Steam Requirement of Single Effect Evaporator

10,000 kg/hr of an aqueous feed containing 1% dissolved solids is to be concentrated to 20% solids, in a single effect evaporator. The feed enters at 25^oC. The steam chest is fed with saturated steam at 110^oC. The absolute pressure maintained in the evaporator is such that the water will boil at 55^oC. The boiling point elevations are as follows:

Feed: 0.2^oC
20% solution: 15^oC

The overall heat transfer coefficient, under normal operating conditions would be 2500 W/m².^oC

Estimate the steam requirement assuming no sub cooling of condensate, heat load on the condenser, and the heat transfer area.

Calculations: The given data are represented in the following diagram.

Let us have the following notations:

Feed: F

Concentrated product: P

Water vapor: V

Steam: S

Solid balance:

F x 0.01 = P x 0.2

P = 10000 x 0.01 / 0.2 = 500 kg/hr

V = F - P = 10000 - 500 = 9500 kg/hr

Temperature of Water vapor, leaving from the evaporator

= 55^oC + Boiling point elevation = 55^oC + 15^oC = 70^oC

Enthalpy of feed at 25^oC (H_F)= 104.8 kJ/kg (the data for water at 25^oC - from Steam Tables)

Enthalpy of product at 70^oC (H_P)= 293.0 kJ/kg (the data for water at 70^oC - from Steam Tables)

Pressure in the evaporator vapor space = 15.74 kPa(abs) (saturation pressure of water vapor at 55^oC - from Steam Tables)

Enthalpy of water vapor leaving at 70^oC and 15.74 kPa(abs) (H_V)= 2640 kJ/kg (from Mollier Diagram)

Enthalpy of saturated water at 15.74 kPa(abs) = 230.2 kJ/kg

Latent heat of steam at 110^oC (λ_S)= 2230 kJ/kg (from Steam Tables)

F H_F + S λ _S = V H_V + P H_P

10000 x 104.8 = S x 2230 = 9500 x 2640 + 500 x 293

S = 10842.4 kg/hr

Steam requirement = 10842.4 kg/hr = 3.01178 kg/sec

Heat load on Condenser (assuming condensate water leaves as saturated liquid corresponding the vapor space pressure) = 9500 x (2640 - 230.2) = 22893100 kJ/hr = 6359.2 kW

Q = Rate of heat transfer through heating surface from steam = U A DT

i.e.,

S λ_S = U A ΔT

3.01178 x 2230 = 2.5 x A x (110 - 70)

A = 67.2 m²

Heat transfer area = 67.2 m²