57201283 37468203 hukum termodinamika ii dan siklus carnot

8/10/2019 57201283 37468203 Hukum Termodinamika II Dan Siklus Carnot

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Hukum Termodinamika II


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Hukum pertama membahas mengenaikonservasi energi.

Hukum kedua mengatur bagaimanakemungkinan konversi.

Semua proses aktual hanya bergerak pada

satu arah Energi panas tidak dapat dikonversi

seluruhnya menjadi energi mekanik


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Kegunaan hukum termo II mengidenfikasi arah dari suatu proses

mengetahui kualitas energi (hukum Iberhubungan dengan kuantitas energi danperubahan bentuk energi)

menentukan batas toeritis unjuk kerja suatu

sistem memperkirakan kelangsungan reaksi kimia

(degree of completion of chemical reaction )


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Mesin Kalor ( heat engines )

Adalah mustahil untukmembuat sebuah mesin

kalor yang bekerja hanyadengan mengambil panasdari Source dan menyerap

seluruhnya menjadienergi


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Efesiensi Untuk semua proses cyclic, U = 0

dimana W = Q h - Q c Efesisensi termal adalah

perbandingan antara kerjayang dilakukan dengan energipanas yang diperoleh darireservoir panas ( source ).

hQW

e h

ch

QQQ

h

c

QQ

e 1


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Konsekwensi hukum termodinamika II;mustahil membuat mesin denganefesiensi 100%.

Automobile engine < 20% Diesel engine < 40% Turbine uap < 50 %


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Refrigerators

refrigerator adalahmesin kalor yangbekerja secaraterbalik


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Refrigerators Alat ini bekerja dalam

sebuah siklus, dimanakerja dilakukan untukmengambil energi panas

| Q L| dari dari reservoirtemperatur dingin

Coefficient of performance

C dari refrigeratordidefinisikan :


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Pompa kalor mentransfer panas dari media

temperatur rendah ke media temperaturtinggi

Coefficient of performance pompa kalor :

Pompa Kalor


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Proses Reversible

Proses reversible adalah proses yang dapatdibalik tanpa meninggalkan jejak pada

lingkungannya Proses irreversible adalah kebalikan dari

proses reversible Proses reversible memiliki efesiensi

maksimum yang mungkin dari suatu mesinkalor.


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The Carnot Cycle and Ideal Heat Engines: Siklus carnot adalah siklus reversible yang dikemukakan

pertama kali oleh sadi carnot tahun 1824 Mesin yang menggunakan siklus karnot disebut mesin carnot Kenyataannya tidak ada mesin kalor yang beroperasi

diantara dua reservoir panas yang efesiensinya melebihiefesiensi mesin carnot


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The Carnot Cycle Idealisasi siklus thermodinamik yang terdiri dari empat proses reversible :

Reversible isothermal expansion (1-2, T H=constant)

Reversible adiabatic expansion (2-3, Q=0, T H TL) Reversible isothermal compression (3-4, T L=constant) Reversible adiabatic compression (4-1, Q=0, T L TH)

1-2 2-33-4

4-1


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The Carnot CycleKerja dilakukan oleh gas = PdV, daerahdibawah kurva proses 1-2-3.

12

3

32

1

Kerja dilakukan padagas = PdV, daerahdibawah kurva proses 3-4-1

subtract

Net work1

2

34

dV>0 dari 1-2-3PdV>0

sejak dV


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The Carnot Principles

The efficiency of an irreversible heat engine is always lessthan the efficiency of a reversible one operating between thesame two reservoirs. hth, irrev < hth, rev

The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. ( hth, rev )A= (hth,rev)B

Both Can be demonstrated using the second law (K-P

statement and C-statement). Therefore, the Carnot heat enginedefines the maximum efficiency any practical heat engine canreach up to.


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Efficiency of a Carnot Engine


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The Carnot Principles

Thermal efficiency hth=W net /Q H=1-(Q L /Q H)=f(T L,TH) and itcan be shown that hth=1-(Q L /Q H)=1-(T L /T H). This iscalled the Carnot efficiency.

For a typical steam power plant operating betweenTH=800 K (boiler) and T L=300 K(cooling tower), themaximum achievable efficiency is 62.5%.


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Let us analyze an ideal gas undergoing a Carnot cycle between twotemperatures T H and T L.

1 to 2, isothermal expansion, DU 12 = 0

QH = Q 12 = W 12 = PdV = mRT Hln(V2/V1)

2 to 3, adiabatic expansion, Q 23 = 0(TL/TH) = (V 2/V3)k-1 (1)

3 to 4, isothermal compression, DU 34 = 0QL = Q 34 = W 34 = - mRT Lln(V4/V3)4 to 1, adiabatic compression, Q 41 = 0

(TL/TH) = (V 1/V4)k-1 (2)

From (1) & (2), (V 2/V3) = (V 1/V4) and (V 2/V1) = (V 3/V4)hth = 1-(Q L/Q H )= 1-(T L/TH) since ln(V 2/V1) = ln(V 4/V3)

It has been proven that hth = 1-(Q L/Q H )= 1-(T L/TH) for all Carnotengines since the Carnot efficiency is independent of the workingsubstance.



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Karena siklus carnot bekerja pada duadaerah isothermal dan adiabatis maka

h

c

h Q

Q

Q

W e 1

h

c

h

c

T T

QQ

h

cCarnot

T

T e 1

Ideal, Carnot efficiency:


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Carnot Efficiency

A Carnot heat engine operating between a high-temperature source at 900 K and reject heat to alow-temperature reservoir at 300 K.

(a) Determine the thermal efficiency of the engine.(b) If the temperature of the high-temperature source

is decreased incrementally, how is the thermalefficiency changes with the temperature.


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Carnot Efficiency

h

h

h

th

L

H

th H

H

th H

L

T T

K

T T

K

T T

1 1300900

0 667 66 7%

300

1300

900

1900

. .

( )

( )

( )

( )

Fixed T and lowering T

The higher the temperature, the higher the "quality"

of the energy: More work can be done

Fixed T and increasing T

L H

H L

200 400 600 800 10000

0.2

0.4

0.6

0.8

1

Temperature (TL)

E f f i c i e

n c y

TH( )TL

TL

200 400 600 800 10000

0.2

0.4

0.6

0.8

1

Temperature (TH)

E f f i c i e

n c y

Th( )T

T

Lower T H


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Carnot Efficiency Similarly, the higher the temperature of the low-temperature sink, the more difficult for a heatengine to transfer heat into it, thus, lower thermal

efficiency also. That is why low-temperaturereservoirs such as rivers and lakes are popular forthis reason.


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Carnot Efficiency

To increase the thermal efficiency of a gaspower turbine, one would like to increase the

temperature of the combustion chamber.However, that sometimes conflict with otherdesign requirements. Example: turbine bladescan not withstand the high temperature gas,thus leads to early fatigue. Solutions: bettermaterial research and/or innovative coolingdesign.


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Carnot Efficiency

Work is in general more valuable compared toheat since the work can convert to heat almost100% but not the other way around. Heatbecomes useless when it is transferred to a low-temperature source because the thermalefficiency will be very low according to hth=1-

(TL /T H). This is why there is little incentive toextract the massive thermal energy stored in theoceans and lakes.


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Refrigerator dan Pompa kalorCarnot

COP R f i i D P K l


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COP Refrigerasi Dan Pompa KalorCarnot


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Kesimpulan Hukum I dan IITermodinamika

Hukum Pertama mengatakan : U = W + q ; Anda tidak dapat memperoleh lebih banyak

energi dari sistem melebihi yang telah andaberikan pada sistem ( you cant win ) Hukum kedua mengatakan :efesiensi

57201283 37468203 hukum termodinamika ii dan siklus carnot

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