IChO 2025 · Theory problem 5

Solar-powered multi-stage flash desalination

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5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 Total
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Freshwater is scarce in the arid climate of the UAE. The country relies on solar-powered desalination plants to produce freshwater. A Dubai desalination plant uses a multi-stage flash (MSF) desalination process. In MSF desalination, seawater is heated by passing through a series of heat exchangers to raise its temperature, reaching its saturation temperature at a high pressure. When pressure is reduced, this superheated seawater gives off pure water vapour to become saturated again, a process known as flashing.

Figure 1

Assume seawater in Dubai is at \(T_0 = 25.00\ ^\circ\mathrm{C}\) and is an aqueous solution of 3.45 % mass \(\ce{NaCl}\). Assume complete ionisation of \(\ce{NaCl}\) in water.

The boiling point of seawater is higher than pure water. The boiling point elevation constant, \(K_{\mathrm b} = 0.5120\ \mathrm{K\,kg\,mol^{-1}}\).

5.1 Calculate the boiling point, \(T\), (in °C) of Dubai seawater at atmospheric pressure.

If you couldn't calculate the answer, use a temperature of 373.50 K for further calculations.

5.2 Calculate the percentage mass of \(\ce{NaCl}\), \(w_{\ce{NaCl}}\), in water that has a boiling point, \(T_{\mathrm b} = 378.00\ \mathrm{K}\).

5.3 The boiling point of seawater also increases as pressure increases. Calculate the boiling point, \(T\), (in °C) of the initial Dubai seawater at a pressure, \(p = 2.50\ \mathrm{atm}\). The latent heat of vaporisation of water, \(E_{\mathrm{vap}} = 2260\ \mathrm{kJ\,kg^{-1}}\) (40.716 kJ/mol). Assume this is the same for seawater.

In the plant, a flash chamber (of volume, \(V = 100\ \mathrm{L}\)) contains a mass of 1.00 kg of seawater (l) at \(T_1 = 90.0\ ^\circ\mathrm{C}\).

Another mass of 1.00 kg of seawater is then overheated to \(T_2 = 110.0\ ^\circ\mathrm{C}\) at high pressure before being added to this chamber, where a reduction in pressure causes some water to turn into steam.

After equilibrium was established, the temperature in the chamber was \(T_{\mathrm f} = 97.0\ ^\circ\mathrm{C}\). Assume there was no loss of energy, and that the latent heat of vaporisation is independent of temperature.

The specific heat capacity of seawater, \(C_{\mathrm p} = 3.85\ \mathrm{kJ\,kg^{-1}\,K^{-1}}\), assumed to be independent of temperature. The density of seawater, \(d = 1025\ \mathrm{kg\,m^{-3}}\).

5.4 Calculate the amount, \(n\), (in mol) of water that vapourised, assuming that the latent heat of vaporization does not depend on pressure, and that the heat capacity of water vapor in the chamber is negligible.

If you couldn't calculate the answer, use 0.56 mol for further calculations.

5.5 Assuming the initial vapour pressure before addition of the overheated water in the chamber was \(p_{\mathrm i} = 0.690\ \mathrm{atm}\), calculate the final vapour pressure, \(p_{\mathrm f}\), (in atm) in the chamber assuming the water vapour behaves as an ideal gas.

5.6 Calculate the thermal energy, \(E\), (in kWh) required to heat a mass of 1.00 kg of seawater from \(T_1 = 25.0\ ^\circ\mathrm{C}\) to \(T_2 = 110.0\ ^\circ\mathrm{C}\).

If you couldn't calculate the answer, use 0.1 kWh for further calculations.

5.7 A plant produces 50,000 \(\mathrm{m^3}\) of pure water per day. The plant passes Dubai seawater through 30 flash cycles and overall can extract 85% of the water present. Calculate the mass, \(m\), (in kg) of Dubai seawater needed per day.

If you couldn't calculate the answer, use \(6.0 \times 10^7\) kg for further calculations.

Plants make efficient use of heat exchangers to save energy. We can assume the total energy required by the plant is the same as the energy needed to heat all the seawater from \(T_1 = 25.0\ ^\circ\mathrm{C}\) to \(T_2 = 110.0\ ^\circ\mathrm{C}\).

5.8 Assuming a 10 \(\mathrm{m^2}\) solar panel has a power of 2.00 kW averaged during the 12 operational hours every day and the efficiency of using energy is 67%, calculate the number of panels, \(N\), required for the plant.

Periodic table and data sheet

Physical constants and formulae

Avogadro constant \(N_{\mathrm A}\) \(6.022\times10^{23}\ \mathrm{mol^{-1}}\)
molar gas constant \(R\) \(8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\)
Faraday constant \(F\) \(96485\ \mathrm{C\,mol^{-1}}\)
Planck constant \(h\) \(6.626\times10^{-34}\ \mathrm{J\,s}\)
speed of light in vacuum \(c\) \(2.998\times10^8\ \mathrm{m\,s^{-1}}\)
mass of electron \(m_{\mathrm e}\) \(9.109\times10^{-31}\ \mathrm{kg}\)
charge of electron \(e\) \(1.602\times10^{-19}\ \mathrm{C}\)
atmospheric pressure \(p_{\mathrm{atm}}\) \(101325\ \mathrm{Pa}\)
ionic product of water at 298 K \(K_{\mathrm w}\) \(10^{-14}\)
\(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}\)
\(1\ \text{Å}=1\times10^{-10}\ \mathrm{m}\)
\(1\ \mathrm{pm}=1\times10^{-12}\ \mathrm{m}\)
\(0\ ^\circ\mathrm{C}=273.15\ \mathrm{K}\)
\(1\ \mathrm{kWh}=3600\ \mathrm{kJ}\)
\(1\ \mathrm{W}=1\ \mathrm{J\,s^{-1}}\)
\(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}\)
\(1\ \mathrm{bar}=10^5\ \mathrm{Pa}\)
area of triangle \(A=\frac12 ab\cdot\sin C\)
volume of cube \(V=a^3\)
volume of sphere \(V=\frac43\pi r^3\)
surface area of sphere \(S=4\pi r^2\)
ideal gas equation \(pV=nRT\)
frequency of light \(\nu=\frac{c}{\lambda}\)
energy of a photon \(E=\frac{hc}{\lambda}\)
Beer-Lambert law \(A=\log\left(\frac{I_0}{I}\right)=\varepsilon cd\)
spin-only magnetic moment \(\mu=\sqrt{n(n+2)}\ \mathrm{BM}\)
Henderson–Hasselbalch equation \(\mathrm{pH}=\mathrm{p}K_{\mathrm a}+\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\)
\(0^{\mathrm{th}}\) order integrated rate law \([\ce{A}]=[\ce{A}]_0-kt\)
\(1^{\mathrm{st}}\) order integrated rate law \(\ln[\ce{A}]=\ln[\ce{A}]_0-kt\)
Arrhenius equation \(k=A\cdot e^{-E_{\mathrm a}/RT}\)
enthalpy \(H=U+pV\)
Gibbs energy \(G=H-TS\)
standard Gibbs energy change \(\Delta_{\mathrm r}G^\circ=-RT\ln K=-nFE^\circ_{\mathrm{cell}}\)
Clausius-Clapeyron equation \(\ln\left(\frac{P_1}{P_2}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac1{T_2}-\frac1{T_1}\right)\)
specific heat \(Q=mC_{\mathrm p}\Delta T\)
boiling point elevation \(\Delta T=iK_{\mathrm b}m_i\)
freezing point depression \(\Delta T=iK_{\mathrm f}m_i\)

Consider all gases ideal unless told otherwise.

In equilibrium constant calculations, all concentrations are referenced to a concentration of \(1\ \mathrm{mol\,dm^{-3}}\).

Periodic table

Figure 2

Official English text of the problem.

Answers

  • 5.1: \(T=100.63\ ^\circ\mathrm{C}\).
  • 5.2: \(w_{\ce{NaCl}}=21.7\%\) by mass.
  • 5.3: \(T=128.7\ ^\circ\mathrm{C}\).
  • 5.4: \(n=0.567\ \mathrm{mol}\).
  • 5.5: \(p_{\mathrm f}=0.886\ \mathrm{atm}\), accounting for the volume occupied by liquid.
  • 5.6: \(E=0.0909\ \mathrm{kWh}\).
  • 5.7: \(m\approx6.1\times10^7\ \mathrm{kg}\) of seawater per day.
  • 5.8: \(N\approx3.4\times10^5\) panels; the nominal calculation gives a minimum of 344,422 panels.

Parts 5.7–5.8 use \(\rho_{\text{pure water}}=1000\ \mathrm{kg\,m^{-3}}\).

Solution

The relevant species are water, \(\ce{H2O}\), SMILES O; sodium ion, \(\ce{Na+}\), SMILES [Na+]; and chloride ion, \(\ce{Cl-}\), SMILES [Cl-]. Sodium chloride, \(\ce{NaCl}\), has the ionic SMILES representation [Na+].[Cl-].

5.1 — Boiling point at atmospheric pressure (2 points)

Complete ionisation gives two ions per formula unit:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)}. \]

Thus the van ’t Hoff factor is \(i=2\). From the periodic table,

\[ M(\ce{NaCl})=22.99+35.45=58.44\ \mathrm{g\,mol^{-1}}. \]

Take \(100.00\ \mathrm{g}\) of seawater. It contains \(3.45\ \mathrm{g}\) of salt and \(96.55\ \mathrm{g}=0.09655\ \mathrm{kg}\) of water. Its molality is

\[ b=\frac{3.45\ \mathrm{g}/(58.44\ \mathrm{g\,mol^{-1}})} {0.09655\ \mathrm{kg}} =0.611444\ \mathrm{mol\,kg^{-1}}. \]

The boiling point elevation is

\[ \Delta T_{\mathrm b}=iK_{\mathrm b}b =2(0.5120)(0.611444) =0.626119\ \mathrm{K}. \]

Pure water boils at \(100.00\ ^\circ\mathrm{C}\) at atmospheric pressure, so

\[ \boxed{T=100.63\ ^\circ\mathrm{C}}. \]

The corresponding absolute temperature, retained for part 5.3, is \(373.776119\ \mathrm{K}\).

5.2 — Salt mass percentage for a boiling point of 378.00 K (2 points)

At atmospheric pressure,

\[ \Delta T_{\mathrm b}=378.00-373.15=4.85\ \mathrm{K}. \]

Therefore,

\[ b=\frac{\Delta T_{\mathrm b}}{iK_{\mathrm b}} =\frac{4.85}{2(0.5120)} =4.736328\ \mathrm{mol\,kg^{-1}}. \]

For \(1.000\ \mathrm{kg}\) of water, the required salt mass is

\[ m_{\ce{NaCl}} =(4.736328\ \mathrm{mol})(58.44\ \mathrm{g\,mol^{-1}}) =276.791\ \mathrm{g}. \]

The mass percentage uses the total solution mass:

\[ w_{\ce{NaCl}} =\frac{276.791}{1000+276.791}\times100\% =21.6786\%. \]

\[ \boxed{w_{\ce{NaCl}}=21.7\%}. \]

5.3 — Boiling point at 2.50 atm (3 points)

Use the boiling point of the initial seawater from part 5.1 as the reference:

\[ P_{\mathrm{ref}}=1.00\ \mathrm{atm}, \qquad T_{\mathrm{ref}}=373.776119\ \mathrm{K}. \]

With the given constant molar enthalpy of vaporisation,

\[ \Delta H_{\mathrm{vap}}=40716\ \mathrm{J\,mol^{-1}}, \]

the Clausius–Clapeyron equation gives

\[ \ln\left(\frac{P}{P_{\mathrm{ref}}}\right) =\frac{\Delta H_{\mathrm{vap}}}{R} \left(\frac{1}{T_{\mathrm{ref}}}-\frac{1}{T}\right). \]

Rearranging,

\[ T= \left[ \frac{1}{T_{\mathrm{ref}}} -\frac{R}{\Delta H_{\mathrm{vap}}} \ln\left(\frac{P}{P_{\mathrm{ref}}}\right) \right]^{-1}. \]

Substitution yields

\[ T= \left[ \frac{1}{373.776119} -\frac{8.314}{40716}\ln(2.50) \right]^{-1} =401.881381\ \mathrm{K}. \]

Thus,

\[ \boxed{T=128.7\ ^\circ\mathrm{C}}. \]

5.4 — Amount of water vaporised (2 points)

The hot seawater releases heat as it cools from \(110.0\ ^\circ\mathrm{C}\) to \(97.0\ ^\circ\mathrm{C}\):

\[ Q_{\mathrm{hot}} =m_2C_{\mathrm p}(T_2-T_{\mathrm f}) =(1.00)(3.85)(110.0-97.0) =50.05\ \mathrm{kJ}. \]

Heating the initial seawater from \(90.0\ ^\circ\mathrm{C}\) to \(97.0\ ^\circ\mathrm{C}\) consumes

\[ Q_{\mathrm{cold}} =m_1C_{\mathrm p}(T_{\mathrm f}-T_1) =(1.00)(3.85)(97.0-90.0) =26.95\ \mathrm{kJ}. \]

With no heat loss and negligible heat capacity of the vapour, the remaining energy causes vaporisation:

\[ \ce{H2O(l) -> H2O(g)}. \]

Consequently,

\[ n\Delta H_{\mathrm{vap}} =Q_{\mathrm{hot}}-Q_{\mathrm{cold}} =23.10\ \mathrm{kJ}. \]

\[ n=\frac{23.10\ \mathrm{kJ}}{40.716\ \mathrm{kJ\,mol^{-1}}} =0.567345\ \mathrm{mol}. \]

\[ \boxed{n=0.567\ \mathrm{mol}}. \]

The corresponding mass evaporated is

\[ m_{\mathrm{evap}} =\frac{23.10}{2260} =0.0102212\ \mathrm{kg}. \]

5.5 — Final vapour pressure (2 points)

The initial vapour remains in the chamber, and the newly evaporated water adds to it. Both the temperature change and the change in vapour-space volume must be included.

Initially,

\[ T_{\mathrm i}=363.15\ \mathrm{K}, \qquad V_{\mathrm{g,i}} =V-\frac{m_1}{d} =0.100-\frac{1.00}{1025} =0.0990244\ \mathrm{m^3}. \]

The initial amount of vapour is therefore

\[ n_{\mathrm i} =\frac{p_{\mathrm i}V_{\mathrm{g,i}}}{RT_{\mathrm i}} =\frac{(0.690\times101325)(0.0990244)} {(8.314)(363.15)} =2.29304\ \mathrm{mol}. \]

The final liquid mass is

\[ m_{\mathrm{l,f}} =1.00+1.00-m_{\mathrm{evap}} =1.989779\ \mathrm{kg}. \]

Using the given liquid density,

\[ V_{\mathrm{g,f}} =0.100-\frac{1.989779}{1025} =0.0980588\ \mathrm{m^3}. \]

At \(T_{\mathrm f}=370.15\ \mathrm{K}\),

\[ n_{\mathrm f}=n_{\mathrm i}+n =2.29304+0.567345 =2.86039\ \mathrm{mol}. \]

Thus,

\[ p_{\mathrm f} =\frac{n_{\mathrm f}RT_{\mathrm f}}{V_{\mathrm{g,f}}} =\frac{(2.86039)(8.314)(370.15)}{0.0980588} =8.9769\times10^4\ \mathrm{Pa}, \]

and

\[ \boxed{p_{\mathrm f}=0.886\ \mathrm{atm}}. \]

If the liquid volume is neglected and the vapour volume is approximated as \(100\ \mathrm{L}\) throughout, the corresponding approximation is

\[ p_{\mathrm f} =p_{\mathrm i}\frac{T_{\mathrm f}}{T_{\mathrm i}} +\frac{nRT_{\mathrm f}}{V} =0.876\ \mathrm{atm}. \]

5.6 — Energy required to heat 1.00 kg of seawater (2 points)

The temperature increase is

\[ \Delta T=110.0-25.0=85.0\ \mathrm{K}. \]

Therefore,

\[ E=mC_{\mathrm p}\Delta T =(1.00\ \mathrm{kg}) (3.85\ \mathrm{kJ\,kg^{-1}\,K^{-1}}) (85.0\ \mathrm{K}) =327.25\ \mathrm{kJ}. \]

Using \(1\ \mathrm{kWh}=3600\ \mathrm{kJ}\),

\[ E=\frac{327.25}{3600}=0.0909028\ \mathrm{kWh}. \]

\[ \boxed{E=0.0909\ \mathrm{kWh}}. \]

5.7 — Daily seawater requirement (1 point)

Using the usual approximation \(\rho_{\text{pure water}}=1000\ \mathrm{kg\,m^{-3}}\), the daily product mass is

\[ m_{\mathrm{product}} =(50000\ \mathrm{m^3})(1000\ \mathrm{kg\,m^{-3}}) =5.0000\times10^7\ \mathrm{kg}. \]

The water mass fraction of the feed is

\[ w_{\mathrm{water}}=1-0.0345=0.9655. \]

Since \(85\%\) of this water is recovered,

\[ m_{\mathrm{product}}=0.85(0.9655)m. \]

Hence,

\[ m=\frac{5.0000\times10^7}{0.85(0.9655)} =6.09255\times10^7\ \mathrm{kg}. \]

\[ \boxed{m\approx6.1\times10^7\ \mathrm{kg\ per\ day}}. \]

The stated \(85\%\) is the overall recovery across all 30 cycles, so it is applied once.

5.8 — Number of solar panels (2 points)

Under the stated heat-recovery assumption, the daily thermal requirement is the energy needed to heat the entire daily seawater feed once:

\[ E_{\mathrm{day}} =m\left(0.0909028\ \mathrm{kWh\,kg^{-1}}\right) =5.53829\times10^6\ \mathrm{kWh}. \]

Each panel supplies useful energy per day of

\[ E_{\mathrm{panel}} =(2.00\ \mathrm{kW})(12\ \mathrm{h})(0.67) =16.08\ \mathrm{kWh}. \]

The given \(2.00\ \mathrm{kW}\) is already the power of one complete \(10\ \mathrm{m^2}\) panel.

Therefore,

\[ N=\frac{E_{\mathrm{day}}}{E_{\mathrm{panel}}} =\frac{5.53829\times10^6}{16.08} =344421.23. \]

Rounding upward to a whole number gives the nominal minimum:

\[ \boxed{N_{\min}=344422\ \text{panels}}. \]

At the precision of the supplied efficiency,

\[ \boxed{N\approx3.4\times10^5\ \text{panels}}. \]

The solution was checked against the official solution.