Quality point of view - In terms of “Entropy”

1

VS 2

Laws of Thermodynamics

•

Direction of a process

•

Quality point of view - In terms of “Entropy”

•

Entropy generation always increases

- If not, it violates 2

law of thermodynamics

•

A process increase Entropy high à Irreversibility high

•

A process increase Entropy low à Irreversibility low

•

Hot à Cold (O); Cold à Hot (X)

•

Energy conservation

•

Quantity point of view - In terms of “Energy”

•

Energy cannot be created or destroyed, but it always

conserves

- If not, it violates 1

law of thermodynamics

•

Energy input – Energy output = Energy stored

•

100% output w/o any loss

The first Laws The second Laws

1 ^st law of Thermodynamics

Control mass (Closed System)

Control volume (Opened System)

W Q U KE P E

D + D = D + D + D

W Q KE PE

U H

E E D

D

D + D +D = + D + D

D + D = + D + D D

If your system is a stationary system

U W Q

D + D = D W Q E

(

) U

H W

E Q

D + D + =

D

D + D

D

= D

D

E_mass=PV

Moving boundary (closed System)

Some Remarks about Entropy

1. Processes can occur in a certain direction only, not in anydirection. A process must proceed in the direction that complies with the increase of entropy

principle, that is, S_gen≥0.

2. Entropy is non-conserved property, and there is no such thing as the conservation of entropy principle.

3. The performance of engineering systems is degraded by the presence of irreversibility, and entropy generationis a measure of the magnitudes of the irreversibility's present during that process

Week 1. Gas Power Cycles I

Objectives

1. Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle

2. Develop simplifying assumptions applicable to gas power cycles 3. Discuss both approximate and exact analysis of gas power cycles 4. Review the operation of reciprocating engines

5. Solve problems based on the Otto, Diesel, Stirling, and Ericsson cycles 6. Solve problems based on the Brayton cycle; the Brayton cycle with

regeneration; and the Brayton cycle with intercooling, reheating, and regeneration

7. Analyze jet-propulsion cycles

8. Identify simplifying assumptions for second-law analysis of gas power cycles

9. Perform second-law analysis of gas power cycles

Combustor And Cycle

Chemical Energy Thermal Energy Mechanical Energy Fuel

Low Heat Value

Combustion Mechanical linkage

Heat Power Output

Temperature rise Pressure rise

Rotational torque

Rankine Cycle Stirling Cycle

Otto Cycle

Diesel Cycle Brayton Cycle

Jet-propulsion Cycle Combustor

External Combustor

Internal Combustor

Stirling Engine Steam Engine

Diesel Engine

(Compression-ignition) Gasoline Engine

(Spark-ignition)

Gas Turbine Jet Engine

Basic Considerations in the Analysis of Power Cycles

The analysis of many complex processes can be reduced to a manageable level by utilizing some idealizations

The idealizations and simplifications commonly employed in the analysis of power cycles can be summarized as follows:

1. The cycle does not involve any friction. Therefore, the working fluid does not experience any pressure drop as it flows in pipes or devices such as heat exchangers.

2. All expansion and compression processes take place in a quasi-equilibrium manner. 3. The pipes connecting the various components of a system are well insulated,

and heat transfer through them is negligible.

4. Neglecting the changes in kinetic and potential energies of the working fluid is another commonly utilized simplification in the analysis of power cycles.

The Carnot Cycle And Its Value in Engineering

• The Carnot cycle is the most efficient cycle that can be executed between a heat source and a sink

• It is composed of four totally reversible processes:

Isothermal heat addition, Isentropic expansion,

Isothermal heat rejection, and Isentropic compression

• Its thermal efficiency

•Example 9-1(show that the thermal efficiency of a Carnot cycle operating between the temperature limits of T_H and T_Lis solely a function of these two temperature)

• The real value of the Carnot cycle comes from it being a standard against which the actual or the ideal cycles can be compared

P-v and T-s diagrams of a Carnot cycle

H L

T - T

Carnot

= 1

h

Air-Standard Assumptions

• The working fluid is air, which

continuously circulates in a closed loop and always behaves as an ideal gas

• All the processes that make up the cycle are internally reversible

• The combustion process is replaced by a heat-addition process from an external source

• The exhaust process is replaced by a heat- rejection process that restores the working fluid to its initial state

• Air has constant specific heats whose

values are determined at room temperature (25^oC) ß cold air standard assumption

Entropy Change of Ideal Gases

T vdP T

dh

T Pdv T

ds du

-

=

+

=

dT c

dh

dT c

du

RT Pv

p v

=

=

=

v

p

dT dv

ds c R

T v

dT dP

c R

T P

= +

= -

2 2 1

1 2 2

1 1 2

ln )

(

ln )

(

P R P

T T dT c

v R v

T T dT c

s s

p v

-

=

+

= -

ò ò

The differential entropy change of an ideal gas

The entropy change for a process obtained by integrating

Constant Specific Heats (Approximate Analysis)

2 2

2 1 avg

1 1

2 2

,avg

1 1

, ln ln

ln ln (kJ/kg K)

v

p

T v

s s c R

T v

T P

c R

T P

- = +

= - ×

2 2

2 1 avg

1 1

2 2

,avg

1 1

, ln ln

ln ln (kJ/kmol K)

v u

p u

T v

s s c R

T v

T P

c R

T P

- = +

= - ×

Entropy changes can also be expressed on a unit mole basis

The entropy change relations for ideal gases under the constant specific heat assumption

Isentropic Processes of Ideal Gases (Approximate Analysis)

cv

R

v

v

v T

T v

v c

R T

T

÷÷ ø ö çç è

= æ Þ

-

=

2 1 1

2 1

2 1

2

ln ln ln

ln

1

, = Þ = -

-

= k

Rc c

k c c c R

v v

p v

p

1

2 1

1 const. 2

(ideal gas)

k

s

T v

T v

-

=

æ ö æ ö

ç ÷ =ç ÷

è ø è ø

1^stisentropic relation

2 2

2 1 avg

1 1

2 2

,avg

1 1

, ln ln

ln ln

v

p

T v

s s c R

T v

T P

c R

T P

- = +

= -

( ¹)

2 2

1 const. 1

(ideal gas)

k k

s

T P

T P

-

=

æ ö æ ö

ç ÷ = ç ÷

è ø è ø

2^nd isentropic relation

2 1

1 const. 2

(ideal gas)

k

s

P v

P ₌ v

æ ö æ ö

ç ÷ = ç ÷

è ø è ø

3^rd isentropic relation

1 1

constant

constant (ideal gas) constant

k

( -k) k k

Tv TP

Pv

-

=

=

=

Compact forms

Variable Specific Heats (Exact Analysis)

K) (kJ/kg

ln

1 2 1

2 1

2

- = - - ×

P R P

s s

s

s

It is expressed on a unit-mole basis

The entropy change relations for ideal gases under the variable specific heat assumption

ò

=

T T dT c

s

( )

o o

1 2

2

1

( ) s s

T T dT

c

= -

ò

K) (kJ/kmol

ln

1 2 1

2 1

2

- = - - ×

P R P

s s

s

s

Isentropic Processes of Ideal Gases (Exact Analysis I)

0 ln

1 2 1

2 1

2

- = - - =

P R P

s s s

s

) exp(s R P_r = ^o

1 2 1

2

ln

P R P

s s^o

=

+

÷ ø ç ö

è æ

÷ ø ç ö

è æ - =

=

R s

R s R

s s P

P

o o o

o

1 2 1

2 1

2

exp exp exp

Relative pressure

1 2 const.

1 2

r r

s P

P P

P

÷÷ = ø

ö çç è

æ

=

2 1 1 2 1

2 2

2 2 1

1 1

P P T T v

v T

v P T

v

P = ® =

1 1

2 2

2 1 1

2 1

2

r r r

r

P T

P T P

P T T v

v = =

1 2 const

1 2

r r

s

v

v v

v ÷÷ = ø

ö çç è æ

=

Relative specific volume v_r=T/P_r

• Strictly valid for isentropic processes of ideal gases only

• The values of P_r and v_r are listed for air in Table A-17

Isentropic Processes of Ideal Gases (Exact Analysis II)

• The values of P_r and v_r listed for air in Table A-17 are used for calculating the final temperature during an isentropic process