4a_Proses Politropik
Transcript of 4a_Proses Politropik
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Termodinamika
Abdi H Sby, ST, MT 11
PROSES
POLITROPIC
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Termodinamika
Abdi H Sby, ST, MT 22
Proses Politropik
Merupakan proses ekspansi atau kompresinyata yang hubungan antara p dan V
diberikan
pV
n
= constann adalah indek proses politropik
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Termodinamika
Abdi H Sby, ST, MT 33
Diagram p-V
Hubungan p,V dan T
Dimana pV n = c
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Termodinamika
Abdi H Sby, ST, MT 44
p1V 1n = p2V 2
n
Dari beberapa proses diperoleh :
pV = mRT
)........(1 12
21
1
2
2
22
1
11
T xV
T xV
p
p
T
V p
T
V p
=
=
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Termodinamika
Abdi H Sby, ST, MT 55
Dari persamaan diatas :
Sustitusi ke (1)
n
n
n
V
V
V
V
p
p
==
2
1
2
1
1
2
n
V
V
T
T x
V
V
=
2
1
1
2
2
1
1
2
2
1
1
2
V
V x
V
V
T
T n
=
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Termodinamika
Abdi H Sby, ST, MT 66
1
11
1
2
1
21
1
1
2
1
2
−+
−
=
=
n
n
T
T
T
T x
T
T
p
p
1
2
1
2
1−
= V
V
xV
V n
Maka,
atau pers. Dibuat,
Subtitusi ke (1)
Sehingga :
1
2
1
1
2−
=
n
V
V
T
T
1
1
1
2
2
1 −
= n
T
T
V
V 1
1
2
1
2 −
=
nn
T
T
p
p
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Termodinamika
Abdi H Sby, ST, MT 77
Kerja
Dimana pV n = c , maka p =c/V n
Subtitusi
∫ =2
1
v
v
dV pW [ ]
( )111
2
1
1
11
2
1
2
1
2
1
+−+−
+−
−
−+−
=+−
=
=
=
∫
∫
nn
v
vn
v
v
n
v
vn
V V cn
n
V c
dV cV
dV V
cW
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Termodinamika
Abdi H Sby, ST, MT 88
c = p1V 1n = p2V 2
n , maka
( )
( )
( )
( ),1
1
:
1
1
1
1
21
2211
1122
1111
1222
T T n
mRW
lainealternativ
n
V pV pW
atau
V pV pn
W
V V pV V pn
W nn
−−
=
−−
=
−+−
=
−+−
= +−+−
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Termodinamika
Abdi H Sby, ST, MT 99
Panas
Q – W = U2- U1
Q = mc v (T 2- T 1 ) +
= - mc v (T 1- T 2 ) +
Dari, p1V 1 – p2V 2 = mR(T 1- T 2 ), maka
Q = - mc v (T 1-T 2 ) +
= m(T 1- T 2 )
( )
1
2211
−−n
V pV p
( )
1
2211
−−n
V pV p
( )
1
21
−−
n
T T mR
−− vcn
R
1
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Termodinamika
Abdi H Sby, ST, MT 1010
R = c p – c v , maka harga Q diperoleh
Q = m (T 1- T 2 )
Dimana,
Disebut panas spesifik politropik (cn), maka
Q = mc n (T 1 – T 2 )
−
−
1n
ncc v p
nv p c
nncc =
−−
1
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Termodinamika
Abdi H Sby, ST, MT 1111
contoh
Satu kg gas ideal pada temperatur dan tekanan awalmasing-masing 45oC dan 98 kPa, kemudiandikompresikan secara politropik (n=1,2) hingga tekanan980 kPa (gage). Tentukan perbandingan kompresi mesindan temperatur akhir juga kerja yang dilakukan. R = 287
J/kg K
Jawab :
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Termodinamika
Abdi H Sby, ST, MT 1212
After ignition of the fuel mixture at the top of the stroke,an internal combustion engine cylinder contains 0,1 L
of hot gas at a temperature of 1500o
C and apressure of 7 MPa (absolute). The hot gas expandspolytropically (n = 1,5) to the bottom of the stroke.The compression ratio 10 : 1, cp = 1,0 kJ/kg K and cv
= 0,72 kJ/kg K
Determine the following :1. Temperature and pressure at the bottom of the
stroke
2. Work transfer during the stroke
3. Internal energi change during the stroke
4. Heat flow from the cylinder during the stroke
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Termodinamika
Abdi H Sby, ST, MT 1313
Proses Adiabatik
Pada proses adiabatik, interaksi kalortidak terjadi antara sistem danlingkungan. Q = 0
secara general proses adiabatik samadengan politropik.
Dari proses politropik diperoleh :
Q = mc n
(T 1
– T 2
)
dimana :
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Termodinamika
Abdi H Sby, ST, MT 1414
proses adiabtik Q = 0
Q = m (T 1- T 2 ) = 0
(T 1- T 2 ) tidak sama dengan 0 tetapi :
−
−
1n
nccv p
nv p c
nncc =
−−
1
01 =
−−n
nccm v p
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Termodinamika
Abdi H Sby, ST, MT 1515
c p = n c v
n = c p /c v
c p /c v = γ disebut indeks Adiabatik
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Termodinamika
Abdi H Sby, ST, MT 1616
Hubungan p, V dan T
p1V 1γ =
p2V 2
γ
Kerja
Dimana n = γ ( )
( ),1
1
21
2211
T T mR
W
lainealternativ
V pV pW
−−
=
−−=
γ
γ
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Termodinamika
Abdi H Sby, ST, MT 1717
Kalor, Q = 0
Energi dalam U,
∆U = – W
U 2– U
1= – W
atau
W = U 1– U
2
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Termodinamika
Abdi H Sby, ST, MT 1818
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Termodinamika
Abdi H Sby, ST, MT 1919
Proses Hubungan p V & T Kerja W Energi Dalam Kalor Q
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Termodinamika
Abdi H Sby, ST, MT 2020
Proses Hubungan p, V & T Kerja, W Energi Dalam
(U2-U1)
Kalor, Q
TekananKonsatan
p = c
V1 /T1 = V2 /T2
p (V2-V1) mcv (T2-T1) mcp(T2-T1)
Volumekonstan V = cp1 /T1 = p2 /T2
0 mcv (T2-T1) Q = U2 – U1
mcv (T2-T1)
TemperaturKonstan
T = c
p1V1 = p2V2
p1V1 ln(V2 /V1) 0 Q = W
= p1V1 ln(V2 /V1)
Politropic pVn = c
T2 /T1 = (V1 /V2)n-1
p2 /p1 = (T2 /T1)n/(n-1)
P2 /P1 = (V1 /V2)n
mcv (T2-T1) Q = W + U2-U1
= mcn (T1-T2)
Dimana
Adiabatic Sama denganpolitropic n = γ Sama denganpolitropic n = γ
mcv (T2-T1) 0
1−−
=n
cncc
v p
n
1
2211
−−n
V pV p
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Termodinamika
Abdi H Sby, ST, MT
Penyelesaian contoh soal
n = 1,2
p1 = 98 kPa
T1 = 45+273 = 318 K
p2 = 980 kPa (gage)= 1081 Mpa (absolut)Di minta Perbandingan V 1 /V 2 :
Dari p1V 1n = p2V 2
n
21
n
p
p
V
V atau
p
p
V
V /1
1
2
2
1
1
2
n
2
1
==
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Termodinamika
Abdi H Sby, ST, MT
Sehingga perbandingan kompresi 7,393 : 1
Temperatur akhir
T2 = 318 x (7,393)1,2-1
= 474 K (201 oC)
22
393,798
10812,1/1
2
1
=
=V
V
1
2
112
1
2
1
1
2−−
=→∴
=
nn
V
V T T
V
V
T
T
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Termodinamika
Abdi H Sby, ST, MT
Kerja yang dilakukan
Dimana massa = 1 kg
23
( )211
T T n
mRW −−
=
( ) kJ J W 224474318
12,1
)287(1−=−
−=
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Termodinamika
Abdi H Sby, ST, MT 24
V1 = 0,1 L = 0,1 x 10-3 m3
p1 = 7 Mpa = 7000 kPa
T1 = 1500+273 = 1773 K
V 2 /V 1 = 10 (perbandingan Kompresi)
V 2 = 1 L = 1 x 10-3 m3
a)Dari persamaan :
T2 = 1773 x (1/10)1,5-1
= 561 K (288 oC)
1
2
1
12
1
2
1
1
2
−−
=→∴
=
nn
V
V T T
V
V
T
T
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Termodinamika
Abdi H Sby, ST, MT
Sehingga dari p1V 1n = p2V 2
n
b) Kerja
25
kPa x p
V V p p
n
22110
17000
5,1
2
2
112
=
=
=∴
( )
kJ x x x x
W
n
V pV pW
957,015,1
)101221()101,07000(
133
2211
=−−
=
−−
=
−−
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Termodinamika
Abdi H Sby, ST, MT
c) U2 – U1 = mcv (T2-T1)
R = cp-cv = 0,28 kJ/kg K
U2 – U1 = 0,00141 (0,72) (561-1773) == -1,230 kJ
d) Q – W = U2 – U1
Q = W + U2 – U1
Q = 0,957 – 1,23 = -0,273 kJ Kalor Keluarsistem
26
kg x
x xm 00141,0
1773280
0001,0107 6
==