If a machine uses LESS effort to overcome a given resistance force (if Fe is less than FR), it has an actual mechanical advantag
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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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Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
<u><em>Deflection angle at B</em></u>
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
<u><em>slope at B </em></u>
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
<u><em>Deflection angle at C </em></u>
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
<u><em>Slope at C </em></u>
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
