Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 9 (Q.No. 39)
39.
A gaseous reactant is introduced in a mixed reactor of 3 litres volume at the rate of 1 litre/second. The space time is __________ seconds.
Discussion:
1 comments Page 1 of 1.
Waqas said:
3 years ago
The Spacetime, τ, is obtained by dividing the reactor volume by the volumetric flow rate entering the reactor.
So, 3/1=3.
So, 3/1=3.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers