Physical Separation Tuto

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 1. A feed containing 2 wt% dissolved organic solids in water is fed to a double-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling- point rise can be considered negligible. Each evaporator has a 1000 ft² surface area, and the heat transfer coefficients are U1=500, and U2=700 btu/h.ft.F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure in the vapour space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate Lj of a solution containing 25 % solids. (Hint: Assume a feed rate of, say F=1000 lbm/h and a value of T). Calculate the area and T1. Then calculate the actual feed rate by multiplying 1000 by 1000/calculated area.)

Answer :

V at Ti 2 at 0.98 Psi T2 at 100F Steam Ti Pq=0.98psi = OF 0.02 at 100 psia my 3q=0.25 stood +42 9 Le at 0.98 psi Li at Ti botat 100°F) H$2 = 1037 btuleb 700 X 1000 X 94.155 - V2 X 10 34 = in = 6.358103 lbl for 7.358104 do for va V+V2 Nor, Favtli , 12

2.

Answer :

As per the given data:

Solids in feed= 2%

Solids in concentrate= 25%

HTA1 = HTA2 = 1000 ft^2

U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F

Overall material balance: Feed= Distillate + concentrate ----> Eq-1

Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25

Feed = 12.5 * concentrate ---> Eq-2

Boiling point rise = negligible, so solution & solvent vapor temperature will be same.

Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).

As per the data:

Latent heat @ 212F = 300 Btu/lb

Latent heat @ 100F = 320 Btu/lb

As per material balance:

Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT

1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr

2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr

Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr

Substituting the above in Eq-1

Feed = 435,000 + concentrate

Substitute Eq-2 in the above

12.5 * concentrate = 435,000 + concentrate

Concentrate, L1 = 435,000/11.5 = 37826 lb /hr

Feed, F = 435,000 + 37826 = 472,826 lb/hr

1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.

M2- Vapour @ 100F M1- Vapour @ 212F Condensate @ 212F Effect-1 Effect-2 Steam @ 326F 14.5 PSI 0,98 PSI Feed @ 100F Concentrat

3.

Question :

A single-effect evaporator is concentrating a feed of 9072 g/h of a 10 wt% solution of NaOH in water to a product of 50% solids. The pressure of the saturated steam used is 42 kPa (gage) and the pressure in the vapor space of the evaporator is 20 kPa (abs). The overall heat-transfer coefficient is 1988 W/m2*K. a) calculate the steam use, steam economy and area if the feed temperature at 288.8 k (15.6 c) b) predict the new product composition if the feed rate is reduced to 4536 kg/h

Answer :

@ Single effect evaporados may Howrates TIP P condensate v (reglam) Yv, hv To alp, he peed f F, LV, S are Kolm) So vapour spaGIVEN DATA Single effect evaporator > N, Y F = 9072 Hbr Xp z Oil (NaOH) P, T 14 4= 0.5 FNE Ps, Is So As steamin 1 p sad z 126FNE Leta STEPTO Using Energy balance Sts = F (p (T-Tp) + Vi= Voto (To-T) (110 Sasa UoAo (ToT) (17) (Where To z Temp of sat stSep use steam table Temp of Vapour cewater) space Tz 60°c or. 3334 at zokpa Latent heat of steam As a 2260 이 steam As a 22600Av = 0.027m2-7 J767 (VTT Sts z U, A. (TS-T) 3 x 2260 1988 x36wy 0.027 (383-333) S 2 4275 g/h] = S[HP (VMI Economy ( S 7257.6

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