Thank you so much for this series of videos! I have a question regarding the 3rd question. It's said that the specific growth rate is under 0.3 per hr. If the dilution rate is equal to the specific growth rate, shouldn't the answer be D < 0.3?
A batch reactor is loaded with 40 liters of media. Following sterilization, the reactor is inoculated with four grams of bacteria. Assuming that the lag phase is negligible, determine the cell concentration (g/l) in the reactor after three hours of growth. The specific growth A batch reactor is loaded with 40 liters of media. Following sterilization, the reactor is inoculated with four grams of bacteria. Assuming that the lag phase is negligible, determine the cell concentration (g/l) in the reactor after three hours of growth. The specific growth rate (μg) of the organism is 0.25 h-1 under these conditions.rate (μg) of the organism is 0.25 h-1 under these conditions. ?
Good afternoon , i keep getting 285per litre for questions number 2 , since we’re supposed to find the flow rate , i made dilution rate subject if the formulae 0.1D=0.8(0.5-D) and making D subject of the formalae i got it as 0.57 and I found my flow rate by saying 500*0.57 which gave me 285 as my final answer instead of 222.22 What am I doing wrong? I’ve tried to do it in all possible ways the nearest answer i got was 220 which is still not correct
For question no 2 we are calculating the feed flow rate so by using this formula f/v=umax.s/ks+s this is what i was using but after calculating the answer i am getting the answer 222.22 L/hr rather than the answer given in the description box that is 227l/hr so can you help me with this question
Hi Advesh, The numericals in the description the video are entirely based on the video content itself, have a good look on the video and the formulas and you will be definitely able to solve the questions and if you have any problem in a particular questions get back to me I'll definitely help.
The growth of an organism on glucose is described by the following Monod model parameters: μm = 0.5 h-1 and Ks =0.1 g.l-1, if the concentration of glucose in the feed is 10 g.l-1 and the dilution rate is set to 0.4 h-1, then the steady state concentration of glucose in the effluent will be A.0 g.l-1B.0.5 g.l-1C.1.0 g.l-1D.10 g.l-1 Answer: Option B Sir plzz guide me how to solve this question
Thank you bro. Appreciate all your teachingsss. More success!
How to solve 1? My answer is not matching with yours
Sir when we will get more videos on bioprocess, fermentation n all
Thank you so much for this series of videos! I have a question regarding the 3rd question. It's said that the specific growth rate is under 0.3 per hr. If the dilution rate is equal to the specific growth rate, shouldn't the answer be D < 0.3?
Hey, i guess it's just a grammatical thing
Specific growth rate is 0.3 .. its written "under is 0.3".. not "under 0.3".
I hope it clarifies.
@@manmohanmitruka9596 Ah, yes, it makes sense now. Thank you for your feedback!
Is this lecture is also useful for chemical engineering newly added topics?
Sir please make videos on topics which are newly added to the syllabus of gate biotechnology 2021
Sir iam not understanding how to do the first problem....as the answer I got is different from what is mentioned can u please explain asap
Hey what answer did you get?
As i got answer mentioned by sir
A batch reactor is loaded with 40 liters of media. Following sterilization, the reactor is inoculated with four grams of bacteria. Assuming that the lag phase is negligible, determine the cell concentration (g/l) in the reactor after three hours of growth. The specific growth A batch reactor is loaded with 40 liters of media. Following sterilization, the reactor is inoculated with four grams of bacteria. Assuming that the lag phase is negligible, determine the cell concentration (g/l) in the reactor after three hours of growth. The specific growth rate (μg) of the organism is 0.25 h-1 under these conditions.rate (μg) of the organism is 0.25 h-1 under these conditions.
?
Mass consumed in this continuous stirred tank reactor is ZERO ,am I crt sir
In mass balance eqn why u take input zero plz explain
@Vemulapalli Chanukya Pavan bs17b034 anyhow we used to add some amount of biomass for the further process. Then how it become zero
Good afternoon , i keep getting 285per litre for questions number 2 , since we’re supposed to find the flow rate , i made dilution rate subject if the formulae 0.1D=0.8(0.5-D) and making D subject of the formalae i got it as 0.57 and I found my flow rate by saying 500*0.57 which gave me 285 as my final answer instead of 222.22
What am I doing wrong?
I’ve tried to do it in all possible ways the nearest answer i got was 220 which is still not correct
For question no 2 we are calculating the feed flow rate so by using this formula f/v=umax.s/ks+s this is what i was using but after calculating the answer i am getting the answer 222.22 L/hr rather than the answer given in the description box that is 227l/hr so can you help me with this question
222.22 is correct
(0.5×0.8)/(0.1+0.8)×500
lovely
Sir how to solve the numericals which you given in the description please help...
Hi Advesh,
The numericals in the description the video are entirely based on the video content itself, have a good look on the video and the formulas and you will be definitely able to solve the questions and if you have any problem in a particular questions get back to me I'll definitely help.
great video
Glad you liked it.
The growth of an organism on glucose is described by the following Monod model parameters: μm = 0.5 h-1 and Ks =0.1 g.l-1, if the concentration of glucose in the feed is 10 g.l-1 and the dilution rate is set to 0.4 h-1, then the steady state concentration of glucose in the effluent will be
A.0 g.l-1B.0.5 g.l-1C.1.0 g.l-1D.10 g.l-1
Answer: Option B
Sir plzz guide me how to solve this question
Plzz rply sir??
Thank you Sir!
Don't forget to share it with your friends Jd
Thank you sir
Question1. answer is 0.6g/l.....if answer is correct pls response
Yes that's what I have mentioned.
Should not the third formula be S = (KsD)/(µm - D). It's µ + D in the video.
Yes that is a mistake and the mistake is also acknowledged in the description
great video