HVAC system - Charnnarong Assavatesanupapcharnnarong.me.engr.tu.ac.th/charnnarong/My...
Transcript of HVAC system - Charnnarong Assavatesanupapcharnnarong.me.engr.tu.ac.th/charnnarong/My...
HVAC
system
by
Asst. Prof. Channarong Asavatesanupap
Mechanical Engineering Department
Faculty of Engineering
Thammasat University
What is HVAC?
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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
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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%
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HVAC
System
in
Building
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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
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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
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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.
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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.
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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.)
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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.
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i
i
iconditionspowerInput
iconditionscapacityCooling
SEER@
@
Energy Label (cont.)
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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
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][]//[
]/[/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?
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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.
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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.
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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
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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
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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
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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.
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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
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kW/TR = Cooling capacity [TR]
Required chiller input [kW]
kW/TR = 3.517 / COP
Energy consumption estimation* from kW/TR
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][][/ 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
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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.
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