Test Index

GATE Civil Engineering (CE) 2019 Shift 2 Solved Paper

© examsnet.com

Question : 43 of 65

Marks: +1, -0

Consider the reactor shown in the figure. The concentration (in

\mathrm{mg/l})

of a compound in the influent and effluent are

C_{0}

C

, flow rate through the reactor is

Q\,\mathrm{m}^3/\mathrm{h}

respectively. The compound is degraded in the reactor following the first order reactions. The mixing condition of the reactor Complete Mix Flow Reactor (CMFR) or a plug-flow reactor (PFR). The length of the reactor can be adjusted in these two mixing conditions to L

\text{CMFR}

L_{\text{PER}}

while keeping the cross-section of the reactor constant. Assuming steady state and for

\frac{C}{C_{0}}=0.8

\frac{L_{\text{CMFR}}}{L_{\text{PER}}}

(round off to 2 decimal places) is ___________

Your Answer:

Solution:

A=

constant

Q=AL

For plug flow

C=C_{0} e^{-k t_{d_1}}

[t_{d_1}=

detention time for plug flow reaction]

For completely mix

C=\frac{C_{o}}{1+ k t_{d_{2}}}

[t_{d_{2}}=

detention time for completely mix reaction]

\Rightarrow 0.8=e^{-k d_{1}}

\Rightarrow k t_{d_1}=0.223=k \cdot \frac{A l_{\text{plug}}}{Q}

Also

0.8=\frac{1}{1+ k t_{d_{1}}}

\Rightarrow k t_{d_2}=0.25=k \cdot \frac{A L_{\text{com}}}{Q}

\Rightarrow \frac{k t_d}{k t_{d_1}} = \frac{L_{\text{plug}}}{L_{\text{com}}} = \frac{0.223}{0.25} = 0.892 = 0.89

\Rightarrow \frac{L_{\text{com}}}{L_{\text{plug}}} = \frac{1}{0.89} = 1.12

© examsnet.com

Go to Question:

Prev Question

Next Question