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BALANCES- MODELAMIENTO EVAPORADOR DE EFECTO SIMPLE

F3(t) CB3(t) hF3(t) Qvapor

Fs (t)

Fs(t)

Ξ½ (t)

T0 (t)

Ts(t)

F2(t) F1(t)

CA2(t)

CA1(t)

hF2(t)

hF1(t)

BALANCE GLOBAL ο‚· F1 (t) – F2(t) – F3(t) =

𝒅 𝒗(𝒕)

ο‚· F1D(t) – F2D(t)- F3D(t) =

𝒅𝒕 𝒅𝝂𝑫 (𝒕) 𝒅𝒕

(1)

BALANCE COMPONENTE A ο‚· CA1(t) F1 (t) – CA2(t) F2(t) = ο‚·

𝒅 𝒅𝒕

[Ξ½(t). CA2(t)]

a1 CA1D(t) + a2CA2D(t) + a3 CA2D(t) + a4 F2D(t) = a5 𝑑π‘ͺ𝑫 π‘¨πŸ (𝒕) 𝑑𝑑

(2)

𝒅𝝂𝑫 (𝒕) 𝒅𝒕

+ a6

ο‚· ο‚· ο‚· ο‚· ο‚· ο‚·

a 1 = π‘­πŸ a2 = π‘ͺπ‘¨πŸ a3 = -π‘­πŸ a4 = -π‘ͺπ‘¨πŸ a5 = π‘ͺπ‘¨πŸ a6 = 𝑽

REEMPLAZANDO (1) EN (2) ο‚·

a1CA1D(t) + a2 F1D(t) + a3 CA2D (t) + a4F2D(t) = a5 F1D(t) - a5 F2

D(t)

- a5 F3

D(t)

+ a6

𝒅 π‘ͺ𝑫 π‘¨πŸ (𝒕) 𝒅𝒕

DESPEJANDO CA2D(t) ο‚·

βˆ’π’‚πŸ π’‚πŸ‘

CA1D(t) +

π’‚πŸ” 𝒅π‘ͺ𝑫 π‘¨πŸ (𝒕) βˆ’π’‚πŸ‘

ο‚·

ο‚·

𝒅𝒕

G1(s) =

G2(s) =

ο‚· G3(s) =

(π’‚πŸ βˆ’π’‚πŸ“ ) βˆ’π’‚πŸ‘

F1D(t) +

(π’‚πŸ’ +π’‚πŸ“) βˆ’π’‚πŸ‘

+ CA2D (t)

π‘ͺ𝑫 π‘¨πŸ (𝒔) π‘ͺ𝑫 π‘¨πŸ (𝒔) π‘ͺ𝑫 π‘¨πŸ (𝒔) 𝑭𝑫 𝟏

π‘ͺ𝑫 π‘¨πŸ (𝒔) 𝑭𝑫 𝟐 (𝒔)

=

=

=

π‘²πŸ

=

𝟏.πŸ“πŸ

π‰πŸ 𝒔+𝟏 𝟏𝟎.πŸ” 𝑺+𝟏 π‘²πŸ π‰πŸ 𝑺+𝟏

π‘²πŸ‘

=

=0

𝝉𝑺+𝟏

βˆ’πŸ’.πŸ–πŸ“π‘ΏπŸπŸŽβˆ’πŸ‘ 𝟏𝟎.πŸ” 𝑺+𝟏

F2D(t) +

π’‚πŸ“ βˆ’π’‚πŸ‘

F3D(t) =

ο‚·

G4(s)=

ο‚·

K1 =

ο‚·

K2 =

ο‚·

K3 =

ο‚·

K4 =

ο‚·

Ο„1=βˆ’π’‚πŸ”

π‘ͺ𝑫 π‘¨πŸ (𝒔) 𝑭𝑫 πŸ‘ (𝒔)

=

π‘²πŸ’ 𝝉𝑺+𝟏

=

𝟎.πŸŽπŸπŸ“ 𝟏𝟎.πŸ” 𝑺+𝟏

π’‚πŸ

βˆ’π’‚πŸ‘ (π’‚πŸ βˆ’π’‚πŸ“ ) βˆ’π’‚πŸ‘

(π’‚πŸ’ +π’‚πŸ“ ) π’‚πŸ‘ π’‚πŸ“

βˆ’π’‚πŸ‘ 𝒂

πŸ‘

BALANCE DE ENERGIA ο‚· F1(t)hF1(t) – F2(t)hF2(t) – F3(t)hF3(t) + Qch(t) =

𝒅 𝒅𝒕

[𝝆 𝑽(𝒕) 𝑼(𝒕)]

ο‚· F1(t)Cp1T1(t) - F2(t)Cp2T2(t)- F3(t)Cp3Teb(t) + UA[T2(t)- Ts(t)] 𝒅

= [𝝂(𝒕)𝑼(𝒕)] 𝒅𝒕

REEMPLAZAR U(t) ο‚· a7F1D(t) + a8T1D(t) + a10F2D(t) + a11T2D(t)+a12F3D(t)+a13TebD(t) + a14

TsD(t)

ο‚· a7F1

D(t)

=a15

𝒅𝝂𝑫 (𝒕) 𝒅𝒕

+ a 8 T1

D(t)

𝒅

+a16Cv [T2D(t)] 𝒅𝒕

+ a10F2

D(t)

+ a11T2D(t)+a12F3D(t)+a13TebD(t)+ 𝒅

a14TsD(t) = a15 F1D(t)- a15F2D(t)- a15 F2D(t) + a16 Cv 𝑻𝑫 𝟐 (𝒕) 𝒅𝒕

ο‚· (a7 – a14 )F1D(t)- a8 T1D(t)- (a9+a14)F2D(t) – a10T2D +( a11+a14) F3D(t) + a12 T3D(t)= a15Cv ο‚·

π’‚πŸ• βˆ’π’‚πŸπŸ“ βˆ’π’‚πŸπŸ

F1D(t) + (

TebD(t) +

οƒΌ οƒΌ οƒΌ οƒΌ

π’‚πŸπŸ’ βˆ’π’‚πŸπŸ

π’‚πŸ– βˆ’π’‚πŸπŸ

TsD(t) =

a7 = Cp1𝑻1 a8 =Cp1 π‘­πŸ a10 = - Cp2𝑻2 a11 =βˆ’π‘­2Cp2 + UA a13= βˆ’π‘­3Cp3 a14 = UA a15= Cv 𝑻2 a16= Cv 𝑽

οƒΌ K5 = οƒΌ K6 =

π’‚πŸ• βˆ’π’‚πŸπŸ“ βˆ’π’‚πŸπŸ π’‚πŸ– βˆ’π’‚πŸπŸ

= 𝟎. πŸŽπŸ•πŸ’ =

= 𝟏. πŸ“πŸ’

𝒅𝒕

T2D(t)

)T1D(t) +

οƒΌ a12= - Cp3𝑻eb οƒΌ οƒΌ οƒΌ οƒΌ

𝒅

π’‚πŸπŸŽ +π’‚πŸπŸ“ βˆ’π’‚πŸπŸ

π’‚πŸπŸ” π‘ͺ𝒗 𝒅 𝑻𝑫 𝟐 (𝒕) βˆ’π’‚πŸπŸ

𝒅𝒕

F2D(t) + + T2D(t)

π’‚πŸπŸ +π’‚πŸπŸ“ βˆ’π’‚πŸπŸ

F3D(t) -

π’‚πŸπŸ‘ βˆ’π’‚πŸπŸ

οƒΌ K7 = οƒΌ K8 = οƒΌ K9 =

π’‚πŸπŸŽ +π’‚πŸπŸ“ βˆ’π’‚πŸπŸ π’‚πŸπŸ +π’‚πŸπŸ“ βˆ’π’‚πŸπŸ π’‚πŸπŸ‘ βˆ’π’‚πŸπŸ π’‚πŸπŸ’

= βˆ’πŸŽ. πŸπŸ‘ = 𝟐. πŸ‘π’™πŸπŸŽβˆ’πŸ‘

= βˆ’πŸŽ. πŸ“πŸ

οƒΌ K10 =

= βˆ’πŸŽ. πŸŽπŸπŸ”

οƒΌ Ο„2 =

π‘ͺ𝒗 = 𝟏𝟏. πŸπŸ’

βˆ’π’‚πŸπŸ π’‚πŸπŸ”

βˆ’π’‚πŸπŸ

ο‚· G5(s)= ο‚· G6(s)= ο‚· G7(s)= ο‚· G8(s)= ο‚· G9(s)=

𝑻𝑫 𝟐 (𝒔)

=

𝑭𝑫 𝟏 (𝒔) 𝑻𝑫 𝟐 (𝒔) 𝑻𝑫 𝟏 (𝒔) 𝑻𝑫 𝟐 (𝒔) 𝑭𝑫 𝟐 (𝒔) 𝑻𝑫 𝟐 (𝒔) 𝑭𝑫 πŸ‘ (𝒔) 𝑻𝑫 𝟐 (𝒔) 𝑫 𝑻𝒆𝒃 (𝒔) 𝑻𝑫 𝟐 (𝒔) 𝑻𝑫 𝒔 (𝒔)

=

π‰πŸ 𝑺+𝟏

=

𝟎.πŸŽπŸ•πŸ’πŸ

=

π‰πŸ 𝑺+𝟏 𝟏𝟏.πŸπŸ’ 𝑺+𝟏 π‘²πŸ”

=

ο‚· G10(s)=

π‘²πŸ“

π‘²πŸ• π‰πŸ 𝑺+𝟏 π‘²πŸ– π‰πŸ 𝑺+𝟏

=

𝟏𝟏.πŸπŸ’ 𝑺+𝟏

= =

π‘²πŸ— π‰πŸ 𝑺+𝟏

=

𝟏.πŸ“πŸ’

=

π‘²πŸπŸŽ π‰πŸ 𝑺+𝟏

βˆ’πŸŽ.πŸπŸ‘ 𝟏𝟏.πŸπŸ’ 𝑺+𝟏 𝟐.πŸ‘π’™πŸπŸŽβˆ’πŸ‘

𝟏𝟏.πŸπŸ’ 𝑺+𝟏

= =

βˆ’πŸŽ.πŸ“πŸ 𝟏𝟏.πŸπŸ’ 𝑺+𝟏 βˆ’πŸŽ.πŸŽπŸπŸ” 𝟏𝟏.πŸπŸ’ 𝑺+𝟏

ο‚· Qch(t) = Qs1(t) + Qvap(t) ο‚· UA [T2(t)-Ts(t)] = F1(t) Cp1 [Teb(t) – T1(t)] + Ξ»F3(t) 𝒂 𝒂 𝒂 𝒂 𝒂 ο‚· F3D(s) = πŸπŸ• T2D(s) + πŸπŸ–TsD(s) + πŸπŸ—F1D(s) + 𝟐𝟎 TebD(t) + 𝟐𝟏 T1D(t) 𝝀

𝝀

𝝀

𝝀

𝝀

ο‚· ο‚· ο‚· ο‚· ο‚·

a17 = 𝑼𝑨 a18 = βˆ’π‘Όπ‘¨ a19 = βˆ’π‘ͺπ’‘πŸ (𝑻𝒆𝒃 βˆ’ π‘»πŸ ) a20 = - π‘­πŸ π‘ͺπ’‘πŸ a21 = π‘­πŸ π‘ͺπ’‘πŸ

οƒΌ K11 = οƒΌ K12 = οƒΌ K13 = οƒΌ K14 = οƒΌ K15 =

π’‚πŸπŸ• 𝝀 π’‚πŸπŸ– 𝝀 π’‚πŸπŸ— 𝝀 π’‚πŸπŸŽ 𝝀 π’‚πŸπŸ 𝝀 𝑭𝑫 πŸ‘ (𝒔)

π’‚πŸπŸ•

ο‚· G11(s)=

=

ο‚· G12(s)=

=

ο‚· G13(s)=

=

ο‚· G14(s)=

=

ο‚· G15(s)=

=

𝑻𝑫 𝟐 (𝒔) 𝑭𝑫 πŸ‘ (𝒔) 𝑻𝑫 𝒔 (𝒔) 𝑭𝑫 πŸ‘ (𝒔) 𝑻𝑫 𝟏 (𝒔) 𝑭𝑫 πŸ‘ (𝒔) 𝑫 𝑻𝒆𝒃 (𝒔) 𝑭𝑫 πŸ‘ (𝒔) 𝑻𝑫 𝟏 (𝒔)

𝝀 π’‚πŸπŸ– 𝝀 π’‚πŸπŸ— 𝝀 π’‚πŸπŸŽ 𝝀

π’‚πŸπŸ 𝝀

= 𝟎. πŸŽπŸ’πŸ“ = βˆ’πŸŽ. πŸŽπŸ’πŸ“ = βˆ’πŸŽ. πŸπŸ’πŸ‘ = βˆ’πŸ’. πŸ‘πŸ = πŸ’. πŸ‘πŸ

Teb(t) = Teb + 0.15 Keb CA2(t) TebD(t) = a22 CA2D(t) a22 = 0.15Keb 𝑻𝑫 𝒆𝒃 (𝒔) = π’‚πŸπŸ π‘ͺ𝑫 (𝒔) π‘¨πŸ

οƒΌ K16 = π’‚πŸπŸ = 𝟎. πŸπŸ“ Keb ο‚· G16(s)=

𝑻𝑫 𝒆𝒃 (𝒕)

π‘ͺ𝑫 π‘¨πŸ (𝒕)

= K16 = 0.078

BALANCE DE MASA PARA LA CHAQUETA ο‚· FS1(t) - FS2(t) = 0 ο‚· FS1(t) = FS2(t) BALANCE DE ENERGÍA PARA LA CHAQUETA ο‚· FS(t) Cp TO(t) – FS(t) CP TS(t) – Q1(t) = ρ Vc Cv

𝒅 𝑻𝑺 (𝒕) 𝒅𝒕

ο‚· FS(t) Cp TO(t) – FS(t) CP TS(t) – UA [Ts(t) – T2(t)] = ρ Vc Cv ο‚· FS(t) [Cp TO(t) - CP TS(t)] – UA [Ts(t) – T2(t)] = ρ Vc Cv ο‚· a23FsD(t) + a23ToD(t) + a25TsD(t) + a26T2D(t) = a25

𝒅 𝑻𝑫 𝒔 (𝒕) 𝒅𝒕

𝒅 𝑻𝑺 (𝒕)

𝒅𝒕 𝒅 𝑻𝑺 (𝒕) 𝒅𝒕

ο‚· ο‚· ο‚· ο‚· ο‚·

a23 = π‘ͺ𝒑 π‘»πŸŽ - π‘ͺ𝒑 𝑻𝒔 a24 = π‘ͺ𝒑 𝑭𝒔 a25 = -UA - π‘ͺ𝒑 𝑻𝒔 a26 = UA a27 = ρ Vc Cv

οƒΌ K17 = οƒΌ K18 = οƒΌ K19 = οƒΌ Ο„3 =

π’‚πŸπŸ‘ βˆ’π’‚πŸπŸ“ π’‚πŸπŸ’ βˆ’π’‚πŸπŸ“ π’‚πŸπŸ” βˆ’π’‚πŸπŸ“ π’‚πŸπŸ•

βˆ’π’‚πŸπŸ“

ο‚· G17(s)=

𝑻𝑫 𝒔 (𝒔)

𝑭𝑫 𝒔 (𝒔) 𝑫 π‘»πŸ (𝒔) 𝑻𝑫 𝟎 (𝒔) 𝑻𝑫 𝒔 (𝒔) 𝑻𝑫 𝟐 (𝒔)

=

ο‚· G18(s)=

=

ο‚· G19(s)=

=

π‘²πŸπŸ• π‰πŸ‘ 𝑺+𝟏 π‘²πŸπŸŽ

π‰πŸ 𝑺+𝟏 π‘²πŸπŸŽ π‰πŸ 𝑺+𝟏

= = =

𝟎.πŸŽπŸ’πŸ• 𝟎.πŸ—πŸ“π‘Ί+𝟏 𝟎.πŸ—πŸ“ 𝟎.πŸ—πŸ“ 𝑺+𝟏 𝟎.πŸŽπŸ“πŸ‘ 𝟎.πŸ—πŸ“ 𝑺+𝟏

a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16 a17 a18 a19 a20 a21 a22 a23 a24 a25 a26 a27

5 [lb/min] 0,032 -3,3 [lb] -0,048 0,048 35 [ lb] 1247 9493,85 -1599,68 -1590,68 -6165,94 -776,2 -3113,5104 -100 790,4 33022,08 100 -100 -314,7 -9493,85 9493,85 0,078

K1 T0 T1 T2 TEB Cp Cv F1 F2 F3 UA Ξ»

1,52 423 298Β°k 100,17 Β°C 4,186 [J/gΒ°K] 2,08 [J/gΒ°K] 5,5 [lb/min] 3,3 [lb/min] 1,7 [lb/min] 100 2204,586

EVAPORADOR DE EFECTO SIMPLE

NATHALIA FALK URIBE JÚLIAN GONZALEZ JEREZ JUAN CAMILO MORENO SEGOVIA FERMÍN ORTEGA GARCIA

GIOVANNY MORALES MEDINA

UNIVERSIDAD INDUSTRIAL DE SANTANDER ESCUELA DE INGENIERIA QUÍMICA BUCARAMANGA

29 DE MARZO DE 2019 OBJETIVOS  Aplicar los conceptos obtenidos durante la asignatura control de procesos, para poner en practima en el tema asignado evaporador de efecto simple  Modelar un evaporador de efecto simple bajo condiciones especiales para un funcionamiento Γ³ptimo sin inconvenientes a futuro  Manejar herramientas como Simulink para realizar el diagrama de bloques y sus respectivos cΓ‘lculos, ecuaciones y condiciones que permite el programa para el modelamiento y lineralizaciΓ³n del proceso.  Se muestra el desarrollo del sistema de control y modelo matemΓ‘tico para un evaporador de efecto simple utilizado para dos compuestos

DESCRIPCIΓ“N DEL PROCESO

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