Chemistry Labs

Problem 5

A Dubai desalination plant uses multi-stage flash (MSF) desalination: seawater is heated at high pressure then flashed, releasing pure water vapour. Dubai seawater is at T0=25.00T_0 = 25.00 °C and contains 3.45 % NaCl by mass (assume complete ionisation); the boiling point elevation constant is Kb=0.5120K_b = 0.5120 K kg mol−1^{-1} and the latent heat of vaporisation of water is Evap=2260E_{\text{vap}} = 2260 kJ kg−1^{-1} (40.716 kJ mol−1^{-1}). (a) Calculate the boiling point of Dubai seawater at atmospheric pressure. (b) Calculate its boiling point at p=2.50p = 2.50 atm using the Clausius–Clapeyron equation. (c) A 100 L flash chamber holds 1.00 kg of seawater at 90.0 °C; a second 1.00 kg portion overheated to 110.0 °C is added, pressure is reduced, and the chamber equilibrates at Tf=97.0T_f = 97.0 °C. With cp=3.85c_p = 3.85 kJ kg−1^{-1} K−1^{-1}, calculate the amount nn (mol) of water that vaporised. (d) The plant produces 50 000 m³ of pure water per day, extracting overall 85 % of the water present in the seawater; calculate the mass of Dubai seawater needed per day.
Step 4 of 4: Daily seawater demand
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msea=5.0×104×1030.85×0.9655=6.1×107 kg per daym_{\text{sea}} = \dfrac{5.0\times 10^{4}\times 10^{3}}{0.85\times 0.9655} = 6.1\times 10^{7}\ \text{kg per day}
Analysis

50 000 m³ of product water is 5.0×1075.0\times 10^{7} kg. Seawater is 96.55 % water and only 85 % of that water is recovered, so the feed is 5.0×107/(0.9655×0.85)=6.1×1075.0\times 10^{7}/(0.9655\times 0.85) = 6.1\times 10^{7} kg per day.