A completely mixed chemical reactor has an in fluent flow with conc. 150 mg/L of

Question

A completely mixed chemical reactor has an in fluent flow with conc. 150 mg/L of A and a flow rate of 100 gal/min (380 L/min). The reaction is first order, The rate equation : - dC/dt = KCa= -ra ( and the rate constant is 0.4 h^-1) Determine: a- The required detention time and volume of the reactor if the effluent contains 20 mg/L of A. Express the volume as gallons and liters. b- A plot of percent Ca removed or converted versus the detention time. c- How many times larger a completely mixed reactor must be than a plug flow reactor for 90% removal or conversion. Please solve the question by numbers and steps and don`t give me links

Explanation / Answer

The required detention time and the volume (gallons) of the reactor if the effluent contains 20 mg/L of A.

= (CO- CF) / KCF = (150-20) / (0.40*20) = 16.3 hrs

Volume V= Q* = (100gal/min)* (60 min/hr) *16.3 hr = 97800 gals

How many times larger a completely mixed reactor must be than a plug flow reactor for 80% removal or conversion.

80% removal = (CO- CO * 80/100) = (150 -150* (80/100)) = 150-120 = 30

Therefore, CF = 30mg/L

PFR

CF/CO = e– (K * )

30/150 = e– (0.4 * )

= 4.02 hrs

CSTR

= (CO - CF)/ (K * CF )

= (150-30) / (0.4 * 30)

= 10 hrs

The ratio of CSTR to PFR

CSTR / PFR = 10 / 4 = 2.5

. How many times larger a completely mixed reactor must be than a plug flow reactor for 90% removal or conversion.

90% removal = (CO- CO * 90/100) = (150 -150* (90/100)) = 150-135 = 15

Therefore, CF = 15 mg/L

PFR

CF/CO = e– (K * )

15 /150 = e– (0.4 * )

= 5.76 hrs

CSTR

= (CO - CF)/ (K*CF )

= (150-15) / (0.4*15)

= 22.5 hrs

The ratio of CSTR to PFR

CSTR / PFR = 22.5 / 5.76 = 3.91

Therefore, CSTR = 3.91 times larger than PFR

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