Question

A single-effect evaporator generates a distillate product at a flow rate of 2 kg/s. The system operating temperatures are as follows: • The boiling temperature, Tb is 90 ºC. • The feed temperature, Tf, is 85 ºC. • The steam temperature, Ts, is 100 ºC. • Boiling Point Elevation is neglected. • The Temperature of cooling Water, Tcw, is 25 ºC.

Determine the heat transfer areas in the evaporator and the condenser, the thermal performance ratio, the flow rates of feed seawater and reject brine, and the flow rate of cooling seawater.

Assume that the specific heat of seawater is constant and equal to 4.2 kJ/kg.ºC, and assume the over all heat transfer coefficeint for evapporater and condenser is 2 kW/m2 .ºC ,and assume the salinity of the feed is 42,000 ppm and rejected brine salinity is 84,000 ppm.

Answer 1

Solution: The solution proceeds with evaluation of the vapor temperature. This requires calculation of the boiling point elevation (BPE) using the correlation :

BPE = (0.0825431 +0.0001883×Tb+0.00000402(Tb^2) (Ts-Tb) +0.0007625 +0.0000902 (Tb) - 0.00000052 (Tb^2)((Ts-Tb)^2)+0.0001522-0.000003 (Tb) - 0.00000003 (Tb^2 ) ((Ts-Tb)^3)

= 0.90 C

The resulting temperature of the vapor formed in the evaporator (Tv) is calculated from Eq.

Tv= Tb - BPE = 90 -0.903 = 89.1°C

The temperatures of the heating steam and vapor are used to calculate the latent heat for the steam and distillate vapor, and are calculated from the correlations. hee resulting values are:

= 2501.897149 – 2.407064037 Ts +1.192217 x10^-3 (Ts^2) -1.5863x10^-5 (Ts^3) = 2501.897149 -2.407064037 (100) +1.192217 x10^-3 (100^2)-- 1.5863x10^-5 (100^3)

= 2264.1 JK/kg

= 2501.897149 – 2.407064037 Tv+1.192217 x10^-3 (Tv^2) - 1.5863x10^-5 (Tv^3 )

= 2501.897149- 2.407064037 (89.1) +1.192217 x10^-3 (89.1^2) - 1.5863x 10^-5(89.1^3

=2284.6 kJ/kg

The overall heat transfer coefficients in the evaporator and condenser are calculated using the correlations . The resulting values for the two coefficients are:

Ue = 1.9695+1.2057x10^-2Tb-8.5989x10^-5(Tb^2)+2.5651x10^-7(Tb^3)  

= 3.054 KJ)s m^2 C

Uc = 1.7194+3.2063x10^-3 Tv +1.5971x10^-5(Tv^2)-1.9918x10^-7(Tv^3)

= 1.986 kJ/s m^2 C

The system performance parameters are calculated . The thermal performance ratio is given by

PR = Md/Ms