When fermentation units are operated with high aeration rates, significant amounts of water can be evaporated into the air passi
ng through the fermentation broth. Since fermentation can be adversely affected if water loss is significant, the air is humidified before being fed to the fermenter. Sterilized ambient air is combined with steam to form a saturated air–water mixture at 1 atm and 90°C. The mixture is cooled to the temperature of the fermenter (35°C), condensing some of the water, and the saturated air is fed to the bottom of the fermenter. For an air flow rate to the fermenter of 10 L/min at 35°C and 1 atm, estimate the rate at which steam must be added to the sterilized air and the rate (kg/min) at which condensate is collected upon cooling the air–steam mixture.
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Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate =
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.
Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is
As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%