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Estimates of Dry Air Requirements of the P0D

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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 references

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.

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).

- Estimates assuming steady state flow and ;
- 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.

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 mda, enthalpy h1, and

humidity ratio W1. Water with enthalpy hw is absorbed at a mass rate of mw. The

resulting moist air mass rate is still mda, but with enthalpy h2 and humidity ratio W2.

h1, W1

h2, W2

hw

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