A Comprehensive Course on Energy Management for FM ... · 1. Concepts of energy calculation in...

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A Comprehensive Course on Energy

Management for FM Professionals

Energy Efficiency in Air-conditioning

Installations

Contents

1. Concepts of energy calculation in Air-conditioning Installations.

2. Applications and Calculation of energy saving on Air Conditioning System

3. COP of chiller, COP of unitary air-conditioner, air distribution system fan

power, duckwork leakage limit, thermal insulation, piping system

frictional loss, energy metering and system control etc.

4. Brief review on respective Building Energy Code from perspective of FM

5. Introduction of latest technologies in energy efficient HVAC system

6. Experience sharing and updates

7. Case studies

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Concepts of energy calculation in Air-

conditioning Installations.

Concepts of energy calculation

Method 1

Annual Cost Saving (kwh) = (Eold – Enew) x top x Energy Tariff Charge

Method 2

Annual Cost Saving (kwh) = (Eold x Percentage of Saving) x top x Energy Tariff

Charge

Simple Payback = ��

��

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Performance Terms and Definitions

Tons of refrigeration (TR): One ton of refrigeration is the amount of cooling obtained by one ton of ice melting in one day: 3024 kCal/h, 12,000 Btu/h or 3.516 thermal kW.

Net Refrigerating Capacity. A quantity defined as the mass flow rate of the evaporator water multiplied by the difference in enthalpy of water entering and leaving the cooler, expressed in kCal/h, tons of Refrigeration.

kW/ton rating: Commonly referred to as efficiency, but actually power input to compressor motor divided by tons of cooling produced, or kilowatts per ton (kW/ton). Lower kW/ton indicates higher efficiency.

Coefficient of Performance (COP): Chiller efficiency measured in Btu output (cooling) divided by Btu input (electric power).

Energy Efficiency Ratio (EER): Performance of smaller chillers and rooftop units is frequently measured in EER rather than kW/ton. EER is calculated by dividing a chiller's cooling capacity (in Btu/h) by its power input (in watts) at full-load conditions. The higher the EER, the more efficient the unit.

energy efficiency parameters conversion

COP = 0.293 EER EER = 3.413 COP

kW/Ton = 12 / EER EER = 12 / (kW/Ton)

kW/Ton = 3.516 / COP COP = 3.516 / (kW/Ton)

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Applications and Calculation of energy

saving on Air Conditioning System

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Key Energy Efficiency Requirements

Topic 9 Energy Audit

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Energy Utilization Intensity (EUI)

Annual energy and EUI (3 years)

Monthly EUI (recent 12 months)

MJ/m2/annum


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Energy Consumption Distribution


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Components of an HAVC system

A typical commercial heating, ventilation and air-conditioning (HVAC) system consists of 5 components:

Airside loop (yellow) Chilled-water loop (blue)

Refrigeration loop (green) Heat-rejection loop (red)

Controls loop (purple)


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Components of an HAVC system (Airside Loop)

Airside loop:

Cooling coil is used to cool and dehumidify the

supply air before it is delivered to the space

A typical cooling coil includes rows of tubes passing

through sheets of formed fins. A cold fluid, either

water or liquid refrigerant, enters one header at the

end of the coil and then flows through the tubes,

cooling both the tubes and the fins

Air-handling system draws outdoor (fresh) air intake,

transfers the cooling effect to the air and

distributes cooled air in the indoor environment


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The airside loop responds to changing cooling loads in

the conditioned space by varying wither the temperature

or the quantity of air delivered to the space

A constant-volume system provides a constant quantity

of supply air and varies the supply-air temperature in

response to the changing cooling load in the space

A thermostat compares the temperature in the

conditioned space to a set point. It then modulates

cooling capacity until the space temperature matches the

set point


Constant supply-air quantity, variable supply-air temperature

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A variable-air-volume (VAV) system varies the quantity of

constant-temperature supply air in response to the

changing load in the space

Each space has a separate VAV terminal unit that varies

the quantity of supply air delivered to that space

A thermostat compares the temperature in the

conditioned space to a set point. It then modulates the

quantity of supply air delivered to the space by changing

the position of the airflow modulation device (a rotating-

blade damper) in the VAV terminal unit.


Variable supply-air quantity, constant supply-air temperature

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Components of an HAVC system (Chilled Water Loop)

Chilled water loop:

Why chilled water?

Water is an effective medium for

transfer of thermal energy

Specific heat:

For water = 4.2 kJ/kg/K

For air = 1 kJ/kg/K

Density:

For water = 1000 kg/m3

For air = 1.2 kg/m3


Water is abundant

Water is chemically stable

Water is nontoxic

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Chilled water flows through the cooling

coil, absorbing heat from the air

Evaporator, is one component of the

refrigeration equipment (chiller), is used

to cool the chilled water back to desired

supply water temperature

Pump is used to move water around the

loop.

This pump needs to have enough power

to move the water through the piping,

the evaporator, the tubes of the coil etc.

installed in the chilled water loop


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Similar to the airside loop, the chilled-water loop responds to changing cooling

loads by varying either the temperature or the quantity of water delivered to

the cooling coil

The most common method is to vary the quantity of water flowing through

the cooling coil by using a control valve

As the cooling load decreases, the modulating control valve reduces the rate

of chilled water flow through the coil, decreasing its cooling capacity


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Components of an HAVC system (Refrigeration Loop)

In the chilled water loop, the evaporator

allows heat to transfer from the water to

cold liquid refrigerant

As heat is transferred from the water to the

refrigerant, the liquid refrigerant boils

The resulting refrigerant vapor is further

warmed (superheated) inside the

evaporator before being drawn to the

compressor

The compressor is used to pump the low-

pressure refrigerant vapor from the

evaporator and compress it to a higher

pressure


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This increase in pressure also raises the temperature of the refrigerant vapor

Common types of compressors includes: reciprocating, scroll, helical-rotary

(screw), and centrifugal

The hot, high pressure refrigerant vapor enters a condenser


The condenser is a heat exchanger that transfers

heat from the hot refrigerant vapor to air, water, or

some other fluid that is at a colder temperature

As heat is removed from the refrigerant, it condenses

and returns to the liquid phase

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The final step of the refrigeration cycle is for this hot liquid refrigerant to pass

through an expansion device

This device creates a large pressure drop that reduces the pressure, and

correspondingly the temperature, of the refrigerant


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1 → 2

Constant-pressure heat absorption

(cooling effect)

2 → 3

State 2 is a saturated vapor;

isentropic (reversible, adiabatic)

compression

3 → 4

Constant-pressure heat rejection;

stage 4 is a saturated liquid

4 → 1

Constant-enthalpy expansion

22



Cooling effect:

=

Isentropic compression:

=

Heat rejection:

=

Throttling,

=

( )1212

hhmq −=•

( )2323

hhmw −=•

( )4334

hhmq −=•

14hh =

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Coefficient of Performance (COP)


COP = Cooling effect/ Power input

The objective of a refrigeration cycle is to

remove heat (q12) from the refrigerated

space. To accomplish this objective, it

requires a work input (w23)

COP is an index of performance of a

thermodynamic cycle or a thermal system

23

12

hh

hh

−

−=

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Example:

Consider an ideal refrigeration cycle using R-

134a as the refrigerant. The gage pressure of

the refrigerant in the evaporator is 0.192 MPa.

In the condenser, the gage pressure of the

refrigerant is 1.217 MPa. Estimate the

coefficient of performance of the air-

conditioning system

Solution:

Absolute pressure

= Gage pressure + atmospheric pressure

p1 = p2 = 0.192 + 0.101 = 0.293 MPa

p3 = p4 = 1.217 + 0.101 = 1.318 MPa

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Components of an HAVC system (Heat Rejection Loop)

In the refrigeration loop, the condenser

transfer heat from the hot refrigerant to

air, water, or some other fluid.

In a water-cooled condenser, water

flows through the tubes while the hot

refrigerant vapor enters the shell space

surrounding the tubes,

Heat is transferred from the refrigerant

to the water, warming the water


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Components of an HAVC system (Heat Rejection Loop)

A heat exchanger is required to cool the

water that returns from the condenser

back to the desired temperature (29.4 oC) before it is pumped back to the

condenser

When a water-cooled condenser is used,

this heat exchanger is typically either a

cooling tower


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Components of an HAVC system (Control Loop)

Each of the previous four loops contains several components

Each component must be controlled in a particular way to ensure proper

operation

Many building operators want to monitor the system, receive alarms and

diagnostics at a central location, and integrate the HVAC system with other

systems in the building. These are some of the ructions provided by a

Building Automation System or Building Management System


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Energy Management Opportunity (EMO)

During the audit, the air-cooled chillers are found to be operated for over 20

years. Would chiller replacement help in reducing the energy consumption?

What is the COP stated in the manual?

Any record of electricity consumption?

Any record of chilled water supply and return temperature?

Any record of the chilled water / refrigerant pressure?

How frequent of those record?


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What is the COP stated in the manual?

The rated COP is usually the for a specific condition:

Specify loading

Specify chilled supply temperature

Specify outdoor temperature


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For example:

Annual electricity consumption for the aged chiller = 273,480 kWh

If the COP of the aged chiller = 2

Cooling load = kWh = 546,960 kWh

COP of the new chiller = 3

Annual electricity consumption after chiller replacement =

=182,320 kWh

About 33% energy saving per year, that is 91,160 kWh


InputPower

Load Cooling COP =

480,2732×

3960,546 ÷

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Where can I get the COP of the existing chiller?

Method 1:

By recording the refrigerant pressure of the evaporator ad the condenser

By knowing the type of refrigerant

Refer to the pressure – enthalpy diagram


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Method 2:

Measure the chilled water flow rate

Measure the supply and return chilled water temperature

Cooling load =

m = mass flow rate of the chilled water

C = specific heat capacity

= temperature difference between supply and return chilled water

Measure the power consumption or current


TmC∆

InputPower

Load Cooling COP =

T∆

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.


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Yes, but I do not have the electricity consumption of chiller


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Make reasonable assumption:

For example: 60% electricity consumption is due to chiller installation


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Monthly cooling load:


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Monthly energy saving:


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Cost saving:


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What is the payback for this chiller replacement?

Simple payback =

Total annual saving = Electricity cost x Annual electricity consumption

Assume the electricity cost = $1 per kWh

Assume the cost of chiller replacement = $400,000

Simple payback = = 4.4 years


Saving AnnualTotal

Investment Total

$91,160

$400,000

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Pump

In a pumping station, it may be

advantageous to install two or more

pumps together – either in series or

parallel

Pumps connected in series cause an

increase in pressure but no increase in

discharge

If two identical pumps are connected

in series, output pressure will

approximately double that afforded by

one pump while discharge remains

unchanged


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Pump

Pumps connected in parallel cause an increase in discharge but no increase in pressure

If two identical pumps are connected in parallel and are discharging to the atmosphere, discharge will approximately double that afforded by one pump while pressure remains unchanged


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Pump

System pressure loss vs. flow rate

Cv

is a cumulative effect of fluid

friction, valves, fittings and other

flow resisting devices (e.g. heat

exchangers)


pCV v ∆=•

rateflow system =•

V

constant system =vC

system in drop pressure=∆p

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Pump


Pump Curve and System Curve

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Pump


Parallel-pump operation

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Pump


Power of circulating water:

pVpm

Pw ∆=∆

=•

•

ρ

waterof rateflow mass =•

m

waterof rateflow volumetric=•

V

increase pressure =∆p

waterof density =ρ

input Power

power Water efficiency Pump =

50


Replacement of Pump

In case of a chiller plant using an oversized pump, the chilled water flow rate

will exceed the design condition. The chilled water flow rate can be lowered by

using a partially closed valve.

A possible EMO is to replace the existing oversized pump with a pump that has

an appropriate lower rating.


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Reset the temperature in air-conditioned zone

The Electrical & Mechanical Services Department (EMSD) of HKSAR

recommends setting the indoor air temperature to 25.5 oC for an air-

conditioned zone to save energy. If the indoor air temperature is raised, heat

gain of the building from the ambient environment will decrease.

The amount of heat gain Q can be estimated by:

i = type of construction material (concrete wall, glass window, roof etc.)

A = area of building envelop

K = thermal conductance

Text_surf = temperature of the exterior surface

Tin = indoor air temperature


( )∑ −=i

insurfextii TTKAQ_

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Given measurement for the existing HVAC operation:

Indoor air temperature, Tin = 21 oC

Total cooling load, Q1 = 350 kW

Electric power consumption, P1 = 100 kW

COP1 = Q1/ P1 = 350/100 = 3.5

Building envelop area = 6,600 m2

Average thermal conductance, K = 1 W/m2K

Building exterior surface temperature, Text_surf = 32 oC

If indoor air temperature is raised to 25.5 oC, chilled water temperature can be

raised and the COP becomes, COP2 = 3.8

Estimate the energy saving by raising the indoor air temperature to 25.5 oC


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Cooling load due to heat gain through the envelop for the existing operation,

Qe1 = AK(Text_surf – Tin) = 6,600 x 1 x (32 – 21) = 72.6 kW

Cooling load due to heat gain through the envelop after indoor air temperature

is raised to 25.5 oC,

Qe2 = AK(Text_surf – Tin) = 6,600 x 1 x (32 – 25.5) = 42.9 kW

Total cooling load after temperature adjustment,

Q2 = 350 – (72.6 – 42.9) = 320.3 kW

Electrical power consumption after temperature adjustment,

P2 = Q2/ COP2 = 320.3/ 3.8 = 84.3 kW

Energy saving = P1 – P2 = 100 – 84.3 = 15.7 kW (or 15.7%)


VAV Box Minimum Airflow Reset

• VAV box manufacturer specify minimum airflow for each VAV

box

• Actual acceptable minimum set point usually lower than that

of this minimum value

• Minimum shall reset to fit actual minimum flow requirement

• CO2 concentration level monitoring system shall be installed

to supplement the flow control system

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VAV Box Minimum Airflow ResetExisting annual electrical energy consumption for AHUs

= 7,200,00MJ

= 2,000,000kWh

Existing annual electrical energy consumption for chillers

= 10,800,000MJ

= 3,000,000kWh

Total cooling air flow reduced percentage

= percentage of VAV boxes x reduced airflow percentage

= 30% x 20%

= 6% of total cooling airflow

Fan energy saving percentage for AHUs

= 1- (1-6%)3

~ 15%

Annual energy saving

= AHU fan energy saving + cooling energy saving

= AHU annual energy consumption x energy saving percentage x time percentage + chiller energy x saving percentage x time percentage

= 2,000,000kWh x 15% x 20% + 3,000,000kWh x 6% x 20%

= 96,000Wh

Annual cost saving

= Annual energy saving x Electricity unit charge

= 96,000kWh x HK$0.97 / kWh

= HK$ 93,120

Heat Pump

• Heat sources : air, chilled water return

• Output temperature : max 60oC

• COP : ~3 with output temperature at 45oC

• Need hot water storage tanks

• Potential uses : central domestic hot water, HP dehumidification,

pool heating

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Heat Pump

Heating capacity : 150kW – 2000kW

COP : > 5 (temperature rise 40oC)

Temperature of

water source

: 7 to 40oC

Max. output temp. : ~ 60oC

Heat Source : Condensing water

Potential uses : • Hot Water / Dehumidification

• Customers with simultaneous high air-

conditioning & hot water demand, e.g. Hotels,

Hospitals, etc

Heat Pump

COP : > 3.5

Refrigerant : Carbon Dioxide (CO2)

Max. output temp. : ~ 85oC

Heat Source : Condensing water / Geothermal heat /

Waste heat generated from

manufacturing process

Application : Hot water supply / Dehumidification /

Industrial process drying

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COP of chiller, COP of unitary air-conditioner,

air distribution system fan power, duckwork

leakage limit, thermal insulation, piping system

frictional loss, energy metering and system

control etc

Air-conditioning System Load Design

Conditions

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Air Leakage Limit of Ductwork

Minimum Insulation Thickness for

Chilled Water Pipework

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Minimum Insulation Thickness for Refrigerant Pipework

(Suction)

Minimum Insulation Thickness for Ductwork and

AHU Casing

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Minimum Coefficient of Performance for Chiller


Air Cooled Chiller

Water Cooled Chiller

Brief review on respective Building

Energy Code from perspective of FM

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Building Energy Efficiency Ordinance

Building Energy Efficiency Ordinance (BEEO) (Cap. 610)

http://www.legislation.gov.hk/blis_pdf.nsf/6799165D2FEE3FA94825755E003

3E532/4F82C8740A5D73DA482577ED005330F2?OpenDocument&bt=0

3 December 2010 Gazetted

21 March 2011 Subsidiary regulations in operation

21 September 2012 Full Operation

Subsidiary regulations: Registered Energy Assessors (REA)

Codes of Practice: Energy Audit Code (EAC)

Building Energy Code (BEC)


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Building Energy Code (BEC)

Commercial building

Industrial building: common area

Residential building: common area

Composite building: common area, portion not for residential or industrial

use

For example:

Hotel and guest house Municipal services

Educational building Medical and health

Community building Government building

Railway station Airport passenger building


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Energy Audit Code (EAC)

Commercial building

Composite building (commercial portion)

BEEO does not govern:

Small building

Building with main electrical switch with approved load ≤ 100 A

Historical building

BS installations, with specific operational and technical natures such as fire

protection, lift safety, industrial undertaking etc.

Building that will cease to fall within 12 months


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For existing building:

OA (occupation approval) on or after 1 Jan 1988 (<24 years)

� 1 year from 21 Sept 2012

OA after 31 Dec 1977 but before 1 Jan 1988 (25 – 35 years)

� 2 years from 21 Sept 2012

OA after 31 Dec 1969 but before 1 Jan 1978 (35 – 43 years)


OA on or before 31 Dec 1969 (>43 years)


For newly constructed building:

� 10 years after issue of Cert of Compliance Registration


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Building Energy Audit

Responsibility of building owner:

Engage REA to carry out energy audit according to prescribed time frame

Obtain from REA the Energy Audit Form and energy audit report

Exhibit Energy Audit Form at building main entrance

Energy audit totally independent from BEC compliance – not a checking for

BEC compliance

Implementation of Energy Management Opportunity (EMO) not mandatory


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Building Energy Audit

Energy Audit Form

http://www.beeo.emsd.gov.hk/en/mi

bec_forms.html

Name of building

Address of building

Registered Energy Assessor (REA)

Commencement and completion

dates of energy audit

Energy Utilization Index

(MJ/m2/annum)

Valid for 10 years from the date of

completion of energy audit


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Building Services Installations

BS installations required by Ordinance to comply with BEC?

For newly constructed building:

YES

For existing building:

Only for major retrofitting works completed on or after 21 Sept 2012

Major retrofitting works:

Addition/ replacement of BS installation:

- a total floor area covered by the works under the same series of work

within 12 months ≥ 500 m2

Addition/ replacement of a main component:

A complete electrical at rating ≥ 400 A

A chiller or a unitary air-conditioner at rating ≥ 350 kW capacity


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Penalties

Fines at various levels from HK$2,000 to HK$1,000,000 to be imposed for

offences, e.g. (not exhaustive):

Reference: Section 22

Building owner failing to cause an energy audit to be carried out within the

time frame: Level 5 (Max. Penalty: HK$50,000)

REA failing to send a copy of Energy Audit Form and energy audit report to the

Director, within 30 days of issue: Level 3 (Max. Penalty: HK$10,000)


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Registered Professional Engineer (RPE) under Engineers Registration

Ordinance – electrical, mechanical, building services or environmental

discipline

� at least 2 years practical experience

� has the knowledge for the performance of the duties and functions under

the BEEO

Corporate member of HKIE in electrical , mechanical, building services or

environmental discipline or equivalent qualification

� at least 3 years practical experience

� has the knowledge for the performance of the duties and functions under

the BEEO


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Newly constructed buildings:

�Certify BS installations comply with BEC for developers to apply for

Certificate of Compliance Registration (CORC)

�Certify BS installations comply with BEC for owners to apply for COCR

renewal

�Issue Form of Compliance (FOC) for major retrofitting works

Existing building:

� Issue the FOC for major retrofitting works


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Central BS installations in commercial buildings/ composite building

�Conduct energy audit

�Issue Energy Audit Form

�Prepare and issue Energy Audit Report


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Energy Audit Code

Objective

Systematic review: Energy consuming equipment/ system

Information for decision making – accounting for

environmental and economic benefits

Key steps

1. Collection of information of the building

2. Review of energy consuming equipment/ systems

3. Identification of Energy Management Opportunity (EMO)

4. Cost benefit analysis of EMO

5. Recommendations

6. Compiling energy audit report/ Energy Audit Form


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Energy Audit Code

1. Collection of Building Information

�Energy consuming equipment inventories, brochures, manuals, drawings

�Building’s internal floor areas

�Energy consumption data past 36 months (or since operation)

�Building’s O&M programmes

�Operation records hours, temperature, flow, pressure vs. settings

�Past audit/ EMO


80

Energy Audit Code

2. Review of Energy consuming Equipment

Compile records based on findings

Calculate power and energy consumptions

� Utilization pattern: operation hours, occupant density etc

� Indicative parameters, absolute and changes – temperature, flow rate etc.

� Control mechanism

� AC systems and components: chillers, AHUs, AC water pumps etc.

� Luminaires

� Lifts and escalators

� Other BS equipment/ system: plumbing and drainage pumps

� Electrical circuit

� Power quality


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Energy Audit Code

2. Review of Energy consuming Equipment

� Types, capacity rating, rating conditions

� Metering in-situ/ external

� Measurement at representative instants and time

� Other characteristics affecting energy consumption (e.g. ext shading,

glazing shading coefficient etc.)


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Energy Audit Code

3. Identification of Energy Management Opportunity (EMO)

Reference to codes, guidelines and practices of established international/ local

standards

� Energy performances vs. operation conditions

� Identify deviations from efficient operation

� Comparison with original design

� Applicable operating conditions and system configurations

� Behaviors of responsible persons affecting energy consumption

Lighting W/m2 AHU/ Fan system fan power W/L/s

Pump W/L/s EUI – MJ/m2/annum

Chiller/ heat pump kwh/yr


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Energy Audit Code

4. Cost Benefit Analysis of EMO

Category I – Housing keeping measures

� Involving housekeeping measures which are improvements with practically

no cost investment and no disruption to building operation

Example: Turning off A/C or lights when not in use, revising A/C temperature

set-points, etc.

Category II – Low cost measures

� Involving changes in operation measures with relatively low cost investment

Example: Installing timers to turn off equipment, replacing T8 fluorescent

tubes with T5 fluorescent tubes, etc.


84

Energy Audit Code

4. Cost Benefit Analysis of EMO

Category III – Higher cost measures

� Involving relatively high capital cost investment to attain efficient use of

energy

Example: Adding variable speed drives, installing power factor correction

equipment, replacing chillers, etc.

5. Recommendations

Need of further studies


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Energy Audit Code6. Energy audit report

a) Executive Summary

b) Energy audit scope

c) Building characteristics

d) General description of equipment/ systems audited

e) Energy consumption and performance evaluation

f) Air-conditioning systems/ equipment information – chiller/ air-conditioner capacity & type, system type (CAV, VAV etc)

g) Lighting installations total lighting power

h) Analysis of historical energy consumption

i) Indication of energy supply to units from central BS installations

j) Findings from information review and site inspections

k) Evaluations of potential EMO

l) Referencing to past energy audit (if any)

m) EMO - recommendations


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Energy Audit Code

6. Energy audit report

Building characteristics

Internal floor area of building Internal floor area of common area

Number of floors Occupant density

Hours of operation Days of operation

Chiller/ unitary air-conditioner

Type, capacity and quantity

Total lighting power


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Energy Audit Code

6. Energy audit report

Energy consumption analysis

Annual energy and EUI (3 years) Monthly EUI (recent 12 months)

Energy supply to unit (e.g.. Chilled water supply)

Breakdown into air-conditioning, lighting, lift and escalator, others (recent 12

months)

Energy management opportunities

Categorized


Introduction of latest technologies in

energy efficient HVAC system

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Chiller selection

COP AnalysisOil Free centrifugal chiller(150RT)Normal screw chiller25% 50% 75% 100%

024681012

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Chiller selection

• COP, IPLV, Initial cost, life cycle cost

• Variable speed and dual compressor control deliver significant energy benefits, when applied to a wide range of building types and climates.

• Oil-free design eliminates oil fouling of flooded heat exchangers, which sustains energy efficiency over the life of the system.

• Oil-free, magnetic bearing design has lower annual maintenance and no periodic bearing inspections. Vibration monitoring is built- in to the compressor as standard.

• Every project is unique, and Life Cycle Costing is a valuable tool that can be used to look at the particular chiller options available

Chiller Plant Demand Flow Technology

• Manages chiller plant chilled & condenser water demand flow with specialized control algorithms.

• These algorithms require the conversion of constant speed for all pumps and fans adopted in chller plant to variable speed through the installation of Variable Frequency Drives (VFDs).

• The VFDs allow the Demand Flow algorithms to maintain optimal differential system pressure, reduce excessive pumping energy, reduce equipment runtime and increase system deliverable cooling capacity to end users with require thermal comfort

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Energy Management System

• ISO 50001 is a framework that helps companies manage their energy systems

and plan better to save energy and to reduce pollution as well as costs

• Benefit

1. Resolves energy efficiency problems

2. Improves energy usage of energy-consuming assets

3. Estimates environmental impact of greenhouse gases;

4. Improves energy management and communication;

5. Provides best practices for energy efficiency;

6. Prioritizes new energy-saving technology;

7. Improves energy efficiency of supply chains; and

8. Details greenhouse gas reduction plans

• An Energy Management System (EnMS) is a systematic process for continually

improving energy performance.

Energy Management System

• Assess current state (energy audit)—identify all energy management

opportunities (baseline)

• Use energy audit recommendations to prioritize, plan and allocate

resources

• Create plan to engage organization and drive energy awareness (energy

culture)

• Sub-metering system to identify energy wastage and trigger optimal plant

operation model

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Case Study


Average electricity unit charge = HK$0.97/kWh

The total annual energy consumption of air conditioners installed = 300,000kWh

Air conditioners could be replaced with energy efficient air conditioners = 80%

Saving percentage by adopting Grade 1 energy label air conditioners = 15%


= The total annual energy consumption of air conditioners with a cooling capacity 7.5kW or below x saving percentage

= 300,000kWh x 80% x 15%

= 36,000kWh

Annual cost saving

= Annual energy saving x Average electricity unit charge

= 36,000Wh x HK$0.97/kWh

= HK$34,920

Payback Period

Estimated capital cost

= (HK$ 126,500 ~ HK$ 235,000) / (23 sets)

Payback period

= Estimated capital cost / Annual cost saving

= (HK$ 126,500 ~ HK$ 235,000) / HK$34,920

= 3.6~6.7 years

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Chiller efficiency could be improved by 3% when applying BMS control for chiller start &stop sequence

Average electricity unit charge = HK$0.97/kWh

Total energy consumption of chiller =3,000,000kW


= energy consumption of chillers x saving percentage

= 3,000,000kW x 0.5%

= 15,000kWh

Annual cost saving

= Annual energy saving x Average electricity unit charge

= 15,000kWh x HK$0.97/kWh

= HK$14,550

Payback Period

Unit rate for calibration of flow meter

= (HK$ 10,000 ~ HK$ 20,000) / unit (including installation cost) x 6 flow meters

= HK$ 60,000 ~ HK$ 120,000

Payback Period

= (HK$ 60,000 ~ HK$ 120,000) / HK$ 14,550

= 4.1 ~ 8.2 years


Damaged thermal insulation area = 10m2

Temperature difference between cold surface to ambient = 22oC

Thermal conductivity improved by proper thermal insulation = 30W/m2oC

Total cooling energy saved = Damaged thermal insulation area (m2) x Temperature difference between cold surface to ambient (oC) x

Thermal conductivity improved

=10m2 x 10oC x 30W/m2oC

=3,000W


= (Total cooling energy saved / COP of air-cooled chiller) x Annual operating hours of chilled water system

= (3,000W/ 5.2) x 8,760 hours

= 5,053.8kWh

Annual cost saving

= Annual electrical energy to be saved / electricity unit charge

= 5,053.8kWh/ year x HK$ 0.97 / kWh

= HK$ 4,902.2

Payback period

Unit rate for pipe work rectification

= (HK$ 25,000 ~ HK$ 35,000) / unit (including installation cost)

Payback Period

= (HK$ 25,000 ~ HK$ 35,000) / HK$4,902.2

= 5.0 ~ 7.1years

A Comprehensive Course on Energy Management for FM ... · 1. Concepts of energy calculation in...

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