Termodinamika

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1

Effisiensi Isentropis Turbin, Nozzle, Kompresor Dan Pompa

Termodinamika

Abdi H Sby, ST, MT

2

Efisiensi isentropis meliputi perbandingan antara performance

actual dari alat dan performance yang akan dicapai dibawah

keadaan ideal untuk kondisi masuk yang sama dan tekanan

keluar yang sama. Misalnya turbin, keadaan bahan masuk turbin

dan tekanan keluar ditentukan. Heat transfer antara turbin dan

lingkungannya diabaikan seperti pengaruh energi kinetik dan

potensial. Dengan asumsi ini, kesetimbangan laju massa dan

energi berkurang pada keadaan steady, untuk memberikan kerja

yang dibangkitkan per unit massa aliran melalui Turbin :

)(21 aktualaa

cvWhh

m

W

Termodinamika

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Keadaan berlabel 2s pada gambar 1 hanya

dicapai pada batas reversible internal. Hal

ini berhubungan pada ekspansi isentropis

melalui turbin. Untuk tekanan keluar

ditentukan, entalpi spesifik h2 menurun

sehingga entropy spesifik menurun.

Karena itu harga yang dicapai semakin

kecil h2 berhubungan pada keadaan 2s

dan harga maksimum kerja turbin adalah

)(21 isentropisss

s

cvWhh

m

W

Gambar 1

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Pada ekspansi aktual melalui turbin h2a > h2s maka kerja

berkurang dari maksimum yang dibangkitkan.

Perbandingan dapat diukur dengan efisiensi isentropis

didefenisikan dengan :

s

a

teoritis

aktual

s

cv

cv

t hh

hh

W

W

m

W

m

W

21

21

Termodinamika

Abdi H Sby, ST, MT

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EXAMPLE Isentropic Efficiency of a Steam Turbine

Steam enters an adiabatic turbine steadily at 3 MPa and 400°C and leaves at 50 kPa and 100°C. If the power output of the turbine is 2 MW, determine (a) the Isentropic efficiency of the turbine and (b) the mass flow rate of the steam flowing through the turbine.

Solution Steam flows steadily in a turbine between inlet and exit states. For 'specified power output, the isentropic efficiency and the mass flow rate are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The changes in kinetic end potential energies are negligible

Analysis A sketch of the system and the T-s diagram of the process are g in Fig. 7-50.

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The enthalpies at various states are

P1 = 3 MPaT1 = 400 oC

P2 = 50 kPaT2 = 100 oC

2 MWSTEAM TURBINE

400

100

3 MPa

50 kPa

1

2

2s

s2s = s1

. .

.

T, oC

s

Actual Prosess

Isentropic Prosess

Gambar 7.50

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a) The enthalpies at various states areP1 = 3 MPa h1 = 3231,7 kJ/kg

State 1 (TABLE A-6)T1 = 400 oC s1 = 6,9235 kJ/kg

P2a = 50 kPaState 2 h2a = 2682,4 kJ/kg (TABLE A-6)

T2a = 100 oC

The exit enthalpy of the steam for the isentropic process h2s is

determined the requirement that the entropy of the steam

remain constant (s2s = s1)

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P2s = 50 kPa sf = 1,0912 kJ/kg. KState 2s (TABLE A-5)

(s2s = s1) sg = 7,5931 kJ/kg. K

Obviously, at the end of the isentropic process steam exists as saturated mixture since sf < s2s < sg. Thus we need to find the

quality at state 2s first :

And :

h2s = hf + x2s hfg = 340,5 + 0,897 (2304,7) = 2407,9 kJ/kg

897,05019,6

0912,19235,622

fg

fss s

ssx

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By substituting these enthalpy values into Eq. 7-61. the

isentropic efficie of this turbine is determined to be

b) The mass flow rate of steam through this turbine is

determined from energy balance for steady-flow systems

%7,66667,09,24077,3231

4,26827,3231

21

21 orhh

hh

s

at

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10

kg/s64,3

kJ/kg)4,26827,3231(1

kJ/s10002

)( 21

21

m

mMW

MW

hhmaW

hmaWhm

EE

aout

aout

outin

Termodinamika

Abdi H Sby, ST, MT

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Effisiensi Nozzel

Pendekatan yang sama untuk turbin dapat digunakan untuk

mengantarkan efisiensi isentropis dari nozzle yang beroperasi

pada keadaan steady. Efisiensi isentropis nozzle didefinisikan

sebagai rasio energi kinetik spesifik

actual dari gas meninggalkan nozzle terhadap energi

kinetik pada keluar yang dicapai pada ekspansi isentropis

antara keadaan inlet sama dan tekanan keluar sama :

2

2

2V a

2

2

2V s

Termodinamika

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s

aN

as

hh

hh

hh V

21

21

2

221 2

VV

s

anozzle exitnozzleatKEisentropic

exitnozzleatKEActual2

2

2

2

h1

h2a

h2s2s

2a

P1

ActualProcess

Isentropic Process

1

h

ss2s = s1

.

.

.P2

Inlet State

2

22aV

2

22sV

Dengan kesetimbangan energi untuk steady, maka

Sehingga :

Termodinamika

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EXAMPLE Effect of Efficiency on Nozzle Exit Velocity

Air at 200 kPa and 950 K enters an adiabatic nozzle at low velocity and is

discharged at a pressure of 80 kPa. If the isentropic efficiency of the

nozzle is 92 percent, determine (a) the maximum possible exit velocity, (b)

the exit temperature, and (c) the actual exit velocity of the air. Assume

constant specific heats for air.

Solution The acceleration of air in a nozzle is considered. For specified

exit pressure and isentropic efficiency, the maximum and actual exit

velocities and the exit temperature are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3

the inlet kinetic energy is negligible.

Analysis A sketch of the system and the T-s diagram of the process are

given in Fig. 7—55.

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Gambar 7-55

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The temperature of air will drop during this acceleration

process because some of its internal energy is converted to

kinetic energy. This problem can be solved accurately by

using property data from the air table. But we will assume

constant specific heats (thus sacrifice some accuracy) to

demonstrate their use. Let us guess the average temperature

of the air to be about 800 K. Then the average values of cp

and k at this anticipated average temperature are determined

from Table A—2b to be cp = 1.099 kJ/kg.K and k = 1.354.

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The exit velocity of the air will be a maximum when the process in

the nozzle involves no irreversibilities. The exit velocity in this case

is determined from the steady-flow energy equation. However, first

we need to determine the exit temperature. For the isentropic

process of an ideal gas we have:

Or :

kk

ss

P

P

T

T/)1(

1

2

1

2

KKP

PTT

kk

ss 748

kPa200

kPa80)950(

354,1/354,0/)1(

1

212

Termodinamika

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This gives an average temperature of 849 K, which is

somewhat higher than the assumed average temperature

(800 K). This result could be refined by reevaluating the k

value at 749 K and repeating the calculations, but it is not

warranted since the two average temperatures are

sufficiently close (doing so would change the temperature

by only 1.5 K, which is not significant). Now we can

determine the isentropic exit velocity of the air from the

energy balance for this isentropic steady-flow process

Termodinamika

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22

22

2

21

1s

s

outin

Vh

Vh

ee

m/s666

kJ/kg1

/sm1000K]748)[(950kJ/kg.K)2(1,099

)(2)(2

22

21,212

savgpss TTchhV

Termodinamika

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b) The actual exit temperature of the air is higher than the

isentropic exit temperature evaluated above and is

determined from

KTT

TTc

TTc

hh

hh

aa

savgp

aavgp

s

aN

764748950

95092,0

)(

)(

22

21,

21,

21

21

Termodinamika

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That is, the temperature is 16 K higher at the exit of the actual

nozzle as a result not irreversibility such as friction. It represents a

loss since this rise in temperature comes at the expense of kinetic

energy (Fig. 7-56). The actual exit velocity of air can be determined

from the definition of isentropic efficiency of a nozzle

m/s639)m/s666(92,0 2222

22

22

sNa

s

aN

VV

V

V

Termodinamika

Abdi H Sby, ST, MT

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950 KAIR

Actual Nozzle

Isentropic Nozzle

764 K, 639 m/s

748 K, 666 m/s

Termodinamika

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Effisiensi Kompresor

Bentuk efisiensi isentropis kompressor,

dengan mengacu gambar 2. Dimana

proses kompresi pada diagram Mollier.

Keadaan massa masuk dan tekanan

keluar ditentukan. Untuk heat transfer

diabaikan dengan lingkungan dan energi

potensial dan kinetik diabaikan, kerja

input per unit massa aliran melalui

kompresor :

12 hhm

Wa

cv

Gambar 2

..

.s2s = s1 s

h

h2a

h2s

h1

1

2s

2a

Isentropic Prosess

Actual Prosess

ws

wa

Inlet state

P2

P1

Termodinamika

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Dari keadaan 1 ditentukan, entalphi spesifik h1 diketahui. Dengan

demikian harga kerja input tergantung pada entalpi spesifik keluar

h2a. Persamaan diatas menunjukkan bahwa jumlah kerja input

menurun jika h2a menurun. Kerja input minimum berhubungan

terhadap semakin kecil harga yang dicapai entalpi spesifik pada

saat keluar kompresor. Kerja input minimum yang diberikan :

12 hhm

Ws

cv

Termodinamika

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Pada kompresi actual h2a > h1, maka kerja lebih besar dari pada

minimum yang diperlukan. Perbedaan ini dapat diukur dengan

efisiensi isentropis kompresor:

12

12

hh

hh

W

W

m

W

m

W

a

s

a

s

cv

s

cv

c

Termodinamika

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EXAMPLE 7-15Effect of Efficiency on Compressor Power Input

Air is compressed by an adiabatic compressor from 100 kPa and 12°C to a pressure of 800 kPa at a steady rate of 0.2 kg/s. If the isentropic efficiency of the compressor is 80 percent, determine (a) the exit temperature of air and (b) the required power input to the compressor.

Solution Air is compressed to a specified pressure at a specified rate. Fora given isentropic efficiency, the exit temperature and the power input are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The changes in kinetic and potential energies are negligible.

Analysis A sketch of the system and the T-s diagram of the process are given in Fig. 7-53.

Termodinamika

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P2 = 800 kPa

P1 = 100 kPaT1 = 285 K

..

.s2s = s1 s

T, K

T2a

T2s

2851

2s

2a

Isentropic Prosess

Actual ProsessAir

Compresorm = 0,2 kg/s

Gambar 7.53

Termodinamika

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(a) We know only one property (pressure) at the exit state, and we

need to know one more to fix the state and thus determine the exit

temperature. The property that can be determined with minimal

effort in this case is hi, since the isentropic efficiency of the

compressor is given. At the compressor inlet

T1 = 285 K h1 = 285.14 kJ/kg (Table A-17) (Pr1 = 1.1584)

The enthalpy of the air at the end of the isentropic compression

process is determined by using one of the isentropic relations of

ideal gases

Termodinamika

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And

Pr2 = 9,2672 h2s = 517,05 kJ/kg

Substituting the known quantities into the isentropic efficiency

relation, we have

Thus :

h2a = 575,03 kJ/kg T2a = 569,5 K

2672,9kPa100

kPa8001584,1

1

212

P

PPP rr

14,285

14,28505,51780,0

212

12

aa

sc hhh

hh

Termodinamika

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The required power input to the compressor is determined from

the energy ce for steady-flow devices

kW0,58

kJ/kg)14,28503,575(kg/s2,0

)( 12,

2,1

hhmW

hmWhm

EE

aina

aina

outin

Termodinamika

Abdi H Sby, ST, MT

30

Effisiensi Pompa

Sama dengan kondisi dengan kompresor :

12

12 )(

hh

PPv

W

W

aa

s

p