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Estimates of Dry Air Requirements of the P0D Lets assume a simple model of input dry air (from the membrane filter) with a relative humidity that mixes with water vapor from leaking from the P0D water target bags and creates a final mixture of output moist air with an increased relative humidity. The final mixture of output moist air should have a relative humidity that is less than the saturated dew point of the temperature of the cooling tubes in the P0D to prevent dripping. We use a psychrometric model of humidifier with adiabatic mixing of water inject into moist air from the 2001 ASHRAE handbook on Fundamentals (HVAC), pg 6.18. In this model input air is mixed with water and creates moist output air. The temperatures and humidity ratios (ratio of mass of water vapor to mass of dry air mass) of the airs will determine the enthalpy which is read off the psychrometric charts. For background web material see http://en.wikipedia.org/wiki/Psychrometrics. Other useful referenceshttp://en.wikipedia.org/wiki/Psychrometrics are Schaum’s outline on “Thermodynamics for Engineers”, chapter 12, example 12.12, pg 294. Since the processes are isobaric (constant pressure) we can set up energy equations using enthalpy and mass rate eqns. These eqns are usually sufficient to solve for the relevant parameters in the problem. The problem is similar to an evaporative water cooler, but the input and output temperatures are the same.

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Psychrometric Chart at sea level pressure. The horizontal axis is air temperature, the vertical axis is humidity ratio. the black lines are enthalpy, the red lines are relative humidity, the green lines are temperature, and the purple lines are humidity ratio (water mass/air mass).

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Estimates assuming steady state flow and ; 1)Lets assume the worst case measured leakage of an Auger water bag at 137cc/day. this corresponds to 9.5x10 -5 kg water/min. 2) Assume 30liters of dry air at 20C and 10% humidity. The mass rate is 3.9x10 -2 kg/min and the humidity is 0.0015 kg water/kg of air. 3) The P0D is a huge heat sink, so the temperature will be constant. Assume 20C. 4) the mixture of water vapor and air to be removed is 0.0024 kg water/kg of air. hence the humidity ratio will increase to 0.004. Note this is well below the saturation point for 15C (assume cooling pipe temp) which is 0.011. 5) From the Psychrometric tables, we find the enthalpy increases by ~7kJ/kg of air so multiplying this by air rate 3.9x10 -2 kg/min we find 0.27 JK/min of power must is provided by the P0D heat sink to keep temperature constant. This corresponds to 4.5W of cooling effect. Remarks; I have had two meeting with our thermal engineer, Pat Burns, who also teaches HVAC and he gave me the ASHRAE standard reference to look up how to model our dry air questions. I have not yet cross checked the above with Burns.

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h 1, W 1 h 2, W 2 hwhw We use a psychrometric model of of adiabatic mixing of water inject into moist air from the 2001 ASHRAE handbook on Fundamentals (HVAC), pg 6.18. A schematic diagram is given below. Dry air enters a volume at an air mass rate of m da, enthalpy h 1, and humidity ratio W 1. Water with enthalpy h w is absorbed at a mass rate of m w. The resulting moist air mass rate is still m da, but with enthalpy h 2 and humidity ratio W 2. The energy and mass rate equations are These two eqns can be readily solved to eliminate the mass rates. The interesting number we seek is the input air mass rate

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