Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H
( YH
)
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH
= molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
Answer:
The level of the service is loss and the density is 34.2248 pc/mi/ln
Explanation:
the solution is attached in the Word file
Answer:
attached below
Explanation:
SO AS WE CAN SEE FOR GASES THE CONVECTIVE HEAT TRANSFER COEFFICIENT HAS VERY LOW VALUE WITH RESPECT TO LIQUID THATS WHY WE USE FIND TO MAXIMIZE HEAT TRANSFER RATE WITH GASES AS A CONVECTIVE MEDIUM.
Answer:
i) Heat transfer coefficient (h) = 7 w/m²k
ii) Heat transfer per meter width of wall
= h x L x 1 x (Ts - T₆₀)
= 7 x 0.3048 x (505.4 - 322) = 414.747 w/m
Explanation:
see attached image