Procedura de calcul al fluxului de caldura necesar prepararii apei calde de consum

The procedure for calculating the heat flow necessary for preparing hot water

a )Regim de baza - vara

a)Basic scheme-summer

Φ_acb  = G_ac x C_2b x (  t_ac – t _{21 b} ) / 3600

G_ac = 0.8 x  3600 / 0.9832 = 2929 kg/h

Φ_acb  = 2929 x 4198.6 x (  60 - 10 ) / 3600

Φ_acb  = 170800 w

b)Regim de varf - iarna

b) Mode  peak - winter

Φ_acv  = G_ac x C_2v x (  t_ac – t _21v ) / 3600

Φ_acv  = 2929 x 4198.6 x (  60 - 7  ) / 3600 = 181048

Φ_acv  = 181048

Debitul de agent termic primar

Agent termic primar apa

Primary heat flow

Primary water heating

Regim de baza – vara

a)Basic scheme-summer

Φ : flux de caldura  [  w ]

Φ: heat flux [w]

acb : apa calda de consum [ regim de baza , varf ]

acb : hot water [the regime basic, top]

G : debit de fluid [ kg/h ]

G: flow of fluid [kg / h]

C : caldura specifica a fluidului [  j/kg x grd ]

C: specific heat of fluid [J / kg x degrees]

2b : circuitul primar in regim de baza - valoarea finala

2b: the primary circuit in the scheme - final value

t  :  temperatura [ ° C ]

t: temperature [° C]

ac : apa calda de consum in regim de baza

Action: hot water in the basic regime

2.1b : circuit secundar in regim de baza - valoarea initiala

2.1B: secondary circuit in the scheme - initial value

Ψac : greutate specifica  [ kg/m³ ]

Ψac: specific weight [kg/m3]

                                   TABEL  VALORI DE STARE PENTRU APA

VALUES OF STATE FOR WATER TABLE

Tm		Ψac	Cp
° K                 ° C		kg/m3	j/kg x °K
303	30	995.6	4198.6
313	40	992.2	4198.6
323	50	988	4198.6
333	60	983.2	4202.7
343	70	977.7	4211.1
353	89	971.8	4215.3
363	90	965.3	4223.7
373	100	958.3	4227.9
383	110	951.7	4238.3
393	120	943.1	4248.8

                                      

t _ac    = 60° C

t _{21 b}= 10° C

C_2b = 4198.6 j/kg °K pentru 35 ° C

C_1b = 4200    j/kg °K pentru 55 ° C

C_1v = 4220    j/kg °K pentru  85 ° C

Calculul debitului

Flow Calculation

CALCULUL VOLUMULUI DE ACUMULARE

STORAGE VOLUME CALCULATION

Q_A = 3600 x Φ_acb x 0.8 = 3600 x 170800 x 0.8 = 491.904.000

Volum de apa de consum ce trebuie acumulat

Volume of water consumption be accumulated

V = Q_A x 45 / ( t_ac - 10  ) x C_2b x Ψac x 60

V = 491.904.000 x 45/( 60 -10  ) x 4198.6 x 983.2 x 60

_{V = 1.49 mc}

_{Boiler vertical                                                                                                                                              Vertical Boiler}

_{QA = 3600 170800 x 0.57 = 350481600}

_{V =} 350481.600 x 45/( 60 -10  ) x 4198.6 x 983.2 x 60 =

_{V =1.2mc}

_{Se alege un boiler vertical cu capacitatea de 1600 litri}

_{Choose a vertical boiler with a capacity of 1600 liters}