Download - Chap.5 thermodynamic cycles

CHAPTER 5Thermodynamic

cycles

Missam Raza Mechanical Dept LJ Polytechnic

Thermodynamic cycles Introduction

Is a series of processes which form a closed path.

The initial and the final states are coincident

Thermal engines work in a cyclic process.

A Thermal engines draws heat from a hot source and rejects heat to a cold source producing work


Thermodynamic cycles Heat Engine Power Cycles

Hot body or source

Cold body or sink

System, or heat engine

Qin

Qout

Wcycle


Thermodynamic cycles Energy analysis of cycles

4

1

3

2

For the cycle, E1 E1 = 0, or

0ΔE cycle

0 W Q ΔE cycle cyclecycle


Thermodynamic cycles formula

Q Wcycle cycle

Qcycle and Wcycle represent net amounts

which can also be represented as:

Qcycle = Wcycle


Thermodynamic cycles Carnot Cycle

The Carnot cycle is a reversible cycle that is composed of four internally reversible processes.

Two isothermal processes

Two adiabatic processesMissam Raza Mechanical Dept

LJ Polytechnic

Thermodynamic cycles The Carnot cycle for a gas might occur as visualized below.


Thermodynamic cycles Introduction

This is a Carnot cycle involving two phases -- it is still two adiabatic processes and two isothermal processes.

It is always reversible -- a Carnot cycle is reversible by definition.Missam Raza Mechanical Dept

LJ Polytechnic

Thermodynamic cycles Otto Cycle

P-V diagram (Work)

T-S diagram (Heat Transfer)


Otto cycle Performance of cycle

H

L

H

net

Q

Q1

Q

w

Thermal Efficiency:

Need to know QH and QL


Otto cycle Process

• Heat addition 2-3 QH = mCV(T3-T2)• Heat rejection 4-1 QL = mCV(T4-T1)

• or in terms of the temperature ratios

23V

14V

H

L

TTmC

TTmC1

Q

Q1

1TTT

1TTT1

Q

Q1

232

141

H

L

qout

qi

n


Otto cycle Process

• 1-2 and 3-4 are adiabatic process, using the adiabatic relations between T and V

4

3

1

2

4

3

1

3

4

1

2

1

1

2

T

T

T

T

T

T

V

V

V

V

T

T

RATIOVOLUMESAME

2

11

2

1

V

V r ;

r

11

T

T1

qout

qin


Otto cycle Process

• Heat addition 2-3 QH = mCV(T3-T2)• Heat rejection 4-1 QL = mCV(T4-T1)

• or in terms of the temperature ratios

23V

14V

H

L

TTmC

TTmC1

Q

Q1

1TTT

1TTT1

Q

Q1

232

141

H

L

qout

qi

n


Otto cycle

Cycle performance with cold air cycle assumptions

1k2

1Otto,th r

11

T

T1

This looks like the Carnot efficiency, but it is not! T1 and T2 are not constant.

What are the limitations for this expression?


THANK YOU
