6/2/22

Chapter 19A: ALL-AIR SYSTEMS: SINGLE

ZONE AND SINGLE DUCT

1) Introduction and common system types

2) Single zone systems- CAV and VAV operation
3) CAV systems without reheat- summer design
4) Air stream heating due to fans
5) CAV systems with reheat under summer design
6) CAV systems- winter design
7) Part-load operation of CAV systems
8) Part-load operation of VAV systems
9) Control issues and benefits of VAV systems

Introduction
– Large number of all-air system variations to fit different requirements for
comfort in buildings, process applications and also for special
applications requiring close control of temperature and humidity (clean
rooms, computer rooms, hospital operating rooms, …)
– An all-air system provides complete sensible and latent cooling,
preheating, and humidification capacity to the building supply air
– No additional cooling or dehumidification required at zone (but heating
may be required)

– Basically two categories:

(i) single-duct systems: for single zone and multiple zone buildings
all heating, cooling and dehumidification is done within a common
duct distribution system
(ii) dual- duct systems: for multiple buildings only (being phased out)
Two ducts- one carrying cold air stream and another hot air stream

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Common Types of All-Air Systems

1) Single duct
(a) Constant Air Volume (CAV): - single space
- multiple-zone reheat
(b) Variable Air Volume (VAV): - variable volume only
- reheat
- induction
- fan powered
2) Dual duct/stream
(a) Dual duct: - constant air volume
- variable air volume
(b) Multizone: - constant air volume
- variable air volume
- three-deck or “Texas” multi-zone

Single Zone Systems

- Operating principle of CAV and VAV under peak design cooling and heating conditions
- Operation of CAV under part load
- Operation of VAV under part load

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All Air Systems

Fig. 11.18. Pneumatic actuator on

Fig. 11.16. Single duct AHU
outdoor air dampers

Single Zone Systems

Exhaust Return
air air
!̇# , %#
Air from room
Damper is recycled
Damper
for energy
efficiency
!̇' Outdoor !̇& Supply
%' air %& air

Supply
fan

Fig. 19.2 The three sets of dampers and their locations

which are meant to inversely vary the return and
exhaust air flows in unison, and also the outdoor air
ventilation amount as needed.

3
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Fig. 19.6 Cutaway drawing of a packaged air-handler unit (AHU).

Single Zone Single Duct CAV Systems

- Summer Peak
sized so that no reheat is required
Return fan
8 7 (optional) 6
Exhaust Return
air air

Thermostat
9 Recycled
air Space Qspace,cool

Outdoor 5
Supply
air 3 4 air
0 1 2
Supply
PH CC HC fan

Fig. 19.1 Schematic diagram of the basic single zone HVAC system with
some of the major energy equipment. The important psychrometric state
points (0-9) and the return, supply, exhaust and ventilation flows are also
shown.
8

4
 6/2/22

Example illustrating analysis during peak or

design conditions of CAV system
Example 19.1: Building air conditioning

A space is to be maintained at 78° F (25.5° C) dry-bulb temperature and 50% RH. The total
cooling load (heat to be removed from the space to maintain comfort) is 120,000 Btu/h (35
kW) of which 70% is sensible heat. Ventilation air at 1000 ft3/min (472 L/s) is required on
the peak day when the outdoor conditions are 95° F (35° C) and 55% RH. The room supply
air temperature is 20o F (11.1o C) below that of the room set point temperature (a typical
value for commercial buildings at peak operating conditions). What are the space air supply
flow rate and the cooling coil rating (total cooling capacity and SHR)?

Figure: See Fig. 19.1 and Fig. 19.3.

Assumptions: The location is at sea level. The supply air temperature to the space is 58°
F. The duct heat transfer and the fan air temperature rise are ignored for simplicity.
Given: SHRspace = 0.7, Q! space,cool = 120,000 Btu/h, V! 0= 1000 ft3/min = 60,000 ft3/h
RH=
Tdb,6 = 78 o F, Tdb,o = 95 o F, fo = 55%, Tdb,5 = 58° F (outdoor and indoor air conditions)
Find: m! a, Q! cc,tot, SHRcc
Lookup values: specific volume vo = 14.4 ft 3 /lba , Wo = 0.0197 lbw /lba , W6 = 0.0103 lbw /lba

Fig. 19.3. The air-conditioning process diagram on a psychrometric chart. The state
points correspond to those in Fig. 19.1 while the numerical values apply to Example
19.1.
10

5
 6/2/22

Solution
1. Locate specified points 0 and 6 on the psychometric chart
1. Locate specified points 0 and 6 on the psychometric chart
2. Locate supply air point 5. The conditioned space inlet and outlet conditions lie on line
2. Locate supply
3-6 in air Condition
Fig. 19.3. point 5. The6 (78° conditioned
F, 50% humidity)space isinlet and known
already outlet from
conditions lie on line
the problem
3-6 in Fig. 19.3.
statement. TheCondition 6 (78°
slope of line 3-6 isF,determined
50% humidity) from theis inner
already
scaleknown
of the from the problem
protractor as
statement.
shown forTheSHR
slope of line
= 0.7. The 3-6 is determined
location of point 3 isfrom the inner scale
the intersection of the of
linethe
3-6protractor
and a as
shown for SHR
vertical = 0.7.
line drawn The
from ,3 = 58° F.of
Tdblocation Thispoint 3 is thetointersection
corresponds a RH value ofof80%.the line 3-6 and a
vertical
3. Findline
the drawn from airflow58°
space supply rate.F.TheThis corresponds
first to a RHisvalue of 80%.
law for this process
3. Find the space supply airflow Q! = m!
rate. The
space,cool a ( h - h
6 first)
5 law for this process is
(19.6)
The enthalpies can be read from Fig. 13.6: h5 = h3 = 23 Btu / lba , h6 = 30 Btu / lba
(19.6)
The air mass flow rate is thus
The enthalpies can 120,000
be read Btu/h
from Fig. 13.6:
m! a = = 17,140 lba /h (3770 ft 3 /min)
The air mass flow rate-is23)thus
(30 Btu/lba
4. Locate mixed air point 1. The end points 0 and 9(6) are known, and so the inverse
relation between mass flows and line segment lengths [Eq. (13.29)] is used to locate
point 1 (see sec. 13.6.2). The outdoor air mass flow is
4. Locate mixed air point 1. The end 3 points 0 and 9(6) are known, and so the inverse
relation between mass V!0 60,000 ft / segment
h
m! a,,0 =flows= and line3
lengths
= 4,170 lb a /h
[Eq. (13.29)] is used to locate
point 1 (see sec. 13.6.2). The v0 14.4 ft / lb
outdoor aira mass flow is
The ratio of air mass flows is (4,170/17,140) = 0.243. Therefore, point 1 is located 24.3
percent of the distance from point 6 along line segment 6-0. The properties at point 1
can be read from the psychrometric chart. They are Tdb = 82° F, Twb = 69° F, and moist
The air
ratio of airhmass
enthalpy flows
a =33.4 Btu/lbisa. (4,170/17,140) = 0.243. Therefore, point 1 is located 24.3
percent of the distance from point 6 along line segment 6-0. The properties at point 1
This allows point
Alternatively,
1use
to the
be fixed
can be read fromwethecan weighted
psychrometric average
chart. Theyrule
areto calculate
82°the
F, mixed air
69°humidity
F, and moist
11
ratio:
air enthalpy ha =33.4 Btu/lba.
W1 =[0.243 × 0.0197 lbw/lba + (1 – 0.243) × 0.0103 lbw/lba] = 0.01216 lbw/lba
Alternatively, we can use the weighted average rule to calculate the mixed air humidity
ratio:Similarly, assuming constant specific heats, the mixed air dry-bulb temperature is:
T
db,1 =[0.243 × 95° F + (1 – 0.243) ×78° F)] = 82° F (consistent with the chart
W1 =[0.243
reading).
× 0.0197 lbw/lba + (1 – 0.243) × 0.0103 lbw/lba] = 0.01216 lbw/lba

Similarly, assuming constant specific heats, the mixed air dry-bulb temperature is:
This allows point 1 to be fixed on the psychrometric chart.
Tdb,1 =[0.243 × 95° F + (1 – 0.243) ×78° F)] = 82° F (consistent with the chart
5. reading).
Determine cooling coil load. Line 1-3 can now be constructed by connecting points 1
and 3. The slope of the resulting line is transposed to the protractor, and the coil's
sensible heat ratio SHRcc can be read off as 0.55. The corresponding relative humidity is
80% and the humidity ratio is 0.0084 lbw/lba. Finally, the coil heat removal rate (or “coil
Thiscooling
allowsload”)
point can
1 tobebefound.
fixed on the psychrometric chart.

5. Determine cooling coil load. Line 1-3 can now be constructed by connecting points 1
and 3. The slope of the resulting line is transposed to the protractor, and the coil's
sensible heat ratio SHRcc can be read off as 0.55. The corresponding relative humidity is
80% and the humidity ratio is 0.0084 lbw/lba. Finally, the coil heat removal rate (or “coil
cooling load”) can be found.
Q! cc ,tot = m! a ( h1 - h3 ) = 17,140 lba /h ´ (33.4 - 23) Btu/lba
= 178,300 Btu/h or 14.9 Tons (52.2 kW)

12

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Air Stream Heating Due to Fans

Due to fan inefficiency, there is a temperature rise across a fan which can impose an additional
cooling requirement.
Dp(kPa )
DTdb ( oC ) = 0.807 ´ (19.9 SI)
h fan
Dp(in - WG)
Or DTdb ( o F ) = 0.363 ´
h fan (19.9 IP)

For a pressure rise in a fan of 250 Pa (1.0 in-WG), the temperature rise with a 100%
efficient fan is 0.2o C (0.36° F).

For a 70% efficient fan, the temperature rise is 0.29o C (0.52° F).

In commercial building HVAC systems, a pressure rise of several inches water gauge is common.
The resulting temperature rise can be 1.0 to 2.0o C (2 to 3° F) and must be considered when cooling coils
and central cooling plants are sized.

13

For spaces with high latent loads- reheat may be

required even under peak summer design conditions

6(7,8,9)
4(5)

Fig. 19.7 Process diagram for single duct CAV system to air-condition a single space with
high latent loads. The state points correspond to Fig. 19.1 while the numerical values apply
to Example 19.3. Note that the supply fan reheat is neglected.

14

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Example 19.3: Peak design for high latent load spaces

Example 19.1 illustrated the fact that the CAV system for a one-
zone space can be designed so as to only require cooling energy (without
any heating) under peak load conditions. This is not always possible. One
important instance where heating is needed even under peak load
conditions arises when the peak space latent load fraction is high such as
in hot and humid locations. This solved example will discuss why this
occurs, and illustrate how to analyze such cases.

We shall use the same design specifications as that of the CAV

system analyzed under peak load condition but with SHR = 0.5

Given: SHRspace = 0.50, Q! space,tot = 120,000 Btu/h, Q! space,sen = Q!

space,lat= 60,000 Btu/h outdoor air conditions: To =95 F, fo
o

! 3
=0.55, V 0= 1000 ft /min
Cooling coil leaving air conditions: Tdb,3 = 58°F, and fand
Tdb,3 = 58°F, f3 = 0.8.
3 = 0.8.
Cooling coil leaving air conditions:
o f
SpaceSpace condition Tdb,6 =T78
condition F 78
db,7 = andF and
o = 0.5. 7 = 0.5.

Find: Find:a, m ! a, ,Q! cc,tot

cc,tot , Q!and
cc,sen cc,sen and ! hc,tot
hc,totQ

Lookup values: specific volume v0 = 14.4 ft3/lb, humidity ratio W0 = 0.0197 lbw/lba,
h0 = 44.52 Btu/lba, W3 = 0.0082 lbw/lba, h3 = 22.85 Btu/lba, W7 = 0.0103 lbw/lba, h7 = 30 Btu/lba.

15

SOLUTION
1. Locate specified points 0, 3 and 7 on the psychrometric chart (see Fig. 19.7) which are
specified from the problem statement.
2. Locate supply air point 5.
As previously, the slope of line 5-7 can be determined from the inner scale of the protractor.
The location of point 6 is the intersection of the line drawn from point 7 with a slope equal
to 0.5 and the horizontal line drawn from point 3. This is found to be
Tdb,5 = 69.1o F and h5 = 25.5 Btu/lba and W5 = 0.0082 lbw /lba .
3. Determine space supply airflow rate:
The enthalpy balance equation (Eq. 19.7) is re-arranged to yield:

Q! space,cool
120,000 Btu/h
m! a = = = 26,667 lba / h
h7 - h5 (30 - 25.5) Btu/lba
4. Determine mixed air condition point 1.
The outdoor air mass flow is the same as before: m! a,0 = 4,170 lba / h

The ratio of air mass flows is (4,170/26,667) = 0.156. Therefore, point 1 is located 15.6 percent
of the distance from point 7 along line segment 7-0.

Using the weighted average rule:

- the mixed-air temperature Tdb,1 = [0.156 × 95° F + (1 – 0.156) x 78° F)] = 80.65° F, and
- the humidity ratio W1 = 0.0118 lbw/lba and enthalpy h1 = 32.31 Btu/lba.

16

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5. Determine cooling coil load (Eq. 19.7):

Q! cc,tot = m! a ( h1 - h3 ) = 26,087 lba /h ´ (32.31 - 22.85) Btu/lba = 246, 783 Btu/h (20.6 Tons)

6. Determine reheat coil load:

Q! hc = m! a ( h4 - h3 ) = 26,087 lba /h ´ (25.5 - 22.85) Btu/lba = 69,131 Btu/h

Note:
-substantial increase in supply air flow rate
(from 17,140 lba/h to 26,667 lba/h

-substantial additional cooling

(from 178 kBtu/h to 246.7 kBtu/h)

- The increase in cooling is equal to reheat = 69 kBtu/h

17

Single Zone Single Duct CAV System Winter Operation

Cooling coil is inactive
9 8
7
Local
exhaust
fan
10
Space

0 1 2 3 4 5
PH CC HC
Steam

Fig. 19.5a Schematic diagram of the complete single-duct single zone CAV system
with ducted return. The preheater and the steam humidifier would operate during
winter while the cooling coil and the reheat coil would be active when cooling
loads are to be met.

18

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Example 19.4: Winter peak design

Consider the CAV system analyzed under peak cooling load condition in
o
previous
Example example..
19.1. The same space is to be maintained at 72 F and 50% RH. The
total heating load of the space is 150,000 Btu/hr with a sensible heat ratio of 0.8.
The outdoor design condition is 40 0F dry-bulb and 40% RH. The same amount of
ventilation air (1000 ft3 /min) is needed. Design the necessary heating and
humidification equipment assuming saturated steam at 200 0F is available and that
the supply air flow rate is kept at the summer design peak value of 17,140 lba/h.
The supply air dry-bulb temperature should not exceed 105 0F.

Assumptions: The location is at sea level. The duct heat transfer and the fan
air temperature rise are ignored for simplicity.
!
Given: ma = 17,140 lba /h , SHRspace = 0.80, Q" space,tot = 150,000 Btu/h,
3
outdoor air conditions: Tdb,o = 40 oF, fo =0.40, V" 0= 1000 ft /min,
space condition T7 = 72 F and f7 = 0.5, Tdb,6 < 105 F.
o o
•
Find: m steam , Q" ph and Q" hc
Lookup values: specific volume v0 = 12.65 ft3/lb, humidity ratio W0 = 0.0021
lbw/lba, W7 = 0.0084 lbw/lba, h7 = 26.4 Btu/lba, specific heat of air ca =
0.24 Btu/lba oF, and enthalpy of steam hsteam= 1146 Btu/lbw.

19

Fig. 19.5

(a) Schematic
diagram

(b) Psychrometric
process
diagram

HCB 3-Chap 19A: All-Air Systems_Single

20
Zone

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 6/2/22

SOLUTION
We will illustrate the use of the graphical procedure as previously.
1. Locate specified points 0 and 7 on the psychrometric chart (see Fig. 19.5b) which are
specified from the problem statement.
2. Determine outdoor air mass flow rate.
The outdoor air mass flow is given by:
V!0 1,000 ft 3 / min´ 60 min/h
m! a,0 = = = 4,743 lba / h
v0 12.65 ft 3 / lba
Note that this mass flow rate is over 13% higher than the previous case which is
attributed to the difference in the specific volumes of outdoor air between summer and
winter conditions.
3. Locate mixed air condition point 1.
The inverse relation between mass flows and line segment lengths (Eq. 13.34) is used to
locate 1. The ratio of air mass flows is (4,743/17,140) = 0.277. The dry-bulb
temperature and humidity ratios can be calculated from the chart or just as simply using
the weighted average rule; e.g.,

- the mixed-air temperature: Tdb,1 = [0.277 ×40° F + (1 – 0.277) × 72° F] = 63.15° F

- the humidity ratio : W1 = [0.277 × 0.0021 lbw /lba + (1 – 0.277) × 0.0084 lbw /lba] =
0.00665 lbw /lba.

The corresponding moist-air enthalpy is read from the psychrometric chart, h1 = 22.4
Btu/lba.
21

1.
4. Determine supply air condition point 5.
As previously, the slope of line 5-7 is equal to the inner scale of the protractor
corresponding to SHRspace = 0.80. We determine enthalpy h5 of the specified supply air
mass flow rate as follows:
Q! 150,000 Btu/h
h5 = h7 + space,heat = 26.4 Btu/ lba + = 35.15 Btu/ lba
m! a 17,140 lba /h
The intersection of the enthalpy line corresponding to 35.15 Btu/lba and the line 6-7
yields a point whose corresponding dry-bulb value is Tdb,5 = 102o F (this is acceptable
since it is lower than the stipulated maximum value of 105o F). The corresponding
humidity ratio W5 = 0.0096 lbw/lba.

5.
2. Determine the amount of steam needed
This is easily calculated since the entering and leaving humidity ratio of the air streams
are known:
"
msteam = m
! a (W5 - W4 ) = 17,140 lba /h ´ (0.0096 - 0.00665) lb w /lb a = 50.56 lbw /h

3. Compute sensible heating needed.

We can use different combinations of preheat and reheat amounts to attain the desired
effect. One option is to assume that all heating is provided by the pre-heater. This is then
22
a simple heating and humidification problem

We draw a line from point 6 parallel to the outer scale of the protractor corresponding to
1,146 ( since enthalpy of steam hsteam= 1,146 Btu/lbw.) which intersects the horizontal
line from point 1 at point 2. This is determined to be: Tdb,2= 100.8o F and h2 = 31.6
Btu/lba.

Finally, the sensible heat provided by the heating coil is found:

11
 1. Determine supply air condition point 5.
As previously, the slope of line 5-7 is equal to the inner scale of the protractor
corresponding to SHRspace = 0.80. We determine enthalpy h5 of the specified supply air
mass flow rate as follows:

6/2/22
The intersection of the enthalpy line corresponding to 35.15 Btu/lba and the line 6-7
yields a point whose corresponding dry-bulb value is Tdb,5 = 102o F (this is acceptable
since it is lower than the stipulated maximum value of 105o F). The corresponding
humidity ratio W5 = 0.0096 lbw/lba.

2. Determine the amount of steam needed

This is easily calculated since the entering and leaving humidity ratio of the air streams
are known:

6.3. Compute sensible heating needed.

We can use different combinations of preheat and reheat amounts to attain the desired
effect. One option is to assume that all heating is provided by the pre-heater. This is then
a simple heating and humidification problem

We draw a line from point 6 parallel to the outer scale of the protractor corresponding to
1,146 ( since enthalpy of steam hsteam= 1,146 Btu/lbw.) which intersects the horizontal
line from point 1 at point 2. This is determined to be: Tdb,2= 100.8o F and h2 = 31.6
Btu/lba.

Finally, the sensible heat provided by the heating coil is found:

Q! hc = m! a ( h2 - h1 ) = 17,140 lba /h ´ (31.6 - 22.4) Btu/lba = 157,688 Btu/h

23

Alternative system:
Not requiring a
steam boiler

Fig. 19.8
(a) Schematic of the winter
heating and humidification
system with injection of
liquid water. The cooling
coil is not shown.

(b) Psychrometric process

diagram with state points.

HCB 3-Chap 19A: All-Air Systems_Single

24
Zone

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 6/2/22

Part-load operation of CAV

Example 19.5: Analysis of CAV systems under part-load operation
(Example 19.1)
Consider the CAV system which was analyzed under peak load condition. This
example will serve to illustrate how the CAV system needs to be operated under
part-load condition so as to maintain satisfactory indoor space conditions.
Given: SHRspace = 0.70, Q! space,tot = 70,000 Btu/h, Q! space,sen = 70,000 x
0.7= 49,000 Btu/h outdoor conditions: Tdb,o =80 oF, fo =0.6, V! 0=
1000 ft3/min
f
Cooling coil conditions: Tdb,3 = 58°F, and 3 = 0.8.
0
Space condition Tdb,6 = 78 F and supply air mass flow rate = 17,140
lba/h
Find: Q! cc,tot, Q! cc,sen and Q! hc,tot
Lookup values: specific volume v0 = 13.9 ft3/lb, humidity ratio W0=0.01325
lbw/lba, W3=0.0082 lbw/lba, specific heat of air ca= 0.24 Btu/lb.oF, and
latent heat of vaporization hv =1075 Btu/lbw.

25

Return fan
8 7 (optional) 6
Exhaust Return
air air

Thermostat
9 Recycled
air Space Qspace,cool

Outdoor 5
Supply
air 3 4 air
0 1 2
Supply
PH CC HC fan

Fig. 19.1 Schematic diagram of the basic single zone

HVAC system with some of the major energy equipment.

26

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 6/2/22

SOLUTION
1. Calculate outdoor air mass flow rate
V! 1, 000 ft 3 / min´ 60 min/h
m! a,0 = 0 = = 4,320 lba / h
v0 13.9 ft 3 / lba
2. Calculate supply air temperature point 5 using sensible heat balance
since the room thermostatic control is based on dry-bulb temperature.
Using sensible heat balance equation:
" "
Q space, sen = ma .ca .(Tdb,6 - Tdb,5 )

"
Q space, sen 49,000 Btu/h
Tdb,5 = Tdb,6 - "
= 78 o F - = 66.10 F
ma ´ c a 17,140 lba /h × 0.24 Btu/lba . o F
3. Verify indoor comfort.
Calculate humidity ratio of air leaving room using a latent heat balance
SOLUTION
" "

,lat = m a .hv .(W6 - W5 )

1. Calculate outdoor air mass
Q space flow rate
"
Qspace,lat 70,000 Btu/h × (1-0.7)
W6 = W5 + ×
= 0.0082 lb w /lb a + =0.00934 lb w /lb a
2. Calculate supply air temperature
ma h v 17,140heat
point 5 using sensible lb a /h × 1075
balance Btu/lb
since the wroom
thermostatic control is based on dry-bulb temperature.
Using sensible
This corresponds heatindoor
to an balance equation:
relative humidity of fo =45%, which along with
a dry-bulb temperature of 78 0F is satisfactory for indoor human comfort.

4. Calculate mixed air condition point 1.

3. Verify indoor comfort.
Calculate humidity ratio of airwhich
leaving room using 27
A common assumption simplifies theaanalysis
latent heat
is balance
to assume a
constant specific heat of moist air ca. The energy balance equation can
then be expressed as:

or Tdb,1 = 78.5 0F
This corresponds to humidity
Similarly the an indoor isrelative humidity
determined of ==45%,
as: W 0.0103which
lbw/lbalong with a dry-bulb
1 a
temperature of 78 0F is satisfactory for indoor human comfort.

4. 5. Determine
Calculate mixedcooling coil loads
air condition point(process
1. 1-3)
Using the simplified, we have
A common assumption which simplifies the analysis is to assume a constant specific heat
- Sensible load:
of moist air ca. The energy balance equation can then be expressed as:
! ! ! !
ma ,1 Tdb,1 = (m a ,1 - mo )Tdb,7 + mo Tdb,o
17,140 lba /h×Tdb,1 =(17,140-4,320) lba /h ×78 o F+4,320 lba /h × 80 o F
or
or Tdb,1 = 78.5 0F
Similarly the humidity is determined as: W1 = 0.0103 lbw/lba

1. Determine cooling coil loads (process 1-3)

5. 1-3)
Using the simplified Eq.(13.39), we have
- Sensible load:
- Sensible
! load:
Qcc,sen =17,140 lba /h × 0.24 Btu/(lba . o F) ´ (78.5-58) o F = 84,329 Btu/h =7.03 Tons

- Latent load:
!
Qcc,lat =17,140 lba /h × 1075 Btu/lb w × (0.0103-0.0082) lbw /lba =38,931 Btu/h =3.24 Tons
!
Q cc ,tot = 84,329 + 38,931 = 123,260 Btu/h =10.27 Tons
- Total load:
Note:
6.2. Determine reheat coil load
Using the sensible heat equation: Energy penalty
! due to reheat
Qrc,tot =17,140 lb/h×0.24 Btu/lba . o F ×(66.1-58) o F =73,320
33,320Btu/h
Btu/h
Reheat load:

28

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 6/2/22

Variable Air Volume Space

(VAV) Systems
VAV box
with reheat coil
PC CC
Fig. 19.10 (a) Terminal reheat configuration for a VAV system for a single zone ducted
return system for a small laboratory

Great energy savings compared to CV systems under part-load operation:

VAV systems modulate air flow as load reduces while keeping supply air
temperature constant at cooling coil leaving air temperature. This reduces both
cooling load on coil, fan power and avoids reheat

Supply air flow modulation is achieved by having terminal boxes

at each zone AND having a variable speed supply fan

29

Part-load operation of VAV System

Example 19.6: VAV analysis under part-load operation

Example
Consider previous 19.5 where a CAV system was assumed. We will solve this
example
problem for the case when a VAV system is used.

Given: Q! space,tot = 70,000 Btu/h, Q! space,sen = 49,000 Btu/h

"
3
outdoor conditions: To =80 oF, fo =0.6, V! 0= 1000 ft /min or m 0 =
4,320 lba/h
f
Cooling coil conditions: Tdb,3 = 58°F, and 3 = 0.8.
o
Space condition Tdb,6 = 78 F.
Find: Q! cc,tot, Q! cc,sen and Q! hc,tot
Lookup values: W0=0.01325 lbw/lba, W3=0.0082 lbw/lba, specific heat of air ca=
0.24 Btu/(lba. oF), and latent heat of vaporization hv=1075 Btu/lbw.

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SOLUTION
1. Calculate supply air mass flow rate with supply air temperature
1. assumed to be equal to that of the cooling coil set temperature. From a
sensible heat balance equation :

!
! Qspace,sen 49,000 Btu/h
ma = = =10,208.3 lba /h
(Tdb,6 -Tdb,5 )×ca (78-58) o F × 0.24 Btu/(lba . o F)
2. Verify indoor comfort, calculate humidity ratio of air leaving room using
a latent heat balance

!
Q 70,000 Btu/h×(1-0.7)
W6 = W5 + ! space ,lat =0.0082 lb w /lb a + =0.0101 lb w /lb a
SOLUTION m a ´ h 10,208.3 lb a /h×1075 Btu/lb w
v
1. Calculate supply air mass flow rate with supply air temperature
1. assumed to be equal to that of the cooling coil set temperature. From a
fo =49%, which along
sensible heat balance equation :
This condition corresponds to an indoor relative humidity of
with a dry-bulb temperature
!
of 75 0F which is within the indoor human comfort
! Qspace,sen 49,000 Btu/h
range. ma = = =10,208.3 lba /h
(Tdb,6 -Tdb,5 )×ca (78-58) o F × 0.24 Btu/(lba . o F)
2. Verify indoor comfort, calculate humidity ratio of air leaving room using
a latent heat
3. Calculate balance
mixed air condition
! ! ! !
ma,1´! Tdb,1 = (ma,1 - mo ) ´ Tdb,7 + mo ´ Tdb,o 31
Q space ,lat 70,000 Btu/h×(1-0.7)
OrW6 = W5 + !
=0.0082 lb w /lb a + =0.0101 lb w /lb a
m a ´ hv 10,208.3 lb a /h×1075 Btu/lb w
10,208.3 lba /h × Tdry,1 = (10,208.3-4,320) lba /h × 78 o F+ 4,320lba /h × 80 o F
or Tdb,1 = 78.85 0F fo =49%, which along
Similarlycorresponds
This condition the humidity to is
andetermined as: humidity
indoor relative W1 = 0.0114
of lbw/lba
with a dry-bulb temperature of 75 0F which is within the indoor human comfort
range.
4. Determine cooling coil loads
Using the simplified equation, we have
3.- Calculate
Sensible load
mixed air condition
! ! ! ! !
ma,1´ Tdblb,1a=/h(m
Qcc , sen = 10,208.3 ,1 - moBtu/(lb
×a0.24 ) ´ Tdb,7 a+. om o ´ Tdb ,o
F)×(78.85-58) o
F=51,082 Btu/h =4.26 Tons (down from 7.03 T)
Or !
Qcc,lat = 10,208.3 lba /h×1075 Btu/lb w ×(0.0114-0.0082)lb w /lba
-
Latent load:lba /h × Tdry,1 = (10,208.3-4,320) lba /h × 78 o F+ 4,320lba /h × 80 o F
10,208.3 =35,116 Btu/h =2.93 Tons (down from 3.24 T)
or Tdb,1 = 78.85 0F
- Similarly
Total
theload:
humidity is determined as: W1 = 0.0114 lbw/lba
!
Q cc,tot = 51,082 + 35,116 = 86,198 Btu/h =7.18 Tons (down from 10.27 T)
4. Determine cooling coil loads
Using the simplified equation, we have
- Sensible load
!
Qcc , sen = 10,208.3 lba /h × 0.24 Btu/(lba . o F)×(78.85-58) o F=51,082 Btu/h =4.26 Tons (down from 7.03 T)
!
For CAV = 7.03 T
Qcc,lat = 10,208.3 lba /h×1075 Btu/lb w ×(0.0114-0.0082)lb w /lba
-
Latent load:
=35,116 Btu/h =2.93 Tons (down from 3.24 T)
For CAV = 3.24 T
- Total load:
!
Q cc,tot = 51,082 + 35,116 = 86,198 Btu/h =7.18 Tons (down from 10.27 T)
For CAV =10.27 T
No reheat needed
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Comments:

In the above example, the reduced airflow has eliminated the need for
reheat but has somewhat compromised occupant comfort due to the
slightly elevated humidity in the space (RH increased from 45% to
49%).

The flow reduction of (10.208.3/17,140) = 0.595 is right at the lower

end of the norm allowed.

Had it been lower, the proper operating strategy would be to provide a

small amount of reheat at the entrance to the space

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(c) Heating (b) Mixed (a) Cooling Design peak

only mode only air flow rate

Supply
air
flowrate Minimum air
flowrate
(~ 60% design)
Hot air Reheat Cold air

100% 100%
Heating load Cooling load
Fig. 19.9 Modulation of room supply air flow for a single zone VAV system as space
loads vary over the year. Note the three modes of operation: (a), (b) and (c).

Design flow: range 0.8 - 1.2 cfm/ft2

Different types of VAV Operation:

-Cooling only
- VAV reheat (because minimum supply air flow reached)
-Heating only
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Single Zone CAV

+ Used for simple rooms with similar conditioning Used in:

requirements over year (interior rooms)
Pros:

• Small commercial
+ Simple to build and operate • Strip malls
+ Lower initial cost • Elementary
• Small health clinic
- Higher energy costs during operation • Auditoriums
Cons:

- Not energy efficient for rooms whose loads • Theaters

vary over year

35

Single-zone VAV

Can be used for single

zone (many rooms)

+ Used for simple rooms whose conditioning Used in:

requirements are similar but varies during day
Pros:

• Small commercial
+ Simple to build (inexpensive) and operate • Strip malls with
different types of
stores
- Requires reheat if rooms with different conditioning • Small health clinic
requirements since supply air cannot be reduced
Cons:

beyond a certain amount

with different
specialization
- Total amount of outside air may be difficult to
maintain due to variable volume
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Outcomes
• Understanding of the common all-air system types used for air
distribution in buildings
• Understanding the operation of the basic CAV and VAV systems for
single zone buildings
• Knowledge of air stream heating across fans
• Familiarity with different configurations of CAV and VAV systems
• Be able to analyze single zone CAV with reheat, CAV without reheat,
and VAV under summer and winter design conditions
• Be able to analyze the performance of CAV and VAV under part-load
operation
• Understanding of the various control issues and energy efficiency
benefits of using VAV system

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