Conditioning of Moist Air l Properties of moist air l Equations – Conservation of mass – Conservation of energy l Classic moist air processes – Sensible.

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

23

1 1

2

12a TTm02.1q

12a hhmq

W

hhw

ww

hm

q

W

h

31

23

m

m

2a

1a

21

23

m

m

3a

1a

21

31

m

m

3a

2a

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:8.1/)32F(C oo

s/L472cfm1000

C8.37F100 oo

sL4719.0cfm1

C9.23F75 oo C6.15F60 oo C10F50 oo

s/L944cfm2000

3

1

2

kg/m9.0v 31

Adiabatic mixing - Solution

kg/m825.0v 32

s/kg05.19.0/944.0m 1a

s/kg57.0825.0/472.0m 2a

kg/g3.109.123.562.1

57.00.13

C30T o3

123a

2a13 WW

m

mWW

C9.19T o3wb %383

21

32

m

m

3a

1a

510

0

20

30

40

50

units4.31

units5.4862.1

05.132

method graphical partial

method graphicalfully

Typical air handling system

Outdoor air

Exhaust air

filter

heating coil

cooling coil

Return air

energyrecoverysystem

recirculation air

damper

-20°C

22°C

23°C

15°C - 35°C

Supply air

economizer

Sensible heat factor

a2a1a mmm

q

q

qq

q

ferheat trans total

ferheat trans sensibleSHF s

ls

s

1

2

)hh(mq 2zas

)hh(mq z1al

z

sensib

le

laten

t

1 2

coil q 2222a h,W,T,m1111a h,W,T,m

www h,T,m

Sensible heat factor - Example

1

2

kW6.6)3.395.52(5.0

)hh(mq 12a

kW0.2qqq sl

– Conditioned air is supplied to a space at T=15C and Twb=14C 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 24C . Find the sensible and latent cooling loads for the space.

condition lineskg

1 5.0m C15T o

1 C14T o

1wb

1

2

C24T o7.0SHF

kW6.47.06.6SHFqqs

parallel

If the cooling load is 6.6kW,what is m1 if state 1 is here?

(i.e., the supply air is heated and the space air is cooled)

1

3

Cooling coil bypass

2

4

41

34

m

mb

1a

6a

bypass factor

condition line

1 3

333a W,T,m111a W,T,m

www h,T,m55 ,T

5

T4 = apparatus dew point

1 3

333a W,T,m111a W,T,m

www h,T,m

6

4

55 ,T

Evaporative cooling - 1

1 2

1111a h,W,T,m

ww h,m

1

2

– Energy balance

– Given air at outdoorconditions (1) and theSHF of the space we canuse evaporative cooling tocool from state 1 to state 2

– The flow rate of air thatis required to cool thespace depends on cooling load

2aww1a hmhmhm negligible for water

2222a h,W,T,m

21 hh

3

condition line

3

6.0SHF

Is this a more favourable outdoorcondition for evaporative cooling?

sq

lq

Evaporative cooling - 2

OA

OAdesign target

good potential

no potential

– Low cost alternative to refrigerant systems Requires less energy for cooling Requires less capital investment

– Cooling potential (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 subjectto mold and bacteria growth

– Freezing may be a problemduring swing seasons(spring/fall) in colder climates

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 & ) to properly control

OA1 2 3

4

5

– Known:1. Outdoor design conditions2. Indoor design conditions3. Loads -> SHF -> condition line

4. Often T3,min -> state 3 -> ma3 5. Knowing the ventilation rate

ma1, mix 1 and 4 to get 26. Cool (+bypass) from 2 to 3

– Increase ventilation rate for “off-design”conditions of outdoor air

Economizer cycle

OA1 2 3

4

4

1

4

2

condition line

1

3

Td

1

2

Td

C8Twant od

Effect of fans

– Large HVAC systems have both a supply and exhaust air fan to keep the building at a desired P 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)

OA1 2 3’

4

5mf1000

PQPower

(L/s) rate flowair volumetricQ

(Pa) rise pressurefan P

80%) to60( efficiencyfan f

airstream theoutside ismotor if 100%

90%) to80( efficiencymotor m

3

4’

1

4

2

condition line3

Td

1 2 3’

4

5

3

4’

3’

4’

Supply fan increases Td or reduces reheat

Where are state points 3’ and 4’ ?

Exhaust fan increases theload on the cooling coil

Effect of fans - summer

1

condition line

1 2 4

6

7

5

Both fans reduce the heating coil load

Effect of fans - winter

qh

qs

ql

3

64

7

2 3

5

condition line

Designing indoor RH - cooling

1 2

4

qcqs

ql

3

1

– Given: qs - load calculation

ql - load calculation

T4 - comfort

T3 - comfort, energy

m1 - ventilation

T1 - weather data/design conditions

1 - design conditions

T1

1

T4T3

ls

s

qq

qSHF

Only state point 1 and the slope of the load line are specifiedThe condition line is not fixed vertically

condition line

Humid climate

1 2

4

qcqs

ql

3

1

– Assume condensation at the coil

– We can now design 4 and control it with the apparatus dew point

T1

1

T4T3

4

4

2

3

d

d

4

condition line

Dry climate

1 2

4

qcqs

ql

3

1

– Assume no condensation at the coil

– We must use outdoor air to control the indoor humidity (4)

– More difficult because we may not be able to control the outdoor air flow rate over a large range

T1

1

T4T3

4

23Horizontal line for no condensation

fgwl hmq fglw h/qm

Since there is no condensation at the coil, this moisture must be removed by outdoor air

141w WWmm

11

w4 W

m

mW

4

Dry climate - 2

1 2

4

qcqs

ql

3

1

– 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

T1

1

T4T3

4

23

fgwl hmq fglw h/qm

Since there is no condensation at the coil, this moisture must be removed by outdoor air

141w WWmm

11

w4 W

m

mW

Solving for W4

This locates state point 4 and determines 4

W4