Question

In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 25. 0×10

3

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. 00×10

5

ft

3

/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. Physical Property Tables Input Air Check significant figures throughout the calculation. Recheck conor pressure calculations/tabulations Make sure that you understand the definition of relative humidity. Estimate the intake of air at ambient conditions in cubic feet per minute. Estimate the hours of operation required to fill the cistern. Hours

Answer 1

Here are the step-by-step calculations to solve this problem:

1) Calculate the density of air at the intake (ambient) conditions using the ideal gas law:

P = ρRT

Where:

P = atmospheric pressure = 1 atm

R = gas constant for air = 0.3704 ft3·atm/lb·°R

T = ambient temperature = 27.0°C = 298.15 K

Plugging in:

ρ = P/(RT)

= (1 atm)/(0.3704 ft3·atm/lb·°R)(298.15 K)

= 0.0764 lb/ft3

2) Calculate the relative humidity of the intake air:

Given:

Relative humidity = 70.0%

Relative humidity is defined as the ratio of the partial pressure of water vapor in the air to the saturated vapor pressure of water at that temperature.

At 27°C, the saturated vapor pressure of water is 3.756 kPa (from steam tables).

The partial pressure of water vapor is:

(Relative humidity)(Saturated vapor pressure)

= (0.700)(3.756 kPa) = 2.629 kPa

3) Use the ideal gas law to calculate the density of water vapor in the intake air:

ρ = P/(RT)

Where:

P = partial pressure of water vapor = 2.629 kPa

T = 298.15 K

R = 461.5 J/kg·K for water vapor

Plugging in:

ρ = (2.629 kPa)/(461.5 J/kg·K)(298.15 K)

= 0.01907 kg/m3

= 0.001191 lb/ft3

4) Calculate the density of dry air:

ρ_dry air = ρ_total - ρ_water vapor

= 0.0764 - 0.001191

= 0.0752 lb/ft3

5) Calculate the intake volume flow rate of air:

Given:

Intake air flow rate = 1.00x105 ft3/min

At ambient conditions:

Volume flow rate = 1.00x105 ft3/min

6) Calculate the mass flow rate of dry air:

ṁ = ρV

Where:

ρ = density of dry air = 0.0752 lb/ft3

V = volume flow rate = 1.00x105 ft3/min

Plugging in:

ṁ = (0.0752 lb/ft3)(1.00x105 ft3/min)

= 7520 lb/min of dry air

7) Calculate the mass flow rate of water vapor:

ρ = 0.001191 lb/ft3

V = 1.00x105 ft3/min

ṁ = (0.001191 lb/ft3)(1.00x105 ft3/min)

= 119.1 lb/min of water vapor

8) Calculate the total hours of operation to fill the cistern:

Given:

Cistern volume = 25.0x103 L = 6612 ft3

Converting the mass flow rates to volumetric flow rates:

Dry air:

7520 lb/min / 0.0752 lb/ft3 = 100,000 ft3/min

Water vapor:

119.1 lb/min / 0.001191 lb/ft3 = 100,000 ft3/min

Total flow rate = 100,000 + 100,000 = 200,000 ft3/min

Operating hours needed = Cistern volume / Flow rate

= 6612 ft3 / (200,000 ft3/min)

= 0.03306 hours

Therefore, the total hours of operation required to fill the cistern is 0.03306 hours.