Liquid Plug Flow Reactor + 1st and 2nd Order Reactions // Reactor Engineering - Class 71

[@zhiqkhir102]

6:31 why does the integrate of 1/(1-x) equal to - ln ? Why is there negative sign?

[@ChemicalEngineeringGuy]

Hi, that is the way the integral is given, those can be found on the integral tables

[@LaytonWrigth]

Hello I dont really understand the term of the v with a dot over it, could you tell me what is it exactly please?

[@ChemicalEngineeringGuy]

Volumetric Flow Rate, M3/h or cm3/s or L/min etc

[@quickscopic]

I think you made a mistake when in the second order reaction you put (-kCa^2) in the integral instead of kCa^2.

[@ChemicalEngineeringGuy]

+Lawrence Yes; i should fix it! There are other comments as well; you may review them =) thanks for the tip

[@nicholasfisher1609]

@@ChemicalEngineeringGuy I was just about to say the same thing!! -ra = KCa2, so kCa2 goes in the denominator, NOT -kCa2

[@MultiRebond]

dx/(1-x)^2 = x/(1-x) right? there is no negative sign, -x/(1-x) is incorrect, I checked is.

[@ChemicalEngineeringGuy]

+Remon Balzer you are correct; I must fix it!

[@robindb4994]

Do you happen to have intermediate steps for the 'just solve for X, its easy' at 8.32? Can't figure it out. Thanks anyway, amazing videos :)

[@ChemicalEngineeringGuy]

You mean the mathematics?

[@mo7ammedksa10]

in 2ed order u don't change the sing of (K) u but (-k) I think u must but (v=(v/KCA0)

[@ChemicalEngineeringGuy]

+mohammed al-amoudi depends on how you use rate of reaction.. I used -ra therefore I need to use K

[@MultiRebond]

+Chemical Engineering Guy yeah I think Mohammed is right, there is the lost minus haha