89
Chapter 1: Introductory concepts and definitions Thermodynamics Thermo: Heat Dynamis: Force System Surroundings Boundary Closed System Control Volume Macroscopic approach: Classical thermodynamics Microscopic approach: Statistical thermodynamics Property: intensive, extensive State Process o Isothermal process o Adiabatic process Cycle Equilibrium Phase

All Lectures

Embed Size (px)

Citation preview

Page 1: All Lectures

Chapter 1: Introductory concepts and definitions

Thermodynamics

Thermo: Heat

Dynamis: Force

• System

• Surroundings

• Boundary

• Closed System

• Control Volume

• Macroscopic approach: Classical thermodynamics

• Microscopic approach: Statistical thermodynamics

• Property: intensive, extensive

• State

• Process

o Isothermal process

o Adiabatic process

• Cycle

• Equilibrium

• Phase

Page 2: All Lectures

• Specific Volume

• Density

• Pressure (units)

• Temperature scale

o Kelvin scale

o Rankine scale

Page 3: All Lectures

Chapter 2: Energy and First Law of Thermodynamics

Work-Energy 2

1

2 22 1

1 ( )2

S

SFds m V V= −∫

Conservation of energy 2 21 1 2

1 12 2

mV mgZ mV mgZ+ = + 2

Expansion /Compression work

Work is a path function

(Depends on process)

Sign convention:

w>0 work done by the system

w<0 work done on the system

Page 4: All Lectures

A gas in a piston-cylinder undergoes an expansion for which . constantnPV =

Evaluate the work

done by the gas if:

a) n=1.5

b) n=1.0

c) n=0

2

1

v

vw P= ∫ dv Expansion work

c) 2

1

5 3 42 1 2( ) 3*10 (0.2 0.1) 3*10 30KJ

v

v

Nw Pdv P v v m Jm

= = − = − = =∫

b) 2 2

1 1

52

1

0.2ln 3*10 (0.1) ln( ) 20.79KJ0.1

v v

v v

vdvw P dv c PVv v

= = = = =∫ ∫

2

1

1.512 1

2

1 12 1

2 2 1 1

2 2 1 1

0.1) ( ) 3( ) 1.060.2

1

But

17.61

n

n nV

nV

n n

Va P P BarV

c V Vw dV cV n

C PV PV

PV PVW KJn

− −

= = =

−= =

= =

−∴ = =

Page 5: All Lectures

Energy transfer by heat:

Q>0 heat transfer to the system

Q<0 heat transfer from the system

Heat transfer modes:

Conduction: xdTQ KAdx

= −

Convection: 2 1( )Q hA T T•

= −

Radiation: Stefan-Boltzman Law4 42 1(Q A T Tεσ

= − )

Page 6: All Lectures

1st Last of Thermodynamics for Closed System:

Q W U KE PE= + Δ + Δ + Δ

Energy analysis of cycles:

net netQ W=

(a) Power cycle:

in

in out

in

WQQ Q

Q

η

η

=

−=

(b) Refrigeration cycle:

in

cycle

in

out in

QW

QQ Q

β

β

=

=−

Heater:

out

cycle

QW

γ =

Page 7: All Lectures
Page 8: All Lectures

Chapter 3: Evaluating Properties

P-v-T Relation

P=P(T,v)

The graph of such a function is a surface called (P-v-T) surface

Page 9: All Lectures

Projecting the liquid, two phase, and vapor regions of P-v-T surface onto the temperature-

specific volume Plane results in a (T-v) diagram

Two Phase Flow:

Divide by m

( ) (

(1 )

vap g

tot f g

f g

f f g g

f g

f g

m mx

m m m

V V V

vm v m v m

mf mgv v vm m

v x v xv

= =+

= +

= +

= +

= − +

)

The concept of independent properties:

(1 )

(1 )f g

f g

H V Pvh u Pvu x u x

h x h x

= += += − +

= − +

u

h

Page 10: All Lectures

Ideal gas model

( )( ) ( )( ) ( )

Pv RTu u Th h T u T Pvh h T u T RT

=== = += = +

Internal energy, enthalpy, and specific heats of ideal gases

For a gas obeying the ideal gas model, specific energy depends only on temperature.

Hence, the specific heat is also a function of temperature alone. That is, vc

( )vduc TdT

= (ideal gas)

This is expressed as an ordinary derivative because depends only onT . u

By separating variables

( )vdu c T dT=

On integration

2

12 1( ) ( ) ( )

T

vTu T u T c T dT− = ∫ (ideal gas)

Similarly, for a gas obeying the ideal gas model, the specific enthalpy depends only on

temperature, so the specific heat defined by Eq. 3.9, is also a function of temperature

alone. That is

pc

( )pdhc TdT

= (ideal gas)

An important relationship between the ideal gas specific heats can be developed by

differentiating the enthalpy equation with respect to temperature

dh du RdT dT

= +

Page 11: All Lectures

Therefore

( ) ( )p vc T c T R= + (ideal gas)

On a molar basis, this is written as

( ) ( )p vc T c T R= + (ideal gas)

Although each of the two ideal gas specific heats is a function of temperature, the above

equations show that the specific heats differ by just a constant. Knowledge of either

specific heat for a particular gas allows the other to be calculated by using only the gas

constant. The above equations also show that and pc c> v pc c> v , respectively.

For an ideal gas, the specific heat ration, k, is also a function of temperature only

( )( )

p

v

c Tk

c T= (ideal gas)

Since , it follows that . Combining the above two equations results in pc c> v 1k >

v

( )1

c ( )1

pkRc T

kRT

k

=−

=−

Page 12: All Lectures

Polytropic process of an ideal gas

Recall that a polytropic process of a closed system is described by a pressure-volume

relationship of the form

constantnpV =

Where is a constant. n

For a polytropic process between two states

1 1 2 2n np V p V=

or

2 1

1 2

( )np Vp V

=

The exponent may take on any value from n −∞ to +∞ , depending on the particular

process. When , the process is an isobaric (constant-pressure) process, and when

, the process is an isometric (constant –volume) process.

0n =

n = ±∞

For a polytropic process

22 2 1 1

1( 1

1p V p VpdV n

n)−

= ≠−∫

For any exponent n except . When 1n = 1n = ,

22

1 111

ln ( 1)VpdV p V nV

= =∫

The above equations apply to any gas (or liquid) undergoing a polytropic process.

When the additional idealization of ideal gas behavior is appropriate, further relations

can be derived. Thus, when the ideal gas equation of state is introduced, the following

expressions are obtained, respectively

Page 13: All Lectures

( 1) / 12 2 1

1 1 2

( ) ( ) (idealgas)n n nT p VT p V

− −=

22 1

1

( ) (ideal gas, n 1)1

mR T TpdVn−

= ≠−∫

22

11

ln (ideal gas, n=1)VpdV mRTV

=∫

For an ideal gas, the case n=1 corresponds to an isothermal (constant-temperature)

process, as can readily be verified. In addition, when the specific heats are constant, the

value of the exponent corresponding to and adiabatic polytropic process of and ideal

gas is the specific heat ratio k

n

Example 3.11 illustrates the use of the closed system energy balance for a system

consisting of an ideal gas undergoing a polytropic process.

Linear Interpolation:

Page 14: All Lectures

Chapter 4: Control volume analysis

A region of space through which mass flows

Conservation of mass cvi e

dm m mct

• •

= −∑ ∑

Since m VAρ•

=

VAmv

=

Then

cv i i e e

i e

dm V A V Adt v v

= −∑ ∑

Page 15: All Lectures

Most engineering systems can be assumed to be at steady state, meaning that all

properties are unchanged with time.

For a control volume then: 0cvdmdt

=

Therefore eim m• •

=∑ ∑

Page 16: All Lectures

A stream of water vapor mixes with a liquid water stream to produce a saturated liquid

stream at the exit. The states at the inlets and exit are specified. Mass flow rate and

volumetric flow rate data are given at one inlet and at the exit, respectively.

Determine the mass flow rates at inlet 2 and at the exit, and the velocity . 2V

Analysis:

The principle relations to be employed are the mass rate balance and the expression

. At steady state the mass rate balance becomes: /m VA v•

=

tvdmdt

0

1 2m m m• • •

= + − 3

1

Solving for 2m•

2 3m m m• • •

= −

Page 17: All Lectures

The mass flow rate is given. The mass flow rate at the exit can be evaluated from the

given volumetric flow rate

1m•

33

3

( )AVmv

=

Where , is the specific volume at the exit. In writing this expression, one-dimensional

flow is assumed. From Table A-3, .

3v

3 33 1.108*10 m /kgv −=

Hence

3

3 3 3

0.06 / 54.15kg/s(1.108*10 / )

m smm kg

−= =

The mass flow rate at inlet 2 is then

2 3 1 54.15 40 14.15kg/sm m m• • •

= − = − =

For one-dimensional flow at 2, , so 2 2 2 2/m A V v•

=

22 2 /V m v A•

= 2

State 2 is a compressed liquid. The specific volume at this state can be approximated by

2 2( )fv v T= .

Page 18: All Lectures

From Table A-2 at 40°C, . 3 32 1.0078*10 m /kgv −=

Therefore

3 3 4 2

2 2 2

(14.15kg/s)(1.0078*10 m /kg) 10 cm 5.7m/s25 1

Vcm m

= =

Page 19: All Lectures

Conservation of energy for a control volume

2 2

( ) (2 2

cv i ei ei i e

dE V VQ W m u gz m u gzdt

• • • •

= − + + + − + + )e

( ) ( )cv e e e i i iW W p A V p A V• •

= + −

2 2

( ) (2 2

cv i ecv i ei i i i e e e ecv

dE V VQ W m u p v gz m u p v gzdt

• • • •

= − + + + + − + + + )

2 2

( ) (2 2

cv i ecv i ei i ecv

dE V VQ W m h gz m h gzdt

• • • •

= − + + + − + + )e

2 2

( ) (2 2

cv i ecv i ei i ecv

i e

dE V VQ W m h gz m h gzdt

• • • •

= − + + + − + +∑ ∑ )e

Page 20: All Lectures

Analyzing control volume at steady state

0cvdEdt

=

2 2

0 ( ) (2 2i e

cv i ei i ecvV VQ W m h gz m h gz

• • • •

= − + + + − + +∑ ∑ )e

Since i em m• •

=∑ ∑

Many applications involve one inlet and one exit

Therefore:

i em m m• •

= =

2 21 2

1 2 1 2( )0 [( ) (

2cvcv

V VQ W m h h g z z• • • −

= − + − + + − )]

2 21 2

1 2 1 2( )0 ( ) (

2cvcvQ V VW h h g z z

m m

• •

• •

−= − + − + + − )

2 21 2

1 2 1 20 ( ) (2

V Vq w h h g z z−= − + − + + − )

Or:

2 22 1

2 1 1 2( ) (2

V Vq w h h g z z−= + − + + − )

This is the statement of first law of thermodynamics for controlled volume

Page 21: All Lectures

Application of the first law to various engineering devices:

Nozzles and diffusers:

2 22 1

2 1 2V Vq h h −

= − +

Page 22: All Lectures

Turbines:

2 1( )q w h h= + −

Page 23: All Lectures

Compressors:

2 1( )q w h h= + −

Page 24: All Lectures

Pump:

2 22 1

2 1 2 1( )2

V Vq w h h g z z−= + − + + −

2 22 1

2 2 2 1 1 1 2 10 (2

V Vw u P v u Pv g z z−= + + − − + + − )

Since water has a low viscosity 2 1 0u u− ≈

And incompressible flow 2 1v v v= =

Then

2 22 1

2 1 2 10 ( ) (2

V Vw v P P g z z−= + − + + − )

)

The above formula is used to find the power used by a pump

2 1(v P P− is known as Pressure head

2 22 1

2V V− is known as Velocity head

2 1(g z z− ) is known as Elevation head

Page 25: All Lectures

Throttling Devices:

2 22 1

2 1 2 1( ) (2

V Vq w h h g z z−= + − + + − )

There is usually no significant heat transfer with the surrounding and the change in

potential energy from inlet to exit is negligible. With these assumptions, the first law

Of thermodynamics reduces to:

2 21 2

1 22 2V Vh h+ = +

Although velocities may be relatively high in the vicinity of the restriction, measurements

made upstream and downstream of the flow area show in most cases that the change in

the specific kinetic energy of the fluid between these locations can be neglected. With

this further simplification, the last equation reduces to

1 2h h=

Page 26: All Lectures

System integration (putting it all together):

Page 27: All Lectures

Heat exchangers:

Devices that transfer energy between fluids at different temperatures by heat transfer

modes such as conduction, convection, and radiation are called heat exchangers. One

common type of heat exchanger is a vessel in which hot and cold streams are mixed

directly as show in figure (a). An open feedwater heater is an example of this type.

Another common type of heat exchanger is one in which a gas or liquid is separated from

another gas or liquid by a wall through which energy is conducted. These heat

exchangers, known as recuperators, take many different forms. Counterflow and parallel

tube-within-a-tube configurations are shown in Figures (b) and (c), respectively. Other

Page 28: All Lectures

configurations include cross-flow, as in automobile radiators, and multiple-pass shall-

and-tube condensers and evaporators. Figure (d) illustrates a cross-flow heat exchanger.

The only work interaction at the boundary of a control volume enclosing a heat

exchanger is flow work at the places where matter enters and exits, so the term of

the energy rate balance can be set to zero. Although high rates of energy transfer may be

achieved from stream to stream, the heat transfer from the outer surface of the heat

exchanger to the surroundings is often small enough to be neglected. In addition, the

kinetic and potential energies of the flowing streams can often be ignored at the inlets and

exits.

cvW•

Page 29: All Lectures

Transient analysis:

Page 30: All Lectures

Chapter 5: The second law of thermodynamics

Although the first law of thermodynamics allows unrestricted convertibility from one

from of energy to another, as long as the overall quantity is conserved, experimental

evidence tells that in certain types of energy conversion, restriction must be placed to the

direction end extent of transformation. For example, it is a fact that although work can be

completely converted into heat, heat cannot be completely converted into work, no matter

how ideal the process may be. For example, rubbing two pieces of sandpaper together,

work used to overcome friction causes an increase in internal energy of the paper and

heat flow to surroundings, but heat addition will not result in doing mechanical work. It

is the second law of thermodynamics that imposes restrictions to the direction and extent

of energy transformation processes. In this chapter we sturdy the second law of

thermodynamics and some consequences of it.

In summary second law provides means for

1. Predicting the direction of processes.

2. Establishing conditions for equilibrium.

3. Determining the best theoretical performance of cycles, engines, and other

devices.

4. Evaluating quantitatively the factors that preclude the attainment of the best

theoretical performance level.

Page 31: All Lectures

Additional uses of the second law include its roles in

5. Defining a temperature scale independent or the properties of any thermometric

substance.

6. Developing mans for evaluating properties such as u and h in terms of properties

that are more readily obtained experimentally.

It is well known that physical processes in nature process toward equilibrium

spontaneously. Liquids flow from a region of high elevation to one of low elevation;

gases expand from a region of high pressure to one of low pressure; heat flows from a

region of high temperature to one of low temperature; and material diffuses from a region

of high concentration to one of low concentration. A spontaneous process can proceed

only in a particular direction. Energy from an external source is required to reverse such

processes. The second law of thermodynamics epitomizes out experiences with respect

to the unidirectional nature of thermodynamics processes.

Page 32: All Lectures

Kelvin – Planck statement of the second law (Heat Engine)

Clausius statement of the second law (Heat pump)

Page 33: All Lectures

Reservoir – A large container that is unaffected in temperature as heat goes into and out

of it.

Page 34: All Lectures

Reversible process: Is a process that can be reversed and leave no resultant change in

either the system or surroundings. Reversible processes are fictitious, they are ideal

operations that represent the only means by which engineering calculations can be made.

Irreversible process: The system and the surrounding can not be exactly restored to

their initial states. All real processes are irreversible, since they involve dissipative

effects such as friction, heat transfer, etc...

The process is irreversible because of heat transfer to the surroundings. To reverse the

process, heat must be added to the block and an equal amount of mechanical work must

be done, which is impossible.

Page 35: All Lectures

The most common irreversible processes are:

• Heat transfer across a finite temperature difference

• Unrestrained expansion of a fluid

• Mixing of fluids of different compositions and states

• Solid and fluid friction

• Spontaneous chemical reaction

• Inelastic deformation

• Electric resistance

• Hysteresis effects in magnetization and polarization

Page 36: All Lectures

Thermodynamic cycles:

1 – Power cycle

Thermal efficiency 1cycle H c c

H H H

W Q Q QQ Q

η −= = = −

Q

2 – Refrigeration/Heat pump cycle

Coefficient of performance

c c

cycle H c

H H

cycle H c

Q QW Q

Q QW Q Q

β

γ

= =Q−

= =−

Page 37: All Lectures

Maximum thermal efficiency of a power cycle:

For a reversible cycle

( )( )

c c

H H

Q f TQ f T

=

Can use number of temp functions.

Kelvin suggested to use ( )( )

c c

H H

f T Tf T T

==

Therefore c c

H H

Q TQ T

=

max 1 c

H

TT

η = −

Maximum performance of a refrigeration cycle:

max

max

c

H c

H

H c

TT T

TT T

γ

Β =−

=−

Page 38: All Lectures

Carnot cycle

What is the most efficient heat engine? What heat pump provides the highest coefficient

of performance? All processes in the cycle must be reversible.

Page 39: All Lectures

Chapter 6: Using Entropy

• Clasius inequality

• Principle of entropy increase

• Entropy is a non conserved property

• Entropy change for pure substances

• Entropy change for ideal gases

• Isentropic process

• Isentropic efficiencies of various engineering devices

Page 40: All Lectures

Clasius inequality

0QTδ

≤∫ for reversible & irreversible cycles

Sum of all differential heat transfers divided by absolute temp at the boundary.

' cQ W dEcδ δ= + I

Where 'cW W Wδ δ δ= +

Since the cyclic device is a reversible one

'

res

QT T

Qδ δ= II

Eliminating from I & II 'SQ

c resQW T dE

T cδδ = −

Let the system undergo a cycle while the cyclic device undergoes one or more cycles.

c resQW T dE

T cδδ = −∫ ∫ ∫

Page 41: All Lectures

Since net change of energy during a cycle is zero 0cdE =∫

c resQW T

= ∫

For the heat engine, based on Kelvin-Planck statement is a negative quantity and

is positive

cW resT

0QTδ

≤∫

= reversible

< irreversible

Definition of entropy change

A quantity whose cyclic integral is zero depends on the state only and not on the process

path therefore the quantity ( )revQ

Tδ must represent a property in differential form.

( )revQds

=

Units of entropy:

KJKg.K

BtuLbm.°R

Integrating both sides:

2

2 1 1( )

rev

Qs s sTδ

Δ = − = ∫

Engineers are often concerned with change in entropy.

Page 42: All Lectures

The principle of entropy increase:

0QTδ

≤∫ Clasius inequality

2 1

1 2( )

A C

revQ Q

T T0δ δ

+ ≤∫ ∫

2

1 210Q s s

+ − ≤∫

Therefore 2

2 1 1

Qs sTδ

− ≥ ∫

Or QdsTδ

≥ statement of second law of thermodynamics

Entropy generation ( 0 irreversible

) 0 reversible0 impossible

ds>⎧⎪=⎨⎪<⎩

Conclusion:

During a thermodynamic process the entropy of the universe can either increase or

remain the same (Reversible process). It is not possible for a process to occur which

involves decrease in entropy of the universe.

( ) ( ) ( )univ system surrS S SΔ = Δ + Δ

The universe is filling up with entropy

Page 43: All Lectures

1Lbm of saturated R-22 vapor at 155°F is condensed at constant pressure to saturated

liquid at 155°F by transfer of heat to surrounding air at 95°F

112.495 58.04 54.45g fBtuQ h hLbm

= − = − =

54.45( ) 0.09811(95 460)sur

BtuSLbm R

Δ = =+ °

Defined reservoir: no change in temp as heat goes into it.

( ) 0.11036 0.19895 0.08862system f gS s sΔ = − = − = −

( ) ( ) ( ) 0.0095univ system surBtuS S S

Lbm RΔ = Δ + Δ =

°

Find the highest temp of air permissible under 2nd law

ANS: must be a reversible process:

( ) ( ) 0sys surS SΔ + Δ =

54.450.08862 0460T

− ++

=

155T F= °

reversible heat transfer

Page 44: All Lectures

Entropy change of pure substances:

(1 ) f gs x s x= − + s

Numerical evaluation of entropy:

dQdsT

= for reversible process

dQ=dU+PdV 1st law in differential form for closed system

TdS dU PdV= +

For unit mass I Tds du Pdv→ = +

Units of s: Btu KJ Lbm°R Kg K

h u Pv= +

dh du Pdv vdP= + + plug in equation I

Tds dh vdP= − II

Equations I & II for open, closed, rev systems

Page 45: All Lectures

Entropy change for ideal gases:

(I) for ideal gasvRTTds c dT dvv

= +

vdT dvds C RT v

= +

2 22 1

1 1

ln lnvT vs s c RT v

− = +

(II) pRTTds c dT dpP

= − for ideal gas

pdT dPds c RT P

= −

2 22 1

1 1

ln lnPT Ps s c RT P

− = −

Page 46: All Lectures

Isentropic relations for an ideal gas:

2 22 1

1 1

ln lnvT v

s s c RT v

− = +

Since 2 1s s=

2 2

1 1

0 ln lnvT vc RT v

= +

2 2

1 1

0 ln ( ) lnv p vT vc c cT v

= + −

Divide by 2 2

1 1

0 ln ( 1) lnvT vc kT v

→ = + −

2 2

1 1

ln (1 ) lnT vkT v

= −

12 2

1 1

( ) kT vT v

−=

2 22 1

1 1

ln lnpT Ps s c RT P

− = −

2 2

1 1

0 ln ( ) lnp p vT Pc c cT P

= − −

Divide by vc 2 2

1 1

0 ln ( 1) lnT Pk kT P

= − −

2 2

1 1

1ln lnT PkT k

−=

P

12 2

1 1

( )kkT P

T P

=

Page 47: All Lectures

kPv c=

1 1 2 2k kPv P v=

2 1

1 2

( )kP vP v=

Isentropic process = reversible + adiabatic

Enables us to compare the actual performance of engineering devices to the performance

under idealized condition.

2 1s s=

Page 48: All Lectures

Isentropic efficiencies of various engineering devices:

Turbine:

s s

w Ww W

η•

•= =

Page 49: All Lectures

Compressor & Pump:

ssw Ww W

η•

•= =

Nozzle:

222

2

( / 2)( / 2)s

VV

η =

Page 50: All Lectures

More on ideal gas model

For two states having the same specific entropy,

22 2 1 1 2 1

1

( , ) ( , ) ( ) ( ) ln ps T p s T p s T s T Rp

− = ° − ° −

Reduces to

22 1

1

0 ( ) ( ) ln ps T s T Rp

= ° − ° − (I)

The above equation involves four property values: 1 1 2 2, , , and p T p T . If any three are

known, the fourth can be determined. If, for example, the temperature at state 1 and the

pressure ratio 2 / 1p p are known, the temperature at state 2 can be determined from

22 1

1

( ) ( ) ln ps T s T Rp

° = ° + (II)

Since is known, would be obtained from the appropriate table, the value of

would be calculated, and temperature would then be determines by

interpolation. If

1T 1( )s T°

2( )s T° 2T

1p , , and are specified ad the pressure at state 2 is the unknown,

equation I would be solved to obtain

1T 2T

2 12 1

( ) ( )exp s T s Tp pR

° − °⎡ ⎤= ⎢ ⎥⎣ ⎦ (III)

These equations can be used when (or s° s°) data are known, as for the gases of Tables

A-22 and A-23

Page 51: All Lectures

Air

For the special case of air modeled as an ideal gas the above equation provides the basis

for an alternative approach for relating the temperatures and pressure at two states having

the same specific entropy. To introduce this, rewrite the equation as

[ ][ ]

22

1 1

exp ( ) /exp ( ) /

s T Rpp s T

°=

° R (IV)

The quantity [ ]exp ( ) /s T R° appearing in this expression is solely a function of

temperature, and is given the symbol ( )rp T . A tabulation of rp versus temperature for

air is provided in Tables A-22. In terms of the function pr, the last equation becomes

2 21 2

1 1

(s , air only)r

r

p p sp p

= = (V)

Where 1 2( )r rp p T= and 2 ( )r r 2p p T= . The function rp is sometimes called the relative

pressure. Observe that rp is not truly a pressure, so the name relative pressure is

misleading. Also, be careful not to confuse rp with the reduced pressure of the

compressibility diagram.

A relation between specific volumes and temperature for two states of air having

the same specific entropy can also be developed. With the ideal gas equation of state,

, the ratio of the specific volumes is /v RT p=

2 2

1 2

( )(v RT pv p RT= 1

1

) (VI)

Then, since the two states have the same specific entropy, equation V can be introduced

to give

Page 52: All Lectures

2 2

1 2

( )( )

r

r

v RT p Tv p T RT

1

1

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(VII)

The ratio / ( )rRT p T appearing on the right side of the last equation is solely a function

of temperature, and is given the symbol . Values of for air are tabulated versus

temperature in Tables A-22. In terms of the function , the last equation becomes

( )rv T rv

rv

2 21 2

1 1

( , air only)r

r

v v s sv v= = (VIII)

Where and . The function is sometimes called the relative

volume. Despite the name given to it, is not truly a volume.

1 ( )r rv v T= 1 22 ( )r rv v T= rv

( )rv T

Page 53: All Lectures

Chapter 8: Cycles

A series of processes in which a system starts at a given state and is returned to that same

state.

• Closed loop cycle: Same working fluid.

• Open loop cycle: New working fluid continuously furnished.

Since in a cycle then 0uΔ =

net netQ W=

Power cycle: Thermal efficiency can not be greater than thermal efficiency of a Carnot

cycle.

Refrigeration cycle: Coefficient of performance.

We will study the application of 1st and 2nd law to analyze the performance of power and

refrigeration cycles.

Page 54: All Lectures

Vapor power systems – Rankine cycle

turb pumpth

in

w wq

η−

=

bwr = Back work ratio

pump

turb

wbwr

w=

The back work ration is an alternative way of measuring the performance of a Rankine

cycle.

Page 55: All Lectures

Increasing Rankine cycle efficiency

1 – Lowering the condenser temp

2 – Superheating the steam (before entering turbine)

Note: The boiler and super heater taken together is called steam generator.

3 – Increasing the boiler pressure (while maintaining the peak temp)

4 – Reheat of the steam to maintain the 90% quality of better in turbine to prevent blade

damage.

Page 56: All Lectures

Principle sources of irreversibilities & losses in turbine and/or pump:

Ideal reheat cycle:

Page 57: All Lectures

Reheat cycle with turbine irreversibilities:

Page 58: All Lectures

Regenerative vapor power cycle, open feedwater heater:

Page 59: All Lectures

Chapter 9: Gas power systems

Internal combustion engine:

Net work for one cycleMean effective pressure = Displacement volume

Page 60: All Lectures

Air standard Otto cycle:

Cycle analysis:

1-2 isentropic compression

2-3 constant volume heat addition

3-4 isentropic expansion

4-1 constant volume heat rejection

Page 61: All Lectures

Air standard analysis

34122 1 3

23 413 2 4

,

,

WW u u u um m

Q Qu u u um m

= − = −

= − = −

4

1

(I)

34 123 4 2 1( ) (cycleW W W u u u u

m m m= − = − − − ) (II)

23 413 2 4 1( ) (cycleW Q Q u u u u

m m m= − = − − − ) (III)

3 2 4 1 4

3 2 3 2

( ) ( ) 1u u u u u uu u u u

η − − − −= =

− −1− (IV)

22 1

1

( ) rr r

v vv vv r

= = 1 (V)

44 3

3

( )r rvv v rvv

= = 3r (VI)

Cold air standard analysis

12 1

1 2

( )k kT V rT V

−= = 1− (VII)

1341

3 4

1( )kk

VTT V r

−−= = (VIII)

Page 62: All Lectures

Effect of compression ratio on performance of Otto cycle

By referring to the T-s diagram of the previous figure, we can conclude that the Otto

cycle thermal efficiency increases as the compression ratio increase. An increase in the

compression ratio changes the cycle from 1-2-3-4-1 to 1-2’-3’4-1. Since the average

temperature of heat addition is greater in the latter cycle and both cycles have the same

heat rejection process, cycle 1-2’-3’-4-1 would have the greater thermal efficiency. The

increase in thermal efficiency with compression ratio is also brought out simply by the

following development on a cold air-standard basis. For constant , Eq. IV becomes vc

4 1

3 2

(1( )

v

v

c T Tc T T

η )−= −

− (IX)

On rearrangement

1 4 1

2 3 2

/ 11 (/ 1

T T TT T T

η )−= −

− (X)

Knowing 14 4

3 3

( ) kT vT v

−= and 11 1

2 2

( ) kT vT v

−=

Since and therefore the right hand sides are equal so the left hand sides

must be equal.

4v v= 1 23v v=

4

3 2

T TT T

= 1 or 34

1 2

TTT T

=

1

2

1 TT

η = − (XI)

Finally, introducing Eq. VII

1

11 krη −= − (XII)

Page 63: All Lectures

Air Standard Diesel Cycle

Cycle analysis:

1-2 isentropic compression

2-3 constant pressure heat addition

3-4 isentropic expansion

4-1 constant volume heat rejection

In the diesel cycle the heat addition takes place at constant pressure. Accordingly,

Process 2-3 involves both work and heat. The work is given by

3232 3 22(W pdv p v v

m= = −∫ ) (XIII)

Page 64: All Lectures

The heat added in Process 2-3 can be found by applying the closed system energy

balance

3 2 23 23( )m u u Q W− = −

Introducing Eq. XIII and solving for the heat transfer

233 2 3 2 3 3 2 2 3( ) ( ) ( ) ( )Q u u p v v u pv u pv h h

m= − + − = + − + = − 2 (IXV)

Where the specific enthalpy is introduced to simplify the expression. As in the Otto

cycle, the heat rejected in process 4-1 is given by

414 1

Q u um

= −

The thermal efficiency is the ration of the net work of the cycle to the heat added

41 4 1

23 23 3 2

/ /1 1/ /

cycleW m Q m u uQ m Q m h h

η −= = − = −

− (XV)

As for the Otto cycle, the thermal efficiency from Eq. XV requires values for

or equivalently the temperatures at the principle states of the cycle. Let

us consider next how these temperatures are evaluated. For a given initial temperature

and compression ratio r , the temperature at state 2 can be found using the following

isentropic relationship and data

1 4 2 3, , , and u u h h

1T

rv

32 2

2r c

Vv T rV

= = 2T

1

Where , called the cutoff ratio, has been introduced. Since , the volume

ratio for the isentropic process 3-4 can be expressed as

3 2/cr v v= 4V V=

4 4 2 1 2

3 2 3 2 3 c

V V V V V rV V V V V r

= = = (XVI)

Page 65: All Lectures

Where the compression ratio and cutoff ratio have been introduces for conciseness.

Using Eq. XVI together with at , the temperature can be determines by

interpolation once is found from the isentropic relationship

r cr

3rv 3T 4T

4rv

44 3

3r r

c

V rv vV r

= = 3rv

In a cold air-standard analysis, the appropriate expression for evaluating is provided

by

2T

12 1

1 2

( )k kT V rT V

1− −= = (constant k)

The temperature is found similarly from 4T

1 134

3 4

( ) ( )k kc

r

V rTT V r

− −= =

Where Eq. XVI has been used to replace the volume ratio.

Page 66: All Lectures

Effect of compression ratio on performance

1

111( 1)

kc

kc

rr k r

η −

⎡ ⎤−= − ⎢ −⎣ ⎦

⎥ (constant k) (XVII)

Page 67: All Lectures

Air Standard Brayton Cycle (Gas Turbine Cycle)

Cycle analysis:

1-2 isentropic compression

2-3 constant pressure heat addition

3-4 isentropic expansion

4-1 constant pressure heat rejection

Air standard analysis

3tW h h

m

• = − 4 (XVIII)

2cW h h

m

• = − 1 (IXX)

3inQ h h

m

• = − 2 (XX)

Page 68: All Lectures

4outQ h h

m

• = − 1 (XXI)

3 4 2 1

3 2

( ) (/ /

/

t c

in

h h h hW m W mh hQ m

η• • • •

• •

− − −−= =

−) (XXII)

22 1

1r r

pp pp

= (XXIII)

44 3 3

3 2r r r

pp p pp p

= = 1p (XXIV)

Cold air standard analysis:

14 4

3 3

( )kkT P

T P

=

11 1

2 2

( )kkT P

T P

=

Page 69: All Lectures

Effect of compressor pressure ratio on the performance of Brayton cycle

3 4 2 1 4 1

3 2 3 2

( ) ( ) ( )1( ) ( )

p p

p

c T T c T T T Tc T T T T

η− − − −

= =− −

1 4 1

2 3 2

/ 11 (/ 1

T T TT T T

η −= −

−)

Knowing 1

4 4

3 3

( )kkT P

T P

= and 1

1 1

2 2

( )kkT P

T P

=

Since therefore right hand sides are equal and the left hand side must be equal. 4

3 2

P PP P==

1

4

3 2

T TT T

= 1 or 34

1 2

TTT T

=

1

2

1 TT

η = −

( 1) /2 1

11( / ) k kp p

η −= − (k constant) (XXV)

Page 70: All Lectures

Comparison with Rankine Cycle

The working fluid of Brayton cycle is a gas while the Rankine cycle involves a phase

change (condenser, evap). The pump work is only a small fraction of turbine work,

however in Brayton cycle the compressor work is a large fraction of turbine work this is

why the Rankine cycle has better efficiency. But Brayton cycle is used for Jet engines

because it has a better thrust to weight ratio than Rankine cycle.

T-s Diagram of Ideal Rankine and Brayton Cycles

Page 71: All Lectures

The Actual Cycle

The deviation of ideal cycle is due to the irreversibilities in turbine and/or compressor.

Effect of Component Efficiencies on Brayton Cycle Performance (Pressure Ratio = 5).

Page 72: All Lectures

Brayton Cycle With Regenerator

The gases at turbine exit are relatively high (500-600°C). These hot gases are discharged

and not used and represent losses. It is good to recover some of this energy. Such heat

recovery can be accomplished by the use of a ‘regenerator’ or heat exchanger

Hot gases from turbine exhaust are used to heat the cooler air before it enters the

combustion chamber. Therefore less heat (fuel) needs to be supplied in the combustion

chamber to reach a certain max turbine inlet temp.

Disadvantage: Additional ducting and space (extra weight) “not practical for jet engine”.

Effectiveness of a heat exchanger (Regenerator)

2

max 4 2

Actual enthalpy increase of the air flowing through the compressor side of regeneratorMaximum theoretical enthalpy increase

actualq hx heq h h

−= = =

2

4 2

'( )''( )

p x

p

c T Te

c T T−

=−

2

4 2

if ' '' (constant specific heat)p p

x

c c

T TeT T

=

−=

Page 73: All Lectures
Page 74: All Lectures

Example 9.7: Brayton Cycle with Regeneration

Air enters the compressor of an ideal Brayton cycle with regenerator at 100 kPa, 300 K

with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. The turbine

inlet temperature is 1400 K. The regenerator effectiveness is 80%. Determine the thermal

efficiency.

Solution

Known: A regenerative gas turbine operates with air as the working fluid. The

compressor inlet state, turbine inlet temperature, and compressor pressure ratio are

known.

Find: For a regenerator effectiveness of 80%, determine the thermal efficiency. Also

plot the thermal efficiency versus the regenerator effectiveness ranging from 0 to 80%.

Schematic and Given Data:

Page 75: All Lectures

Assumptions:

1. Each component is analyzed as a control volume at steady state. The control

volumes are shown on the accompanying sketch by dashed lines.

2. The compressor and turbine processes are isentropic.

3. There are no pressure drops for flow through the heat exchangers.

4. The regenerator effectiveness is 80% in part (a).

5. Kinetic and potential energy effects are negligible.

6. The working fluid is air modeled as an ideal gas.

Page 76: All Lectures

Analysis:

(a) The specific enthalpy values at the numbered states on the T-s diagram are the

same as those in Example 9.4: kg1 300.19 /h kJ= , kg2 579,9 /h kJ= ,

kg , kg . 3 1515.4 /h kJ= =4 808.5 /h kJ

To find the specific enthalpy xh , the regenerator effectiveness is used as follows: By

definition

2

4 2

xreg

h hh h

η −=

Solving for xh ,

4 2 2( )

(0.8)(808.5 579.9) 579.9 762.8 /x regh h h h

kJ kg

η= − +

= − + =

With the specific enthalpy values determined above, the thermal efficiency is

3 4 2 1

3

( ) (( / ) ( / )( )( / )

(1515.4 808.5) (579.9 300.19)(1515.4 762.8)

0.568(56.8%)

t c

xin

h h h hW m W mh hQ m

η• • • •

• •

− − −−= =

− − −=

−=

)

Page 77: All Lectures

Gas Turbines for Jet Propulsion:

Gas turbines are particularly suited for aircraft propulsion because of their favorable

power-to-eight ratios. The turbojet engine is commonly used for this purpose. As

illustrated in the figure below, this type of engine consists of three main sections: the

diffuser, the gas generator, and the nozzle. The diffuser placed before the compressor

decelerates the incoming air relative to the engine. A pressure rise known as the ram

effect is associated with this deceleration. The gas generator section consists of a

compressor, combustor, and turbine, with the same functions as the corresponding

components of a stationary gas turbine power plant. In a turbojet engine, the turbine

power output need only be sufficient to drive the compressor and auxiliary equipment,

however. The gases leave the turbine at a pressure significantly greater than atmospheric

and expand through the nozzle to a high velocity before being discharged to the

surroundings. The overall change in the velocity of the gases relative to the engine gives

rise to propulsive force, or thrust.

Page 78: All Lectures

Turbojet engine with afterburner:

Some turbojets are equipped with an afterburner, as show below. This is essentially a

reheat device in which additional fuel is injected into the gas exiting the turbine and

burned, producing a higher temperature at the nozzle inlet than would be achieved

otherwise. As a consequence, a greater nozzle exit velocity is attained, resulting in

increased thrust.

Page 79: All Lectures

Cycle analysis

The T-s diagram of the process in an ideal turbojet engine is show on the previous page.

In accordance with the assumptions of an air-standard analysis, the working fluid is air

modeled as an ideal gas. The diffuser, compressor, turbine, and nozzle processes are

isentropic, and the combustor operated as constant pressure.

Process a-1 shows the pressure rise that occurs in the diffuser as the air decelerates

isentropically through this component.

Process 1-2 is an isentropic compression.

Process 2-3 is a constant-pressure heat addition.

Process 3-4 is an isentropic expansion through the turbine during which work is

developed.

Process 4-5 is an isentropic expansion through the nozzle in which the air accelerates and

the pressure decreases.

Owing to irreversibilities in an actual engine, there would be increases in specific entropy

across the diffuser, compressor, turbine, and nozzle. In addition, there would be a

pressure drop through the combustor of an actual engine. The subject of combustion is

discussed in Chap. 13.

Page 80: All Lectures

In a typical thermodynamic analysis of a turbojet on an air-standard basis the

following quantities might be known: the velocity at the diffuser inlet, the compressor

pressure ratio, and the turbine inlet temperature. The objective of the analysis would be

to determine the velocity at the nozzle exit. Once the nozzle exit velocity is determined,

the thrust is determined by applying Newton’s second law of motion in a form suitable

for a control volume. All principles required for the thermodynamic analysis of turbojet

engines on an air standard basis have been introduces.

Redo example 9.12 pg 441 for homework.

Page 81: All Lectures

Combined Gas Turbine Vapor Power Cycle

A combined power cycle couples two power cycles such that the energy discharged by

heat transfer from one cycle is used partly or wholly as the input for the other cycle.

The stream exiting the turbine of a gas turbine is at a high temperature. One way this

energy can be used, thereby improving overall fuel utilization, is by a regenerator that

allows the turbine exhaust gas to preheat the air between the compressor and combustor,

which we already discussed.

Another method is the combined cycle. The two power cycles are coupled so that the heat

transfer to the vapor cycle is provided by the gas turbine cycle, called the Topping cycle.

Page 82: All Lectures

Reheat and Intercooling

Reheat between turbine stages and intercooling between compressor stages provide two

important advantages: The net work output is in creased, and the potential for

regeneration is enhanced. Accordingly, when reheat and intercooling are used together

with regeneration, a substantial improvement in performance can be realized. One

arrangement incorporation reheat, intercooling, and regeneration is shown. This gas

turbine has two stages of compression and two turbine stages. The accompanying T-s

diagram is drawn to indicate irreversibilities in the compressor and turbine stages. The

pressure drops that would occur as the working fluid passes through the intercooler,

regenerator, and combustors are not shown.

Page 83: All Lectures

Chapter 10: Vapor compression refrigeration systems:

Cycle analysis:

Process 1-2: Isentropic compression of the refrigerant from state 1 to the condenser

pressure at state 2.

Process 2-3: Heat transfer from the refrigerant as it flows at constant pressure through

the condenser. The refrigerant exists as a liquid at state 3.

Process 3-4: Throttling process from state 3 to a two-phase liquid-vapor mixture at

state4.

Page 84: All Lectures

Process 4-1: Heat transfer to the refrigerant as it flows at constant pressure through the

evaporator to complete the cycle.

1 4inQ h h

m

• = −

1 2cW h h

m

• = −

2 3outQ h h

m

• = −

4 3h h=

/

/in

c

Q m

W mβ

• •

• •=

Therefore:

1 4

1 2

h hh h

β −=

− Which is the coefficient of performance of the ideal cycle

Page 85: All Lectures

Actual cycle with irreversibilities:

Page 86: All Lectures

Example:

An automobile air conditioning system uses R-22 as the working fluid. The evaporator

pressure is 80 Lbf/in2, and the condenser outlet pressure is 250 Lbf/in2. The temperature

of the R-22 leaving the evaporator is 5 0F above saturation. Assume an isentropic

compressor with no changes of pressure or temperature in the lines connecting the

components of the cycle. The maximum heat load is 22,000 Btu/h; that is, 22,000 Btu/hr

must be removed to lower the air temperature of the interior of the car to a comfortable

level. Determine the compressor horsepower required and the coefficient of performance

of the cycle. Then repeat the problem for a compressor efficiency of 85 percent.

T-s and p-h diagrams of Vapor Compression Cycle with Nonisentropic Compression

Page 87: All Lectures

Solution

1 2

1 1 12

1 2 2o 2

2 2

3 2

4 3

80 37.71

Btu42.71 and 80 108.9Lbm

Btu0.2222 and 250Lbm.

Btu 150.8 and 121.3Lbm

Btu at 250 43.46Lbm

Btuand 43.46L

sat

f

LbfP T Fin

LbfT F P hin

Lbfs s PR in

T F h

Lbfh hin

h h

= ∴ = °

= ° = ∴ =

= = =

∴ = ° =

= =

= =

1 4

1 2

bmBtu65.44Lbm

Btu12.40Lbm

5.277

Btu22000 Lbmhr 336.18Btu hr65.94Lbm

1336.18 (12.40 )2545

1.638

Nonisentropic Case:

1.638 10.85

in

comp

in

comp

comp comp

comp

comp

comp

q h h

w h h

qw

m

W m w

Lbm BtuWhr Lbm

W HP

W

β

• •

= − =

= − = −

= =

= =

=

=

=

= = .927 The actual work into compressor

1.927 * 2545 4904

4.486

comp

in

comp

HP

BtuWhr

Q

= =

= =

Page 88: All Lectures

Refrigerant properties

Page 89: All Lectures

Selecting refrigerants