Answer:
rate of water condensation in cistern = 2,604.628L/ min
the hours of operation required to fill the cistern 0.128hr
Explanation:
Given,
At 22°C, the properties of conditioned air are-
Flowrate = 2830m^3/min ; [1 m^3= 1000 L]
= 2830 x (1000L) / min
= 2.830 x 10^6 L
Consider:
intake at 31°C = X liters/ min.
Therefore
X liters = volume of air flowing per minutes
Moisture content (relative humidity)
= 70.0 % of X L = 0.70X L
Dry (some moisture removed) air content
= X L - 0.70X L = 0.30 L
Used charles' law to determine the Volume of released air at 31°C
(V1/ T1) = (V2/ T2) - equation 1
Where,
V1 = 2830m^3
T1 = 22°c = 295k
V2 = ?
T2 = 31°c = 304k
2830m^3/ 295k = V2/304k
V2 = 2830m^3 × 304k
--------------------------
295k
= 2916.339m^3
Therefore,
The volume of same exaled air (2830m^3/min at 22°c) is equal to 2916.339m^3 at 31°c
During condensation, only water is removed
Therefore
The volume of dry gas and 50% relative humidity is equal to 2916.339m^3 at 22°c
Now, only water is removed during condensation. After removing water, the volume of dry gas and 40% RH is equal to 2916.339m^3
So,
Dry air content + 50% of dry air content
=0.30xm^3 + 50% of 0.30xm^3
=0.30xm^3 + 0.15xm^3
= 0.45xm^3
Intake per minute = x = ?
Let
0.45xm^3 = 2916.339m^3
X = 2916.339m^3
-----------------------
0.45
X = 6480.754m^3
Therefore, intake per minute at 31°C = 6480.754m^3
Volume of moisture removed = volume of intake air at 31°C - Volume of exhausted air at 31°C
= 6480.754m^3 - 2916.339m^3
= 3,564.415m^3
Convert m^3 to L (1m^3 =1000L)
= 3,564.415 × 10^3L
= 3.564 × 10^6L
Assuming moisture behave ideally at 27°C, calculate the moles of water vapor using ideal gas equation-
PV = nRT ....equation 2
Where,
P = pressure in atm = 1.00atm
V = volume in L = 3.564 × 10^6L
n = number of moles = ?
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K) = 300k
Putting the values for amount of moisture removed at 22°C-
1.00 atm x 3.564 × 10^6L = n x (0.0821 atm L mol-1K-1) x 300 K
n = 3.564 × 10^6atmL/ 24.63 atm L mol-1
n = 144,701.583 mol
Thus,
during conditioning, 144,701.583 mol of water was removed.
Mass of water removed = moles x molar mass
=144,701.583 mol x (18.0 g/ mol)
= 2,604,628.44g
= 2,604.62844 kg
= 2,604.628kg
Let density of water be 1.00 kg/ L at through the temperatures (31°C to 22°C), the volume of liquid water condensed in the cistern is given by-
Volume of water condensed = Mass of moisture removed x density of water
= 2,604.628kgx (1.00 kg/ L)
= 2,604.628L
A. Therefore, rate of water condensation in cistern = 2,604.628L/ min
B. Time required to fill the cistern = Capacity of cistern/ rate of water condensation
Given
Capacity of cistern = 20000L
= 20000 L/ (2,604.628L/ min)
= 7.679min
Equivalent to 0.128hr
