Question

a single effect evaporator is used to concentrate a solution of 10% of total solids to 30% of total solids at a rate of 250 kg / h. if the evaporator pressure is 77 kpa, and if steam is available at 200 kpa gauge, the amount of steam needed per hour and the area of ​​heat transfer area if the overall heat transfer coefficient is 1700 jm -22s-1 ° C. the feed temperature is 18 ° C and the boiling point of the solution below 77 kpa is 91 ° C. the specific heat of the solution is the same as water, which is 4,186 x 103 jkg-1, and the latent heat of the evaporation of the solution is the same as the water under the same conditions
a. the amount of steam needed (kg/hour)
b. amount of heat moved area (m²)

Answer 1

From steam tables, the condensing temperature of steam at 200 kPa (gauge)[300 kPa absolute] is 134°C and latent heat 2164 kJ kg^-1; the condensing temperature at 77 kPa (abs.) is 91°C and latent heat is 2281 kJ kg^-1.

Mass balance (kg h^-1)

	Solids	Liquids	Total
Feed	25	225	250
Product	25	58	83
Evaporation			167

Heat balance
Heat available per kg of steam
                                       = latent heat + sensible heat in cooling to 91°C
                                       = 2.164 x 10⁶ + 4.186 x 10³(134 - 91)
                                       = 2.164 x 10⁶ + 1.8 x 10⁵
                                       = 2.34 x 10⁶ J

Heat required by the solution
                                       = latent heat + sensible heat in heating from 18°C to 91°C
                                       = 2281 x 10³ x 167 + 250 x 4.186 x 10³ x (91 - 18)
                                       = 3.81 x 10⁸ + 7.6 x 10⁷
                                       = 4.57 x 10⁸ J h^-1

Now, heat from steam       = heat required by the solution,
Therefore quantity of steam required per hour = (4.57 x 10⁸)/(2.34 x 10⁶)
                                       = 195 kg h^-1

Quantity of steam/kg of water evaporated = 195/167
                                       = 1.17 kg steam/kg water.

Heat-transfer area
Temperature of condensing steam = 134°C.
Temperature difference across the evaporator = (134 - 91) = 43°C.
Writing the heat transfer equation for q in joules/sec,
                                    q = UA DT

              (4.57 x 10⁸)/3600 = 1700 x A x 43
                                     A = 1.74 m²

Area of heat transfer surface = 1.74 m²