Chapter 7. Application of Thermodynamics

Chapter 7. Application of Thermodynamics

to Flow Processes

Introduction

 Thermodynamics of flow

 Based on mass, energy, and entropy balance on open system

 Flow  also fluid mechanics problem

 Thermodynamics of flow

- Mass conservation, laws of thermodynamics

 Fluid mechanics

- Momentum balance

 This chapter

 Duct flow – pipe, Nozzle, Throttle

 Turbines (expanders)

 Compressors, Pumps

7.1 Duct Flow of Compressible Fluids (1)

 Adiabatic, steady state, one-dimensional flow of compressible fluid

 No shaft work and no change in potential energy

 Energy Balance Equation – 1^st law

W Q

m zg 2 u

H 1 dt

) mU (

d

fs

CV 2

     



 



 



 



  





0 0 0 0

0 2 u

H 1



 

  

 dH   udu

Changes in enthalpy directly go to changes in velocity

Steady state

7.1 Duct Flow of Compressible Fluids (2)

 Mass balance equation – Continuity Equation

  ^{m 0}

dt dm

fs

cv

    ) 0

V ( uA d )

m (

d   

0

A 0 dA u

du V

dV   

 equation 2.27 (page 47)

7.1 Duct Flow of Compressible Fluids (3)

 Thermodynamic Relations

 Replace V in terms of S and P VdP TdS

dH  

P P

P

P s

S T T

V S

V

S dS dP V

P dV V

 

 







 



 







 

 

 









 

 







 

 

 







 

T

V V

1 



 







 

 (eqn 3.2)

T C T

S

P

 



 







 (eqn 6.17)

P

P

C

VT S

V   



 









 Relation from physics

7.1 Duct Flow of Compressible Fluids (4)

S 2

2

V V P

c 



 







 



is related with pressure derivative w.r.t volume

with constant enthalpy (S)

S dS dP V

P dV V

P s

 

 







 

 

 







 

c dP dS V

C T V

dV

2 P

 



7.1 Duct Flow of Compressible Fluids (5)

 Variables : dH, du, dV, dA, dS, dP

 Equations : four udu

dH  

A 0 dA u

du V

dV   

VdP TdS

dH  

c dP dS V

C T V

dV

2 P

 



dS, dA : treat as independent Can develop equations of other derivatives with dS and

dA

7.1 Duct Flow of Compressible Fluids (6)

0 A dA

u M

1 TdS 1

M 1

C M u udu

0 A dA

TdS u C

1 u VdP

) M 1

(

2 2

2 2 p

2

2

p 2 2

 



 





 



 









 









 





 







 



 







M : Mach number = u/c

Derive yourself!

7.1 Duct Flow of Compressible Fluids (7)

dx 0 dA A

u M

1 1 dx

T dS M

1 C M

u dx

u du

dx 0 dA A

u dx

T dS C

1 u dx

V dP ) M 1

(

2 2

2 2 p

2

2

p 2 2

 



 





 



 









 









 





 







 



  





According to second law, (dS/dx) ≥ 0

Pipe Flow

 Pipe Flow : constant cross sectional area (dA/dx=0)

dx dS M

1 C M u dx T

u du

dx dS )

M 1

(

C 1 u

V T dx

dP

2 2 p

2

2 p

2



 









 









 





 









 









 





Subsonic flow :

Implies :

0 )

M 1

( 



dx 0 dP 

dx 0 du 

Pressure drops

in the direction of flow Velocity increases

in the direction of flow

Pipe flow

 The velocity does not increase indefinitely.

 If the velocity exceeds the sonic value,

dx 0

dP  0

dx du  Supersonic flow

Shock wave and turbulence

Unstable flow

Sonic Boom?

 What is happening around the aircraft?

Example 7.1

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.

Example 7.1

(a) From the continuity equation,

1 2

1 2

1 2

1 1 1 2

2

2 A A , V V ,so u u

V A u V

A 0 u

V ) (uA

d       

(b) For the entropy of incompressible fluid,

T 0 ln T C S T VdP

C dT dS

1 2 P

P     



(c) From dH = -udu, dH = 0 since du = 0





dT C

dH  _P      ₂  ₁  _{P 2}  ₁  ₂  ₁ 0

) T T

( C )

P P

( V 0

H   ₂  ₁   _{P 2}  ₁ 



G G

j j

cv j

fs S 0 S S

T Q dt

) mS ( ) d

m S

(       

 ^



Nozzles

 Flow within a pipe or a duct (variable cross-sectional area)

 Assume isentropic flow (dS = 0)  reversible flow

dx dA M

1 1 A

u dx

du

dx dA M

1 1 VA

u dx

dP

2 2 2

 

 





 



 

 





 

Nozzles

Subsonic (M<1) Supersonic(M>1) Converging Diverging Converging Diverging

dA/dx - + - +

dP/dx - + + -

du/dx + - - +

Converging Diverging

Converging Nozzle

 For subsonic flow in converging nozzle

 Pressure Velocity

 Maximum obtainable velocity = speed of sound, c

 Converging subsonic nozzle can be used to deliver constant flow into a region of variable pressure

 As P₂ decreases, velocity increases and maximum value at sonic velocity.

 Ultimately, the pressure ratio P₂/P₁ reaches a critical value at which the velocity at the nozzle exit is sonic!

 Further decrease in P₂ has no effect on velocity.

Converging / Diverging Nozzle

 Speed of sound is attained at the throat of converging/diverging nozzle only when the pressure at the throat is low enough that critical value of P₂/P₁ is reached.

Value of critical pressure ratio

VdP TdS

dH   udu dH  

VdP udu  

 _ ^





 





 

 



 





 











2 

1

P

P

/ ) 1 (

1 2 1

2 1 1 2

2

P

1 P 1

V P VdP 2

2 u

u

dS=0  Adiabatic

 





 

 



 





/ 1 1 1

1 1

P V V P

V P const

PV

Value of critical pressure ratio

 _ ^





 





 

 



 





 











2 

1

P

P

/ ) 1 (

1 2 1

2 1 1 2

2

P

1 P 1

V P VdP 2

2 u

u

Critical value  u = c

2 2 2

2

S

S 2

2 2

2

V P u

V P V

P

V V P

c u





 

 



 









 

 







 





1

1 2

2 2 2

2 1

1 2 P

P

V P u

0 u







 

 







 







Example 7.2

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

oC. At the nozzle inlet, the velocity is 30 m/s. Calculate values of the ratio A/A₁ (A₁ is the cross-sectional area of the nozzle inlet) for the sections where the pressure is 600 kPa. Assume that the nozzle operates isentropically.

From the continuity equation,

2 1

2 1 1

2 1

1 1 2

2 2

u V

V u A

A V

A u V

A

u   

From dH = -udu,

) H H

( 2 u

u²₂  ₁²  ₂  ₁ Compare the unit of u² and H

Example 7.2

The initial values for entropy, enthalpy, and specific volume can be found from steam table.

 S₁ = 7.2997 kJ/kg∙K, H₁ = 3,059.8×10³ J/kg, V₁ = 371.39 cm³/g Then,

) 10 059 , 3 H

( 2 30

u u and

V 39 . 371

30 A

A ₃

2 2

2 2 2

2 1

2     

Since S₂ = S₁ = 7.2997 kJ/kg∙K (isentropic), other values, H₂ and V₂, can be found from steam table (at 600 kPa and S₁ = 7.2997 kJ/kg∙K)

From page 727,

S₂ = 7.2997 kJ/kg∙K, H₂ = 3,020.4×10³ J/kg, V₂ = 418.25 cm³/g Then, A₂/A₁ and u₂ can be calculated!

Example 7.3

Consider again the nozzle in Ex. 7.2, assuming now that steam behaves as an ideal gas. Calculate the critical pressure ratio and the velocity at the throat.

Let  = 1.3 for steam,

55 . 1 0

3 . 1

2 1

2 P

P ^{1.3 1}

3 . 1 1

1

2 





 



 







  ^{ }



mol / J 17 . 765 , 4 K 15 . 573 K

mol / J 314 . 8 RT V

P , gas ideal For

obtained be

should V

P P 1 P

1 V P VdP 2

2 u

u

1 1

1

1 1 P

P

/ ) 1 (

1 2 1

2 1 1 2 2

2

1











 















 









 















2 2 1

1V 4,765.17 J/mol 296,322J/kg 296,322m /s

P   

   

s / m 35 . 544 u

322 , 296 ]

55 . 0 1 1 [

3 . 1

511 , 264 3

. 1 900 2

P 1 P

1 V P u 2

u

2

3 . 1 / ) 1 3 . 1 ( /

) 1 (

1 2 1

2 1 1 2

2







 



 















 









 

 ^







Throttling Process

 Throttling Process : Orifice , Partly closed valve, porous plug, …

 Primary result : pressure drop

 For ideal gases, H = H(T) and no change in T

 For real gases, slight change in T

1

2

H

H , 0

H  



Example 7.4

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.

! solver or

iteration with

T for solve can

we C or

H T H

T

0 H

H ) T T

( C

H H

dT C H

H H

2 H

ig P

R 2 R

1 1

2

R 1 R

2 1

H 2 ig P R

1 R

2 T

T

ig P 1

2

2

1

 



























R 1 R

2 1

T 2 T

ig

P S S

P ln P T R

C dT S ²

1













Example 7.5 Joule-Thompson Coefficient

 Temperature change resulting from a throttling a real gas.

 Joule-Thompson coefficient

 Temperature increase/decrease as a result of throttling process

P

T  

 







 



Joule/Thomson Coefficient and other properties

P

T  

 







 



T P

T P

H

P

H T

H P

H H

T P

T 



 







 



 







 

 



 







 



 







 

 



 









P

P

H

C 



 







 

 1



J-T coeff. comes from the pressure dependence of H

Joule/Thomson Coeff. from PVT relation

P P

2

P 2

T

P T

T Z P

C RT

T Z P

RT P

H

T T V

P V H

 

 







 



 

 







 

 



 









 

 







 

 



 









With C

and PVT relation, any property can be

predicted.

7.2 Turbines (Expanders)

 Expansion of gas  Production of Work

 Internal Energy  Kinetic Energy  Shaft Work

Turbines (Expanders)

 Heat effects are negligible, Inlet and outlet velocity changes are small

 Normally T₁, P₁ and P₂ are given

 If turbine expands reversibly and adiabatically…

 isentropic process (adiabatic process): S = 0

 Maximum work

) H H

( H W

) H H

( m H

m W

1 2

s

1 2

s















  



S s

( isentropic ) ( H )

W  

Turbines (Expanders)

 Turbine Efficiency

 Turbine efficiency of properly designed turbine : 0.7 to 0.8

S s

s

) H (

H

) isentropic (

W

W



 







Turbines (Expanders)

Adiabatic expansion process in a turbine or expander

Example 7.6

A steam turbine with rated capacity of 56,400 kW operates with steam at inlet conditions of 8,600 kPa and 500 ^oC, and discharges 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 of flow of the steam.

At the initial condition of 8,600 kPa and 500 ^oC,

H₁ = 3,391.6 kJ/kg, S₁ = 6.6858 kJ/kg∙K (from steam table, page 752) For isentropic process, S₂ S₁  6.6858 at P₂ 10kPa

8047 .

0 x

1511 .

8 x

6493 .

0 ) x 1 ( 6858 .

6

S x S

) x 1 ( S

x S

x S

, Then

) liquid vapor

(

! phase two

is it 6493

. 0 S

and 1511 .

8 S

, kPa 10 P

at

v 2 v

2 v

2

v 2 v 2 l

2 v 2 v

2 v 2 l

2 l 2 2

l 2 v

2 2























 











Example 7.6

 

kg / kJ 4 . 117 , 2 8 . 584 , 2 8047 .

0 8 . 191 )

8047 .

0 1 ( H

kPa 10 P

at 8 . 191 H

and 8 . 584 , 2 H

, table steam

From

1 2

S 2

2 l

2 v

2









 















 







Now, a turbine efficiency is  = 0.75,

    

kg / kJ 0 . 436 , 2 6 . 955 6

. 391 , 3 H H

H

kg / kJ 6 . 955 2

. 274 , 1 75

. 0 H

H

1 2

S































The composition of steam is

s / kg 02 . 59 m

) 6 . 391 , 3 0 . 436 , 2 ( m 400

, 56 W

), H H

( m H

m W

From

K kg / kJ 6846 .

7 1511 .

8 9378 .

0 6493 .

0 ) 9378 .

0 1 ( S

...

and

9378 .

0 x

8 . 2584 x

8 . 191 )

x 1 ( 0 . 436 , 2

s

1 2

s 2

v 2 v

2 v

2



















































 



 

Example 7.7

A stream of ethylene gas at 300 ^oC and 45 bar is expanded adiabatically in a turbine to 2 bar. Calculate the isentropic work produced. Find the properties of ethylene, assuming an ideal gas.

1 2 1

2 S

ig P 1

T 2 T

ig P

1 H 2

ig P T

T

ig P 1

2

P ln P T R

ln T P C

ln P T R

C dT S

) T T

( C

dT C H

H H

2

1

2

1



























Example 7.7

For isentropic process (S = 0),

) T T

( C

dT C

) H ( ) isentropic (

W

, T calculate we

Once

P 0 ln P T R

C dT S

from T

for solve or

C g calculatin by

T Find

45 0 ln 2 15 R

. 573 ln T

C P 0

ln P T R

ln T C

S

1 H 2

ig P T

T

ig P S

s

2

1 T 2

T

ig P S 2

ig P 2

2 S

ig P 1

2 1

2 S

ig P

2

1

2

1





































7.3 Compression Processes

 Compression Devices : Rotating blades, Reciprocating pistons

 Turbine operating in reverse

Compressors

 Compressor efficiency : 0.7 to 0.8

H ) H (

W

) isentropic (

W

S s s



 







Compressors

Adiabatic compression process

H ) H (

W

) isentropic (

W

S s s



 







Isentropic Compression of Ideal Gas

 For ideal gas,

Assuming S = 0 (isentropic process),

Remember that S is not zero for real compression process!

1 2 1

2 P S

1 T 2

T ig P

P P

ln P T R

ln T P C

ln P T R

C dT P S

R dP T

C dT

dS ²

1



















 

) T T

( C

dT C

) isentropic (

W H

, P then

T P T

1 '

H 2 T '

T ig P S S

C / R

1 2 1 '

2

P 2

1 S

' P









 



 



 



Example 7.8

Saturated-vapor steam at 100 kPa (T_sat = 99.63 ^oC) 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,

S₁ = 7.3598 kJ/kg∙K and H₁ = 2,675.4 kJ/kg For isentropic compression to 300 kPa,

For steam at 300 kPa with this entropy,

K kg / kJ 3598 .

7 S

S₂  ₁   kg

/ kJ 8 . 888 , 2 H₂ 

   

kg / kJ 5 . 284 H

W ,

And

) 723 page

, table steam

from (

K kg / kJ 5019 .

7 S

and C

1 . 246 T

kg / kJ 9 . 959 , 2 5 . 284 4

. 675 , 2 H H

H , Then

kg / kJ 5 . 75 284

. 0

4 . 213 H H

kg / kJ 4 . 213 4

. 675 , 2 8 . 888 , 2 H

S

2 o

2

1 2

S S





























 

 











Example 7.9

If methane is compressed adiabatically from 20 ^oC and 140 kPa to 560 kPa, estimate the work requirement and the discharge temperature of the methane.

The compressor efficiency is 0.75. Assume that methane is an ideal gas.

 

dT C or

) T T

( C

W H

) and isentropic (

W W

then is

work actual

The

dT C or

) T T

( C

) isentropic (

W H

, Then

K 37 . 397 T

, T for Solve

140 0 ln 560 T R

C dT S

or

140 0 ln 560 15 R

. 293 ln T

P C ln P T R

ln T C

S

2

1 '

2

P 1 2

T T

ig P 1

H 2 P S

S S

T T

ig P 1

' H 2 ' S

S 2 2 T

15 . 293

ig P

2 P S

1 2 1

2 P S













 











 

















 





Pumps

 Liquids are usually moved by pumps, generally rotating equipment.

 Same assumption (adiabatic process) as compressor

 For isentropic process (adiabatic pump),

 In general process,

  ^{ } 





1

1

P S P

S

( isentropic ) H VdP W

VdP dH

P T V

ln T C S

P ) T 1

( V T

C H

T VdP C dT

dS dP

) T 1

( V dT

C dH

1 2 P

P

P P









































Example 7.10

Water at 45 ^oC and 10 kPa enters an adiabatic pump and is discharged at a pressure of 8,600 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 the water.

For saturated liquid water at 45 ^oC,

V = 1,010 cm³/kg,  = 425×10^-6 K^-1, C_p = 4.178 kJ/kgK

W_s(isentropic)=(H)_S=1,010×(8,600-10)=8.676×10^-6 kPa  cm³/kg=8.676 kJ/kg

P T V

ln T C S P

) T 1

( V T C H

calculated be

can S and T

kg / kJ 57 . 75 11

. 0

676 . H 8

H , ) H Since (

1 2 P

P

S

































 

 



Homework

 Problems

 7.20, 7.23, 7.34

 7.16

- Just prove that

- You don’t have to work on parts (a) and (b)!

 Due:

 Recommend Problems

 7.18, 7.27, 7.28, 7.32, 7.49, 7.52

T

Z T

T Z 



 









 

 



 









