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.

can you show how you calculated the latent heat of water and the latent heat of the steam

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 (from steam tables)

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 (from steam tables)

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