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Flow Problems dan Pendinginan~ 1 ~ Tugas 2 Pemodelan Matematika
Tugas β 3 Mata Kuliah: PEMODELAN MATEMATIKA
HUKUM PENDINGINAN NEWTON
DAN
FLOW PROBLEMS
Dikerjakan Oleh:
Nama : YUSRI
Nim : 809715022
Kelas : A - Reguler
Prodi : Pendidikan Matematika
PROGRAM STUDI PENDIDIKAN MATEMATIKA
PROGRAM PASCA SARJANA
UNIVERSITAS NEGERI MEDAN
2010
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Flow Problems dan Pendinginan~ 2 ~ Tugas 2 Pemodelan Matematika
1. Sebuah adonan kue mula-mula bersuhu 300C, dimasukkan kedalam oven yang bersuhu tetap 1400C. Setelah 5 menit suhu adonan menjadi 400C. Pada menit ke berapakah suhu adonan menjadi 1000C. Penyelesaian: Diketahui: Suhu adonan mula-mula = T (0) = 30 0C Suhu lingkungan (oven) tetap =π = 140 0C Suhu adonan setelah 5 menit =T (5) = 40 0C Ditanya: Suhu adonan menjadi 1000C saat t =.....? Atau: T (t) = 1000C pada t =...? Jawab:
ππ
ππ‘= π (π β π) ......................................................................................... (1)
βΊππ
ππ‘= π (π β 140)
βΊ βππ
ππ‘= βπ(π β 140)
βΊ βππ
ππ‘= πβπ β β140)
βΊ π π π β π 0 = π. π π β 140
π . π
βΊ π π π β 30 = π. π π β 140.π
π
βΊ π π π β π. π π = 30 β 140.π
π
βΊ (π β π)π π = 30 β 140.π
π
βΊ π π =30
π β πβ
140. π
π (π β π)
βΊ π π‘ = ββ1 30
π βπβ
140.π
π π βπ
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Flow Problems dan Pendinginan~ 3 ~ Tugas 2 Pemodelan Matematika
βΊ π π‘ = ββ1 30
π β π β ββ1
140. π
π π β π
βΊ π π‘ = ββ1 30
π β π β ββ1
π΄
π +
π΅
π β π
Pencarian A dan B:
140π
π π β π =
π΄
π +
π΅
π β π
140π
π π βπ =
π΄π βπ΄π+π΅π
π (π βπ)
140π
π π βπ =
π΄+π΅ π βπ΄π
π (π βπ)
βπ΄ = 140 βΊ π΄ = β140
π΄ + π΅ = 0 βΊ π΄ = βπ΅
ππππ π΅ = 140
βΊ π π‘ = ββ1 30
π β π β ββ1
β140
π +
140
π β π
βΊ π π‘ = ββ1 30
π β π + ββ1
140
π β ββ1
140
π β π
βΊ π π‘ = ββ1 140
π βββ1
110
π β π
βΊ π π‘ = 140. ββ1 1
π β110.ββ1
1
π β π
βΊ π π‘ = 140 β 110ππ.π‘ ................................................................. (1-a)
Untuk solusi khusus t = 5, T (5) = 40 maka
40 = 140 β 110ππ .5
βΊ 110ππ .5= 100
βΊ ππ .5= 100
110
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Flow Problems dan Pendinginan~ 4 ~ Tugas 2 Pemodelan Matematika
βΊ πΏπ(ππ .5)=Ln ( 100
110)
βΊ 5π= β0,095311179
βΊk = -0, 019062235........................................................................ (1-b)
Untuk solusi khusus T (t) = 100 maka
π π‘ = 140 β 110ππ.π‘
βΊ 100 = 140 β 110πβ0,019062235 .t
βΊ 110πβ0,019062235 .t= 140 β 100
βΊ πβ0,019062235 .t = 40
110
βΊ πΏπ πβ0,019062235 .t = πΏπ 40
110
βΊ β0,019062235. t = β1,011600922
βΊ π‘ =β1,011600922
β0,019062235
βΊt = 53, 06832707............................................................................ (1-b)
Maka pada menit ke β 53 suhu adonan menjadi 1000C
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Flow Problems dan Pendinginan~ 5 ~ Tugas 2 Pemodelan Matematika
2. A tank initially holds 10 galloons of fresh water. At t=0, brine solution containning Β½ lb of salt per gallon is poured in to the tank off rate of 2 gallon per minute while the well stined mixture leaves the tank at the same. Find the amount of salt in the tank at anytime t.
Penyelesaian:
Diketahui:
Volume tangki = V0 = V (0) =10πππππ.
Jumlah garam dalam tangki mula-mula=Q (0) = 0 (karena Fresh water)
et = 2 πππππ
πππππ‘.
ft = 2πππππ
πππππ‘
a untuk fresh water = 0
b = 1
2
ππ
πππππ
Ditanya:
Q (t) = jumlah garam dalam tangki setiap saat?
Jawab:
βΊππ
ππ‘ = b.e β f. [
π
π£0+ππ‘βππ‘]
βΊππ
ππ‘ =
1
2
ππ
πππππ.2
πππππ
πππππ‘ β 2
πππππ
πππππ‘ [
π
10πππππ +2 πππππ
πππππ‘β2
πππππ
πππππ‘
]
βΊππ
ππ‘ =
1
2
ππ
πππππ .2
πππππ
πππππ‘ β 2
πππππ
πππππ‘ [
π
10galon]
βΊππ
ππ‘ = 1
ππ
πππππ‘ β 2
πππππ
πππππ‘ [
π
10galon]
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Flow Problems dan Pendinginan~ 6 ~ Tugas 2 Pemodelan Matematika
βΊππ
ππ‘ = 1
ππ
πππππ‘ β[
π
5 menit]
βΊππ
ππ‘ = 1 β
π
5 .......................................................................................... (2)
βΊππ
1β π
5
= ππ‘
ππ
1β π
5
= ππ‘
βΊ-5 ln (1 β π
5) = t +c
βΊLn (1 β π
5) =
t +c
β5
βΊ 1 β π
5= π
t +cβ5
βΊπ
5= 1 β πβ
t 5 . ππ
βΊ π = 5(1 β Cπβt
5) ....................................................................... (2-a)
Untuk solusi khusus Q t=0=0 maka
0 = 5 (1 β C. πβ0
5 )
βΊ0 = 5 (1 β C. π0)
βΊ0 = 5 (1 β C. π0)
βΊC = 1.................................................................................................. (2-b)
Maka diperoleh solusi umum jumlah garam dalam tangki setiap saat adalah:
Q (t) = 5(1 β πβt
5 )
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Flow Problems dan Pendinginan~ 7 ~ Tugas 2 Pemodelan Matematika
3. A metal bar at a temperature of 100 0F is placed in a room at a constant temperature of 00 F. If after 20 minutes the temperatures of bar is 50 0F. Find an expression for temperature of the bar at anytime! Penyelesaian: Diketahui: Suhu batangan mula-mula = T (0) = 100 0F Suhu lingkungan (ruangan) =π = 0 0F Suhu batangan setelah 20 menit =T (20) = 50 0F Ditanya: Suhu batangan setiap saat = T (t) Jawab:
ππ
ππ‘= π (π β π) ........................................................................................ (3)
βΊππ
ππ‘= π (π β 0)
βΊππ
ππ‘= π π
βΊππ
π= π ππ‘
ππ
π= πππ‘
βΊ ππ
π=π ππ‘
βΊ Ln T= kt +c
βΊ π = πππ‘+π
βΊ π = πππ‘ . ππ
βΊ π = Cπππ‘ ....................................................................................... (3-a)
Untuk solusi khusus T t=0=100 maka
100= Cππ .0
βΊ100= C. π0
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Flow Problems dan Pendinginan~ 8 ~ Tugas 2 Pemodelan Matematika
βΊ100 = C. 1
βΊC = 100............................................................................................. (3-b)
Untuk solusi khusus T t=20=50 maka
50 = 100ππ .20
βΊ50
100 = π20π
βΊLn 0,5= π20π
βΊ-0, 6934718 = 20k
βΊ π =β0,6934718
20
βΊk = -0, 034657359......................................................................... (3-c)
Maka diperoleh solusi umum jumlah garam dalam tangki setiap saat adalah:
T (t) = 100πβ0,034657359 π‘