Answer:
Explanation:
Given that:
The flow rate Q = 0.3 m³/s
Volume (V) = 200 m³
Initial concentration [tex]C_o[/tex] = 2.00 ms/l
reaction rate K = 5.09 hr⁻¹
Recall that:
[tex]time (t) = \dfrac{V}{Q}[/tex]
[tex]time (t) = \dfrac{200}{0.3}[/tex]
[tex]time (t) = 666.66 \ sec[/tex]
[tex]time (t) = \dfrac{666.66 }{3600} hrs[/tex]
[tex]time (t) = 0.185 hrs[/tex]
[tex]\text{Using First Order Reaction:}[/tex]
[tex]\dfrac{dc}{dt}=kc[/tex]
where;
[tex]t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)[/tex]
[tex]0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)[/tex]
[tex]0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)[/tex]
[tex]1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)[/tex]
[tex]\dfrac{200}{C_e} = 1.942[/tex]
[tex]C_e = \dfrac{200}{1.942}[/tex]
[tex]\mathbf{C_e = 102.98 \ mg/l}[/tex]
Thus; the concentration of species in the reactant = 102.98 mg/l
b). If the plug flow reactor has the same efficiency as CSTR, Then:
[tex]t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ][/tex]
[tex]\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ][/tex]
[tex]\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ][/tex]
[tex]\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ][/tex]
[tex]\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)[/tex]
[tex]\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)[/tex]
[tex]\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)[/tex]
[tex]V_{PFR} =0.196(0.663)*0.3*3600[/tex]
[tex]V_{PFR} = 140.34 \ m^3[/tex]
The volume of the PFR is ≅ 140 m³
