HVAC

system

by

Asst. Prof. Channarong Asavatesanupap

Mechanical Engineering Department

Faculty of Engineering

Thammasat University

What is HVAC?

2

HVAC stands for Heating, Ventilation and Air Conditioning,

the three main functions of a building comfort system. A

complete system can control air temperature, humidity and

fresh air intake and maintain the quality of the air in your

building.

Comfort conditions

3

Thermal comfort is the condition

that expresses satisfaction with

the thermal environment and is

assessed by subjective evaluation

(ANSI/ASHRAE Standard 55).

Maintaining this standard of

thermal comfort for occupants of

buildings or other enclosures is

one of the important goals

of HVAC design engineers.Most preferred conditions:

T = 25 C and RH = 50%

4

HVAC

System

in

Building

5

HVAC system

COOLING COILfor cooling and

dehumidifying process

HEATING COILfor heating process

HUMIDIFIERfor humidifying process

AIR FILER and FRESH AIR INTAKE

for air ventilating and cleaning processes

Processes of HVAC

6

The processes by which effective control of parameters in an air conditioned space is maintained are as follows:

• Heating: To increase the temperature by adding

thermal energy to a space.

• Cooling: To decrease the temperature by removing

thermal energy from a space.

• Humidifying: The process of increasing the relative humidity of a space by addition of water vapor or steam.

Processes of HVAC (cont.)

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• Dehumidifying: The process of removing the water vapor or humidity of a space.

• Cleaning: The process of removing dust, pollens, smoke and contaminants from air inside the space.

• Ventilating: The process of adding external air to freshen up the air and maintaining gas ratio.

• Air movement: To control the movement of the supplied air so that the occupants of the space do not feel discomfort.

Moist air

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Cooling and Dehumidification Process

is the process in which the moist air is cooled

sensibly and at the same time the moisture is removed

from it. This process is obtained when the moist air at the

given temperature and %RH is cooled below the dew

point temperature.

“Moist air” is a mixture of dry air and water vapor. In atmospheric air water vapor content varies from 0 to 3% by mass.

Dry air

Water

vapor

=

Cooling and Dehumidification process in a cooling coil

Warm moist air in(higher temperature and higher humidity ratio)

Cool air out(lower temperature and lower humidity ratio)

+ condensate(condensed water vapor)

Coil surface temperature

lower than dew-point temperature

Heat removed

Example 1 Cooling and Dehumidification process

Given:

Moist air in conditions: Tdb = 25 C and %RH = 50

Cold air out conditions: Tdb = 15 C and %RH = 85

Find:

Amount of water and energy removed from moist air during

the process Air in

Air out

11

Cooling and Dehumidification process

15C

85%RH

TDP = 14C w = 0.010 kg/kgda

w = 0.009 kg/kgda

h = 50kJ/kgda

h = 35kJ/kgda

25C

50%RH

Example 1 (cont.)

From Psychrometric chart

Moist air properties at given conditions:

TDP = 14 C

win = 0.010 kg/kgda

hin = 50 kJ/kgda

Cold air properties at given conditions:

wout = 0.009 kg/kgda

hout = 35 kJ/kgda

Note: Coil surface temperature must be lower than 23C

Cooling and Dehumidification process in a cooling coil

Warm moist air in(higher temperature and higher humidity ratio)

Cool air out(lower temperature and lower humidity ratio)

+ condensate(condensed water vapor)Tdb = 25 C, %RH = 50

Tdb = 15 C, %RH = 85 Coil surface temperature

lower than dew-point temperature

(<14 C)

win = 0.010 kg/kgdahin = 50 kJ/kgda

hout = 35 kJ/kgda wout = 0.009 kg/kgda

Example 1 (Cont.)

Answer:

Water removed = win - wout

= 0.010 – 0.09 kg/kgda

= 0.001 kg/kgda

Energy removed = hin - hout

= 50 – 35 kJ/kgda

= 15 kJ/kgda

Example 2 Evaporative cooler (Cooling and Humidifying)

Given:

Moist air in conditions: Tdb = 30 C and %RH = 50

Find:

%RH of air and the leaving air temperature if the amount of water

content increased by 15% during the process.

16

17

Refrigeration system

is an equipment that moves heat from cool indoor

spaces to warmer outdoor locations. It moves heat by

causing a refrigerant to evaporate and condense.

cool indoor space

warmersurrounding

Liquid refrigerant

Vaporized refrigerant

Heat

in

Heat

out

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Basic components of a refrigeration system

Basic components of a refrigeration system

and typical operating conditions.

1) Evaporator

2) Compressor

3) Condenser

4) Expansion valve

Evaporator Compressor

Condenser

Compressor

Air conditioner

Cooling capacity and Performance

Cooling capacity is the measure of a cooling system's

ability to remove heat. The SI units are watts (W).

Other common units include Btu/h and tons.

20

1 W = 3.412 Btu/h

12,000 Btu/h = 1 ton of refrigeration

Coefficient of Performance, COP

is a ratio of cooling capacity to input power

required. Higher COPs equate to lower operating

costs. The COP may exceed 1.

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COP =Required input power [W]

Cooling capacity [W]

A unit-less measure of energy efficiency,

commonly used in thermodynamics,

Energy Efficiency Ratio, EER

is the ratio of cooling capacity (in Btu/h) to

power required (in W) at a given operating point.

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EER =Required input power [W]

Cooling capacity [Btu/h]

EER = 3.412 COP

(ARI*, 1984)* The Air-Conditioning and Refrigeration Institute

Energy Label

The energy efficiency

of the air conditioner is rated

in terms of a set of energy

efficiency classes from 3-5 on

the label. “5” being the most

energy efficient, “1” the least

efficient.

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Energy Label (cont.)

24

Cooling capacity < 8,000 W ( 27,296 Btu/h)

Rated Label EER

5 11.60

4 11.00 - 11.59

3 10.60 - 10.99

Energy Efficiency Rated Year 2011

Seasonal Energy Efficiency Ratio, SEER

is the cooling output during a typical

cooling-season divided by the total electric energy

input during the same period.

25

i

i

iconditionspowerInput

iconditionscapacityCooling

SEER@

@

Energy Label (cont.)

26

Cooling capacity < 8,000 W ( 27,296 Btu/h)

Rated Label SEER

5 15.00

4 14.20 - 14.99

3 13.40 -14.19

Seasonal Energy Efficiency Rated Year 2015

Energy consumption estimation* from EER/SEER

27

][]//[

]/[/htimeoperting

WhBtuEER

hBtucapacityCAkWh

][]//[

]/[/htimeoperating

WhBtuSEER

hBtucapacityCAkWh

* ASSUMPTION : Compressor Run time = 100%.

Example 3 Power drawn by air conditioner

Initially, a well-sealed house is at 32 C. The air conditioner is turned on and cools the entire house to 20 C in 15 min. If the COP of the air conditioning system is 2.5, determine the power drawn by the air conditioner. Assume the entire mas within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg-C.

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Example 4 Energy consumption estimation

A 2-ton air conditioner has an EER of 11.20 Btu/h/W. If the unit operates for a total of 1,200 hours during the summer, how much will electrical energy be consumed?

29

Example 5 Energy consumption by SEER

Re-calculate the amount of energy consumption in the previous problem, if the energy efficiency of the unit is, now, given by SEER of 14.0 Btu/h/W.

30

Cooling loads calculation

What is the cooling load? :

Cooling load is the amount of heat energy that would need to be

removed from a space (cooling) to maintain the temperature in an

acceptable range.

Why is the cooling load important?

To ensure that the right size of Air-conditioning system is installed

in order to provide consistent indoor comfort for occupants. It's

also important for saving energy, which can be wasted by either a

too-large or too-small system.

Cooling load components

Cooling load components (cont.)

Cooling load components

1) Conduction through roof, walls, glazing

2) Solar radiation through glazing

3) Air exchange (Ventilation and Infiltration)

4) People

5) Lights and Equipment

Building cooling load distribution

Cooling loads estimation

TypesRequired cooling load

(Btu/h/m2)

bedroom 500 – 600

office 750 – 900

cafeteria 1,000

patient rooms 600 – 750

examination room 600 – 750

operating room 1,200 – 1,500

meeting room 800 – 1,000

Rule of thumb

Source: เทคนิคการอนรัุกษ์พลงังานในอาคาร กระทรวงพลงังาน

Example 6 Cooling capacity estimation

Calculate the cooling capacity of an air conditioner for a 5m x 6m bedroom.

36

Compressor run time

37Op

erat

ing

time

On ท ำงำน

Off OffOn

Set-point temperature

Room

tem

pera

ture

Time

Time

On

Temperaturevariation

Compressor usually operates

in cycles of on and off

operation. The compressor

is “On” when the room

temperature is above a

desired temperature (Set-

point). Once the desired

temperature is reached the

compressor turns off until the

room temperature increases

again.

Cycle time of Air conditioner

Compressor run time

38

runtimehtimeopertingWhBtuEER

hBtucapacityCAkWh %][

]//[

]/[/

runtimehtimeoperatingWhBtuSEER

hBtucapacityCAkWh %][

]//[

]/[/

Longer cycle time: - more energy efficient

- better humidity removal

Typically, compressor run time is 80%.

Example 7

Calculate the energy consumption of the air conditioner used in the previous example. Assume

EER = 11.2 Btu/h/W, total operating time = 12h, and run time = 80%.

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HW#3

Problem 1)

Given:

Moist air in conditions: Tdb = 34 C and RH = 40%

Cool air out conditions: Tdb = 18 C and RH = 80%

Find:

Rate of heat removal from moist air if the volume flowrate of moist air

is 2.5 m3/s.

Problem 2) If an SEER 12 air conditioning unit is used in problem 1),

determine

a) the required power consumption

b) the annual electricity cost. Assume 3,500 operating hours, 85%

compressor runtime and 3.50 baht /kWh.

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Types of Air conditioning systems

1. Individual room systems

2. Unitary packaged systems

3. Central hydronic systems

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Individual room systems

Supply air

Return airWindow Unit

Supply air

Return air

Indoor unit

Out doorunit

Refrigerant pipe

Window-type Wall split-type

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Unitary packaged systems

A packaged system is used for cooling more than two rooms or a larger space. The cooled air is thrown by the high capacity blower, and it flows through the ducts laid through various rooms. Return

air

diffuser diffuser

T

Self-containedunitary packaged unit

Outdoor air

Supply ductReturn duct

Conditioned space

Packaged system for larger space

Roof-top packaged system for multiple spaces

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Central hydronic systems

A central system is used for cooling big buildings, offices, factories etc. If the whole building is to be air conditioned, HVAC engineers find that putting individual units in each of the rooms is very expensive making this a better option.

Chiller*Building

[Conditioned Space]

Surrounding[Outdoor air]

Heat rejected tosurrounding air

Heat removed fromConditioned space

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Chillers

is a machine that removes heat from a liquid (water) via a refrigeration system

Condenser

Evaporator

CompressorExpansion

valve

CHWR CHWS

CHWS = Chilled water supply

CHWR = Chilled water return

Chiller

Heat removed fromConditioned space

Heat rejected tosurrounding air

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- Air cooled chiller

??M

Supply Air Duct

Air Handling UnitAHU

ReturnAir Duct

Chilled Water Pump

CHS (45F,~7C)

CHR (55F,~13C)

Space (75F, 24C)

Fresh Air, FA

SA (60F, 3 m/s)

SA (60F, ~15C)

RA(75F, ~24C, 2 m/s)

Diffuser

Grille

(95F, ~35C)

Sun

Air-cooled chiller

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- Water cooled chiller

50

Cooling tower

is a heat rejection device which rejects unwanted heat from a

chiller to the atmosphere through the cooling of a water stream

to a lower temperature.

??M

Water-cooled chiller

Supply Air Duct

Air Handling UnitAHU

ReturnAir Duct

Cooling Tower

Condenser Water PumpChiller

Chilled Water Pump

CHS (45F,~7C)

CHR (55F,~13C)CDS (98F,~37C)

CDR (90F, ~32C)

Space (75F, 24C)

Fresh Air, FA

SA (60F, 3 m/s)

SA (60F, ~15C)

RA(75F, ~24C, 2 m/s)

Diffuser

Grille

(95F, ~35C)

Sun

52

53

Cooled Air Distribution

through AHU

??M

Air Handling Unit (AHU)

AHU

is a device used to regulate and circulate air as part of a heating, ventilating, and air-conditioning (HVAC) system. The Air Handling Units can have several components, depending on the complexity and requirements of each specific building and application.

55

Cooled Air Distribution

through FCU

??M

Fan Coil Unit (FCU)is a device consisting of a heating and or cooling 'coil' and fan. It is

part of an HVAC system found in buildings. A fan coil unit is a diverse device sometimes using ductwork, and is used to control the temperature in the space where it is installed, or serve multiple spaces.

Chiller efficiency

The term kW/ton is commonly used to represent

the energy efficiency of the chiller. It is defined as the ratio

of the chiller power in kW to the cooling capacity in tons at

the rated condition

57

kW/TR = Cooling capacity [TR]

Required chiller input [kW]

kW/TR = 3.517 / COP

Energy consumption estimation* from kW/TR

58

][][/ htimeopertingTRcapacityCoolingTRkWkWh

* ASSUMPTION : Compressor Run time = 100%.

Chiller consumption

][htimeopertingkWkWhi

Plant consumption

towerCoolingpumpchiller

i

kWkWkWkW where

Example 8

The 500-TR Chiller is used for a large commercial building. The operating time is 10h/d. If the COP of the chiller is 4.0, determine

(a) kW/TR

(b) Chiller power consumption

(c) Annual energy consumption

59

Example 9

Estimate the annual energy consumption of the chiller plant of the previous example. Assume the power consumptions of chilled water pump, cooling water pump and cooling tower fan are 35 kW, 15 kW, and 5 kW, respectively.

60