Question 6.3.2: Material Balances Around a Condenser A stream of air at 100°......

1. p_{H_{2}O} = y_{H_{2}O }  P = (0.100)(5260 mm Hg) = 526 mm Hg

p^{*}_{H_{2}O} (100°C) = 760 mm Hg > p_{H_{2}O}  \Longrightarrow the vapor is superheated (see Inequality 6.3-2)

From Equation 6.3-3

p_{i} = \boxed{y_{i}P= p^{*}_{i} (T_{dp})}                  (6.3-3)

p_{H_{2}O}=p^{*}_{H_{2}O} (T_{dp}) = 526  mm  Hg \\\left. \Large{\Downarrow} \right. Table  B.3 \\ \boxed{T_{dp} = 90°C}

and the air has 100°C – 90°C = \boxed{10°C  of  superheat}

2. Since the air becomes saturated at 90°C, further cooling must lead to condensation. Since the products are liquid water in equilibrium with a gas phase, the water vapor in the gas must remain saturated.
On the following flowchart, the symbol BDA stands for bone-dry air, a term used to signify the water-free component of an air–water vapor mixture.
Basis: 100 mol Feed Gas

Let us first do the degree-of-freedom analysis. Three unknown variables appear on the chart—n_{1}, n_{2}, and y. Since only two species are involved in the process, we can only write two independent material balances, leaving us one equation short. If we fail to observe that the gas at the condenser outlet is saturated with water, solution of the problem would be impossible; however, the saturation condition supplies the needed third equation, Raoult’s law.
The solution outline is as follows: apply Raoult’s law at the outlet to determine y, the mole fraction of water in the outlet gas; then use a dry air balance to determine n_{2} and a total mole balance or a water balance to determine the final unknown, n_{1}.

Raoult’s Law at Outlet:           yP = p^{*}_{H_{2}O}(T)

\left. \Large{\Downarrow} \right.\\ y= \frac{p^{*}_{H_{2}O} (80°C)}{P} = \frac{355  mm  Hg}{5260  mm  Hg} = \boxed{0.0675 \frac{mol  H_{2}O}{mol}}

Balance on Dry Air:     \begin{array}{c|c}100  mol &0.900  mol  BDA \\ \hline & mol\end{array}= n_{2} (1  –  y)

\left. \Large{\Downarrow} \right. y= 0.0675 \\ n_{2}= 96.5  mol

Total Mole Balance:            100  mol =  n_{1} + n_{2}

\left. \Large{\Downarrow} \right. n_{2}= 96.5  mol \\ n_{1} = 3.5  mol  H_{2}O  condensed

Percentage Condensation:   \frac{3.5  mol  H_{2}O  condensed}{(0.100 \times 100)  mol  H_{2}O  fed} \times 100\% = \boxed{35\%}

3. Initially y_{H_{2}O } P < p^{*}_{H_{2}O} (100°C). Saturation occurs when P is high enough for the inequality to become an equality, or

P_{saturation} = \frac{p^{*}_{H_{2}O} (100°C)}{y_{H_{2}O}} = \frac{760  mm}{0.100} = 7600  mm  Hg

Any increase in P above 7600 mm Hg must cause condensation, so that the products from the compression to 8500 mm Hg must include a liquid stream.
Basis: 100 mol Feed Gas

Before going through the solution, try to outline it as was done following the flowchart of Part 2.
Raoult’s Law:  y= \frac{p^{*}_{H_{2}O} (100°C)}{P}=\frac{760  mm  Hg}{8500  mm  Hg} = \boxed{0.0894 \frac{mol  H_{2}O}{mol} }

Balance on Dry Air:     (100  mol)(0.900) =n_{2} = (1  –  y)

\left. \Large{\Downarrow} \right. y= 0.0894 \\ n_{2}= 98.8  mol

Total Mole Balance:            100  mol =  n_{1} + n_{2}

\left. \Large{\Downarrow} \right. n_{2}= 98.8  mol \\ n_{1} = 1.2  mol  H_{2}O  condensed

Percentage Condensation:  \frac{1.2  mol  H_{2}O  condensed}{(0.100 \times 100)  mol  H_{2}O  fed} \times 100\% = \boxed{12\%}

4. (a) Experimental error (you should be able to list many possibilities). (b) The condenser was not at steady state when the measurements were made, possibly because the system had not yet leveled out following startup or because water vapor was adsorbing on the walls of the condenser. (c) The emerging gas and liquid streams were not at equilibrium (e.g., condensation actually occurred at a temperature below 100°C and the product streams were separated and reheated before emerging). (d) Raoult’s law does not apply (this is not a likely explanation for the air–water system at the given conditions).