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 1 of 4: Boiling point elevation

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\Delta T_b = i\,K_b\,m = 2\times 0.5120\times 0.6114 = 0.626\ \text{K} \Rightarrow T_b = 373.78\ \text{K} \approx 100.63\ \text{°C}

Analysis

For 100 g of seawater: 3.45 g NaCl ( $M = 58.44$ ) in 96.55 g water gives $m = 3.45/58.44/0.09655 = 0.6114$ mol kg $^{-1}$ ; with $i = 2$ for full dissociation, $\Delta T_b = 2\times 0.5120\times 0.6114 = 0.626$ K and $T_b = 373.15 + 0.626 = 373.78$ K.