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Steam Requirement of Single Effect Evaporator

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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^{o}C. The steam chest is fed with saturated steam at 110^{o}C. The absolute pressure maintained in the evaporator is such that the water will boil at 55^{o}C. The boiling point elevations are as follows:

Feed: 0.2^{o}C

20% solution: 15^{o}C

The overall heat transfer coefficient, under normal operating conditions would be 2500 W/m^{2}.^{o}C

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

Mass balance:

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

Energy balance:

Temperature of Water vapor, leaving from the evaporator

= 55^{o}C + Boiling point elevation = 55^{o}C + 15^{o}C = 70^{o}C

Enthalpy of feed at 25^{o}C (H_{F})= 104.8 kJ/kg (the data for water at 25^{o}C - from Steam Tables)

Enthalpy of product at 70^{o}C (H_{P})= 293.0 kJ/kg (the data for water at 70^{o}C - from Steam Tables)

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

Enthalpy of water vapor leaving at 70^{o}C 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^{o}C (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

Heat transfer area estimation:

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 m^{2}

Heat transfer area = 67.2 m^{2}

[Index]

Last Modified on: 01-May-2024

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