Advanced Thermodynamics

Note 6Applications of Thermodynamics to

Flow Processes

Lecturer: 郭修伯

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

Fig 7.2

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