Persamaan Differensial Parsial

Teknik Perminyakan
TPM 433

Kuliah Minggu-2
Capaian Belajar TPM 433
PDP TP

Kuliah #2

Billal M. Aslam
Institut Teknologi Bandung [2]
Persamaan Diferensial TPM 433
PDP TP

Definisi
• Persamaan yang mengandung suku turunan dari variable dependen
terhadap variable independent

Contoh :
𝒅𝒚
= 𝟎. 𝟐𝒙𝒚
𝒅𝒙

Solusi persamaan diferensial adalah bentuk fungsi yang memenuhi pers tsb pada suatu interval I

𝟎.𝟏𝒙𝟐
𝒚= 𝒆

Billal M. Aslam
Institut Teknologi Bandung [3]
Persamaan Diferensial TPM 433
PDP TP

Jenis Persamaan Diferensial

• Persamaan Diferensial Biasa (PDB) – fungsi 1 var.

• Persamaan Diferensial Parsial (PDP) – fungsi > 1 var

Billal M. Aslam
Institut Teknologi Bandung [4]
 TPM 433
Persamaan Diferensial Biasa (Review) PDP TP

Klasifikasi PDB
• Bentuk PDB
𝑑𝑛 𝑦 𝑑𝑛−1 𝑦 𝑑𝑦
𝑎0 𝑥 𝑛
+ 𝑎1 𝑥 𝑛−1
+ ⋯ + 𝑎𝑛−1 𝑥 + 𝑎𝑛 𝑥 𝑦 = 𝑅(𝑥)
𝑑𝑥 𝑑𝑥 𝑑𝑥
Dependent
Independent
Order of ODE Variable(s)
Variable

– Disebut linear jika :

1) y and all derivatives of y are raised to no power other than 1
2) Each coefficient a(x) is only function of x
3) No y and y-derivative products are allowed

Disebut homogen jika : R(x) = 0

Billal M. Aslam
Institut Teknologi Bandung [5]
Persamaan Diferensial Biasa (Review) TPM 433
PDP TP

Contoh

linear

Billal M. Aslam
Institut Teknologi Bandung [6]
Persamaan Diferensial Biasa (Review) TPM 433
PDP TP

Metode Solusi PDB Orde 1

1 Method of Integrating Factor

2 Separation of Variables
1st Order O.D.E. Solution
𝑑𝑦
𝑎0 𝑥 + 𝑎1 𝑥 𝑦 = 𝑅(𝑥) 3 Laplace Transform*
𝑑𝑥

4 Numerical Methods*

• Solution : any function (φ(x)) that satisfies the D.E. on continuous

interval I
Billal M. Aslam
Institut Teknologi Bandung [7]
Persamaan Diferensial Biasa (Review) TPM 433
PDP TP

Jenis Solusi & Medan Arah

𝑑𝑦 3
• Tinjau suatu PDB = sin y, y 0 =−
𝑑𝑥 2

Direction field (Medan Arah)

Bentuk semua solusi yang memenuhi
(solusi umum)
Kurva Solusi
Solusi Khusus yang memenuhi PD

Billal M. Aslam
Institut Teknologi Bandung [8]
 TPM 433
PDP TP

Metode Solusi

PDB Orde 1

Billal M. Aslam Institut Teknologi

Bandung
Solusi PDB Orde 1 TPM 433
PDP TP

Metode Faktor Integrasi

For 1st Order O.D.E. y ′ + 𝑃 𝑥 𝑦 = 𝑄(𝑥)

Let there be M(x) which if multiplied will reduce L.H.S. into common derivative

𝑀 𝑥 y ′ + 𝑃 𝑥 𝑦 …(1) 𝑀 𝑥 𝑃 𝑥 = 𝑀′(𝑥) …(4)

𝑀′ 𝑥
𝑀 𝑥 𝑦 ′ + 𝑀 𝑥 𝑃 𝑥 𝑦 …(2) 𝑃 𝑥 = …(5)
𝑀(𝑥)
𝑀 𝑥 𝑦 ′ + 𝑀′ 𝑥 𝑦 …(3)
‫ = 𝑥𝑑 𝑥 𝑃 ׬‬ln 𝑀(𝑥) …(6)
Total derivative (M(x)y)’ 𝑒‫𝑃 ׬‬ 𝑥 𝑑𝑥
= M x …(7)

“the integrating factor”

Billal M. Aslam
Institut Teknologi Bandung [10]
Solusi PDB Orde 1 TPM 433
PDP TP

Metode Faktor Integrasi (lanjutan)

Multiply by M(x) y ′ 𝑀(𝑥) + 𝑃 𝑥 𝑦𝑀(𝑥) = 𝑄 𝑥 𝑀(𝑥)

𝑑
(𝑦𝑀(𝑥)) = 𝑄 𝑥 𝑀(𝑥)
𝑑𝑥

𝑦𝑒 ‫𝑃 ׬‬ 𝑥 𝑑𝑥
= න 𝑄(𝑥) 𝑒 ‫𝑃 ׬‬ 𝑥 𝑑𝑥
dx + C

𝑦 = 𝑒− ‫𝑃 ׬‬ 𝑥 𝑑𝑥
‫𝑃 ׬ 𝑒 )𝑥(𝑄 ׬‬ 𝑥 𝑑𝑥
dx + C 𝑒 − ‫𝑃 ׬‬ 𝑥 𝑑𝑥

𝑓𝑜𝑟 𝑄 𝑥 = 0 𝑦 = C 𝑒 − ‫𝑃 ׬‬ 𝑥 𝑑𝑥

Billal M. Aslam
Institut Teknologi Bandung [11]
Solusi PDB Orde 1 TPM 433
PDP TP

Metode Faktor Integrasi (lanjutan)

Billal M. Aslam
Institut Teknologi Bandung [12]
Solusi PDB Orde 1 TPM 433
PDP TP

Metode Pemisahan Variable

• Kumpulkan suku sejenis dan integralkan

𝑑𝑦
𝐵 𝑦 = 𝐴(𝑥)
𝑑𝑥
𝐵 𝑦 𝑑𝑦 = 𝐴 𝑥 𝑑𝑥

න 𝐵 𝑦 𝑑𝑦 = න 𝐴 𝑥 𝑑𝑥 + 𝐶

Billal M. Aslam
Institut Teknologi Bandung [13]
Solusi PDB Orde 1 TPM 433
PDP TP

Metode Pemisahan Variable

• Contoh
𝑑𝑦
= 4𝑥 2 + 2
𝑑𝑥
𝑑𝑦 = 4𝑥 2 + 2 𝑑𝑥

𝑦 = න 4𝑥 2 + 2 𝑑𝑥
4 3
𝑦 = 𝑥 + 2𝑥 + 𝐶
3

Billal M. Aslam
Institut Teknologi Bandung [14]
 TPM 433
PDP TP

Metode Solusi

PDB Orde 2

Billal M. Aslam Institut Teknologi

Bandung
Solusi PDB Orde 2 TPM 433
PDP TP

Metode Persamaan Karakteristik

• PDB Orde 2 linear
𝑑2 𝑦 𝑑𝑦
𝑎0 2 + 𝑎1 + 𝑎2 𝑦 = 𝑅(𝑥)
𝑑𝑥 𝑑𝑥
2𝑦
𝑚𝑥 𝑑𝑦 𝑑
Let 𝑦 = 𝑒 𝑚𝑥 be our “trial solution” 𝑦 = 𝑒 , = 𝑚𝑒 𝑚𝑥 , 2 = 𝑚2 𝑒 𝑚𝑥
𝑑𝑥 𝑑𝑥

𝑎0 𝑚2 𝑒 𝑚𝑥 + 𝑎1 𝑚𝑒 𝑚𝑥 + 𝑎2 𝑒 𝑚𝑥 = 𝑅(𝑥)
For homogenous solution , i.e. 𝑅 𝑥 = 0

𝑎0 𝑚2 + 𝑎1 𝑚 + 𝑎2 = 0 “Characteristic Equation”

Billal M. Aslam
Institut Teknologi Bandung [16]
Solusi PDB Orde 2 TPM 433
PDP TP

Metode Persamaan Karakteristik

• Characteristic Equation (cont’d)
𝑎0 𝑚2 + 𝑎1 𝑚 + 𝑎2 = 0 −𝑎1 ± 𝑎12 − 4𝑎0 𝑎2
𝑚1,2 =
2𝑎0
• Case 1, m1,2 real and distinct …
𝑦ℎ = 𝑐1 𝑒 𝑚1𝑥 + 𝑐2 𝑒 𝑚2𝑥
• Case 2, m = a±ib (complex number)
𝑦ℎ = 𝑐1 𝑒 𝑎𝑥 cos 𝑏𝑥 + 𝑐2 𝑒 𝑎𝑥 sin 𝑏𝑥
• Case 3, m1 = m2 = m
𝑦ℎ = 𝑐1 𝑒 𝑚𝑥 + 𝑐2 𝑥𝑒 𝑚𝑥

Billal M. Aslam
Institut Teknologi Bandung [17]
Solusi PDB Orde 2 TPM 433
PDP TP

Solusi Khusus
• In this case, we let 𝑅(𝑥) ≠ 0
• Combination of homogeneous and particular solution yield “General Solution”

1 Method of Undetermined
Coefficient

y = 𝑦ℎ + 𝑦𝑝
Method of Variation of
2
Parameter

Billal M. Aslam
Institut Teknologi Bandung [18]
Solusi PDB Orde 2 TPM 433
PDP TP

Solusi Khusus – Metode Koefisien Taktentu

• for linear diff eq. 𝐋(𝑦) = 𝑅(𝑥), particular solution have a form
𝑦𝑝 𝑥 = 𝐴1 𝑦1 𝑥 + 𝐴2 𝑦2 𝑥 +… +𝐴𝑛 𝑦𝑛 𝑥

• Case 1 𝑅 𝑥 = 𝑝𝑛 (𝑥) nth degree polynomial

𝑦𝑝 𝑥 = 𝐴𝑛 𝑥 𝑛 + 𝐴𝑛−1 𝑥 𝑛−1 + ⋯ + 𝐴1 𝑥 + 𝐴0
• Case 2 𝑅 𝑥 = 𝑘𝑒 α𝑥 , k and α a known constant

𝑦𝑝 𝑥 = 𝐴𝑒 α𝑥
• Case 3 𝑅 𝑥 = 𝑘1 sin 𝛽𝑥 + 𝑘2 cos 𝛽𝑥 k1, k2, 𝛽 known constant

𝑦𝑝 𝑥 = 𝐴 sin 𝛽𝑥 + B 𝑐𝑜𝑠 𝛽𝑥

Billal M. Aslam
Institut Teknologi Bandung [19]
Solusi PDB Orde 2 TPM 433
PDP TP

Contoh
• Solve 𝑦 ′′ − 𝑦 ′ − 2𝑦 = 𝑒 3𝑥
𝑅 𝑥 = 𝑘𝑒 𝛼𝑥 = 𝑒 3𝑥 (case 2)
−𝑎1 ± 𝑎12 − 4𝑎0 𝑎2
𝑚1,2 =
2𝑎0 𝑦𝑝 = 𝐴𝑒 3𝑥 , 𝑦𝑝 ′ = 3𝐴𝑒 3𝑥 , 𝑦𝑝 ′′ = 9𝐴𝑒 3𝑥

𝑚1 = −1, 𝑚2 = 2 (𝒄𝒂𝒔𝒆 𝟏) …substitute to ODE

9𝐴𝑒 3𝑥 − 3𝐴𝑒 3𝑥 − 2𝐴𝑒 3𝑥 = 𝑒 3𝑥
𝑦ℎ = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 2𝑥
4𝐴𝑒 3𝑥 = 𝑒 3𝑥

1 3𝑥
𝑦𝑝 = 𝑒
4

Billal M. Aslam
Institut Teknologi Bandung [20]
 TPM 433
PDP TP

Contoh Aplikasi
PDB pada Persamaan Aliran dalam
Media Berpori – Teknik Perminyakan

Billal M. Aslam Institut Teknologi

Bandung
 Darcy Equation
h1 − h2
Q = KA
L
Dalam bentuk yang lebih luas:
h1 – h2
k  dp g dz 
Q
vs = −  − 
h1
  ds 1.0133 10 ds 
6

s = jarak searah aliran (selalu postif), cm

vs = volume per luas penampang per satuan waktu
h2 searah aliran, cm/s
z = arah vertical (positif ke bawah), cm
 = densitas fluida, gm/cc
g = percepatan gravitasi, 980.665 cm/sq.s
dp/ds = gradien tekanan sepanjang arah alir, atm/cm
k = permeabilitas, D
 = viskositas fluida, cp

Q
 Horizontal Flow
k dp
vx = −
 dx
Q k dp
Q =−
A  dx
A Q k
dx = − dp
A 
Q
L p
p
2
Q k
L
A0 dx = −  dp
p1

kA( p1 − p2 )
Q=
Bagaimana persamaan fluid rate apabila
fluidanya kompressible?
L
 Vertical Flow: Free Flow
Laju Alir, Q ? Free Flow

k  dp dz 
vs = −  − g 
  ds ds 
dp dp
= =0
ds dz Patm
dz
= 1 (z dan s searah)
ds
k
vs = g

L
kA
Q= g

Porous media
saturated with
liquid Patm
Vertical Flow: Downward with Head
Laju Alir, Q ? Flow under
head h
k  dp dz  Patm Liquid
vs = −  − g 
  ds ds 
dp dp Patm − (Patm + gh ) h
= = = − g h
ds dz L L
dz
= 1 (z dan s searah)
ds
Porous media
k h  saturated with
vs = − − g − g 
 L  L liquid
k h 
vs = g  + 1
 L 

kA h  Patm
Q= g + 1
 L 
 dh kA  h(t )  Persamaan Laju Alir
Lanjutan....... −A =

g  + 1
dt  L 
Ketinggian h di atas media kg k h(t ) =
(c1h0 + c2 )exp(− c1t ) − c2 
c1 = ; c2 = g
berpori akan berkurang L  c1

seiring dengan waktu: dh  k h0 k   kg  k 

− = c1h + c2  g + g  exp − t  − g 
 L    L   
dt h(t ) = 
A(h0 − h(t )) = Q(t ) * t y = c1h + c2 kg
L
(h0 − h(t )) dy = c1dh
A = Q(t )  h   kg  
t dh = (1 / c1 )dy h(t ) = L  0 + 1 exp − t  − 1
 L   L  
1 dy
Limit t → 0 − =y kA  h(t ) 
c1 dt Q(t ) = g + 1
  L 
d (h0 − h(t )) dy
= −c1dt
A = Q(t ) kA   h0   kg  
dt y Q(t ) = g   + 1 exp − t  
y (t )
   L   L 
dh t
− A = Q(t ) dy
dt 
y0
y 0
= − c1dt Waktu dan laju alir saat h(t) = 0

maka, y (t ) = y0 exp(− c1t )  h   kg  

0 = L  0 + 1 exp − t  − 1
(c1h0 + c2 )exp(− c1t ) − c2   L   L  
dh kA  h(t )  h(t ) =
−A = g  + 1 c1 1  kg 
= exp − t 
dt   L   h0   L 
Cek:  + 1
dh k  h(t )   L 
− = g  + 1 (c1h0 + c2 ) − c2 
dt   L  t = 0  h(0) = L  1 
t=− ln 
dh kg k c1 kg  h0 / L + 1 
− = h(t ) + g
dt L  h(0) = h0 Substitusi waktu ini ke dalam persamaan laju alir,
maka
kA
Q(t ) = g Sama dengan vertical free flow

 Vertical Flow: Upward with Head
Laju Alir, Q ? Flow under
head h
Liquid Patm h

Porous media Area, A2

saturated with
liquid L

Area, A1
 Vertical Flow: Upward with Head
Flow under
k  dp dz  head h Patm
vs = −  − g 
  ds ds 
Liquid Patm
dp dp (Patm + gx ) − (Patm + g (h + x + L )) h
=− =
ds dz L
dp h x
= − g − g
ds L
dz Porous media
= −1 (z dan s berlawanan arah) Area, A2
ds saturated with
k h  liquid L
vs = − − g − g + g 
 L 
k h
vs = g  
 L
kA h
Q= g   Area, A1
 L
 Radial-Horizontal Flow
Laju Alir, Q ?

Pe Pw

re rw

h
Pe Pw
 Radial-Horizontal Flow
k  dp dz 
vs = −  − g 
  ds ds 
dp dp dz
=− ; =0
ds dr ds Pe Pw
k  dp 
vs =  
  dr 
kA  dp 
Q=  
  dr 
k 2rh  dp  re rw
Q=  
  dr 
re Pe
dr 2kh
Q r
=

dp  h
rw P w Pe Pw
2kh (Pe − Pw )
Q=
 r 
ln e 
 rw 
 TPM 433
PDP TP

End of Lecture

Terima Kasih

Billal M. Aslam Institut Teknologi

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