tme 221 2nd law

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Termodinamika Teknik

Hukum ke Dua Termodinamika

@2014 Harjadi Gunawan

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Todays main concepts:

Understand the need for and the usefulness of the 2ndlaw Be able to write and use the entropy balance

Be able to predict the maximum possible efficiency and COP of

power and refrigeration or heat pump cycles, respectively.

Be able to provide several different expressions which explain

the 2ndLaw of Thermodynamics.

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The First Law of Thermodynamics in an energy balance and tells us the

magnitudes energy flows.

3Sec 5.1: Introducing the Second Law

2

CV V

2CV CV

dEQ W m h gz

dt

It does not say anything about the spontaneousdirection.

Can we spontaneously cool the

refrigerator by removing heat

from the environment?

Q

out

Qin

Can we spontaneously heat the house by

removing heat from the environment?

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The fridge will warm.

Qout

The house will cool.

We know intuitively, that the spontaneousdirection for heat flow is from

warm to cold.

Qin

If TA> TBTA= TB.

But, we have refrigeration and heating, so what is required to

change the direction of heat flow?

Work needs to be done to move in the non-spontaneousdirection

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The Second Law of Thermodynamics answers the following

Which direction the process will move spontaneously?

What is equilibrium?

What is the maximum efficiency?

What parameters can be changed to move closer to

the maximum efficiency?

What is temperature?

How can we measure uand h?

Since the Second Law answers all these

questions, there is not a single statement

of the second law.

Popular forms of the Second Law include:

Everything degrades to chaos

No perpetual motion machine

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7Sec 5.2: Statements of the Second Law

Clausius Statement:

It is impossible for any system to operate in such a

way that the sole result would be an energytransfer by heat from a colder to a hotter body.

Kelvin-Planck StatementIt is impossible for any system to operate in

a thermodynamic cycle and deliver a net

amount of energy by work to its surroundings

while receiving heat transfer from a single

thermal reservoir. Q Thermal Reservoir:

A body where energy exchange as

heat does not effect the temperature.

Kelvin-Planck Statement tells us that

It is not possible to convert heat completely to work.

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Entropy Statement:

It is impossible for any system to operate in a way that

entropy is destroyed.

entropywithin

system

Net entropy

transferred

into the

system

Entropy

generated

with system

[ ]= +

[ ] [ ]Note: Entropy can be generated (unlike mass)

Most processes do not operate at Ssys = 0, so generally, Ssys

sys genQST

( + 0 - ) ( + 0 - ) ( + or 0 )

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9

Energy Balance:

For a Closed Systems:

sys

sys sys

dEQ W

dt

sysdE

dt

Q W

Entropy Balance:

sys gen

QS

T

Entropy Rate Balance:

sys

gen

dS Q

dt T

sys sys sysE Q W

Energy Rate Balance:

sysdS

dt

Q

T

gen

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Process Entropy

Change

Entropy

Transfer

Entropy

Production

Possible,

Impossible, or

Indeterminate.

A > 0 0 if > 0 Possible

B < 0 if < 0 > 0 Possible

C 0 > 0 Impossible

D > 0 > 0 Possible

E 0 < 0 if > 0 Possible

F > 0 < 0 Impossible

G < 0 < 0 Possible

10

Problem 5: 3 Classify the following processes of a closed system as

possible, impossible, or indeterminate.

--------------------------------------------------------------------------------------------------------

SYS SYS

QS

T

Entropy

production

Entropy

transfer

Entropy

change

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An alternate way of looking at Entropy:Entropy, S, is a measure of the Disorderwithin a system

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12Sec 5.3: Identifying Irreversibility

Reversible process:

Both the system and surroundings can bereturned to the initial state.

One grain of sand

is removed.

Gas

Sand

The grain of sandis replaced.

Irreversible process:

System and surroundings cannot bereturned to the original state.

Gas

Sand

Gas

Sand

Gas

Rock

Removerock

Irreversibility can be

internal to the system

external to the system

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14Sec 5.4: Interpreting the Kelvin-Planck Statement

Kelvin-Planck Statement

It is impossible for any system to operate in a thermodynamic

cycle and deliver a net amount of energy by work to itssurroundings while receiving heat transfer from a single

thermal reservoir.

0cycleW< 0 : Internal irreversibility present

= 0 : No internal irreversibility

For a single reservoir:

The inequality (

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15Sec 5.6: Second Law Aspects of Power Cycles

Efficiency of a Power Cycle interacting with two reservoirs.

H

C

Hot

Cold

in

out

in

outin

in

cycle

QQ

QQ

QQ

QQQ

Q

W

111

Second Law statements for power cycles(Carnot Corollaries):

The thermal efficiency of an irreversible

power cycle is always less than the thermal

efficiency of a reversible power cycle when

each operates between the same two

thermal reservoirs.

All reversible power cycles operating

between the same two thermal reservoirs

have the same maximum thermal efficiency.

If QC 0, then 1.0 = 100%

16S 5 7 S d L A f R f i i d H P

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16Sec 5.7: Second Law Aspects of Refrigeration and Heat Pumps

COP of refrigerationand heat pumps cycles interacting with two reservoirs

CH

H

cycle

in

CH

C

cycle

out

QQ

Q

W

Q

QQ

Q

W

Q

&

As WCycle0, then and

Second Law statements for refrigeration:

The coefficient of performancee of an

irreversible refrigeration cycle is always less

than the coefficient of performance of a

reversible refrigeration cycle when each

operates between the same two thermalreservoirs.

All reversible refrigeration cycles operating

between the same two thermal reservoirs

have the same maximum coefficient of

performance.

Such an arrangement would violate the Clausius Statement.

17S 5 8 T t S l

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17Sec 5.8: Temperature Scales

The Celsius Temperature Scale:

0Cfreezing point of water

(at 1 atm)100Cboiling point of water

(at 1 atm)

The Fahrenheit Temperature Scale:

0Ffreezing point of water &NaCl solution (at 1 atm)100Faverage normal human body

temperature (at 1 atm)

It is desirable to have a scale that is notdependent on a single substance.

Phase changes of many substancesallows extension of the scale based onproperties

18S 5 8 T t S l

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18

The driving force for heat transfer is a TThis causes a heat transfer QT

Sec 5.8: Temperature Scales

HCreversibleH

C TTQ

Q,

Choose

H

C

T

T

H

C

reversibleH

C

T

T

Q

Q

Thus

The ratio of the temperatures is equal to the ratio of the

heat rejected and the heat absorbed.

Used to define the Kelvin temperature scale.

reversibleTPQ

QT

16.273

19S 5 8 T t S l

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19

Consider three heat engines , operating reversibly

Sec 5.8: Temperature Scales

CHH

C TT

Q

Q,1

ICI

C TTfQ

Q,

1

2

3

1

2

3 IHI

H

TTfQ

Q,

2

3

IC

IH

IC

IH

TTf

TTf

QQ

QQ

,

,

C

H

C

H

C

H

T

T

Tf

Tf

Q

Q

Therefore

20Sec 5 9 : Ma im m Performance

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20

With

Sec 5.9 : Maximum Performance

H

C

T

T1max

H

C

reversibleH

C

T

T

Q

Q

The Carnot efficiency For a

reversible cycle

increase TH

decrease TC

Thus, ideally we want to

maximize T= (TH - TC)

TH limited by equipment costs

(highpand high T)

To increase

21Sec 5 9 : Maximum Performance

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21Sec 5.9 : Maximum Performance

Typically, the cold reservoir is the atmosphere and large bodies of water

and thus TC= 298 K, since it will cost too much to use a refrigerated

reservoir.

What is the maximum efficiency of a power cycle operating between

TH= 745 K

and TC = 298 K

298

1 1 60%745C

H

T

T


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22

Example: (5.19) A power cycle operating at steady state receives energy by heattransfer at a rate of QHat TH=1000 K and rejects energy by heat transfer to a coldreservoir at a rate QCat TC= 300 K.

For each of the following cases, determinewhether the cycle operates reversibly,

irreversibly, or is impossible.


a) QH= 500 kW, QC=100 kW

2001 1 0.60

500

Cd

H

Q

Q

b) QH= 500 kW, Wcycle= 250 kW,

and QC= 200 kW

d) QH= 500 kW, QC= 200 kWc) Wcycle= 350 kW, QC= 150 kW

max3001 1 0.701000

C

H

T

T

100

1 1 0.80500

C

a

H

Q

Q cycle H C W Q Q

3500.70

350 150

cycle cycle

c

H cycle C

W W

Q W Q

250 500 200 300kW kW

?b

Impossible: a > max

Impossible

Reversible: c = max Irreversible: a < max

Max system efficiency:


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23

Example: (5.63) The refrigerator shown in the figure operates at

steady state with a coefficient of performance of 4.5 and a power

input of 0.8 kW. Energy is rejected from the refrigerator to the

surroundings at 20C by heat transfer from the metal coils whoseaverage surface temperature is 28C. Determine


W=0.8kW

TH= 20C= 293 K

TC= ?

(a) The rate energy is rejected, in kW

(b) The lowest theoretical temperature

within the refrigerator, in K

(c) The maximum theoretical power, in

kW, that could be developed from a

power cycle operating between the

coils and surroundings.

=4.5


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24

Example: (5.63) Determine


W=0.8kW

TH= 20C= 293 K

TC= ?

(a) The rate energy is rejected, in kW

=4.5max

cycle

in

W

Q

cycle H C W Q Q

max H cycle

cycle

Q W

W

COP of refrigeration cycle

Work Energy Balance:

C H cycleQ Q W

Combining:

max1 0.8 1 4.5 4.4H cycleQ W kW kW

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Example: (5.63) Determine


W=?

TH= 28C= 301 K

TH= 20C= 293 K

c) The maximum theoretical power, in kW, that

could be developed from a power cycle operating

between the coils and surroundings.

max 1 cycleC

H H

WT

T Q

kWQH 4.4

1 Ccycle H H

TW Q

T

Note: Such a power cycle wouldoperate between reservoir

temperatures of 20 oC and 28 oC.

Maximum Possible Power Cycle Efficiency:

Solve for possible Work done:

293

(4.4 ) 1 (4.4 ) 0.0266 0.117

301

kW kW kW

From part a) the rejected heat is

tme 221 2nd law

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