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8/10/2019 Flow Process Lecture
1/27
Chapter 7
Applications of Thermodynamics toFlow Processes
8/10/2019 Flow Process Lecture
2/27
The discipline
Principles: Fluid mechanics and Thermodynamics
Contrast
Flow process inevitably result from pressure gradients withinthe fluid. Moreover, temperature, velocity, and evenconcentration gradients may exist within the flowing fluid.
Uniform conditions that prevail at equilibrium in closedsystem.
8/10/2019 Flow Process Lecture
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Duct flow of compressible fluids
Equations interrelate the changes occurring in pressure, velocity, cross-sectional area, enthalpy,entropy, and specific volume of the flowingsystem.
Consider an adiabatic, steady-state, onedimensional flow of a compressible fluid:
The continuity equation:
02
2
u H ududH
0 AdA
udu
V dV 0)/( V uAd
8/10/2019 Flow Process Lecture
4/27
VdP TdS dH
dP P V
dS S V
dV S P
0
AdA
udu
V dV
P P P S
T
T
V
S
V
P T V
V
1
T
C T S P
P
P P C
VT
S
V
2
2
c
V
P
V
S
From physics,c is the speedof sound in afluid
dP cV
dS C
T V dV
P 2
ududH dP
c
V dS
C
T
V
dV
P
2
011222
dA Au
TdS C u
VdP cu
P
01122
2
dA
Au
TdS C u
VdP P
M
c
uM
The Mach number
VdP TdS dH
01
11
2
22
22
dA Au
TdS C u
udu P MM
M
Relates du to dS and dA
8/10/2019 Flow Process Lecture
5/27
Pipe flow
01
1
1
2
22
22
dA A
u
TdS
C u
udu P
MM
M 011
222
dA
Au
TdS C u
VdP P
M
dxdS C
u
T dxdu
u P
2
22
1 M
M
dxdS C
u
V T
dxdP P
2
2
1
1
M
For subsonic flow, M 2 < 1, , the pressure decreasesand the velocity increases in the direction of flow. For subsonicflow, the maximum fluid velocity obtained in a pipe of constantcross section is the speed of sound, and this value is reached at theexit of the pipe.
0dxdu 0
dxdP
8/10/2019 Flow Process Lecture
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8/10/2019 Flow Process Lecture
7/27
01
1
1
2
22
22
dA A
uTdS
C u
udu P
MM
M 011
222
dA
Au
TdS C u
VdP P
M
Reversible flow
01
1 22
dxdA
Au
dxdu
uM
012
2
dxdA
Au
dxdP
V M
Nozzles:
Reversible flow
Subsonic: M 1Converging Diverging Converging Diverging
dxdA
- + - +
dxdP - + + -
dxdu
+ - - +
For subsonic flow in a converging nozzle, the velocity increases as the cross-sectionalarea diminishes. The maximum value is the speed of sound, reached at the throat.
8/10/2019 Flow Process Lecture
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01
1 22
dxdA
Au
dxdu
uM
012
2
dxdA
Au
dxdP
V M
isentropicVdP udu
2
1
22122 P P VdP uu
.const PV
1
1
21121
22 11
2 P P V P
uu
cu 2S V
P V c
22
V P
V P
S
.const PV
01 u
1
1
2
1
2
P
P
8/10/2019 Flow Process Lecture
9/27
A high-velocity nozzle is designed to operate with steam at 700 kPa and 300
C. At thenozzle inlet the velocity is 30 m/s. Calculate values of the ratio A/A 1 (where A 1 is thecross-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 V u
A A
1
1
1
Energy balance: )(2 1212 H H uu
Initial values from the steam table: K kg
kJ S 2997.71 kg kJ H 8.30591
g cmV
31 39.371
uV
A A
39.37130
1
)108.3059(2900 32 H u
Since it is an isentropic process, S = S 1. From the steam table:
600 kPa: K kg
kJ S 2997.7
kg kJ
H 4.3020 g
cmV
3
25.418 sm
u 3.282 120.01 A
A
Similar for other pressures P (kPa) V (cm3/g) U (m/s) A/A 1
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
8/10/2019 Flow Process Lecture
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Consider again the nozzle of the previous example, assuming now that steam behavesas 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
12
P P 3.1 55.0
1
2
P P
1
1
21121
22 11
2 P P V P
uuWe have u 1, P1, V1, P2/P1,
sm
u 35.5442
(a)
(b)2M
sm
u 7.108835.54422
1
1
21121
22 11
2 P P V P
uu kPa P 0.302
8/10/2019 Flow Process Lecture
11/27
Throttling Process:
When a fluid flows through a restriction, such as an orifice, a partlyclosed valve, or a porous plug, without any appreciable change inkinetic or potential energy, the primary result of the process is a
pressure drop in the fluid.
W Qm zg u H dt
mU d
fs
cv
2
21)( 0Q
0W
0 H
Constant enthalpy
For ideal gas: 0 H 12 H H 12 T T
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 vaporizesor 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.
8/10/2019 Flow Process Lecture
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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 R R
H
ig P H H T T C H
Final state at 1 bar: assumed to be ideal gas and 022 R R S H
11
2 T C
H T
H
ig P
R
082.11 r T 471.01 r P
452.0)152.0,471.0,082.1(
),,(1
1
0
11
HRB
OMEGA PRTR HRB RT H
RT H
RT H
c
R
c
R
c
R
And based on 2nd virial coefficients correlation
263 10824.810785.28213.1 T T C ig P
?? H
ig P C
K T 400 K mol
J C ig P 07.94
K T 2.3852 ??? K T 6.3924005.02.3855.0
K mol J
C C ig P H ig
P 73.92
K T 0.3852
R
S
ig
P S P P
RT T
C S 112
1
2
lnln H ig
P S ig
P C C
2934.0)152.0,471.0,082.1(1 SRB RS R
K mol J
S 80.23
8/10/2019 Flow Process Lecture
13/27
Throttling a real gas from conditions of moderate temperature and pressure usuallyresults in a temperature decrease. Under what conditions would an increase intemperature be expected.
Define the Joule/Thomson coefficient: H P
T
When will < 0 ???
T P T P H P H
C P H
H T
P T
1
Always negative
P T T V
T V P H
T P H
Sign of ???
P ZRT
V P T T
Z P
RT P H
2
P P T Z
P C RT
2
Always positive P T
Z
Same sign
The condition may obtain locally for real gases. Such
points define the Joule/Thomson inversion curve.
0
P T Z
8/10/2019 Flow Process Lecture
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Fig 7.2
8/10/2019 Flow Process Lecture
15/27
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 efficientconversion of the internal energy of a high-
pressure stream into shaft work.
S W
Turbine
8/10/2019 Flow Process Lecture
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S fs
cv W Qm zg u H dt
mU d
2
21)(
)( 12 H H m H mW S
12 H H H W
S
The maximum shaft work: a reversible process (i.e., isentropic, S 1 = S 2)
S S H isentropicW )()(
The turbine efficiency
S S
S
H H
isentropicW W
)()(
Values for properly designed turbines: 0.7~ 0.8
8/10/2019 Flow Process Lecture
17/27
A steam turbine with rated capacity of 56400 kW operates with steamat inlet conditions of 8600 kPa and 500
C, and discharge into acondenser 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 offlow of the steam.
S W
Turbine
K kg kJ S kg
kJ H
C T kPa P
6858.66.3391
5008600
11
11
K kg kJ S kPa P 6858.610 22
K kg kJ x xS xS xS vvvvl v 6858.61511.86493.0)1()1( 222
8047.0v xkg kJ H x H x H vvl v
4.2117)1( 222
kg kJ H H H S 2.127412
kg kJ H H S 6.955
vvl v H x H xkg kJ H H H 2212 )1(0.2436
9378.0v x K kg
kJ S xS xS vvl v 6846.7)1( 222
skJ H mW S 56400
skg m 02.59
8/10/2019 Flow Process Lecture
18/27
A stream of ethylene gas at 300
C and 45 bar is expanded adiabaticallyin a turbine to 2 bar. Calculate the isentropic work produced. Find the
properties of ethylene by: (a) equations for an ideal gas (b)appropriategeneralized correlations.
R R
H
ig P H H T T C H 1212 )( R R
S
ig P S S P
P R
T T
C S 121
2
1
2 lnln
K T bar P bar P 15.573245 121
(a) Ideal gas
1
2
1
2 lnln P P R
T T C S
S
ig P
0S
0S
3511.61135.3exp2
RC
T S
ig P
)0.0,6392.4,3394.14,424.1;,15.573( 2 E E T MCPS RC
S
ig P iteration
K T 8.3702
)()()( 12 T T C H isentropicW H ig
P S S
224.7
)0.0,6392.4,3394.14,424.1;18.370,15.573( E E MCPH R
C H
ig P
mol J
isentropicW S 12153)15.5738.370(314.8224.7)(
8/10/2019 Flow Process Lecture
19/27
8/10/2019 Flow Process Lecture
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Pressure increases: compressors, pumps, fans, blowers, and vacuum pumps.
Interested in the energy requirement
S fs
cv W Qm zg u H dt
mU d
2
21)(
)( 12 H H m H mW S
12 H H H W S The minimum shaft work: a reversible process (i.e., isentropic, S 1 = S 2)
S S H isentropicW )()(
The compressor efficiency H
H W
isentropicW S S
S )()(
Values for properly designed compressors: 0.7~ 0.8
S W
compressor Compression process
8/10/2019 Flow Process Lecture
21/27
Saturated-vapor steam at 100 kPa (t sat = 99.63
C ) is compressedadiabatically to 300 kPa. If the compressor efficiency is 0.75, what isthe work required and what are the properties of the discharge stream?
For saturated steam at 100 kPa: K kg
kJ S 3598.71 kg
kJ H 4.26751
Isentropic compression
K kg kJ
S S 3598.712300 kPa
kg kJ
H 8.28882 kg kJ
H S 4.213
kg
kJ H H S 5.284
kg
kJ H H H 9.295912
300 kPaC T
1.2462
K kg kJ S 5019.72
kg kJ
H W S 5.284
8/10/2019 Flow Process Lecture
22/27
If methane (assumed to be an ideal gas) is compressed adiabaticallyfrom 20
C and 140 kPa to 560 kPa, estimate the work requirement andthe discharge temperature of the methane. The compressor efficiencyis 0.75.
R R
S
ig P S S P
P R
T T
C S 121
2
1
2 lnln
0S
S
ig P C
R
P P
T T
2
212
)0.0,6164.2,3081.9,702.1;,15.293( 2 E E T MCPS RC
S
ig P
iteration 41
2 P P K T 15.2931
K T 37.3972
R R
H
ig P
S s
H H T T C
H isentropicW
1212 )(
)( mol J
isentropicW s 2.3966)(
mol
J isentropicW
W s
s 3.5288
)(
)( 12 T T C
H W
H
ig P
s
)0.0,6164.2,3081.9,702.1;,15.293( 2 E E T MCPH R
C H
ig P
K T 65.4282
8/10/2019 Flow Process Lecture
23/27
Pumps
Liquids are usually moved by pumps. The sameequations apply to adiabatic pumps as to adiabaticcompressors.
For an isentropic process:
With
For liquid,
21
)( P P S s
VdP H isentropicW
)()( 12 P P V H isentropicW S s
dP T V dT C dH P )1( VdP T dT
C dS P
P T V T C H P )1(
P V T
T C S
P
1
2ln
8/10/2019 Flow Process Lecture
24/27
Water at 45
C and 10 kPa enters an adiabatic pump and is dischargedat 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 cm
V 3
1010The saturated liquid water at 45
C: K 1
10425 6 K kg
kJ C P 178.4
)()( 12 P P V H isentropicW S s
kg kJ
kg cmkPa
isentropicW s 676.810676.8)108600(1010)(3
6
kg kJ
H isentropicW
W s s 57.11)(
P T V T C H P )1(
K T 97.0
P V T T
C S P 1
2ln
K kg kJ
S 0090.0
8/10/2019 Flow Process Lecture
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Problem 1
Steam enters a nozzle at 800 kPa and 280
C atnegligible velocity and discharges at pressure
of 525 kPa. Assuming isentropic expansion ofthe steam in the nozzle, what is the exitvelocity and what is the cross-sectional area at
the nozzle exit for a flow rate of 0.75 kg/s?
8/10/2019 Flow Process Lecture
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Problem 2
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Problem 3