Update 12 April 2020
a. FDV = Massa
CDV = h (ketinggian)
b. [Laju Akumulasi]=[Laju Alir massa masuk]-[Laju Alir Massa Keluar]
dm/dt = min - mout
d(A. h. ρ) = qin. ρ – qout. ρ
A. ρ. dh/dt = qf. ρ – q.ρ – qc.ρ
Asumsi larutan homogen
A. ρ. dh/dt = qf.ρ – q.ρ – qc.ρ
A = qf – q – qc
c. Diketahui :
h0 = 10 m
qf = 100 /min
qc = 5 m3/min
q = 10h m3/min
A = 314 m2
T = 10
Ditanyakan :
h = ?
Jawab :
A. dh/dt = qf – q – qc
314. dh/dt = 100 – 10h – 5
dh/dt =
dh/dt = 0,3185 – 0,0318h – 0,0159
dh/dt – 0,0318h = 0,3185 – 0,0159
Misalkan : P = 0,3185
Q = 0,3026
FI = e^{Int {P. dt}}
= e^{Int {0,3185. dt}}
= e^{0,3185. t}
Masukkan ke persamaan
h. FI = int{Q. FI} dt
h. e^{0,3185. t}= int{0,3026. e^{0,3185. t}} dt
h. e^{0,3185. t}= 0,3026. int {e^{0,3185. t}} dt
h. = {0,3026×{1/0,3185 }×{e^0,3185 t}+C}/e^{0,3185. t}
h = {0,3026/0,3185 }+{C /e ^0,3185 t}
h = 9,516+{C /e ^0,3185 t}
Untuk mencari C
h0 = 10 m ; t0=0
10 = 9,516
10 = 9,516
10 = 9,516
C = 0,484
Untuk mencari h ; t=10
h = 9,516 +{C /e ^0,3185 t}
h = 9,516 +{0,484 /e ^0,3185 t}
h =9,868 m
T h 109,5360270209,5168286309,5160342409,5160014509,5160000609,5160000709,516809,516909,5161009,516
d. Kesimpulan : Laju Aliran keluar sama dengan Laju Aaliran masuk karena h konstan 9,516
Soal 2
Diketahui :
V0 = 5 m3
M0 = 100kg
qin1= 0,01 m3/min
cin1= 200 kg/m3
qin2= 0,05 m3/min
cin2= 0 (air minum)
C0 =M0/V0 =100kg/5 m3=20 kg/m3
NM Keseluruhan (Non reacting)
[Laju Akumulasi] = [Laju Alir Masuk] – [Laju Alir Keluar]
dm/dt = min – mout
d(A. h. ρ)/dt= (qin. ρ + qout. Ρ) – qout. ρ
A ρ. dh/dt = (qin. ρ + qout. Ρ) – qout. ρAsumsi larutan Homogen
A. ρ. dh/dt = (qin. ρ + qout. ρ ) – qout. ρ
A. dh/dt = qin + qout – qout
A. dh/dt = qin1 + q2out – q3out
dh/dt = {qin1 + q2out – q3out}/A
dh/dt = 0
NM Keseluruhan (Non reacting)
d.{A. C. h}/dt = qin. c + qin. c – qout. c
d.{A. C. h}/dt = qin1. c + q2in. c – q3out. C
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