Question

In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20.0 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 (31.0°C, 86.0% 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.

Answer 1

#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 31⁰C = X liters/ min.

So, volume of air flowing per minutes = X liters

            Moisture content = 86.0 % of X L = 0.86X L

            Dry (some moisture removed) air content = X L – 0.86 X L = 0.14 L

#3. Volume of released air at 31⁰C can be given by Charles’ law

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

Where, V1 and T1 are for 31⁰C (= 304.15 K); V2 and T2 are for 22⁰C (295.15K).

Putting the values in equation 1-

            (V1 / 304.15 K) = 1.00 x 10⁵ ft³ / 295.15 K

            Or, V1 = (1.00 x 10⁵ ft³ / 295.15 K) x 304.15 K = 1.03049 x 10⁵ ft³

That is, the volume of same exhaled air (1.0 x 10⁵ ft^3, per minute, at 22⁰C) is equal to 1.03049 x 10⁵ ft³ at 33⁰C.

Now, during condensation, only water is removed. So, after removing water, the volume of dry gas and 40% RH is equal to 1.03727 x 10⁵ ft³.

            0.14X ft³ + 40% relative humidity of 0.14X ft³ = 1.03049 x 10⁵ ft³

            Or, 0.14 X ft³ + 0.056 X ft³ = 1.03049 x 10⁵ ft³

            Or, 0.196X ft³ = 1.03049 x 10⁵ ft³

            Or, X = (1.03049 x 10⁵ ft³) / 0.196 = 5.26 x 10⁵ ft³

Therefore, intake per minute at 31⁰C = 5.26 x 10⁵ ft³

#4. Volume of moisture removed =

Volume of intake air at 31⁰C – Volume of exhausted air at 31⁰C

= 5.26 x 10⁵ ft³ – 1.03049 x 10⁵ ft³

                        = 4.23 x 10⁵ ft³

                        = 4.23 x 10⁵ (28.3168 L)

                        = 1.21 x 10⁷ L

Assuming moisture behave ideally at 31⁰C, calculate the moles of water vapor condensed 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 1.21 x 10⁷ L = n x (0.0821 atm L mol^-1K^-1) x 304.15 K

            Or, n = 1.21 x 10⁷ L atm L / 24.970715 atm L mol^-1

            Or, n = 484567.62 mol

Thus, during conditioning, 484567.62 mol water is removed.

Mass of water removed = moles x molar mass

                                    = 484567.62 mol mol x (18.0 g/ mol)

                                    = 8722217.16 g

                                    = 8722.22 kg

Assuming density of water to be 1.00 kg/ L at through the temperatures (31⁰C 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

                                                            = 8722.22 kg x (1.00 kg/ L)

                                                            = 8722.22 L

Therefore, rate of water condensation in cistern = 8722.22 L/ min

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

                                                = 20000 L/ (8722.22 L/ min)

                                                = 2.29 min

                                                = 0.0381 hr