Question

A single effect evaporator is used to increase the concentration of a 44% NaOH solution to 65%. The initial flow rate of the feed solution (40°C) is 50,000 kg/h. Steam at 3 atm is used to heat the system. The temperature in the vapor space is 40°C and the overall heat transfer coefficient is estimated as 2,000 W/m2 ·°C. Calculate the amount of steam consumed and surface area needed for heating.

Answer 1

Overall balance

F = L + V

50000 = L + V

NaOH balance

F * xF = L * xL + V *xV

50000 x 0.44 = 0.65 L + 0

L = 33846.15 kg/h

V = F - L

= 50000 - 33846.15

= 16153.85 kg/h

Vapor space temperature T1 = 40°C = 313 K

Latent heat of water Hv = 2406 kJ/kg

Steam pressure = 3 atm x 101.325 kPa /atm

= 303.975 kPa

Saturation temperature Ts = 133.975°C = 407.12 K

Latent heat of steam = 2162.12 kJ/kg

Enthalpy of feed Hf = Cpf x(Tf - T1) = Cpf x (40-40) = 0

Enthalpy balance

F Hf + S = L HL + V Hv

S = 0 + 16153.85 x 2406/2162.12

Steam rate S = 17975.95 kg/h

Heat transfer

Q = S = 17975.95 x 2162.12 = 38866163.1 W

Q = UA(Ts - T1)

38866163.1 = 2000 x A x (407.12 - 313) K

Area A = 206.47 m2