FE Environmental Mathematics Practice Problems

The current FE Environmental specification assigns 5-8 of the 110 exam questions to Mathematics. Form A contains 6 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A completely mixed lagoon of volume $500 m^{3}$ initially holds a contaminant at $40 mg/L$ . Contaminant-free water enters at $25 m^{3} / h$ and the lagoon discharges at the same rate. Within the lagoon the contaminant also decays by a first-order reaction with $k = 0.030 h^{- 1}$ . The contaminant concentration after 30 h is most nearly:

(A)3.6 mg/L
(B)8.9 mg/L
(C)16.3 mg/L
(D)22.0 mg/L

Show worked solution

Answer: (A)

Washout and reaction combine into a single first-order coefficient of $0.080 h^{- 1}$ , leaving about 3.6 mg/L after 30 h.

Washout and reaction both remove mass in proportion to the instantaneous concentration, so the unsteady mass balance collapses to one lumped rate constant.

V \frac{d C}{d t} = - QC - k V C

\frac{d C}{d t} = - (\frac{Q}{V} + k) C

Hydraulic washout at $Q / V$ and reaction at $k$ are additive because both are first order in $C$ .

\frac{Q}{V} + k = \frac{25}{500} + 0.030 = 0.080 h^{- 1}

Exponential decay from the initial 40 mg/L over 30 h leaves

C = 40 e^{- (0.080) (30)} = 40 (0.0907) = 3.6 mg/L

FE Reference Handbook — Mathematics: Differential Equations, First-Order Linear Homogeneous with Constant Coefficients

Why the other choices appear

(B)Neglects the reaction and uses washout alone: $40 e^{- (0.050) (30)} = 8.9$ mg/L.
(C)Neglects washout and uses the decay constant alone: $40 e^{- (0.030) (30)} = 16.3$ mg/L.
(D)Subtracts the decay rate from the washout rate instead of adding: $40 e^{- (0.020) (30)} = 22.0$ mg/L.