TRIPPLE_EFFECT_EVAPORATOR_DESIGN_CALCULATION

Calculation For Triple Effect Evaporator Forward Feed DESCRIPTION OF PROBLEMYou are required to design a triple effect evaporator to concentrate 25000 kg/h of solution at 38 deg C, containing 10% solids to a product which contains 50% solids. Steam is available at 1.877kg/cm2 and the last efect will be assumed to be operated at vacuum of 66 cm Hg referred to 66 cmhg barometer. Water is available at 29 deg C for use in barometric condenser. Assume negligible BPE . average specific heats of 1 in all the effect. The condensate from each effect leaving at its saturation temp egligible radiation losses, avg pressure drop equal in each effect and ht and dia of each tube is 150 cm and 5 cm respectively. Find out the no of tubes required for the purpose , heat consumption and evaporation per kg of steam, if U1,U2,U3 are 2930 ,1220 and 610 kcal/hr.m2.deg C respectively.

CALCULATIONS (part-1) Applying overall material balance F = E + P3 component balance F*XF = P1*Xp1 P1*XP1=P2*Xp2 P2*XP2 = P3*Xp3 hence: F*Xf = P3 * Xp3 P3 = 5000 E = (F - P3)= 20000 Temperature distribution and calculation ∆ T = Ts - Tv3 -∑Bpr ∑Bpr

=

∆T

0 38.4

(3*10)(Given data)

Enthalpy chart

values (Kcal/Kg)

reff(data book)

hf (feed enthalpy) he1(1st effect) he2(2nd effect) he3(3rd effect)

38 106.5 90 51.6

Table 7 Table 7 Table 7 Table 7

Values obtained after suitable interploation techniques Areas 1st effect 2nd effect 3rd effect Steam

latent heat (Kcal/Kg) (λ)

enthalpy (Kcal/Kg)(E)

Reff(Data book)

534.8

641.5

Table 7

545.3 568.1 527.5

635.3 619.6

Table 7 Table 7 Table 7

CALCULATIONS (part-2) S.λs + F.hf = E1.He1 + P1.H1

S(527.5)+(E2+E3)536=12412500

Material balance for 1st effectP1 = F- (E- E2- E3) P1 = 10000 - (8250 - E2 - E3) on solving E2+E3= 13820 For 2nd effect : p1.hp1 + E.λe1 = E2.HE2 + P2.Hp2 1063.6E2+518.3E3=10708500

For 3rd effect : P3 = 5000 P2.hp2 + E2.λE2 = E3.HE3 + P3.hp3 E2=0.971E3-352.1

On furthur solving the amount steam required will be

S = 9488 kg/hr Steam economy = (E/S) = 2.1709

Notations

Values

Unit

Reference

F E E1 E2 E3 P1 P2 P3 Xf Xp3

Feed rate 25000 Overall Vapour streams E1 + E2 + E3 20000 Triple Effect Evaporator Forward FeedVapourEstream - (E2 +atE3) 1st effect 6180 Vapour stream at 2nd effect 6630 Vapour stream at 3rd effect 7190 Product Stream 1st effect 18820 Product Stream 2nd effect P2 = E3 + P3 12190 Product Stream 3rd effect 5000 Solute composition in feed 0.1 Solute composition in product 0.5

Description

Formules

Kg/hr Kg/hr Kg/hr Kg/hr Kg/hr Kg/hr Kg/hr Kg/hr

given

A1

Area of 1st effect

A1 = A2 = A3

A2

Area of 2nd effect

-

164

m2

A3

area of 3rd effect

-

m2

Ps

Steam pressure

-

152.3 26.69

Ts

Steam temperature

-

117.8

o

C

o

C

151.2

given given 2

m

Psi Table 7

Tw Pv3

Water temperature Pressure at last effect

-

29 1.97

Tv3

temperature at last effect

-

51.6

o

C

Table 7

11.3

o

C

given

16.5

o

C

given

38.4

o

C

∆T1

-

∆T2

-

∆T3 U1 U2

Overall heat transfer coeff Overall heat transfer coeff Overall heat transfer coeff

1st effect 2nd effect

Psi

given 2o

given

2o

given

2o

given

2930 Kcal/hr m C 1220 Kcal/hr m C

3rd effect

610 Kcal/hr m C o C 0

I

II

III

Steam

Ts 0

Te1 106.5

Te2 90

Liquid

Tl1 0

(TE1-∆T2) 106.5

(TE2-∆T3) 90

Vapour

0

106.5

90

U3 Bpr for each

Temperature distribution table Effect

Area Calculations A1 A2

E1.λe1/ (U2.∆T2)

151.2 164.2

A3

E2.λe2/ (U3.∆T3)

154.3

S.λs / (U1.∆T1)

Mass flow rate calculation E3.λe3 = m cp ∆T

m = (E3.λe3)/(cp∆T) 178368.51

NO OF TUBES=(MAX AREA)/(3.14*Do*L)

Number of Tubes (N)= (164.2/(3.14*5*1.5)*100 (N)=697 Evaporation/ kg Steam =

E/S = 20000/9488 =

2.11

Reference

100