Step-by-step solved example
First-order PFR, 90% removal, k = 0.50 d⁻¹. Find t.
1. Fraction remaining
C/C0 = 0.10; ln(C/C0) = ln(0.10) = −2.303.
2. Time
t = −(1/k) ln(C/C0) = −(1/0.50)×(−2.303) = 4.61 d.
Answer: t = 4.61 d
First-order PFR, 90% removal, k = 0.50 d⁻¹. Find t.
1. Fraction remaining
C/C0 = 0.10; ln(C/C0) = ln(0.10) = −2.303.
2. Time
t = −(1/k) ln(C/C0) = −(1/0.50)×(−2.303) = 4.61 d.
Answer: t = 4.61 d
1.50% removal in a first-order PFR needs kt =
2.k = 0.20 d⁻¹, 50% removal. t =
3.99% removal: t =
4.For the same k and 90% removal, a single CMFR needs t =
5.C = C0 e^{−kt} is also the integrated law for a
6.k = 0.40 d⁻¹, C/C0 = 0.20. t =
7.Tanks-in-series (N CMFRs) approach a PFR as N
8.This design t assumes the rate is
9.If k is doubled, PFR volume for the same Q and conversion
10.A dispersion number → 0 in an open-channel contactor means it behaves like a