The residence time distribution (RTD) of a chemical reactor is describes the amo

Question

The residence time distribution (RTD) of a chemical reactor is describes the amount of time a fluid element could spend inside a reactor. The RTD is a concept often used by chemical engineers to characterize the mixing and flow within reactors. Let T denote the residence time for a particular application. (a) Assume T follows an exponential distribution in which it is known that the distribution has an 80^th percentile of 8 mins. Calculate the probability that the residence time exceeds 10 mins. (b) Assume the distribution of T is lognormal with a mean of 5 mins and a variance 50 mins^2. Calculate the probability of residence time being below 5 mins. (c) Calculate the 99^th percentiles of the exponential distribution in (a) and the log-normal distribution in (b).

Explanation / Answer

(a) Suppose T ~ exponential(lambda) and cdf of T is:

F(x) = P(T<=x) = 1- exp(-lambda*x)

Then, if 8 mins is at 80-th percentile , then we get:

1-exp(-lambda * 8) = 0.8

So, exp(-lambda*8) = 0.2

So, lambda = - log(0.2)/8 = 0.20118

So, P(T>10) = exp(-lambda*10) = 0.133748 = 0.134 = 13.4 %

(b) Suppose t = Log(T)

Now, P(T<5) = P(log(T) < log(5) ) = P(t < log(5))

Now, t = log(T) ~ N(mean = 5, variance = 50)

So, P(T<5) = P(t<log(5) | t ~ N(mean = 5, variance = 50) )

= 0.27151 = 27.15 %

(c) Suppose 99-th percentile is x and y for exponential and lognormal here. Then,

1-exp(-lambda*x) = 0.99 giving:

exp(-lambda*x) = 0.01

x = - log(0.01) / lambda = - log(0.01) / 0.20118 = 22.89

and P(T < y ) = 0.99 gives: P(t < log(y) )

This can be found using excel function norminv, which gives:

y = norminv(0.99, mean = 5, std dev = sqrt(50) ) = 21.45

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