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Conditioning of Moist Air

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1 Conditioning of Moist Air
Properties of moist air Equations Conservation of mass Conservation of energy Classic moist air processes Sensible heating and cooling Cooling and dehumidifying Adiabatic humidifying Heating and humidifying Adiabatic mixing Space conditioning Sensible heat factor (SHF) Cooling coil bypass (b) Evaporative cooling Economizer cycle Effect of fans Designing indoor RH Change using bulk properties at inlet and outlet 2 2 3 1 1

2 Adiabatic mixing - Example
Two thousand cubic feet per minute (cfm) of air at 100 F db and 75 F wb (state 1) are mixed with 1000 cfm of air at 60 F db and 50 F wb (state 2). The process is adiabatic at a steady flow rate and at standard sea level pressure. Find the condition of the mixed streams. Convert to SI:

3 Adiabatic mixing - Solution
1 3 5 10 20 30 40 50 2

4 Typical air handling system
Exhaust air 23°C Return air recirculation air economizer energy recovery system damper 22°C filter heating coil cooling coil -20°C Supply air Outdoor air 15°C - 35°C

5 Sensible heat factor 1 2 coil q latent sensible 1 2 z

6 Sensible heat factor - Example
Conditioned air is supplied to a space at T=15°C and Twb=14°C at a flow rate of 0.5 kg/s. The sensible heat factor for the space is 0.70 and the space is to be maintained at 24°C . Find the sensible and latent cooling loads for the space. 2 If the cooling load is 6.6kW, what is m1 if state 1 is here? 1 parallel condition line 2 1 (i.e., the supply air is heated and the space air is cooled)

7 Cooling coil bypass 1 3 1 3 6 4 condition line bypass factor
2 5 3 bypass factor 4 T4 = apparatus dew point

8 Evaporative cooling - 1 Energy balance
3 1 2 Energy balance Given air at outdoor conditions (1) and the SHF of the space we can use evaporative cooling to cool from state 1 to state 2 The flow rate of air that is required to cool the space depends on cooling load negligible for water condition line 2 3 1 Is this a more favourable outdoor condition for evaporative cooling?

9 Evaporative cooling - 2 Low cost alternative to refrigerant systems
Requires less energy for cooling Requires less capital investment Cooling potential µ (Twb,design – Twb,outdoors) If the cooling potential is small, the air flow rates become very large and system is not economical because of fan and duct costs If cooling load is large, air flow rates become large Evaporative systems are subject to mold and bacteria growth Freezing may be a problem during swing seasons (spring/fall) in colder climates no potential design target good potential

10 Economizer cycle Using outdoor air to condition the indoor space
Can be used during “off-design” conditions to save energy (usually cooling) by increasing the amount of outside (ventilation) air (usually at night-time) May be limited in amount of outdoor air if RH must be controlled to a specific value (often RH must remain below some maximum value) must measure the enthalpy (T & f) to properly control 4 5 OA 1 2 3

11 Economizer cycle Known:
4 4 OA 1 2 3 Known: Outdoor design conditions Indoor design conditions Loads -> SHF -> condition line Often T3,min -> state 3 -> ma3 Knowing the ventilation rate ma1, mix 1 and 4 to get 2 Cool (+bypass) from 2 to 3 Increase ventilation rate for “off-design” conditions of outdoor air 1 1 2 2 Td 4 3 condition line Td 1

12 Effect of fans Large HVAC systems have both a supply and exhaust air fan to keep the building at a desired DP compared to outdoors. All of the power is ultimately degraded to sensible energy in the airstream. We will assume all of the energy rise occurs at the fan (most does) 4’ 4 5 3 OA 1 2 3’

13 Effect of fans - summer condition line 1 2 3’ 4 5 3 4’
Where are state points 3’ and 4’ ? 1 2 3’ Exhaust fan increases the load on the cooling coil Td 4’ 4 3 condition line Supply fan increases Td or reduces reheat

14 Both fans reduce the heating coil load
Effect of fans - winter 6 qs 7 ql qh 5 1 2 3 4 Both fans reduce the heating coil load 5 4 6 7 condition line 2 3 1

15 Designing indoor RH - cooling
1 2 4 qc qs ql 3 Given: qs - load calculation ql - load calculation T4 - comfort T3 - comfort, energy m1 - ventilation T1 - weather data/design conditions f1 - design conditions f1 1 Only state point 1 and the slope of the load line are specified The condition line is not fixed vertically condition line T3 T4 T1

16 Humid climate Assume condensation at the coil
1 2 4 qc qs ql 3 Assume condensation at the coil We can now design f4 and control it with the apparatus dew point f1 1 2 f4 d 4 condition line 3 d T3 T4 T1

17 Dry climate Assume no condensation at the coil
1 2 4 qc qs ql 3 Assume no condensation at the coil We must use outdoor air to control the indoor humidity (f4) More difficult because we may not be able to control the outdoor air flow rate over a large range Since there is no condensation at the coil, this moisture must be removed by outdoor air f4 4 Horizontal line for no condensation condition line 3 2 f1 1 T3 T4 T1

18 Dry climate - 2 1 2 4 qc qs ql 3 If we can’t increase the outdoor air flow rate to reach the indoor design humidity, the indoor humidity will decrease Usually ok to allow indoor RH to decrease in summer In this case the indoor humidity can be calculated as follows Since there is no condensation at the coil, this moisture must be removed by outdoor air f4 4 W4 Solving for W4 3 2 f1 1 This locates state point 4 and determines f4 T3 T4 T1

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