Question

Exercise

Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h^-1 of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m^-2 s^-1 °C^-1 in the first, second and third effects respectively. Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect.

Answer. A₁ = 2.4 m² = A₂ = A₃

Steam consumption: 0.35kg steam/kg water evaporated.

Note: Trial and error solution is not required here.

Answer 1

Mass balance (kg h^-1)

	Solids	Liquids	Total
Feed	50	450	500
Product	50	117	167
Evaporation			333

Heat balance

From steam tables, the condensing temperature of steam at 200 kPa (g) is 134°C and the latent heat is 2164 kJ kg ^-1.

Evaporating temperature in final effect under pressure of 60 kPa (abs.) is 86°C, as there is no boiling-point rise and latent heat is 2294 kJ kg^-1.

Equating the heat transfer in each effect:

q₁ = q₂ = q₃

U₁A₁DT₁  = U₂A₂ DT₂ = U₃A₃DT₃

And

DT₁ + DT₂ + DT₃ = (134 - 86) = 48°C.

Now, if

A₁  = A₂ = A₃

then

DT₂  = U₁DT₁ /U₂

and

DT₃ = U₁DT₁ /U₃

so that

DT₁(1 + U₁/U₂ + U₁/U₃) = 48,

DT₁ x [1 + (2270/2000) + (2270/1420)] = 48

3.73DT₁ = 48

DT₁ = 12.9°C,

DT₂ = DT₁ x (2270/2000) = 14.6°C

and

DT₃ = DT₁ x (2270/1420) = 20.6°C

And so the evaporating temperature:

In first effect is          (134 - 12.9) = 121°C; latent heat (from Steam Tables) 2200 kJ kg^-1.
In second effect is     (121 - 14.6) = 106.5°C; latent heat 2240 kJ kg^-1
In the third effect is (106.5 - 20.6) = 86°C, latent heat 2294 kJ kg^-

Equating the quantities evaporated in each effect and neglecting the sensible heat changes, if w₁, w₂, w₃ are the respective quantities evaporated in effects 1,2 and 3, and w_s is the quantity of steam condensed per hour in effect 1, then

w₁ x 2200 x 10³ = w₂ x 2240 x 10³
= w₃ x 2294 x 10³
= w_s x 2164 x 10³

The sum of the quantities evaporated in each effect must equal the total evaporated in all three effects so that:

w₁ + w₂ + w₃ = 333 and solving as above,

w₁ = 113 kg h^-1   w₂ = 111kg h^-1     w₃ = 108kg h^-1   w_s = 115 kg h^-1

Steam consumption

It required 115 kg steam (w_s) to evaporate a total of 333 kg water, that is
0.35kg steam/kg water evaporated.

Heat exchanger surface.

Writing a heat balance on the first effect:

(113 x 2200 x 1000)/3600 = 2270 x A₁ x 12.9

A₁ = 2.4 m² = A₂ = A₃

total area = A₁ + A₂ + A₃ = 7.2 m²