CEE 370 Environmental Engineering Principles Lecture #17 - PDF document

CEE 370 Lecture #17 10/4/2019 Print version Updated: 4 October 2019 CEE 370 Environmental Engineering Principles Lecture #17 Ecosystems IV: Microbiology & Biochemical Pathways Reading: Mihelcic & Zimmerman, Chapter 5 Davis & Masten, Chapter 4 David Reckhow CEE 370 L#17 1 Batch Microbial Growth  Observed behavior Stationary Death Lag Exponential Growth Time 2 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 1

CEE 370 Lecture #17 10/4/2019 Exponential Growth model dN  dX   rN X or dt dt Animals Bacteria where, N X = concentration of organisms at time t t t = time  = proportionality constant or specific r growth rate, [time ─ 1 ] dN/dt dX/dt = organism growth rate, [mass per volume-time] 3 CEE 370 L#16 David Reckhow Exp. Growth (cont.) Higher Organisms Microorganisms dN  dN  dX dX     rN rdt X dt or or dt N dt X     N X        ln rt ln t     N X     o o  N  rt  t N o e X X o e 4 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 2

CEE 370 Lecture #17 10/4/2019 Exp. Growth Example A microbial system with an ample substrate and nutrient supply has an initial cell concentration, X o , of 500 mg/L. The specific growth rate is 0.5 /hour. a) Estimate the cell concentration after 6 hours, assuming log growth is maintained during the period. b) Determine the time required for the microbial population to double during this log growth phase. 5 CEE 370 L#16 David Reckhow Solution I a) To determine the microbial concentration after 6 hours, we substitute into the exp growth model, obtaining, = 500 mg  t (0.5/hr x 6 hr) X = X e L x e o X = 10,000 mg L Thus, in a period of six hours, the microorganisms increase 20 fold. 6 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 3

CEE 370 Lecture #17 10/4/2019 Solution II b) To determine the time for the concentration to double, we use the log form. Also, if the concentration doubles, then ln X X  = t or X = 2 X o o Or, solving for t we obtain,   X ln     X ln 2 o t = =  0.5 / hr t = 1.4 hr. Thus, the microbial population can double in only 1.4 hours. By comparison, the human population is currently doubling about every 40 years. 7 CEE 370 L#16 David Reckhow Limitations in population density  Carrying capacity, K, and the logistic growth model:    X      1 max K      dX X     1 X max dt K   K  X t      K X      t 1 0 e     max X     0 8 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 4

CEE 370 Lecture #17 10/4/2019 Logistic Model (cont.)  In many books they use different terms when applying it to animal dynamics:    dX X     1 X max dt  K     dN N    r 1 N dt  K  K KN   N o   t      rt    N K N e K N   o o   rt 1 0 e     N     0 9 CEE 370 L#16 David Reckhow Logistic Growth Example  Determine 10 day population for:  Initial population: X 0 = 2 mg/L  Max growth rate: 1 day -1  Carrying capacity: 5000 mg/L 10 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 5

CEE 370 Lecture #17 10/4/2019 Substrate-limited Growth  Also known as resource-limited growth  THE MONOD MODEL Similar to enzyme kinetic model introduced in 𝑒𝑌 ≡ 𝜈𝑌 � 𝜈 ��� 𝑇 lecture #13 𝐿 � � 𝑇 𝑌 𝑒𝑢 ������ where,  m = maximum specific growth rate, [day -1 ] S = concentration of limiting substrate, [mg/L] K s = Monod or half-velocity constant, or half saturation coefficient, [mg/L] 11 CEE 370 L#16 David Reckhow Monod Kinetics 0.5*µ m K S 12 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 6

CEE 370 Lecture #17 10/4/2019 Decay term  With microorganisms, when using a substrate-limited growth model, it is also appropriate to consider “decay”  Decay covers the “cost of doing business”, including the energy lost in respiration, cell maintenance & reproduction 𝑒𝑌  The mode is simple first order � �𝑙 � 𝑌 𝑒𝑢 �����  And combining it with the Monod model 𝑒𝑌 � 𝜈 ��� 𝑇 𝐿 � � 𝑇 𝑌 � 𝑙 � 𝑌 𝑒𝑢 ������� 13 CEE 370 L#17 David Reckhow Substrate model 𝑍 ≡ ∆𝑌  Yield coefficient ∆𝑇  And 𝑒𝑢 � � 1 𝑒𝑇 𝑒𝑌 𝑍 𝑒𝑢  So, combining with the overall growth model 𝑒𝑇 𝑒𝑢 � 1 𝑒𝑌 � 1 𝜈 ��� 𝑇 𝐿 � � 𝑇 𝑌 � 𝑙 � 𝑌 𝑍 𝑒𝑢 𝑍 14 CEE 370 L#17 David Reckhow Lecture #17 Dave Reckhow 7

CEE 370 Lecture #17 10/4/2019 Competition for Substrate 15 CEE 370 L#16 David Reckhow World Population Growth 16 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 8

CEE 370 Lecture #17 10/4/2019 Human Population Projection  Arithmetic Model  Exponential Model  Decreasing Rate of Increase Model  Graphical extension  Graphical comparison  Ratio method  Other 17 CEE 370 L#16 David Reckhow Linear Model P = P o + k l  t Population k l Time Good for some types of growth, but not all • Babies growth ~ 2 lb/month • Linear model predicts 1,320 lb by age 55 18 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 9

CEE 370 Lecture #17 10/4/2019 Exponential Model Exponential Functions  r t P = P e o where, P = population at time t, P o = population at time zero, r (k e ) = population growth rate, years -1 ,  t = time, years. 19 CEE 370 L#16 David Reckhow Parameter Estimation: Exponential Model  r t Ln P = Ln P e ( ) ( ) o    Ln P ( ) Ln P ( ) r t 0 Ln Population k e or r Time (delta-t) 20 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 10

CEE 370 Lecture #17 10/4/2019 Decreasing Rate of Increase Model P = P o + (P max -P o )(1-exp(-k e  t)) Where: P max 55000 is the limiting k e = 0.15 yr -1 or saturation 50000 population Population k e = 0.04 yr -1 45000 k e = 0.01 yr -1 40000 35000 Decreasing Rate of Increase P max = 50,000 30000 0 10 20 30 40 50 60 70 80 90 100  t (years) 21 CEE 370 L#16 David Reckhow Comparison of Population Models 50000 Exponential 48000 k e (r) = 0.014 yr -1 46000 Linear 44000 Decreasing Rate of Increase Population k l = 500 yr -1 42000 k e = 0.04 yr -1 P max = 50,000 40000 38000 36000 34000 32000 30000 0 5 10 15 20 25 30  t (years) P o = 33,000 for all 22 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 11

CEE 370 Lecture #17 10/4/2019 Parameter Estimation: Decreasing Rate of Increase Model Ln(P max -P) = Ln(P max -P o ) - k d  t Ln (P max -P) k d Time or delta-t 23 CEE 370 L#16 David Reckhow Comparison of Population models: 100 yrs. 130000 120000 110000 Exponential 100000 k e = 0.014 yr -1 90000 Population Linear 80000 70000 k l = 500 yr -1 60000 50000 Decreasing Rate of Increase 40000 k e = 0.04 yr -1 30000 P max = 50,000 20000 10000 0 0 10 20 30 40 50 60 70 80 90 100  t (years) P o = 33,000 for all 24 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 12

CEE 370 Lecture #17 10/4/2019 Example: World Population Growth In 1950 the world population was estimated to be 2.5 billion. In 1990 it was estimated to be 5.5 billion. Assuming this growth rate will be sustained, a) calculate the world population in the year 2000 b) determine the year the global population will reach 10 billion. 25 CEE 370 L#16 David Reckhow Solution We must first calculate the population growth rate, r. To do this we convert Equation into log form: ln P      = r t   P o 26 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 13

CEE 370 Lecture #17 10/4/2019 Solution, cont. Now, if we know the population at two different times in the past, we can calculate the population growth rate, r. We know that the population in 1950 was 2.5 billion, and that it was 5.5 billion in 1990. The time change,  t, is 40 years. Thus, the population growth rate is,   9 ln 5.5 x 10  P      ln     2.5 x 10 9 Po r = =  t 40 years r = 0.020 / yr 27 CEE 370 L#16 David Reckhow Solution, cont. Population in the year 2000?  P = P e r t = 5.5 x e 0.020 x (2000 - 1990) o 9 P = 6.7 x in year 2000 10 Actual: 6.1 x 10 9 28 CEE 370 L#16 David Reckhow Lecture #17 Dave Reckhow 14