APSC 150 Biological Engineering

3 March 2009 Please answer all five questions. Questions 1 is worth 20 marks (1 mark each part). The marks for questions 2 through 5 are shown. 1. Please answer true (T) or False (F) to each of the following: a) Levoisier did experiments to show that matter (mass) is conserved. (True) b) Biological treatment in a lagoon removes the colour from pulp mill wastewaters. (F) c) A secondary clarifier (in an activated sludge reactor) allows the HRT and the SRT to be

different. (True) d) In the BOD test, a control is used to account for the amount of oxygen used by the bacteria

when they are not exposed to effluent. (True) e) Bacteria can break down hydrogen peroxide and produce oxygen. (True) f) If the level of dissolved oxygen goes below 1 mg/L in a BOD test, we would not count the

results because oxygen would be limiting. (True) g) It would be impossible to get more than 10,000 kg of bacteria produced from 10,000 kg of

BOD. (True) h) In activated sludge reactors, biomass must be wasted at the same rate it is produced to keep

the reactor at a steady state. (True) i) In biological systems, if microbial growth decreases, then less of the substrate is completely

oxidized, and less oxygen is required. (F) j) The rate of oxygen transfer from the atmosphere to a body of water would be increased by

increasing the surface area of contact. (True) k) Lagoons would often have more aerators at the front (inlet) end, than at the back (outlet) end.

(True) l) Pharmaceuticals (drugs) would be found in most municipal wastewaters. (True) m) Activated sludge reactors are less susceptible than lagoons to toxic shocks because of their

recycle stream. (F) n) Dupont has developed a process by which plastics can be made from renewable feedstocks.

(True) o) Design problems are typically over-specified. (F) p) If the flowrate of wastewater to a lagoon increases, then the effluent BOD will increase too.

(True) q) Biological systems, because of microbial growth, do not necessarily obey the law of

conservation of mass. (F) r) Use of a larger volume of effluent in the test should result in a lower final dissolved oxygen

concentration (DO). (True) s) I am reasonably confident that I will not remember any of this biological stuff 10 minutes

after I walk out of this room. Joke question…students get mark whether T or F t) Protease enzymes such as subtilisin are often included in laundry detergents and stain

removers. True

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Question 2 10 marks total a) Calculate ThOD for cellobiose

C12H22O11+ 12 O2 → 12CO2 + 11 H20 2.5 marks (for balanced reaction)

342 mg/L cellotriose uses 12x32 (or 384) mg/L oxygen 20 mg/L cellotriose uses

2/5.2238434220 OLmg=× 2.5 marks for calculated oxygen

demand b) First use data to get kinetic constants µmax and Ks by doing a reciprocal plot

BOD Concentration (mg/L)

Specific Growth Rate (μ) (1/day)

1/BOD (or 1/S) 1/μ

2 0.02927 0.5 34.16 50 0.46160 0.02 2.166

Students can get the equation of the line by using these two points, either with or without a

graph. I have given the graph below.

y = 66.67x + 0.833R² = 1

0

5

10

15

20

25

30

35

40

0 0.1 0.2 0.3 0.4 0.5 0.6

1/µ (days)

1/S  (L/mg BOD)

Using this, and the linear form of the Monod equation

2

maxmax

maxmax

max

max

111

1

1

μμμ

μμμ

μμ

μμ

+⎥⎦⎤

⎢⎣⎡=

+=

+=

+=

SKs

SS

SKs

SSKs

SKS

S

Intercept = 1/μmax= 0.833 days. Therefore μmax= 1.2 days-1 2.5 marks

Slope = Ks/μmax= 66.67 d mg BOD/L. Therefore Ks = 80 mg BOD/L. 2.5 marks Question 3 (15 marks total)

a) 10 marks for calculations in part a

PfBBDDLmgBOD )()()/( 2121 −−−

=

Test 1 Lmg1002

300/5.0)7.820.9()03.720.9()L/mg(BOD =

−−−=

Test 2 Lmg1000

300/2)70.820.9()03.220.9()L/mg(BOD =

−−−=

Test 3 Lmg498

300/5)70.820.9()40.020.9()L/mg(BOD =

−−−=

b) • All three tests should give the same value, since they all came from the

same sample. The lower value of BOD for test 3 indicates a problem. In this case, the problem is most likely that we ran low on oxygen. If oxygen goes below 1 mg/L, it becomes limiting, and slows the material metabolism of the organics. In this case it went down to 0.4 mg/L. 5 marks for explanation

Question 4 (25 marks total) a) First, figure out the MCRT (or SRT) required to get to 50 mg/L BOD in the effluent.

3

S d

C

MAX dC

1K kS

1 k

⎛ ⎞+⎜ ⎟θ⎝ ⎠=

⎛ ⎞μ − +⎜ ⎟θ⎝ ⎠

Solve this equation for SRT.

days

kKkSSK

dSdm

SC

14.7

)1.080()1.02.1(20)2080(

)()(

=

×−−+

=−−

+=

μθ

5 marks for solving for SRT (or MCRT) The use the biomass equation to get volume:

C 0

d C

Y(S S)X

(1 k )θ −

=θ + θ

substituting V/Q for HRT (θ) and rearranging

( )( )

3

1

3

720,35

))14.7)(1.0(1)(4000(

)201000)(5.0)(70000)(14.7(

1

m

ddLmgSS

LmgBOD

mgBODmgbacteria

dmd

V

kXSSQYV

sd

Oc

=

+

−=

++

=

−

θθ

5 marks for solving for volume b)

( )( )

dkg

ddL

kgBODkgBOD

kgbiomassd

m

008,20

))14.7)(1.0(1(

)020.00.1)(5.0)(70000(

θCk1S-SQYP 1

3

d

0X

=

+

−=

+= −

5 marks for solving for Px (could also do material balance first, then add QwXw+QeXe to get the same answer)

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c) Overall Biomass Balance around Clarifier (Q + Qr)X = QeXe + QrXr + QwXw

Overall Volumetric Flow Rate Balance Qe = Q – Qw (Q + Qr)X = QwXw + QrXr + (Q – Qw)Xe Solve for Qw

Qw = 620.7 m3/d 5 marks for solving for Qw

d) To achieve the same performance, lagoon needs same MCRT (SRT) of 7.14 days. For lagoon HRT=SRT. Therefore the lagoon needs a 5 day HRT.

33 800,499)14.7)(/000,70( mdaysdmQVQV

===∴

=

θ

θ

As such, 500,000 m3 could do it. 5 marks for solving for lagoon volume

Students may know this without showing calculations, which is okay. Question 5 (30 marks total) a) Define the system = Atmosphere of Earth (an open system) CO2 to be removed is 350-250=100 ppmv 100 ppmv= 100 m3 in 1×106 m3

Therefore in the atmosphere:

Volume CO2 to be removed (m3) = 31436

18m104m100

101104

×=××

×4 marks

5

kmoles CO2 = n = PV/RT (ideal gas law) = (101.3 kPa)(4×1014 m3)/ ((8.314 KJ/kmol.K)(273 K)) = 1.785×1013 kmole = 7.855×1014 kg 4 marks

Time required= years100days1065.3d/kg1015.2

kg10855.7 410

14=×=

×

× 2 marks

b) Methane used by 1 car in a year

year/kg1000km100

methanekg5yrkm20000 =× 4 marks

Methane produced by process in a year

yr/CHkg1085.2yr

day365

CHkmoleCHkg16

COkmoleCHkmole1

COkg44COkmole

dCOkg

1015.2

412

4

4

2

4

22210

×=×

××××

4 marks Therefore, number of cars fuelled

cars1085.2caryear/CHkg

year/CHkg1000

1085.2 94

412×=

⋅×

(2 marks)

Note students could also arrive at the same answer by converting fuel consumption

to kmoles per year (62.5 kmole/year⋅car) and dividing the kmole of CH4 produced in 1 year (1.785×1011 kmole/year)

c) Balance the stoichiometric relationship

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CO2 + aH2 + bNH3 → 0.019 C5H7O2N + cCH4+ dH2O

Left side Right side Nitrogen b ∴b=0.019 0.019 Carbon 1 (5×0.019)+c ∴c=0.905 Oxygen 2 (2×0.019)+d ∴d=1.96 Hydrogen (2a+3b)

2a+(3×0.019) ∴a=3.81 (7×0.019)+ (4×0.905)+ (2×1.96)=7.67

CO2 + 3.81H2 + 0.019NH3 → 0.019 C5H7O2N + 0.905CH4+ 1.96H2O (10 marks for correct balance)

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