Question 13.1: In a process in which it is used as a solvent, benzene is ev......
In a process in which it is used as a solvent, benzene is evaporated into dry nitrogen. At 297 K and 101.3 kN/m², the resulting mixture has a percentage relative humidity of 60. It is required to recover 80 per cent of the benzene present by cooling to 283 K and compressing to a suitable pressure. What should this pressure be? The vapour pressure of benzene is 12.2 kN/m² at 297 K and 6.0 kN/m² at 283 K.
From the definition of percentage relative humidity (RH):
P_{w}=P_{w0}\left({\frac{R H}{100}}\right)At 297 K: P_{w}=(12.2\times1000)\times\left({\frac{60}{100}}\right)=7320\ \mathrm{N/m^{2}}
In the benzene-nitrogen mixture:
mass of benzene = \frac{P_{w}M_{w}}{\mathrm{R}T}=\frac{(7320\times78)}{(8314\times297)}=0.231{\mathrm{~kg}}
mass of nitrogen = {\frac{(P-P_{w})M_{A}}{\mathrm{R}T}}={\frac{\left[(101.3-732)\times1000\times28\right]}{(8314\times297)}}=1.066\mathrm{~kg}
Hence the humidity is: \mathscr{H} =\left(\frac{0.231}{1.066}\right)=0.217\mathrm{~kg/kg}
In order to recover 80 per cent of the benzene, the humidity must be reduced to 20 per cent of the initial value. As the vapour will be in contact with liquid benzene, the nitrogen will be saturated with benzene vapour
and hence at 283 K:
\mathscr{H} _{0}=\frac{{(0.217\times20)}}{100}=0.0433\ \mathrm{kg/kg}Thus in equation 13.2:
\mathscr{H}_{0}=\frac{P_{w0}}{P-P_{w0}}\left(\frac{M_{w}}{M_{A}}\right) (13.2)
0.0433=\left({\frac{6000}{P-6000}}\right)\left({\frac{78}{28}}\right)from which: \underline{\underline{P=3.92\times 10^{5}\ N/m^{2}=392\ kN/m^{2}}}