Hukum Kedua TermodinamikaTermodinamika

Kerja & Panas• Kerja dapat diubah hampir seluruhnya menjadi

panas, energi potensial, energi kinetik atau energi listrik. Efisiensi bisa mencapai 100% dengan menghilangkan friksi.

• Panas tidak dapat seluruhnya diubah menjadi • Panas tidak dapat seluruhnya diubah menjadi kerja, energi potensial, energi kinetik atau energi listrik. Efisiensi tidak bisa lebih dari 40%.

• Panas mengalir dari suhu tinggi ke suhu rendah, tidak pernah sebaliknya.

Gerakan dengan friksiGerakan dengan friksiReaksi kimia spontanReaksi kimia spontan

Beberapa Proses Tak Dapat Balik

............pencampuranpencampuran

Transfer panasTransfer panas

TT11 > T> T22

Q

Ekspansi Ekspansi

P1 > P2

Statements of The Second Law

• No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system.(Kelvin-Plank)– It is impossible by a cyclic process to convert the heat

absorbed by a system completely into work done by the system.

• No process is possible which consists solely into the transfer of heat from one temperature level to a higher one. (Clausius)

Hukum I dan Hukum II• Hukum I :

- Energi dapat berubah dari suatu bentuk energi menjadi bentuk energi lainnya

- Tidak bisa menjelaskan arah perubahannya dan sifat perubahannya perubahannya dan sifat perubahannya (spontan/tidak spontan)

• Hukum II :- Menjelaskan arah perubahan energi dan

sifat perubahannya, spontan/tidak spontan

Mesin Kaloradalah suatu alat

atau mesin yang menghasilkan kerja dari panas (kalor) dalam suatu proses siklis.

– Essential to all heat-engine cycles are absorption of heat into the system at a high temperature, rejection of heat to the surroundings at a lower temperature, and production of work.

– In operation, the working fluid of a heat engine absorbs heat |QH| from a hot reservoir, produces a net amount of work |W|, discard heat |QC| to a cold reservoir, and returns to its initial state.

Basic Thermodynamic Cycle of Heat Engine

HEAT SOURCE

QinWorking Substance

HEAT SINKPump

Engine W

Qout

Substance

The first law:CH QQW −=

The thermal efficiency of the engine:

CCH

Q

Q

Q

QQ

Q

W−=

−=≡ 1η

HHH QQQ

• If a thermal efficiency of 100% is not possible for heat engines, what determines the upper limits?

Refrigerator and Air Conditioner

High-temperature

SOURCE

Refrigerator

QH

Low-temperature

SINK

RefrigeratorW

QC

CnetH QWQ +=

Carnot’s Theorem

No engine operating between two reservoirs can be more efficient than a Carnot engine can be more efficient than a Carnot engine operating those same two reservoirs

Hot reservoir at TH

Eng E|W|

RefrigeratorC = refrigerator

HQ' HQHH QQ '−

Hot reservoir at TH

Cold reservoir at TC Cold reservoir at TC

WQ H −' WQH −HH QQ '−

If engine E has the greater efficiency,

( ) QQWQWQ '' −=−−−

HHHH

Q' Q and QQ'

W>>

W

For the engine/refrigerator combination, the net heat extracted from the cold reservoir is :

( ) HHHH QQWQWQ '' −=−−−The net heat delivered to the hot reservoir is also HH QQ '−Since this is in violation of statement of the second law, the original premise that engine E has a greater efficiency then the Carnot engine is false. Carnot theorem is proved.

Thus, a corollary to Carnot’s theorem state:

The thermal efficiency of a Carnot engine depends only on the temperature levels and depends only on the temperature levels and not upon the working substance of the engine

• A heat engine operating in a completely reversible manner is called a Carnot engine (N.L.S. Carnot, 1824):– Step 1: A system at the temperature of a

cold reservoir TC undergoes a reversible adiabatic process that causes its temperature to rise to that of a hot reservoir temperature to rise to that of a hot reservoir at TH.

– Step 2: The system maintains contact with the hot reservoir at TH, and undergoes a reversible isothermal process during which heat |QH| is absorbed from the hot reservoir.

– Step 3: The system undergoes a reversible adiabatic process in the opposite direction of step 1 that brings its temperature back to that of the cold reservoir at TC.

– Step 4: The system maintains contacts with the – Step 4: The system maintains contacts with the reservoir at TC, and undergoes a reversible isothermal process in the opposite direction of step 2 that returns to its initial state with rejection of heat |QC| to the cold reservoir.

The cycle traversed by an ideal gas serving as the working fluid in a Carnot engine is shown by a PV diagram.

TH

TC

b

c

P|QH|

a → b adiabatic compressionb → c isothermal expansionc → d adiabatic expansiond → a isothermal compression

For the isothermal steps b → c and d→ a

b

cHH V

VRTQ ln= and

a

dCC V

VRTQ ln=

TC

ad

V

|QC|

bV aV

For adiabatic processes a → b and c→ d

b

aT

T

V

V

V

T

dT

R

CH

C

ln=∫ andc

dT

T

V

V

V

T

dT

R

CH

C

ln=∫

C

H

ad

bc

C

H

C

H

T

T

VV

VV

T

T

Q

Q==

)/ln(

)/ln(

H

C

H

C

H T

T

Q

Q

Q

W−=−== 11η 5.0

600

3001 =−=η

Rough practical limits for η of a Carnot engine; actual heat engines are irreversible and η rarely exceed 0.35.

A central power plant, rated at 800,000 kW, generates steam at 585K and discards heat to a river at 295 K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power?

4957.0585

2951max =−=η 347.04957.07.0 =×=η

H

C

H

C

H T

T

Q

Q

Q

W−=−== 11η

347.0=η kWW 800000=

kWQC 1505500=

Entropy

• A Carnot engine:C

C

H

H

T

Q

T

Q= 0=+

C

C

H

H

T

Q

T

Q

0=+ CH dQdQ0=+

C

C

H

H

T

dQ

T

dQ

0=∫ T

dQrev

T

dQdS revt =∫=∆

T

dQS revt

• There exists a property called entropy S, which is an intrinsic property of a system, functionally related to the measurable coordinates which

∫=∆T

dQS revt

related to the measurable coordinates which characterize the system.

– If a process is reversible and adiabatic, dQrev = 0, dSt = 0. The entropy of a system is constant during a reversible adiabatic process and the process is isentropic.

– When a system undergoes an irreversible – When a system undergoes an irreversible process between two equilibrium states, the entropy change of the system ∆St is evaluated by an arbitrarily chosen reversible process. Since entropy is a state function, the entropy changes of the irreversible and reversible processes are identical.

Entropy changes of an ideal gas• For one mole of fluid undergoing a mechanically

reversible process in a closed system:PdVdQdU rev −=

VdPPdVdUdH ++=

VdPdHdQ −= VdPdHdQrev −=

dTCdH igP= PRTV /=

P

dPR

T

dTC

T

dQdS ig

Prev −==

0

ln0 P

P

T

dT

R

C

R

S T

T

igP −=∆

∫

For an ideal gas with constant heat capacities undergoing a reversible adiabatic process:

0

ln0 P

P

T

dT

R

C

R

S T

T

igP −=∆

∫

0=∆S

1

2

1

2 lnln0P

P

T

T

R

C igP −=

igPCR

P

P

T

T/

1

2

1

2

=

RCC igV

igP += ig

V

igP

CC=γ

γγ /)1(

1

2

1

2

−

=

P

P

T

T

Penyelesaian Integral pada Entropi

( )2

02

200 1

2

1ln

0

T

T

DCTBTA

T

dT

R

CT

T

igp

=

−

+

+++=∫

τ

τττ

τ

0T

T=τ

D)C,B,A,T,ICPS(T0,0

=∫T

T

igp

T

dT

R

C

Dengan Matcad:

=∫

0

ln

0

T

T

T

dTC

C

T

T

igP

S

igP

Kapasitas panas rata-rata untuk perhitungan entropi

0

ln0

T

TC

T

dTC

S

igP

T

T

igp =∫

0

20

22

00 ln

1

2

1

T

T

T

DCTBTA

R

CS

igP

=

−

+

+++=

τ

τττ

τ

D)C,B,A,T,MCPS(T0,=R

CS

igP

Dengan Matcad:

0

ln0 P

P

T

dT

R

C

R

S T

T

igP −=∆

∫

0

lnT

TC

T

dTC

S

igP

Tigp =∫

00

lnlnP

P

T

T

R

C

R

S S

igP

−=∆

00

TT ST∫

Contoh perhitungan dengan matcadmatcad

The 2nd law mathematical statement

• Let a quantity of heat |Q| be transferred from the hotter (TH) to the cooler (TC) reservoir. The entropy changes of the two reservoirs are:

tH T

QS

||−=∆ tC T

QS

||−=∆

• For the process of irreversible heat transfer, ∆Stotal is always positive, approaching zero as the process becomes reversible.

HT CT

−=∆+∆=∆CH

CHtH

tHtotal TT

TTQSSS ||

0≥∆ totalS

A 40 kg steel casting (CP= 0.5 kJ/kgK) at a temperature of 450°C is quenched in 150 kg of oil (CP= 2.5 kJ/kgK) at 25°C. If there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together?

0)25)(5.2)(150()450)(5.0)(40( =−+− TT CT o25.46=No heat losses:

K

kJ

T

TmC

T

dTCm

T

dQS P

Pt 33.16ln1

2 −====∆ ∫∫(a) the casting

K

kJ

T

TmC

T

dTCm

T

dQS P

Pt 13.26ln 2 ====∆ ∫∫(b) the oilKTTT P

1∫∫(b) the oil

(c) total entropy changeK

kJSSS t

oiltcasttotal 80.9=∆+∆=∆

Net rate of change in entropy of flowing streams

Time rate of change entropy in control volume

Time rate of change entropy in surroundings

Total rate of entropy generation

Entropy balance for open systems

0)(

)( ≥=++∆ G

tsurrcv

fs Sdt

dS

dt

mSdmS &&

0)(

)(,

≥=−+∆ ∑ Gj j

jcvfs S

T

Q

dt

mSdmS &

&

&

σ

In a steady-state flow process, 1 mol/s of air at 600 K and 1 atm is continuously mixed with 2 mol/s of air at 450K and 1 atm. The product stream is at 400K and 1 atm. Determine the rate of heat transfer and the rate of entropy generation for the process. Assume that air is an ideal gas with CP = (7/2)R, that the surroundings are at 300K, and that kinetic- and potential-energy changes are negligible.

KTs

moln

A

A

600

1

=

=&

C.V.

KTs

moln

400

3

=

=&

T

QSSnSSn

T

QSnSnSnS

BBAA

BBAAG

)()( −−+−=

−−−=

σ

σ

&

&&

&

&&&&

C.V.

KTs

moln

B

B

450

2

=

=&

Q&

sJTTCnTTCnHnHnHnQ BPBAPABBAA /7.8729)()( −=−+−=−−= &&&&&&

Ks

J

T

Q

T

TCn

T

TCn

BPB

APA

446.10

300

7.8729

500

400ln)2(

600

400ln)1()314.8)(

2

7(

lnln

=

+

+=

−+=σ

σ

&

&&

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

T’ = 200°C

kJQ 2000−=′σσ QQQQ +−=+′= 2000

sWQH +=∆

kgK

kJS

kg

kJH

3554.7

2676

1

1

=

=apparatus

kJQ 2000−=′

Tσ = 0°C

σQ

Saturated steam at 100 °C

Liquid water at 0 °C

0.0

0.0

2

2

==

S

H

σQH +−=−=∆ 200026760.0

kJQ 0.676−=σ

K

kJS 3554.73554.70.0 −=−=∆

K

kJS t 227.4

15.273200

2000 =+

=∆

K

kJS t 4748.2

15.2730

0.676 =+

=∆

K

kJStotal 6536.04748.2227.43554.7 −=++−=∆

Calculation of ideal work

• In any steady-state flow process requiring work, there is an absolute minimum amount which must be expended to accomplish the desired change of state of the fluid flowing through the control volume.

• In a process producing work, there is an absolute maximum amount which may be accomplished as the maximum amount which may be accomplished as the result of a given change of state of the fluid flowing through the control volume.

• The limiting value obtains when the change of state associated with the process is accomplished completely reversibly.

For such a process, the entropy generation is zero:

0)( =−∆σT

QmS fs

&

& fsmSTQ )( && ∆= σ )()(2

1 2 revWmSTmzguH sfs

fs

&&& +∆=

++∆ σ

fs

fs

ideal mSTmzguHW )(2

1 2&&& ∆−

++∆= σ( ) fsfsideal mSTmHW )( &&& ∆−∆= σSTHWideal ∆−∆= σ

s

idealt W

Wrequiredwork

&

&

=)(ηideal

st W

Wproducedwork

&

&

=)(η

What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen (assumed an ideal gas) at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar?

mol

JEEICPHdTCH

T

T

igP 15060)5040.0,0.0,3593.0,280.3;300,800(314.8

0

−=+−×==∆ ∫

For an ideal gas, enthalpy is independent of pressure, and its change is given:

mol

J

P

PR

T

dTCS

T

T

igP 042.3

50

0133.1ln314.815060ln

2

11

2 =−−=−=∆ ∫

JSTHW 15973)042.3)(300(15060 −=−−=∆−∆=

molSTHWideal 15973)042.3)(300(15060 −=−−=∆−∆= σ

Or, Step 1, reversible, adiabatic expansion from initial state to 1.0133 bar, T’Step 2, cooling to the final temperature

HWQ S ∆=+Step 1 )( 1HHHWS −′=∆=

)(dQT

TTdWCarnot

σ−=Step 2 STHHT

dQTQW

T

TCarnot ∆−−=−= ∫ σσ )'( 2'

2

STHWWW CarnotSideal ∆−∆=+= σTotal

An inventor claims to have devised a process which takes in only saturated steam at 100 °C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C. The inventor claims further that, for every kilogram of steam taken into the process, 2000 kJ of energy as heat is liberated at the temperature level of 200°C. Show whether or not this is possible. In order to give the inventor the benefit of any doubt, assume cooling water available in unlimited quantity at a temperature of 0°C.

T’ = 200°C

kJQ 2000−=′

267626760.0 −=−=∆H

K

kJS 3554.73554.70.0 −=−=∆

kgK

kJS

kg

kJH

3554.7

2676

1

1

=

=apparatus

kJQ 2000−=′

Tσ = 0°C

σQ

Saturated steam at 100 °C

Liquid water at 0 °C

0.0

0.0

2

2

==

S

H

K

mol

J

STHWideal

9.666

)3554.7)(15.273(2676

−=

−−−=∆−∆= σ

kJ

TT

TWQ

7.15770200

15.2732009.666

||||

=−

+=

−=

σ

Lost work

• Work that is wasted as the result of irreversibilities in a process is called lost work:

idealSlost WWW −≡

QmzguHWS&&& =

++∆= 2

2

1QmzguHW

fs

S & =

++∆=2

fs

fs

ideal mSTmzguHW )(2

1 2&&& ∆−

++∆= σ

QmSTW fslost&&& −∆= )(σ

Surrounding temperature TσσT

QmSS fsG

&

&& −∆= )(

0≥= Glost STW &&σ

0≥= Glost STW σ

The two basic types of steady-flow heat exchanger are characterized by their flow patterns: cocurrent and countercurrent. Consider the two cases, for each of which the following specifications apply:

The minimum temperature difference between the flowing streams is 10K. Assume that both streams are ideal gases with CP = (7/2)R. Find the lost work for both cases. Take Tσ = 300 K.

smolnKTKTKT HCHH 1300350400 121 ==== &

T400 K

TH

TC

TH

TC

Cocurrent Counter current

350 K

300 K

QmzguHWfs

S&&& =

++∆= 2

2

10)()( =∆+∆ CCHH HnHn &&

0)()( 1212 =−+− CCPCHHPH TTCnTTCn &&CCHHfs SnSnnS )()()( ∆+∆=∆ &&&

Negligible pressure change

+=∆

1

2

1

2 lnln)(C

C

H

C

H

HPHfs T

T

n

n

T

TCnnS

&

&&& QnSTW fslost

&&& −∆= )(σ

Case I, cocurrent: J3403507 Case I, cocurrent:

25.1300340

350400 =−−=

H

C

n

n&

& Ks

JnS fs 667.0

300

340ln25.1

400

350ln)314.8(

2

7)1()( =

+

=∆ &

s

JnSTW fslost 1.200667.0300)( =×=∆= &&

σ

Case II, countercurrent:

5556.0300390

350400 =−−=

H

C

n

n&

&

Ks

JnS fs 356.0

300

390ln5556.0

400

350ln)314.8(

2

7)1()( =

+

=∆ &

s

JnSTW fslost 7.106356.0300)( =×=∆= &&

σ

From thermodynamic point of view, the countercurrent case is much more efficient.

The third law of thermodynamics

• The absolute entropy is zero for all perfect crystalline substances at absolute zero temperature.– When the form is noncrystalline, e.g., amorphous or glassy,

calculations show that the entropy of the more random form is greater than that of the crystalline form.form is greater than that of the crystalline form.

– The absolute entropy of a gas at temperature T:

∫∫∫ +∆++∆

+=T

T

gP

v

vT

T

lP

f

fTSP

v

v

f

f

dTT

C

T

HdT

T

C

T

HdT

T

CS

)()()(0