Thermofluids 3a - Chapter 9 - 2015

ENABLING OBJECTIVESWhat do we aim to achieve in this section?

Extend 2ND Law to more general control volumes Apply 2ND law to steady state devices such as

turbines, compressors, nozzles etc.

Apply thermodynamic theory to real devices usingefficiency

Understand difference between cycle efficiency anddevice efficiency

Strengthen understanding of entropy as a measure oftransformability of heat

THERMOFLUIDS 3A11 (TMS 3A11)

LAWS OF THERMODYNAMICS

1. Zeroth Law : Equality of Temperature

2. First Law : Conservation of Energy

3. Second Law : Conversion of Energy / Feasibility

4. Third Law : Limit of Entropy

5. Fourth Law : Limit of Process



When two bodies are independently in thermal equilibrium with a third body,

then the two bodies are in thermal equilibrium with each other

ZEROTH LAW OF THERMODYNAMICS


When two bodies are independently in thermal equilibrium with a third body, then the two bodies are in thermal

equilibrium with each other

ZEROTH LAW OF THERMODYNAMICS


During any thermodynamic cycle a system undergoes, the cyclic integral of the transferred heat is proportional

to the cyclic integral of the work

WQ

1ST LAW OF THERMODYNAMICS

CONSERVATION BALANCE


Kelvin-Plank

It is impossible to construct a device that will operate in a thermodynamic cycle and produce no effect other than the

raising of a weight while exchanging heat with a single reservoir

Clausius Statement

It is impossible to construct a device that will operate in a thermodynamic cycle and produce no effect other than the

transfer of heat from a cooler body to a hotter body

2ND LAW OF THERMODYNAMICS


Implications of Kelvin-Plank

A perpetual motion machine of the second kind is not possible! Impossible to convert 100% heat into work!

Practical conversion of thermal energy into work requires at least two heat reservoirs.

Amount of work extracted from such a heat engine is given by:

0QQWLH

From this we note that:

0Q



Efficiency of such device is given by:

H

L

H

LH

Q

Q1

Q

QQ

From the definition of thermodynamic temperature, we note that:

H

L

H

L

T

T

Q

Q

Hence thermal efficiency of a heat engine can be re-expressed as:

H

L

T

T1



For a heat pump (refrigerator) the coefficient of performance is:

L

HH L

L

Q 1

TQ Q1

T

From the definition of thermodynamic temperature, we also note that:

constantT

Q

T

Q

H

H

L

L

Hence for any reversible thermodynamic cycle:

0constantT

Q

T

Q

L

L

H

HQ T



Implications of Clausius Statement

In the case of irreversible cycles where QH is the same as that in a reversible cycle, QLirr > QLrev, hence:

Thus in general we would have:

This equation is known as the Clausius Inequality

0T

Q

T

Q

L

Lirr

H

HQ irrT

0Q T



From this analysis it is clear that for a control mass

undergoing a process:

is independent of path taken and depends only on the end

states of the system. It follows that this parameter is a change

of property of the system. This property is termed ENTROPY

with symbol S.

For irreversible processes

TQ

revT

QdS

irrT

QdS



Thus in general we have:

In a process that changes state of the system from state 1 to

state 2, the change in entropy is given as the integral:

This inequality implies entropy generation for irreversible

processes i.e.:

2

1 revT

Q

12SS

TQdS

genT

QSdS

where is the generated entropy which is always

positive.gen

S




What causes generation of entropy?

(loss of potential to do work, level of chaos)

Internal friction during the process

External friction during the process

Unrestrained expansion

Finite temperature differences during heat transfer

And other causes

3RD LAW OF THERMODYNAMICS


The entropy of a pure crystal is zero at zero absolute temperature

4TH LAW OF THERMODYNAMICS


It is not possible to design a Carnot Engine or any other physical heat engine, whose source has a positive

(absolute) temperature and sink has a negative (absolute) temperature

2ND LAW OF CONTROL VOLUME


Control Volume Configuration

Targets processes that are accompanied by flow of mass.



Control Volume Configuration

Typical examples of control volumes.



Control Volume Analysis

We recall the control mass entropy relation presented in Chap. 8 as:

gencm ST

QS

Because control volume includes mass flow, this eqn. can be modified to:

gencm ST

Q

dt

dS

where

dt

dSSand

dt

dQQ

gen

gen




Special note:

1. The masses flowing in and out of the control volume carry an amount

of entropy and do not result in any other effect

2. Although processes will take place in the masses, these are confined

to either inside or outside the control volume

3. Processes that occur inside the control volume will lead to a change

in entropy (increase or decrease)

4. Taking these into account, the entropy balance equation can be

expressed as:

Rate of change of entropy = + in out + generation

(i.e. entropy IN less entropy OUT plus entropy generation)




Thus we have:

geneeiicv ST

Qsmsm

dt

dS

where:

sec)/(

)(

sec)/(

)/(s

(kg/sec)outletatflowmassratem

)/(s

(kg/sec)inletatflowmassratem

e

e

i

i

KkgkJvolumecontrolinsidegenerationentropyofrateS

KboundaryatrightsourceheatofetemperaturT

kJvolumecontrolthetotransferheatofrateQ

KkgkJexitatmassofentropy

KkgkJentryatmassofentropy

gen

cv




Since the entropy term is always positive, we have

where equality applies to reversible processes while the inequality

applies to irreversible processes.

NB: if mi = me = 0, the equation reverts back to the control mass eqn.

T

Qesemisim

dt

cvdS



Control volume remains stationary with respect to reference coordinate system

State of mass at each point in the control volume does not change with respect to time

Rate at which heat and work cross the control volume boundary does not change

Mass flux and state of mass at the entry and discharge points of the control volume do not vary with time

The steady State Process




For the steady state case, there is no change of entropy per unit

mass with time at any point within the control volume, thus:

Hence for the steady state process from eqn.

where all time rates do not vary with time. For common systems

e.g. turbines and compressors, mass flux at inlet is equal to mass

flux at outlet i.e.

0dt

dScv

gencv

iiee ST

Qsmsm

mmm ie

geneeiicv S

T

Qsmsm

dt

dS

we have




It follows that the entropy balance equation reduces to:

Dividing through by m gives the equation in terms of intensive

properties i.e.

gencv

ie SdT

Qssm

genie sdT

qss

implying that

genie sT

qss

and hence

iie sT

qss




Governing equations:

ei mm

1. Continuity equation

2. Energy balance

cve

2e

eei

2i

iicv WgZ2

VhmgZ

2

VhmQ

3. Entropy balance

iie sT

qss



Control Volume Analysis: Example

The first stage of a turbine receives superheated steam at 10MPa

and 900oC with an exit pressure of 600 kPa. Assume that this

process is reversible and adiabatic. Neglecting kinetic and potential

energy changes, determine the exit temperature and the specific

work.

Model Representation

i

e

Tw

T

s

i

e

10 MPa

600 kPa




Solution

From continuity we have

ei

i

sKkJ/kg7.6272s

kJ/kg4361.24h

ei mm

From energy balance (first law) we have

eiT hhw

For a reversible adiabatic process, there is no heat transfer and

hence no entropy generation, therefore

ie ss

For inlet state




For exit state we have:

kJ/kg1143.28

kJ/kg3217.964361.24wT

kJ/kg3217.96h

C375T

KkJ/kg7.6272ss

e

oe

ie

Specific work output:



Transient Process

For a transient process, the following assumptions are made:

Control volume remains constant with respect to the reference coordinates

State of mass will change with time but remains uniform throughout the control volume at any time instant

State of mass entering or leaving the control volume remains constant with time although the flow rates may vary with time.



Transient Process

Given these assumptions, the 2ND Law for control volume under

transient conditions can be expressed as:

gencv

eeiicv ST

Qsmsmsm

dt

d

Integrating over time we have:

gen21

t

o gen

ee

t

0 ee

ii

t

0 ii

cv1122

t

0 cv

SdtS

smdtsm

smdtsm

smsmdtsmdt

d



Transient Process

Putting these integrals together we have:

gen21

t

ocv

cveeiicv1122

SdtT

Qsmsmsmsm

Noting that T is constant over the whole control volume at any

instant, we have:

gen21

t

o

cveeiicv1122

SdtT

Qsmsmsmsm



Transient Process: EXAMPLE

Steam at a pressure of 1.4 MPa and temperature of 300oC is flowing in a

pipe as shown in the Figure below. Connected to this pipe through a

valve is an evacuated tank. The valve is opened and the tank fills with

steam until a pressure of 1.4 MPa is attained and the valve is closed.

The process takes place adiabatically and kinetic and potential energy

changes are negligible. Calculate the specific entropy generated in the

filling tank. T

s

2

i




Solution

Because the process is adiabatic, heat in and work out are equal to

zero, i.e.

0,0 2121 WQ

From continuity of mass flow we have

i21 mmhence0m

Energy balance (1ST Law) gives us

i2ii22 hui.e.hmum

And entropy balance (2ND Law) gives us

genS21ii22 smsm




Solution

Inlet State: Pi = 1.4 MPa, Ti = 300oC, hi = 3040.4 kJ/kg,

si = 6.9533 kJ/kg-K

Final State : P2 = 1.4 MPa, u2 = hi = 3040.4 kJ/kg

T2 = 452oC, s2 = 7.4590 kJ/kg-K

Therefore, specific entropy generated is

KkJ/kg0.5057

KkJ/kg6.9533)(7.4590

s-ss i22gen1



The Reversible Steady State Process

Adiabatic

Certain special relations arise when we consider the reversible

process. For reversible adiabatic process

Hence from the property relation

and if ds = 0, it follows that for an adiabatic process we have

ie ss

dPvdhdsT

e

iih dPvhe

Substituting into the first law equation to determine work done we get

eieie

iZZg

VV

2dPvw

22




Isothermal

For reversible isothermal process, entropy balance equation (2ND Law)

is:

From this, the property equation can be integrated to yield:

Substituting these two equations into the energy balance eqn. yields:

T

qss ie

e

iieihhs dPvsT e

which is the same as that obtained for the adiabatic process.

eieie

iZZg

VV

2dPvw

22




GeneralBecause any reversible process can be represented by a series of

alternating adiabatic and isothermal processes as shown below, it

follows that the derived equations apply to all reversible processes.

T

s

2

i Adiabatic

Isothermal

eieie

iZZg

VV

2dPvw

22




ApplicationsIf we consider the case where changes in kinetic and potential

energies are negligible, we have:

e

idPvw

This equation shows the dependence of work on specific volume.

For a workless process and incompressible fluid, we have:

02

)(

22

eiei

ie ZZgVV

PPv

This is the very important fluid mechanics equation called the Bernoulli

equation.




Applications

For polytropic processes, if

e

idPvw

and

then

constantvP n

ie TT1n

Rnw



Principle of Increase of Entropy

We have already established that for a control volume, the entropy

balance equation is:

For the surroundings of the control volume, we have:

and hence we have:

T

Qsmsm

dt

dSiiee

cv

o

cviiee

surr

T

Qsmsm

dt

dS

dt

dS

dt

dS

dt

dS surrcvnet



Principle of Increase of Entropy

That is:

and because T0>T, it follows that:

This is known as the principle of increase of entropy.

o

cv

cv

cvnet

T

Q

T

Q

dt

dS

0Sdt

dS

dt

dS

dt

dSgen

surrcvnet



Process Efficiency

According to the second law, the thermal efficiency of a heat engine is

given by

where Wnet = net work done

QH = Heat transfer from the high temp. reservoir

At this stage, it is also important to consider process efficiency, i.e.

efficiency of the individual processes that make up the cycle.

H

netth

Q

W

This implies comparing the efficiency of the device under given

conditions to the performance in an ideal process under the same

conditions.



Turbine Efficiency



Turbine Efficiency

T

s

Pi

Pe

ws

wactuali

ea

es



Turbine Efficiency

T

s

Pi

Pe

ws

wactuali

ea

es

Assuming adiabatic process

esi

ei

hh

hh

s

aturbine

W

W

Typical turbine efficiency lies

in the range 0.7 to 0.88



Compressor Efficiency

The purpose of a compressor is to

increase the pressure of a fluid.



Compressor Efficiency

Again assume adiabatic process

ei

esi

a

scompressor

hh

hh

W

W

Typical compressor efficiencies also

lie in the range 0.7 to 0.88

T

s

Pi

Pe

ws

wactual

i

es

e



Nozzle Efficiency

Purpose of a nozzle is to

produce high velocity fluid

stream.



Nozzle Efficiency

Purpose of a nozzle is to produce high velocity fluid stream.

2es

2e

nozzleV

V

where Ve is the actual exit

velocity while Ves is the ideal

exit velocity of the nozzle.

Typical nozzle efficiencies also

lie in the range 0.9 to 0.97

T

s

Pi

Pe

i

es

e

As in the previous two cases

nozzles are also considered to

be adiabatic.



Efficiency: EXAMPLEA compressor is used to bring saturated water vapor at 1MPa to 17.5 MPa where

the actual exit temperature is 650oC. Find the isentropic compressor efficiency

and the entropy generation.

Solution

T

s

Pi = 1MPa

Pe = 17.5 MPa

ws

wactual

i

e

es

T=650oC

ei

esi

a

scompressor

hh

hh

W

W



Efficiency: EXAMPLE

Inlet State: Pi=1MPa, Ti=179.91oC, si=6.5864 kJ/kg-K, hi=2778.08

kJ/kg Table B.1.2 p.708

Solution

Ideal Exit State: Pes=17.5MPa, ses=si=6.5864 kJ/kg-K, hes=3560.1 kJ/kg

Table B.1.3 p.715

Actual Exit State: Pe=17.5MPa, Te=650oC, se= 6.7357 kJ/kg-K,

he=3693.9 kJ/kg Table B.1.3 p.715

Therefore ideal work ws = hes hi = (3560.1 2778.08) kJ/kg = 782.02 kJ/kg

And the actual work wa = he hi = (3693.9 2778.08) kJ/kg = 915.82 kJ/kg

Thus compressor efficiency

%4.8582.915

02.782

W

W

a

scompressor



Efficiency: EXAMPLE

Solution

Entropy generation:

K-kJ/kg 0.1492

KkJ/kg6.5865)(6.7357

sss iegen

Thermofluids 3a - Chapter 9 - 2015

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Transcript of Thermofluids 3a - Chapter 9 - 2015