In an attempt to conserve water and to be awarded leed (leadership in energy and environmental design) certification, a 20,000-liter cistern has been installed during construction of a new building. the cistern collects water from an hvac (heating, ventilation, and air-conditioning) system designed to provide 2830 cubic meters of air per minute at 22°c and 50% relative humidity after converting it from ambient conditions (31°c, 70% relative humidity). the collected condensate serves as the source of water for lawn maintenance. estimate (a) the rate of intake of air at ambient conditions in cubic feet per minute and (b) the hours of operation required to fill the cistern.

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