Energy Balance and Temperature Regulation Energy Balance Temperature Regulation.
Matek-Energy Balance
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Rate of heat in : (y.z.q x ) | x Rate of heat out : (y.z.q x )| x+∆x Rate of thermal energy production : (y.z.∆x).Se Substitusikan ke persamaan energy balance : Rate heat in – reat heat out + energy production = 0 Rate heat out – rate heat in = energy production (y.z.q x )| x+∆x - (y.z.q x ) | x = (y.z.∆x).Se …………………… dibagi dengan (y.z.∆x) dan dilimitkan ∆x -> 0 Jadi : lim ∆x→ 0 qx |x−qx |x+ ∆x ∆x =Se
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Chemical Engineering Dept
Transcript of Matek-Energy Balance
Rate of heat in : (y.z.qx) |x
Rate of heat out : (y.z.qx)|x+∆x
Rate of thermal energy production : (y.z.∆x).Se
Substitusikan ke persamaan energy balance :
Rate heat in – reat heat out + energy production = 0
Rate heat out – rate heat in = energy production
(y.z.qx)|x+∆x - (y.z.qx) |x = (y.z.∆x).Se …………………… dibagi dengan (y.z.∆x) dan dilimitkan ∆x -> 0
Jadi :
lim∆ x→0
qx|x−qx|x+∆ x∆ x
=Se
d (qx )dx
=Se
Lalu di integralkan :
q=Se . x+c…………………… ..(1)
Karena perpindahan panas yang terjadi adalan konduksi maka digunakan rumus
q=−k dTdx
Disubstitusikan ke persamaan (1).
−k∫ dTdx =∫ Se . x+C
−k T=12Se . x2+C1 x+C2..................................... (2)
Menggunakan BC x = 0 , T = 212℉
−50(212)=C 2
C2=¿−1 0600¿
Menggunakan BC x = 1,6405×10−3 ft , T = 347℉
−50(347)=12×109×(1,6405×10−3)2+C1 (1,6405×10−3 )+(−10600)
−17350=1345,62+C 1 (1,6405×10−3 )−10600
C1=−4934849 ,131
Jadi didapatkan persamaan (2)
−k T=12Se . x2−4934849 ,131 x−10600
T=[−1
2Se . x2+4934849 ,131 x+10 600]
k
Temperatur (T) sebagai fungsi x =1.6405 .10−3
2 = 8,2025 x 10-4
T=
−12×109×(8,2025×10−4)2+4934849 ,131×(8,2025×10−4)2+10600
50
k= 50 Btu.F/(ft.hr)
Se = 109 Btu/(ft3.hr)
T=218,794℉
