Residence Time Distribution Questions and Answers

Answer: C

Correct answer: True

In ideal plug flow every element takes exactly the same time to traverse the reactor, so all the tracer emerges at \( t = \tau \) and nowhere else - a Dirac spike. A perfectly mixed tank gives the opposite extreme, a decaying exponential beginning at \( t=0 \), and every real reactor lies somewhere between the two.

Answer: D

Some tracer leaves immediately and some stays a very long time, which is exactly what perfect mixing implies. The exponential form is the widest spread any single well mixed vessel can give. Some tracer leaves almost at once.

Answer: A

The distribution describes the flow and says nothing about the chemistry. The other three are exactly the faults a tracer test is run to find. Chemistry needs a different experiment. The tracer passes through unchanged.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the gap. The distribution records how long each element stayed but nothing about whom it kept company with, and for a non-linear rate law the conversion depends on that. Two reactors with identical distributions but different micromixing give different conversions, which is exactly why the segregation and maximum mixedness limits are computed.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: for a first order reaction the distribution fixes the conversion exactly, and for any other order it gives only the bounds set by the two micromixing limits. Statement 3 claims the general case, which the previous item denies - the distribution is a necessary input but not a sufficient one once the rate law is non-linear.

Answer: D

Correct answer: the two limits of micromixing

They are the two extremes of how fluid elements of different ages mix. Complete segregation treats each element as a little batch reactor that never exchanges material, mixing only at the exit; maximum mixedness has elements mix with all others of the same remaining life as early as possible. Real reactors lie between them, and the two calculations therefore bracket the achievable conversion.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mathematics. A first order rate is linear in concentration, so the average of the reacted concentrations equals the reaction of the averaged concentration - averaging and reacting commute. The micromixing therefore cannot matter, and the two limits coincide. For any other order the operations do not commute and the bounds separate.

Answer: B

Correct answer: an early spike of tracer

Bypassing means part of the feed takes a short route and leaves almost at once, so tracer appears far earlier than \( \tau \) and in a sharp peak, followed by the ordinary response of the remainder. A long tail indicates dead volume slowly exchanging with the main flow, and a delay before any tracer appears indicates a plug flow section in the path.

Answer: C

Reading the shape is the whole point of running the test, and each fault leaves its own mark. A curve that matches an ideal model closely is the reassuring case. Each fault leaves its own mark. Reading the shape is the point of the test.

Answer: D

Correct answer: a catalyst that has lost activity

A tracer test reports on the flow and nothing else - it uses an inert species precisely so that nothing chemical interferes. Bypassing, dead volume and channelling are all flow faults and all show as characteristic distortions of the curve. Loss of catalyst activity is a chemical fault, and a deactivated bed would give exactly the same tracer response as a fresh one.