EXAMPLE $18\mathrm{.}4.$ A gas stream with $6\mathrm{.}0$ percent $N{H}_3($dry basis$)$ and a flow rate of

at $\{$:$0C,1(atm))$ is to be scrubbed with water to lower the concen$-$

tration to $0\mathrm{.}02$ percent. The absorber will operate at atmospheric pressure with inlet

temperatures of $20$ and $25C$ for the gas and liquid, respectively. The gas is saturated

with water vapor at the inlet temperature and can be assumed to leave as a saturated gas

at $25C.$ Calculate the value of $N_{Oy}$ if the liquid rate is $1\mathrm{.}25$ times the minimum.

Solution. The following solubility data are given by Perry. ${?}^{15a}$

The temperature at the bottom of the column must be calculated to determine the mini$-$

mum liquid rate.

Basis. $100$gmol of dry gas in$,$ containing $94$mol of air and $6$mol of $N{H}_3.$ The

outlet gas contains $94$mol of air.

The moles of ammonia in the outlet gas, since $y_a=0\mathrm{.}0002,$ are

$94(\frac{0\mathrm{.}0002}{0\mathrm{.}9998})=0\mathrm{.}0188molN{H}_3$

The amount of ammonia absorbed is then $6-0\mathrm{.}0188=5\mathrm{.}98$mol.

Heat effects. The heat of absorption is $5\mathrm{.}988,310=49,690$ cal. Call this $Q_a.$ Then

$Q_a=Q_{sy}+Q_v+Q_{sx}$

where $Q_{sy}=$ sensible$-$heat change in gas

$Q_v=$ heat of vaporization

$Q_{sx}=$ sensible heat change in liquid

The sensible$-$heat changes in the gas are

$Q_{\text{a}\text{i}\text{r}}=94$mol$7\mathrm{.}0ca\frac{l}{m}ol*C5C=3,290$cal

$Q_{H_{2}O}=2\mathrm{.}48\mathrm{.}05=96$cal

$Q_{sy}=3,290+96=3,390$cal

$\{\begin{array}{c}\text{:}{}_{\text{g}\text{r}\text{a}\text{d}}-=3\end{array}$

then $7$~$=30$ C

for unit operations of chemical engineering $7$th editon example $18\mathrm{.}4$

I want to know how the $0\mathrm{.}0125,0\mathrm{.}00780,$ and $0\mathrm{.}00138$ in the table come out.

I tried to get it right, but the answer is not right