Answer:
Calculate the individual residential lifetime cancer risk.
Risk = Potential factor x CDI …… (1)
Calculate the value of C
C =
t = d / v
t = 150 miles / 2 mile per hrs.
t = 75 hrs
t = 75/24
t = 3.13 days
Substitute the obtained value in (2).
C =
C = 0.72 x e^(-0.1*3.13)
C = 0.72 x 0.7313
C = 0.526 mg/L
Substitute the obtained value in (1).
Risk = Potential factor x CDI
Risk = 0.30kg.d/mg x 0.526mg/L x 2L/d x 350day/365days
Risk = 0.3026
Answer:
Figure 1 shows the neural network that computes the XOR of two inputs.
Two Input unit: X₁ and X₂
One hidden layer with two hidden units: h₁ and h₂
Two Weights for hidden unit 1 = 20, 20
Bias for hidden unit 1 = -10
Two Weights for hidden unit 2 = -20, -20
Bias for hidden unit 2 = 30
Weights for output layer = 20, 20
Bias for output layer = -30
Activation function = sigmoid (σ)
Output at hidden unit and at the output unit Y is shown in Table attached in Fig 2
Answer:
The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.
Explanation:
Data
<u>kLa</u> = 0.17/s
<u>Solubility of oxygen</u> = 8 × 10^-3 kg / m^3
<u>The maximum specific oxygen uptake rate </u>= 4 mmol O2 / g h.
<u>Concentration of oxygen</u> = 0.5 × 10^-3 kg/ m^3
<u>**The maximum cell density</u> = 50 g/l
___________________
The calculated maximum cell concentration:
xmax= kLa · CAL*/ qo
CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate
Replacing the data given
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h
4 mmol O2 / g h to kg O2/ g s
= 3.56 x 10^-3 kg O2/ g s
So then,
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s
xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l
_____________________