJ9.959,25.2844.675,2HHH,Then
kg/kJ5.28475.0
4.213HH
kg/kJ4.2134.675,28.888,2H
S
2
o
2
12
S
S
Example 7.9
If methane is compressed adiabatically from 20 oC and 140 kPa to 560 kPa,
estimate the work requirement and the discharge temperature of the methane.
The compressor efficiency is 0.75. Assume that methane is an ideal gas.
dTCor)TT(CWHand)isentropic(W
W
thenisworkactualThe
dTCor)TT(C)isentropic(WH,Then
K37.397T,TforSolve
0140
560lnR
T
dTCSor
0140
560lnR
15.293
TlnC
P
PlnR
T
TlnCS
2
1
'2
1P
2
T
T
ig
P12HPSS
S
T
T
ig
P1
'
2H
'
SS
22
T
15.293
ig
P
2
SP
1
2
1
2
SP
Pumps
Liquids are usually moved by pumps, generally rotating equipment.
Same assumption (adiabatic process) as compressor
For isentropic process (adiabatic pump),
In general process,
1
1
P
PSS VdPH)isentropic(W
VdPdH
PVT
TlnCSP)T1(VTCH
VdPT
dTCdSdP)T1(VdTCdH
1
2PP
PP
Example 7.10
Water at 45 oC and 10 kPa enters an adiabatic pump and is discharged at
a pressure of 8,600 kPa. Assume the pump efficiency to be 0.75. Calculate
the work of the pump, the temperature change of the water, and the
entropy change of the water.
For saturated liquid water at 45 oC,
V = 1,010 cm3/kg, = 425×10-6 K-1, Cp = 4.178 kJ/kgK
Ws(isentropic)=(H)S=1,010×(8,600-10)=8.676×10-6 kPa cm3/kg=8.676 kJ/kg
PVT
TlnCSP)T1(VTCH
calculatedbecanSandT
kg/kJ57.1175.0
676.8H,
H
)H(Since
1
2PP
S
Homework
Problems
7.20, 7.23, 7.34
7.16
- Just prove that
- You don’t have to work on parts (a) and (b)!
Due:
Recommend Problems
7.18, 7.27, 7.28, 7.32, 7.49, 7.52
T
Z
T
ZT