A wine aerator is a small, in-bottle, hand-held pour-through or decantor top device using the venturi effect for aerating the wi
ne. These devices mix air into the wine as it flows through (or over), increasing exposure to oxygen and causing aeration. Apparently, wine which has been aerated tastes better. You want to design a new aerator that will also chill the wine as it is poured. This aerator will be aventuri-shaped tube with an input diameter of 1 cm and an output diameter of 1 mm. Determine the flow rate of the wine exiting theventure tube while pouring as a function of pouring angle and volume of wine in the bottle. The wine bottle can be modeled as a cylinder of 10 cm in diameter and length of 35 cm. You can assume the pressure at the exit is 1 atm and you can assume the head in the bottle while pouring has a pressure of 1 atm.You will need to determine the pressure at the inlet of the aerator as a function of time at various angles. Assume the bottle starts full. Also, calculate the resultant force needed to keep the device attached to the top of your wine bottle as a function of time for the same angles. Comment on maximum forces. Finally, you need to determine the rate that energy needs to be removed to chill the wine from room temperature to 4 C as it flows through the aerator.
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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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