Free FE Chemical practice problems

Ten real questions from our FE Chemical bank, drawn from 10 of its 17 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

8 with figures10 topicsFull worked solutions

1
medium
A. Molecular & convective diffusion
Components A and B of a binary gas mixture diffuse in opposite directions at equal molar rates (equimolar counter-diffusion) across a gas film 2 mm thick at 100 °C and 101.3 kPa. The partial pressure of A is 20 kPa at one face of the film and 5 kPa at the other, as shown, and the mass diffusivity of A in B is $D_m = 2.0\times10^{-5}$ m²/s. The molar flux of A across the film is most nearly:
Figure — drag or pinch to move around
1. 0.0484 mol/(m²·s)
2. 0.0553 mol/(m²·s)
3. 0.0645 mol/(m²·s)
4. 0.180 mol/(m²·s)
Show the answer & worked solution
Answer: A.0.0484 mol/(m²·s)
For equimolar counter-diffusion of gases the FE Handbook (Chemical Engineering — Mass Transfer, p. 249) gives $N_A = \dfrac{D_m}{RT}\cdot\dfrac{p_{A1}-p_{A2}}{\Delta z}$ . With $T = 100 + 273.15 = 373.15$ K, $\Delta z = 0.002$ m and $p_{A1}-p_{A2} = 15000$ Pa: $N_A = \dfrac{(2.0\times10^{-5})(15000)}{(8.314)(373.15)(0.002)}$ = 0.0484 mol/(m²·s). Distractors: the STAGNANT-B formula (p. 248), which multiplies by $P/(p_B)_{lm} = 101.3/88.6$ : it holds only when B does not diffuse, 0.0553 mol/(m²·s); $p_{A1}$ alone used as the driving force ( $p_{A2}$ ignored), 0.0645 mol/(m²·s); and $T$ left in degrees Celsius (100 instead of 373.15 K), 0.180 mol/(m²·s).
2
medium
A. Process dynamics & transfer functions
A closed-loop pressure controller is tuned so that the loop responds to a set-point change as an under-damped second-order system with damping ratio $\zeta = 0.5$ . The percent overshoot of its response to a step change in set point is most nearly:
Figure — drag or pinch to move around
1. 2.66 %
2. 16.3 %
3. 20.8 %
4. 25.7 %
Show the answer & worked solution
Answer: B.16.3 %
For an under-damped second-order response the FE Handbook (Instrumentation, Measurement, and Control, p. 233) gives the percent overshoot as $\%OS = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ . Here $\sqrt{1-\zeta^2} = \sqrt{1-0.5^2} = 0.866$ , the exponent is $0.5\pi/0.866 = 1.814$ , and $\%OS = 100\,e^{-1.814}$ = 16.3 %: the first peak rises 16.3 % of the total change above the final value. Distractors: the DECAY RATIO (ratio of successive peaks, the overshoot fraction squared) reported, $100e^{-2\zeta\pi/\sqrt{1-\zeta^2}}$ = 2.66 %; the $\sqrt{1-\zeta^2}$ dropped, $100e^{-\pi\zeta}$ = 20.8 %; and the root MULTIPLYING instead of dividing, $100e^{-\zeta\pi\sqrt{1-\zeta^2}}$ = 25.7 %.
3
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
Figure — drag or pinch to move around
1. $8,593
2. $13,333
3. $20,593
4. $25,333
Show the answer & worked solution
Answer: C.$20,593
The capital recovery factor (FE Reference Handbook 10.6 p. 235) converts a present amount into equal end-of-year payments: $(A/P,\ 6\%,\ 15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=\dfrac{0.06\,(1.06)^{15}}{(1.06)^{15}-1}=0.10296$ , printed as 0.1030 in the handbook's 6% factor table. The annual payment is $200,000 × 0.10296 = $20,593. Distractors: $P/n$ , spreading the principal evenly and ignoring interest ($13,333); the sinking-fund factor $(A/F)$ read in place of $(A/P)$ ($8,593), which treats the loan as a future sum; and simple interest on the full principal for all 15 years, spread evenly, $P(1+in)/n$ ($25,333), which charges interest on money already repaid.
4
medium
A. Statics, pipe flow & flow measurement
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 150$ mm. The upstream velocity is $V_1 = 1.5$ m/s. The velocity in the smaller section is most nearly:
Figure — drag or pinch to move around
1. 0.844 m/s
2. 2 m/s
3. 2.67 m/s
4. 4.74 m/s
Show the answer & worked solution
Answer: C.2.67 m/s
Continuity (FE Reference Handbook 10.6 p. 185): the same flow $Q=Av$ passes both sections, so $A_1V_1=A_2V_2$ . With $A=\pi D^2/4$ the $\pi/4$ cancels, leaving $V_2=V_1\left(\dfrac{D_1}{D_2}\right)^2=1.5\left(\dfrac{200}{150}\right)^2=2.67$ m/s, faster in the smaller section, as it must be. Distractors: the diameter ratio used un-squared, $V_1D_1/D_2$ (2 m/s), forgetting that area goes as $D^2$ ; the area ratio inverted, $V_1(D_2/D_1)^2$ (0.844 m/s), which would have the water slow down in the smaller section; and the ratio squared twice, $V_1(D_1/D_2)^4$ (4.74 m/s), the area-ratio squaring applied a second time.
5
medium
A. Conduction, convection & radiation
A brick wall 200 mm thick with thermal conductivity $k=0.8$ W/(m·K) has a face area of 10 m². In steady operation its inside surface is at $T_1=20$ °C and its outside surface at $T_2=-4$ °C. The rate of heat loss through the wall is most nearly:
Figure — drag or pinch to move around
1. 640 W
2. 960 W
3. 1500 W
4. 11900 W
Show the answer & worked solution
Answer: B.960 W
The temperature difference across the wall is $T_1-T_2=20-(-4)=24$ K (a difference is the same in °C and K), and $L=0.2$ m. Conduction through a plane wall (FE Reference Handbook 10.6 p. 209): $\dot Q=\dfrac{kA\,(T_1-T_2)}{L}=\dfrac{0.8\cdot10\cdot24}{0.2}=960$ W. Distractors: dropped the minus sign of the outside temperature, taking $\Delta T=20-4=16$ K (640 W); built the resistance as $kL/A$ instead of the handbook's $R=L/(kA)$ , putting the conductivity where a resistivity belongs ( $A\Delta T/(kL)=1500$ W); and converted only the inside temperature to kelvin, $\Delta T=293-(-4)=297$ K (11900 W), where converting both faces, or neither, gives the same 24 K.
6
medium
A. Steady-state material balances
A single-effect evaporator concentrates 1000 kg/h of an aqueous sugar solution containing 12 wt% sugar to a product of 30 wt% sugar. The vapor leaving the evaporator is pure water. At steady state, the rate at which water is evaporated is most nearly:
Figure — drag or pinch to move around
1. 180 kg/h
2. 600 kg/h
3. 760 kg/h
4. 880 kg/h
Show the answer & worked solution
Answer: B.600 kg/h
Steady state with no reaction: mass in = mass out, for the total and for the solute (sugar), which leaves only in the concentrate $L$ (the same balances the FE Handbook prints for a flash drum, p. 250). Solute: $F w_F = L w_L$ , so $L = 1000(0.12)/0.3 = 400$ kg/h. Total: $V = F - L = 1000 - 400 = 600$ kg/h. Check on the water: $880 = 600 + 400(0.7)$ . Distractors: the water in the concentrate figured on the FEED rate, $F(1-w_L)$ , instead of on $L$ , giving $F(w_L - w_F) = 180$ kg/h; the water leaving in the concentrate written as $L w_L$ (its sugar fraction) instead of $L(1-w_L)$ , $880 - 120 = 760$ kg/h; and ALL the water in the feed taken as evaporated, $F(1-w_F) = 880$ kg/h (the water that leaves with the concentrate forgotten).
7
medium
A. Particle properties, comminution & crystallization
A sieve analysis of a crushed limestone sample gives the cumulative mass percent finer than (undersize) each sieve opening: 0.1 mm → 8%, 0.25 mm → 30%, 0.5 mm → 58%, 1 mm → 82%, 2 mm → 100%. The median particle size d₅₀ is most nearly:
Figure — drag or pinch to move around
1. 0.336 mm
2. 0.375 mm
3. 0.429 mm
4. 0.500 mm
Show the answer & worked solution
Answer: C.0.429 mm
The cumulative undersize is the distribution F of the FE Handbook (Solids Processing, p. 259): the fraction finer than each size. The median d₅₀ is the size where F = 50 %, which falls between the 0.25 mm sieve (30 %) and the 0.5 mm sieve (58 %). Interpolating linearly: d₅₀ = 0.25 + (50 − 30)/(58 − 30) × (0.5 − 0.25) = 0.25 + 0.714 × 0.25 = 0.429 mm. Distractors: the interpolation fraction taken as (50 − 30)/58 instead of (50 − 30)/(58 − 30), 0.336 mm; the MIDPOINT of the two bracketing sizes (the percentages ignored), 0.375 mm; and the NEAREST sieve (58 %) read without interpolating, 0.500 mm.
8
medium
A. Distributions & descriptive statistics
The life of a carbide cutting insert turning steel at a fixed cutting speed is normally distributed with mean $\mu=50$ min and standard deviation $\sigma=8$ min. The probability that a randomly selected insert lasts between 46 min and 62 min is most nearly:
Figure — drag or pinch to move around
1. 0.2417
2. 0.4332
3. 0.6247
4. 0.6915
Show the answer & worked solution
Answer: C.0.6247
Standardize each limit (Normal Distribution, $z=(x-\mu)/\sigma$ ): $z_a=(46-50)/8=-0.50$ and $z_b=(62-50)/8=1.50$ . From the Unit Normal Distribution table, $F(1.50)=0.9332$ and $F(-0.50)=1-F(0.50)=1-0.6915=0.3085$ (the table lists only positive $x$ ). So $P(46<X<62)=F(z_b)-F(z_a)=0.9332-0.3085=0.6247$ . Distractors: the minus sign on $z_a=-0.50$ dropped, so the table value $F(0.50)=0.6915$ was subtracted in place of $F(-0.50)=0.3085$ (0.2417); only the area from the mean to 62 min counted, $F(1.50)-0.5$ , leaving out the part between 46 min and the mean (0.4332); and the upper limit ignored, reporting $P(X>46)=1-F(-0.50)$ (0.6915).
9
medium
A. Inorganic chemistry & electrochemistry
A technician adds 200 mL of 0.050 mol/L hydrochloric acid, a strong acid, to 300 mL of pure water. Taking the volumes as additive, the pH of the resulting solution at 25 °C is most nearly:
1. 1.30
2. 1.48
3. 1.70
4. 3.91
Show the answer & worked solution
Answer: C.1.70
Molarity is gram moles of solute per liter of solution (FE Reference Handbook 10.6 p. 86). The acid brings $0.050\times0.2=0.01$ mol of $\mathrm{H^+}$ (HCl dissociates completely) into $200+300=500$ mL of solution, so $[\mathrm{H^+}]=0.01/0.5=0.02$ mol/L. Then $\mathrm{pH}=\log_{10}(1/[\mathrm{H^+}])=-\log_{10}0.02=1.70$ (p. 87). Distractors: the dilution ignored, giving the pH of the 0.050 mol/L stock itself ( $-\log_{10}0.05=1.30$ ); the 300 mL of water taken as the final volume, leaving out the acid's own 200 mL ( $[\mathrm{H^+}]=0.0333$ mol/L, pH 1.48); and the natural log used in place of $\log_{10}$ ( $-\ln 0.02=3.91$ ).
10
medium
A. PFDs, equipment sizing & cost estimation
The delivered cost of the purchased equipment for a new solid-fluid processing plant was estimated at 1,500,000 USD when the plant cost index stood at 550. The index is now 770. Using the Lang-factor method, the present fixed-capital investment for the plant is most nearly:
1. 6,450,000 USD
2. 8,400,000 USD
3. 9,030,000 USD
4. 10,500,000 USD
Show the answer & worked solution
Answer: C.9,030,000 USD
First bring the equipment cost up to date with the cost-index ratio (FE Handbook, Chemical Engineering — Cost Indexes, p. 262): 1,500,000 USD × 770/550 = 2,100,000 USD. The Handbook's Lang-factor table (Capital Cost Estimation, p. 263) gives a solid-fluid processing plant 4.3 for fixed-capital investment and 5.0 for total-capital investment (fixed capital plus working capital). The factor multiplies the delivered equipment cost and already includes it, so the fixed-capital investment is 4.3 × 2,100,000 USD = 9,030,000 USD. Distractors: the cost-index update skipped (the old equipment cost used), 6,450,000 USD; the solid processing row (4.0) read instead of the solid-fluid processing row, 8,400,000 USD; and the TOTAL-capital column (5.0) read for a fixed-capital estimate, which adds working capital, 10,500,000 USD.

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