Students in a robotics class designed cars to race as their end of the year project. Students collected data on their cars and c
1 answer:
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Explanation:
The given data is as follows.
Temperature of dry bulb of air =
Dew point = = (10 + 273) K = 283 K
At the dew point temperature, the first drop of water condenses out of air and then,
Partial pressure of water vapor = vapor pressure of water at a given temperature
Using Antoine's equation we get the following.
= 0.17079
= 1.18624 kPa
As total pressure = atmospheric pressure = 760 mm Hg
= 101..325 kPa
The absolute humidity of inlet air =
= 0.00735 kg / kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg / kg dry air.
Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.
0.07 kg - 0.00735 kg
= 0.06265 kg for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
= 6.265 kg
Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.
Answer : The volume given to the patient should be, 85.5 mL
Explanation :
As we are given that the suspension contains 10g/15mL.
Now we have to determine the volume should be given to the patient.
As, 10 grams of lactulose syrup needed 15 mL volume of solution
So, 57 grams of lactulose syrup needed volume of solution
Thus, the volume given to the patient should be, 85.5 mL