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The discipline
• Principles: “Fluid mechanics” and “Thermodynamics”• Contrast
– Flow process inevitably result from pressure gradients within the fluid. Moreover, temperature, velocity, and even concentration gradients may exist within the flowing fluid.
– Uniform conditions that prevail at equilibrium in closed system.
• Local state – An equation of state applied locally and instantaneously at any
point in a fluid system, and that one may invoke a concept of local state, independent of the concept of equilibrium.
Duct flow of compressible fluids
• Equations interrelate the changes occurring in pressure, velocity, cross-sectional area, enthalpy, entropy, and specific volume of the flowing system.
• Consider an adiabatic, steady-state, one dimensional flow of a compressible fluid:
• The continuity equation:
02
2
uH ududH
0A
dA
u
du
V
dV0)/( VuAd
VdPTdSdH
dPP
VdS
S
VdV
SP
0A
dA
u
du
V
dV
PPP S
T
T
V
S
V
PT
V
V
1
T
C
T
S P
P
PP C
VT
S
V
2
2
c
V
P
V
S
From physics, c is the speed of sound in a fluid
dPc
VdS
C
T
V
dV
P2
ududH dP
c
VdS
C
T
V
dV
P2
011222
dA
A
uTdS
C
uVdP
c
u
P
01122
2
dA
A
uTdS
C
uVdP
P
M
c
uM The Mach number
VdPTdSdH
01
1
1
2
22
22
dA
A
uTdS
Cu
udu P
MM
M
Relates du to dS and dA
Pipe flow
01
1
1
2
22
22
dA
A
uTdS
Cu
udu P
MM
M 011
222
dA
A
uTdS
C
uVdP
P
M
dx
dSCu
Tdx
duu P
2
22
1 M
M
dx
dSCu
V
T
dx
dP P
2
2
1
1
M
For subsonic flow, M2 < 1, , the pressure decreases and the velocity increases in the direction of flow. For subsonic flow, the maximum fluid velocity obtained in a pipe of constant cross section is the speed of sound, and this value is reached at the exit of the pipe.
0dx
du0
dx
dP
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.
Control volume: a finite length of horizontal pipe, with entrance (1) and exit (2)
The continuity equation:2
22
1
11
V
Au
V
Au
21 AA
21 VV 21 uu
incompressible
const. cross-sectional area
Entropy balance (irreversible): 012 SSSG
incompressible liquid with heat capacity C
02
112
T
TG T
dTCSSS 12 TT
Energy balance with (u1 = u2): 21 HH 0)( 1212
2
1
PPVCdTHHT
T
12 TT
12 PP If reversible adiabatic: T2 = T1; P2 = P1. The temperature and pressure change originates from flow irreversibility.
01
1
1
2
22
22
dA
A
uTdS
Cu
udu P
MM
M 011
222
dA
A
uTdS
C
uVdP
P
M
Reversible flow
01
1 2
2
dx
dA
A
u
dx
duu
M
012
2 dx
dA
A
u
dx
dPVM
Nozzles:
Reversible flow
Subsonic: M <1 Supersonic: M >1ConvergingDivergingConvergingDiverging
dx
dA- + - +
dx
dP- + + -
dx
du+ - - +
For subsonic flow in a converging nozzle, the velocity increases as the cross-sectional area diminishes. The maximum value is the speed of sound, reached at the throat.
01
1 2
2
dx
dA
A
u
dx
duu
M
012
2 dx
dA
A
u
dx
dPVM
isentropicVdPudu
2
1
221
22
P
PVdPuu
.constPV
1
1
21121
22 1
1
2
P
PVPuu
cu 2SV
PVc
22
V
P
V
P
S
.constPV 01 u
1
1
2
1
2
P
P
A high-velocity nozzle is designed to operate with steam at 700 kPa and 300°C. At the nozzle inlet the velocity is 30 m/s. Calculate values of the ratio A/A1 (where A1 is the cross-sectional area of the nozzle inlet) for the sections where the pressure is 600, 500, 400, 300, and 200 kPa. Assume the nozzle operates isentropically.
The continuity equation: uV
Vu
A
A
1
1
1
Energy balance: )(2 121
2 HHuu
Initial values from the steam table:Kkg
kJS
2997.71 kg
kJH 8.30591
g
cmV
3
1 39.371
u
V
A
A
39.371
30
1
)108.3059(2900 32 Hu
Since it is an isentropic process, S = S1. From the steam table:
600 kPa:Kkg
kJS
2997.7
kg
kJH 4.3020
g
cmV
3
25.418s
mu 3.282 120.0
1
A
A
Similar for other pressures P (kPa) V (cm3/g) U (m/s) A/A1
700 371.39 30 1.0600 418.25 282.3 0.120500 481.26 411.2 0.095400 571.23 523.0 0.088300 711.93 633.0 0.091200 970.04 752.2 0.104
Consider again the nozzle of the previous example, assuming now that steam behaves as an ideal gas. Calculate (a) the critical pressure ratio and the velocity at the throat. (b) the discharge pressure if a Mach number of 2.0 is required at the nozzle exhaust.
The ratio of specific heats for steam, 3.1
1
1
2
1
2
P
P 3.155.0
1
2 P
P
1
1
21121
22 1
1
2
P
PVPuu
We have u1, P1, V1, P2/P1, γ
s
mu 35.5442
(a)
(b)2M
s
mu 7.108835.54422
1
1
21121
22 1
1
2
P
PVPuu kPaP 0.302
Throttling Process:
When a fluid flows through a restriction, such as an orifice, a partly closed valve, or a porous plug, without any appreciable change in kinetic or potential energy, the primary result of the process is a pressure drop in the fluid.
WQmzguHdt
mUd
fs
cv
2
2
1)( 0Q
0W0H
Constant enthalpy
For ideal gas: 0H 12 HH 12 TT
For most real gas at moderate conditions of temperature and pressure, a reduction in pressure at constant enthalpy results in a decrease in temperature.
If a saturated liquid is throttled to a lower pressure, some of the liquid vaporizes or flashes, producing a mixture of saturated liquid and saturated vapor at the lower pressure. The large temperature drop results from evaporation of liquid. Throttling processes find frequent application in refrigeration.
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.
Constant enthalpy process:
0)( 1212 RR
H
igP HHTTCH
Final state at 1 bar: assumed to be ideal gas and 022 RR SH
11
2 TC
HT
H
igP
R
082.11 rT 471.01 rP
452.0)152.0,471.0,082.1(
),,(1
1
0
11
HRB
OMEGAPRTRHRBRT
H
RT
H
RT
H
c
R
c
R
c
R
And based on 2nd virial coefficients correlation
263 10824.810785.28213.1 TTC igP
??H
igPC
KT 400 Kmol
JC ig
P 07.94
KT 2.3852 ???
KT 6.3924005.02.3855.0 Kmol
JCC ig
PH
igP
73.92
KT 0.3852
R
S
igP S
P
PR
T
TCS 1
1
2
1
2 lnln H
igPS
igP CC
2934.0)152.0,471.0,082.1(1 SRBR
S R
Kmol
JS
80.23
Throttling a real gas from conditions of moderate temperature and pressure usually results in a temperature decrease. Under what conditions would an increase in temperature be expected.
Define the Joule/Thomson coefficient:HP
T
When will μ < 0 ???
TPTPH P
H
CP
H
H
T
P
T
1
Always negative
PT T
VTV
P
H
TP
H
Sign of ???
P
ZRTV PT T
Z
P
RT
P
H
2
PP T
Z
PC
RT
2
Always positive
PT
Z
Same sign
The condition may obtain locally for real gases. Such points define the Joule/Thomson inversion curve.
0
PT
Z
Turbine (Expanders)
• A turbine (or expander):– Consists of alternate sets of nozzles and
rotating blades– Vapor or gas flows in a steady-state expansion
process and overall effect is the efficient conversion of the internal energy of a high-pressure stream into shaft work.
SWTurbine
S
fs
cv WQmzguHdt
mUd
2
2
1)()( 12 HHmHmWS
12 HHHWS
The maximum shaft work: a reversible process (i.e., isentropic, S1 = S2)
SS HisentropicW )()(
The turbine efficiency
SS
S
H
H
isentropicW
W
)()(
Values for properly designed turbines: 0.7~ 0.8
A steam turbine with rated capacity of 56400 kW operates with steam at inlet conditions of 8600 kPa and 500°C, and discharge 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 rate of flow of the steam.
SWTurbine
KkgkJSkg
kJH
CTkPaP
6858.66.3391
5008600
11
11
KkgkJSkPaP 6858.610 22
KkgkJxxSxSxS vvvvlv
6858.61511.86493.0)1()1( 222
8047.0vxkgkJHxHxH vvlv 4.2117)1( 222
kgkJHHH S 2.127412
kgkJHH S 6.955
vvlv HxHxkgkJHHH 2212 )1(0.2436
9378.0vxKkg
kJSxSxS vvlv
6846.7)1( 222
skJHmWS 56400
skgm 02.59
A stream of ethylene gas at 300°C and 45 bar is expanded adiabatically in a turbine to 2 bar. Calculate the isentropic work produced. Find the properties of ethylene by: (a) equations for an ideal gas (b)appropriate generalized correlations.
RR
H
igP HHTTCH 1212 )( RR
S
igP SS
P
PR
T
TCS 12
1
2
1
2 lnln
KTbarPbarP 15.573245 121
(a) Ideal gas
1
2
1
2 lnlnP
PR
T
TCS
S
igP
0S
0S
3511.61135.3
exp2
RC
TS
igP
)0.0,6392.4,3394.14,424.1;,15.573( 2 EETMCPSRC
S
igPiteration
KT 8.3702
)()()( 12 TTCHisentropicWH
igPSS
224.7
)0.0,6392.4,3394.14,424.1;18.370,15.573(
EEMCPH
RC
H
igP
mol
JisentropicWS 12153)15.5738.370(314.8224.7)(
(b) General correlation
030.21 rT 893.01 rP
234.0)087.0,893.0,030.2(
1
1
0
11
HRB
RT
H
RT
H
RT
H
c
R
c
R
c
R
based on 2nd virial coefficients correlation
097.0)087.0,893.0,030.2(1 SRBR
S R
0806.0116.045
2ln
15.573ln 2 R
TCS
S
igP
Assuming T2 = 370.8 K
314.12 rT 040.02 rP
0139.0)087.0,040.0,314.1(2 SRBR
S R
based on 2nd virial coefficients correlation
iteration
KT 8.3652
296.12 rT 040.02 rP
20262.0)087.0,040.0,296.1(2 HRBRT
H
c
R
mol
JHHTTC
HisentropicW
RR
H
igP
Ss
11920)(
)(
1212
• Pressure increases: compressors, pumps, fans, blowers, and vacuum pumps.
• Interested in the energy requirement
S
fs
cv WQmzguHdt
mUd
2
2
1)()( 12 HHmHmWS
12 HHHWS
The minimum shaft work: a reversible process (i.e., isentropic, S1 = S2)
SS HisentropicW )()(
The compressor efficiencyH
H
W
isentropicW S
S
S
)()(
Values for properly designed compressors: 0.7~ 0.8
SWcompressorCompression process
Saturated-vapor steam at 100 kPa (tsat = 99.63 °C ) 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:Kkg
kJS
3598.71 kg
kJH 4.26751
Isentropic compression
Kkg
kJSS
3598.712
300 kPakg
kJH 8.28882
kg
kJH S 4.213
kg
kJHH S 5.284
kg
kJHHH 9.295912 300 kPaCT 1.2462
Kkg
kJS
5019.72
kg
kJHWS 5.284
If methane (assumed to be an ideal gas) is compressed adiabatically from 20°C and 140 kPa to 560 kPa, estimate the work requirement and the discharge temperature of the methane. The compressor efficiency is 0.75.
RR
S
igP SS
P
PR
T
TCS 12
1
2
1
2 lnln
0S
S
igPC
R
P
PTT
2
212
)0.0,6164.2,3081.9,702.1;,15.293( 2 EETMCPSRC
S
igP
iteration 41
2 PP KT 15.2931
KT 37.3972
RR
H
igP
Ss
HHTTC
HisentropicW
1212 )(
)(
mol
JisentropicWs 2.3966)(
mol
JisentropicWW s
s 3.5288)(
)( 12 TTC
HW
H
igP
s
)0.0,6164.2,3081.9,702.1;,15.293( 2 EETMCPHR
CH
igP
KT 65.4282
Pumps
• Liquids are usually moved by pumps. The same equations apply to adiabatic pumps as to adiabatic compressors.
• For an isentropic process:
• With • For liquid,
– – –
2
1
)(P
PSs VdPHisentropicW
)()( 12 PPVHisentropicW Ss
dPTVdTCdH P )1( VdPT
dTCdS P
PTVTCH P )1(
PVT
TCS P
1
2ln
Water at 45°C and 10 kPa enters an adiabatic pump and is discharged at a pressure of 8600 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 water.
kg
cmV
3
1010The saturated liquid water at 45°C:K
110425 6
Kkg
kJCP 178.4
)()( 12 PPVHisentropicW Ss
kg
kJ
kg
cmkPaisentropicWs 676.810676.8)108600(1010)(
36
kg
kJH
isentropicWW s
s 57.11)(
PTVTCH P )1(
KT 97.0
PVT
TCS P
1
2ln
Kkg
kJS
0090.0