Answer:
Total saving would be of 36.917 $\yr
Explanation:
Given Data:
Degree F
Degree F
A = 30 ft^2
%
Heat loss before previous final value
= 360 Btu/hr
Heat loss after new value
saving would be
= 616.1782 kw hr/yr
$
= 36.917 $\yr
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
_____________________
Answer:
a) I=0 b) 4.17V c) 0.354 A d) 14.5s
Explanation:
a) consider circuit in the attachment
i(t)= E/R (1- e^(-t/RL))
i(0)= 12.5/3×(1-e^(0/RL))
i(0)=0
b) at t⇒∞
i(∞)= 12.5/3× (1- e^(-∞/RL))
= 4.17V
c) 1/RL= 1/(6.95×3)= 0.0479616
i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)
= 0.354A
d) I/2= I (1- e^(-t/RL))
t= - RL ln0.5
t= - 3×6.95 × (-0.693)
t= 14.5 s
