In an attempt to conserve water and to be awarded LEED (Leadership in Energy and

ID: 510818 • Letter: I

Question

In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 15.0 x 103 L cistern has been installed during construction of a new building.

The cistern collects water from an HVAC (heating, ventilation, and air-conditioning) system that is designed to provide 1.00 x 105 ft3/min air at 22.0°C and 40.0% relative humidity after converting it from ambient conditions (27.0°C, 70.0% relative humidity).

The collected condensate serves as the source of water for lawn maintenance. Assume P = 1 atm.

a) Estimate the intake of air at ambient conditions in cubic feet per minute.

b) Estimate the hours of operation required to fill the cistern.

Explanation / Answer

Ans.

#1. Given,

At 220C, the properties of conditioned air are-

Flowrate = 1.00 x 105 ft3/ min ; [1 ft3 = 28.3168 L]

= 1.00 x 105 x (28.3168L) / min

= 2.83168 x 106 L

#2. Let the intake at 270C = X liters/ min.

So, volume of air flowing per minutes = X liters

Moisture content = 70.0 % of X L = 0.70X L

Dry (some moisture removed) air content = X L – 0.70 X L = 0.30 L

#3. Volume of released air at 270C can be given by Charles’ law

(V1/ T1) = (V2/ T2) - equation 1

Where, V1 and T1 are for 270C (= 300.15 K); V2 and T2 are for 220C (295.15K).

Putting the values in equation 1-

(V1 / 300.15 K) = 1.00 x 105 ft3 / 295.15 K

Or, V1 = (1.00 x 105 ft3 / 295.15 K) x 300.15 K = 1.01694 x 105 ft3

That is, the volume of same exhaled air (1.0 x 105 ft3, per minute, at 220C) is equal to 1.0694 x 105 ft3 at 270C.

Now, during condensation, only water is removed. So, after removing water, the volume of dry gas and 40% RH is equal to 1.0694 x 105 ft3.

0.30X ft3 + 40% relative humidity of 0.30X ft3 = 1.01694 x 105 ft3

Or, 0.30X ft3 + 0.12X ft3 = 1.01694 x 105 ft3

Or, 0.42X ft3 = 1.01694 x 105 ft3

Or, X = (1.01694 x 105) / 0.42 = 2.42129 x 105 ft3

Therefore, intake per minute at 270C = 2.42129 x 105 ft3

#4. Volume of moisture removed =

Volume of intake air at 270C – Volume of exhausted air at 270C

= 2.42129 x 105 ft3 – 1.01694 x 105 ft3

= 1.40 x 105 ft3

= 1.40 x 105 (28.3168 L)

= 3.964 x 106 L

Assuming moisture behave ideally at 270C, calculate the moles of water vapor using ideal gas equation-

Using Ideal gas equation: PV = nRT - equation 2

Where, P = pressure in atm

V = volume in L

n = number of moles

R = universal gas constant= 0.0821 atm L mol-1K-1

T = absolute temperature (in K)

Putting the values for amount of moisture removed at 220C-

1.00 atm x 3.964 x 106 L = n x (0.0821 atm L mol-1K-1) x 300.15 K

Or, n = 3.964 x 106 atm L / 24.642315 atm L mol-1

Or, n = 160861.51 mol

Thus, during conditioning, 160861.51 mol water is removed.

Mass of water removed = moles x molar mass

= 160861.51 mol x (18.0 g/ mol)

= 2895507.18 g

= 2895.50718 kg

Assuming density of water to be 1.00 kg/ L at through the temperatures (270C to 220C), the volume of liquid water condensed in the cistern is given by-

Volume of water condensed = Mass of moisture removed x density of water

= 2895.51 kg x (1.00 kg/ L)

= 2895.51 L

Therefore, rate of water condensation in cistern = 2895.51 L/ min

Time required to fill the cistern = Capacity of cistern/ rate of water condensation

= 15000 L/ (2895.51 L/ min)

= 5.18 min

= 0.086 hr

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In an attempt to conserve water and to be awarded LEED (Leadership in Energy and

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