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

T_0 = 25.00

°C and contains 3.45 % NaCl by mass (assume complete ionisation); the boiling point elevation constant is

K_b = 0.5120

K kg mol

^{-1}

and the latent heat of vaporisation of water is

E_{\text{vap}} = 2260

kJ kg

^{-1}

(40.716 kJ mol

^{-1}

). (a) Calculate the boiling point of Dubai seawater at atmospheric pressure. (b) Calculate its boiling point at

p = 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

T_f = 97.0

°C. With

c_p = 3.85

kJ kg

^{-1}

^{-1}

, calculate the amount

n

(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 3 of 4: Water vaporised by flashing

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Intuition

Flash vaporisation is self-cooling: the sensible heat released when the hot brine cools supplies the latent heat of the vapour it produces.

Q_{\text{cool}} = 2\times 3850\times(373.15-370.15) = 23100\ \text{J} = n\,(40716 + 0.018\times 3850\times 3) \Rightarrow n = 0.56\ \text{mol}

Analysis

The 2.00 kg of liquid cooling from 100 °C (the mixed mean temperature between 90 and 110 °C; more precisely the hot portion supplies the heat) down to 97 °C releases $Q \approx 2\times 3.85\times 3.0 = 23.1$ kJ; dividing by the latent heat per mole $2260\times 0.01802 = 40.72$ kJ mol $^{-1}$ gives $n = 0.57$ mol (the official balance accounting for the shrinking liquid mass yields $n = 0.565$ mol).