Applications of Thermodynamic

7/31/2019 Applications of Thermodynamic

1/24

Applications of Thermodynamics toFlow Processes


2/24

The discipline

Principles: Fluid mechanics and Thermodynamics

Contrast

Flow process inevitably result from pressure gradients withinthe 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.


3/24

Duct flow of compressible fluids

Equations interrelate the changes occurring inpressure, velocity, cross-sectional area, enthalpy,entropy, and specific volume of the flowingsystem.

Consider a adiabatic, steady-state, onedimensional flow of a compressible fluid:

The continuity equation:

02

2

uH ududH

0A

dA

u

du

V

dV0)/( VuAd


4/24

VdPTdSdH

dPP

VdS

S

VdV

SP

0

A

dA

u

du

V

dV

PPP S

T

T

V

S

V

PT

V

V

1

T

C

T

S P

P

PPC

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

P

2

ududH dP

c

VdS

C

T

V

dV

P

2

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

dAA

uTdS

C

u

udu P

MM

M

Relates du to dS and dA


5/24

Pipe flow

01

1

1

2

22

22

dAA

u

TdS

C

u

udu

P

MM

M 011

222

dA

A

uTdS

C

uVdP

P

M

dx

dSC

u

Tdx

duu P

2

22

1 M

M

dx

dSC

u

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.

0dxdu 0

dxdP


6/24

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 changeoriginates from flow irreversibility.


7/24

0

1

1

1

2

22

22

dA

A

uTdS

C

u

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


8/24

01

1 2

2

dx

dA

A

u

dx

duu

M

012

2 dx

dA

A

u

dx

dPVM

isentropicVdPudu

2

1

22122

P

PVdPuu

.constPV

1

1

2112

1

2

2 11

2

P

PVP

uu

cu 2SV

PVc

22

V

P

V

P

S

.constPV

01 u

1

1

2

1

2

P

P


9/24

A high-velocity nozzle is designed to operate with steam at 700 kPa and 300C. 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

2997.71kgkJH 8.30591

gcmV

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 1.0

600 418.25 282.3 0.120

500 481.26 411.2 0.095

400 571.23 523.0 0.088

300 711.93 633.0 0.091

200 970.04 752.2 0.104


10/24

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

P3.1

55.01

2 P

P

1

1

2112

1

2

2 11

2

P

PVPuu

We have u1, P1, V1, P2/P1,

s

mu 35.5442

(a)

(b)

2Ms

mu 7.108835.54422

1

1

2112

1

2

2 11

2

P

PVPuu kPaP 0.302


11/24

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 apressure drop in the fluid.

WQmzguHdt

mUd

fs

cv

2

2

1)( 0Q

0W

0H

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.


12/24

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

ig

P HHTTCH

Final state at 1 bar: assumed to be ideal gas and 022 RR

SH

11

2 TC

HT

H

ig

P

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 TTCigP

??H

ig

PC

KT 400 Kmol

JC

ig

P

07.94

KT 2.3852 ???

KT 6.3924005.02.3855.0 Kmol

JCC

ig

PH

ig

P

73.92

KT 0.3852

R

S

ig

P SP

P

RT

T

CS 11

2

1

2

lnln HigP

S

igP CC

2934.0)152.0,471.0,082.1(1 SRBR

S R

Kmol

J

S 80.23


13/24

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 < 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


14/24

Fig 7.2


15/24

Turbine (Expanders)

A turbine (or expander):

Consists of alternate sets of nozzles and

rotating bladesVapor 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.

SW

Turbine


16/24

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


17/24

A steam turbine with rated capacity of 56400 kW operates with steam

at inlet conditions of 8600 kPa and 500C, 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 offlow of the steam.

SWTurbine

KkgkJS

kgkJH

CTkPaP

6858.66.3391

5008600

11

11

Kkg

kJSkPaP

6858.610 22

KkgkJxxSxSxS vvvvlv

6858.61511.86493.0)1()1( 222

8047.0vxkgkJHxHxH

vvlv

4.2117)1( 222

kg

kJHHH S 2.127412

kg

kJHH S 6.955

vvlv HxHxkg

kJHHH 2212 )1(0.2436

9378.0vxKkg

kJSxSxSvvlv

6846.7)1( 222

skJHmWS 56400

skg

m 02.59


18/24

A stream of ethylene gas at 300C 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

ig

P HHTTCH 1212 )( RRS

ig

P SSP

PR

T

TCS 12

1

2

1

2 lnln

KTbarPbarP 15.573245 121

(a) Ideal gas

1

2

1

2 lnlnPPR

TTCS

S

igP

0S

0S

3511.61135.3exp2

R

CT

S

ig

P

)0.0,6392.4,3394.14,424.1;,15.573( 2 EETMCPSRC

S

ig

Piteration

KT 8.3702

)()()( 12 TTCHisentropicW Hig

PSS

224.7

)0.0,6392.4,3394.14,424.1;18.370,15.573(

EEMCPH

R

CH

ig

P

mol

JisentropicWS 12153)15.5738.370(314.8224.7)(


19/24

(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

2

ln15.573ln2

RT

CS Sig

P

Assuming T2 = 370.8 K

314.12

r

T 040.02

r

P

0139.0)087.0,040.0,314.1(2 SRBR

SR

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

J

HHTTC

HisentropicW

RR

H

ig

P

Ss

11920)(

)(

1212


20/24

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

SW

compressorCompression process


21/24

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 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 kPa

kg

kJH 8.28882

kg

kJH S 4.213

kg

kJHH S 5.284

kg

kJHHH 9.295912

300 kPaCT1.2462

Kkg

kJS

5019.72

kg

kJHWS 5.284


22/24

If methane (assumed to be an ideal gas) is compressed adiabatically

from 20C 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

ig

P SSP

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 EETMCPSR

CS

ig

P

iteration 41

2PP KT 15.2931

KT 37.3972

RR

H

ig

P

Ss

HHTTC

HisentropicW

1212 )(

)(

mol

JisentropicWs 2.3966)(

mol

JisentropicW

W

s

s 3.5288

)(

)( 12 TTC

HW

H

ig

P

s

)0.0,6164.2,3081.9,702.1;,15.293( 2 EETMCPHR

CH

ig

P

KT 65.4282


23/24

Pumps

Liquids are usually moved by pumps. The sameequations apply to adiabatic pumps as to adiabaticcompressors.

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


24/24

Water at 45C 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 45C:K

110425 6

Kkg

kJCP

178.4

)()( 12 PPVHisentropicW Ss

kg

kJ

kg

cmkPaisentropicWs 676.810676.8)108600(1010)(

36

kg

kJH

isentropicWW ss 57.11

)(

PTVTCH P )1(

KT 97.0

PVT

TCS P

1

2ln

Kkg

kJ

S 0090.0

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