flow process lecture

8/10/2019 Flow Process Lecture

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Chapter 7

Applications of Thermodynamics toFlow Processes


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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.


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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


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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


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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


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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.


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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


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


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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


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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.


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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


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


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Fig 7.2


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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


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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


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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


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)(


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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


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


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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


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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


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


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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?


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Problem 2


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Problem 3

flow process lecture

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