Explanation:
bad weather usually form when low pressure pushed high pressure ot the way.
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
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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
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Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.
Answer:
Height of the water = √(128)/147456 ft
Explanation:
Given
Radius, r = 1 ft
Height, h = 2 ft
Radius of hole = 1/32in
Acceleration of gravity, g = 32ft/s²
c = 0.6
Area of the hold = πr²
A = π(1/32)² ---- Convert to feet
A = π(1/32 * 1/12)²
A = π/147456 ft²
Area of water = πr²
A = π 1²
A = π
The differential equation is;
dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water
A1 = π/147456, A2 = π
dh/dt = (π/147456)/π √(2*32*2)
dh/dt = 1/147456 * √128
dh/dt = √128/147456 ft
Height of the water = √(128)/147456 ft