bilant bere

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BILANT DE MATERIALE BERE Productia pe an : P an =1110 tone/an = 1110000 kg/an bere Fondul anual de timp: F.A.T = 330 zile; Durata unei sarje: t s = t f + t aux ; Durata fermentatiei: t f = 408 ore; t aux = 40ore. t s = 408+40=448 ore Numarul de sarje: N s = F.A.T t s = 330 448 =17,6 18 sarje Productia pe sarje: P s = P an N s = 1110000 18 =61666,66667 kg / sarja Randamentul global este : η g = i=1 n ∙η i - sedimentare: η g =90% - centrifugare: η g =85% - filtrare : η g =90%

Transcript of bilant bere

Page 1: bilant bere

BILANT DE MATERIALE

BERE

Productia pe an : Pan=1110 tone/an = 1110000 kg/an bere

Fondul anual de timp:

F.A.T = 330 zile;

Durata unei sarje:

ts= tf + taux ;

Durata fermentatiei:

tf = 408 ore; taux = 40ore.

ts= 408+40=448 ore

Numarul de sarje:

Ns=F . A .T

t s=330

448=17,6≅ 18 sarje

Productia pe sarje:

Ps=Pan

N s=1110000

18=61666,66667 kg /sarja

Randamentul global este :

ηg=∏i=1

n

∙ ηi

- sedimentare: ηg=90 %

- centrifugare: ηg=85 %

- filtrare :ηg=90 %ηg=0,9 ∙0,85 ∙0,9=¿ηg=0,6885%

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Productia in fermentatie

Pf =P s

ηg=61666,66667

0,6885=¿P f =89566,69088

Volumul util

V u=Pf

ρ=89566,69088

1030,1=¿V u=86 , 94951061 m3

Calculul compozitiei din must

1,04 kg must……………117,5g∙10-3 kg zaharuri fermentescibile(Z.F)89566,69088……………z

z =10119,31363 kg Z.F

1,04 kg must……………13,5 g∙10-3kg zaharuri nefermentescibile(Z.N)89566,69088……………t

t = 1162,644545kg Z.N

1,04 kg must…………….5 g acizi(Ac)89566,69088…………….u

u = 430,6090908 Ac

Mustul contine 7,4% extract1kg must……………96,602 kg89566,69088……….v

v = 82940,54709 kg H2O

Mmust = MH2O + MZ.F + MZ.N + MAc Mmust = 82940,54709 + 10119,31363 + 1162,644545 + 430,6090908Mmust = 94653,11436 kg must.

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Drojdie: 0,5kg/100 L must100 L must……………0,5 kg drojdie86949,51061………….k

k = 434,7475531 kg drojdie;

BERE100kg must……………4,4 kg compusi secundari10119,31363………….x

x = 445,2497997 kg compusi secundari;

100kg must……………46,8 kg CO2

10119,31363………….yy = 4735,838779 kg CO2

100kg must……………48,8 kg etanol10119,31363………….w

w = 4938,225051 kg etanol

Masa de apa evaporata

100 L must……………0,3 L H2O evaporata86949,51061…………a

a = 260,8485318

Contractia de volum

∆ v=86949,51061 ∙ 0,5

100=434,7475531kg H 2 O

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In bere ramane dizolvat 0,4 kg CO2/ 100L must

100 L must(inmultim cu densitatea 1,04)……………0,4 kg CO2

104 kg must……………0,4 kg CO2

89566,69088…………...bb = 344,4872726 kg CO2

M CO2 degajat = 4735,838779 – 344,4872726 = 4391,351506

Materii intrate Valoare Materii iesite Valoare

MUST 94653,11436 BERE TANARA 89566,69088- zaharuri fermentescibile 10119,31363 - etanol 4938,225051- zaharuri nefermentescibile 1162,644545 - CO2 344,4872726- acizi 430,6090908 - Compusi secundari 445,2497997

DROJDIE 434,7475531 - H2O 82679,69856

TOTAL 95087,86191DROJDIE + PIERDERI 869,4951061CO2 degajat 4391,351506H2O evaporat 260,8485318TOTAL 95088,38602