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Page 1: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

BUILDING SYSTEMS

(PART ONE OF TWO)

February 19, 2013

Page 2: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Presented by AIA-Pittsburgh’s Young Architects’ Forum (YAF)

And Allen & Shariff Engineering,LLC

Exam resources are available at AIA-PGH YAF ARE Review

http://yafpghare.wordpress.com/

Page 3: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

ARE Blackout Period

Due to migration of record data,

an estimated eight week blackout

period begins July 1, 2013.

During this period:

• No exam appointments may be

scheduled

• There will be no exams administered

• New Authorizations to Test will not be

granted

• State boards will not be able to

establish eligibility or update records

The last day to take an exam before the

blackout will be June 30, 2013.

Alpine Testing Solutions, Inc. will take over content and candidate management

for the ARE beginning July 1, 2013.

Prometric will continue to be NCARB’s site management consultant.

Page 4: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Agenda

� 5:30 Introduction/ Division Overview

� 5:40 Graphic Vignette Review

� 6:00 Multiple Choice Content Review

� 7:00 Break (if desired)

� 8:00 Questions?

Page 5: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Division StatementThe evaluation, selection, and integration of mechanical,

electrical, and specialty systems in building design

and construction.

Exam Structure� 95 Multiple-Choice Questions (2 hours)

� Break (15 minutes)

� 1 Graphic Vignette (1 hour)

� Mechanical and Electrical Layout

Page 6: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Test Day…

� Introductory Tutorial 0:15

� Multiple-Choice Questions 2:00

� Scheduled (Mandatory) Break 0:15

� Introductory Tutorial 0:15

� Graphic Vignette 1:00

� Exit Questionnaire 0:15

� TOTAL APPOINTMENT TIME 4:00

Page 7: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Suggested Sequence

� Construction Documents and Services

� Programming Planning and Practice

� Site Planning and Design

� Structural Systems

� Building Systems

� Building Design and Construction Systems

� Schematic Design

Page 8: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Content Areas� CODES & REGULATIONS (5-8% of scored items)

� “…building codes, specialty codes, zoning, and other regulatory requirements…”

� ENVIRONMENTAL ISSUES (10-15% of scored items)

� “…sustainable design principles…”

� PLUMBING (10-15% of scored items)

� Principles

� Materials and Technology

� HVAC (18-23% of scored items)

� Principles

� Materials and Technology

� ELECTRICAL (10-15% of scored items)

� Principles

� Materials and Technology

� LIGHTING (18-23% of scored items)

� Principles

� Materials and Technology

� SPECIALTIES (14-19% of scored items)

� Principles

� Materials and Technology

Page 9: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

By the numbers…� 95 questions…

� 2 hour testing time…

…1 minute, 15 seconds per question

� By content areas…

Codes/ Regulations….…………..……………..…4 - 8 questions

Environmental Issues..……………..……..…..10 - 15 questions

Plumbing……………………………………….....10 - 15 questions

HVAC…………………….…………………….……17 – 22 questions

Electrical ……………………………………….....10 - 15 questions

Lighting..……………….…………………….……17 – 22 questions

Specialties………………………………………...13 – 18 questions

Page 10: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

GRAPHIC VIGNETTE

Page 11: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

MECHANICAL & ELECTRICAL PLAN

Develop a reflected ceiling plan that

integrates ceiling, lighting, mechanical, and

structural systems and incorporates life safety considerations.

PROGRAMA reflected ceiling plan for an architect’s office is to be prepared. The space is in

a multistory building and is enclosed by leasable office space on one side, a

corridor on another, and two exterior walls. The client wants flexibility for

furniture placement, efficient lighting levels, and a comfortable environment.

GRAPHIC VIGNETTE

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The Work Screen

Page 13: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Mechanical and Electrical Layout

The ProblemComplete the partially completed reflected ceiling plan on the work screen

by:

(1) providing a grid for acoustical tile,

(2) locating lighting fixtures to achieve specified light intensity, and

(3) developing a schematic HVAC plan complete with fire dampers and air diffusers, ductwork, and return-air grilles to meet specified air distribution requirements.

It is recommended that the ceiling layout be completed before ducts are added. Your solution must be contained within the perimeter walls of the overall space.

The completed plan should reflect effective coordination and integration of structural, mechanical, and electrical units within the ceiling grid and should provide for maximum flexibility for furniture layouts at the most economical cost.

Page 14: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Program RequirementsComply with the following requirements to develop the reflected ceiling plan:

Suspended Ceiling System

1. Provide a 2 ft X 4 ft grid with lay-in acoustical tiles in all spaces.

2. All ceiling heights are 9 ft above the finished floor.

3. Typical walls terminate 6 inches above the finished ceiling; bearing walls and fire-rated walls extend to

the bottom of the floor deck above.

Lighting System -- should be efficient and should minimize overlighting and underlighting.

1. For all spaces, use only recessed fluorescent fixtures to provide uniform light distribution with a light

level of approximately 50 footcandles measured at desk level (3 ft above the finished

floor).

2. In addition to the fluorescent fixtures required above, provide recessed accent light fixtures.

� Locate the accent light fixtures along the west wall of the Architect’s Office so that the direct light

level on the wall at a height of 5 ft above the floor is 80 footcandles.

� Space the accent light fixtures so that the light level between the fixtures at 5 ft above the

floor is 80 footcandles.

� The accent light fixtures should not be considered in determining the uniform light distribution

levels.

� The recessed fluorescent fixtures should not be considered in determining the accent light levels.

Page 15: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Program Requirements, cont.

HVAC System

The space is served by the supply and return risers within the shaft indicated on the floor plan. The HVAC

system should provide for uniform air distribution with an economical duct layout conforming to the

following restrictions:

1. Provide a minimum of one supply diffuser and one return-air grille in each space. An

acceptable air distribution pattern includes one supply diffuser and one return-air grille for every 144

ft of floor area (or portion thereof ) in each space.

2. Connect each supply diffuser to the rigid supply duct system with flexible duct. Do not

exceed 10 ft for flexible duct lengths.

3. Return-air grilles are open to the ceiling space, which serves as a return-air plenum. Connect the plenum to the return riser with rigid duct.

4. Protect duct openings in fire-rated walls with fire dampers.

5. Flexible ducts fit through joist webs.

6. Rigid ducts fit under beams, in spaces between joists, and in a zone that extends 2 ft on either side of

beams and bearing walls in plan view. Rigid ducts do not fit through joists or

between the bottom of joists and the ceiling.

Page 16: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Program Requirements, cont

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Mechanical & Electrical Layout

Sample PASSING Solution

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Mechanical & Electrical Layout

Sample FAILING Solution

Page 19: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

General Tips…

Clear your head.

Remember – it’s not AutoCAD…

… or design studio.

Practice makes perfect…

… but don’t over-practice!!

Take your time.

Follow all of the instructions!!

Don’t second-guess yourself.

Page 20: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

MULTIPLE CHOICE

Page 21: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)
Page 22: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

ELECTRICAL SYSTEMS REVIEW

Page 23: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Basic Physics

• The three most common units are:

• Potential (V, measured in volts)

• Current (I, measured in amps)

• Resistance (R, measured in ohms)

Page 24: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Water Analogy for

Voltagefrom the NEED Project www.NEED.org

Page 25: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Water Analogy for

Currentfrom the NEED Project www.NEED.org

Page 26: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Water Analogy for

Resistancefrom the NEED Project www.NEED.org

Page 27: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Ohm’s Law

Ohm’s Law is the formula which relates these

factors to each other. There are several

ways to show the formula:

• V = I x R

• I = V / R

• R = V / I

Page 28: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Ohm’s Law Example

• Given: 120 volt outlet and a hair dryer with

a resistance of 8 ohms. The current (I) flow

through the hair dryer when it is turned on

is ____.

• a. 15 amps

• b. 15 ohms

• c. 15 volts

Page 29: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Ohm’s Law Relationships

• V and I are directly proportional to each

other, i.e. the greater the voltage, the

greater the current.

• I and R are inversely proportional to each

other, i.e. the greater the resistance, the

smaller the current.

Page 30: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Series Resistances

• Resistors may be situated in series (one

path)…

Page 31: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Parallel Resistances

• Resistors may be situated in parallel

(multiple paths)…

Page 32: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Parallel + Series

Resistances

• Or in combinations of series and parallel

paths….

Page 33: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Parallel + Series

Resistances - Effective

Resistance Solution

Page 34: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Parallel + Series

Resistances - Effective

Resistance Solution

Given three parallel paths, two paths with a

resistance of 4 ohms and one with a

resistance of two ohms, find REFF.

1/REFF = 1/4 +1/4 +1/2

1/REFF =1/4 +1/4+2/4 = 4/4

REFF = 1 ohm

Page 35: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Transmitting Electricity

• Two main ways of transmitting

electricity:

1. Direct Current (DC)

2. Alternating Current (AC)

Page 36: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Direct Current (DC)

• Direct current flows in one direction with

constant voltage.

• Batteries are a common example of a DC

source.

Page 37: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

DC Power Equation

• P = V x I

• Where:

1. P = Power in watts (W)

2. V = Voltage

3. I = Current in amps

Page 38: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

DC Power Example

• Given: 12 volt battery connected to a 4

ohm resistor. Solve for power (P).

• Start with Ohm’s Law

• I = V / R = 12V / 4 ohms = 3A

• P = 12V x 3A = 36 W

Page 39: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

DC Power Equation

Variations

• Since we know:

V = I x R, I = V / R, and P = V x I

• Then by substitution, we also know:

P = I x I x R = I2 x R

P = V x V / R = V2 / R

Page 40: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Alternating Current (AC)

• Alternating current (AC) is based on the

concept that electricity has nearly no

inertia, and therefore the direction of flow

can be reversed very rapidly by reversing

the voltage.

• Plotting voltage versus time results in a

sine wave.

Page 41: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Single Phase AC Sine Wave60 Hz or 60 cycles per second

Page 42: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Single Phase AC Power

Equation

• P = V x I x PF

• Where:

1. P = Power in watts (W)

2. V = Voltage

3. I = Current in amps

4. PF = Power Factor in decimal form

Page 43: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Lead and Lag

• The AC current sine wave is not always in

sync with the AC voltage sine wave due to

capacitances and inductances.

• The current sine wave leads when it is

ahead of the voltage sine wave.

• The current sine wave lags when it is

behind the voltage sine wave.

• This is difference is represented as a phase

angle.

Page 44: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Power Factor

• The power factor is the cosine of the phase

angle between the voltage wave and the

current wave.

• It ranges from 0 to 1.0 and is often expressed

as a percentage from 0 to 100%.

• When entered into an AC power equation, it is

changed to decimal form.

• For a purely resistive load, the power factor is

1.0

Page 45: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

3-Phase AC Power

• In 3-phase power, there are three voltage

sine waves each 120 degrees apart from

each other.

Page 46: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

3-Phase AC Power

Equation

• P = V x I x PF x 1.732

• Where:

1. P = Power in watts

2. V = Voltage

3. I = Current in amps

4. PF = Power Factor in decimal form

Page 47: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Why is line voltage 1.732

times phase voltage?

Answer:

(http://wiki.answers.com/Q/Why_is_line_voltage_1.732_times_phase_voltage)

Line voltage is stated as "phase to phase". Phase voltage is stated as "phase to

ground". In a three phase system, each phase is 120 degrees out of phase with

respect to the other two, one leading, and the other trailing. Draw the vector

diagram for this and you get three triangles inside a larger triangle, the outer

sides being phase to phase and the inner sides being phase to ground. The outer

triangle is equilateral, with angles of 60 degrees, and the inner triangles are

isosceles with angles at the outer triangle's vertices of 30 degrees. Look at one

of the inner triangles and bisect it with a vector from ground perpendicular to

the vector for phase to phase. You see a right triangle. Now you can do

trigonometry...

The base is one half the phase to phase voltage. Lets call that X. In trig,

cosine(theta) = X (one half phase to phase) over hypotenuse (phase to ground).

Cosine 30 is 0.866, so phase to ground is one half phase to phase over 0.866, or

phase to phase over 1.732.

Page 48: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

So why times 1.732?

It’s not important for this test!

Just memorize the 3-phase power formula!

P = V x I x PF x 1.732

Page 49: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

3-Phase AC Power Example

• Given: A motor draws a current of 7 amps

at 240V-3ph with a power factor of 0.8.

Solve for power (P).

• P = 240V x 7A x 0.8 x 1.732 = 2,325 W

• Large power wattages are expressed in

kilowatts or even megawatts.

• 2,325 W = 2.325 KW

Page 50: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Electrical Equipment

• Motor

– Machine that converts electrical energy into mechanical energy.

– Running a current through a wire loop creates a magnetic field. This is the basic principle behind electric motors and solenoids.

Page 51: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Electrical Equipment

• Generator

– Machine that converts mechanical energy into electrical energy.

– Rotating a wire loop between two magnetic poles will generate a current. This is the basic principle behind generating electricity.

Page 52: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Power Generation

• Single phase alternator is the most basic form of power generation.

• Wire loop rotated between magnetic north and south poles creates a single phase AC current.

• Speed of rotation determines the period (peak to peak) of the AC current sine wave produced.

Page 53: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Single Phase Alternator

• If the speed is 60 rpm, it results in a sine

wave cycle time of 1/60th of a second, or 60

cycles per second, or 60 Hertz (Hz).

• 60 Hz, 120V AC power is typical in the USA.

• 50 Hz, 240V AC power is common in the

rest of the world.

Page 54: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

3-Phase Alternator

• Add two more wire loops spaced evenly to

the single phase alternator, and circuit

them separately.

• If the loops are evenly spaced, the resultant

current sine waves produced by each loop

are shifted by 120 degrees from one

another.

Page 55: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Single Phase and 3-Phase

Alternator Diagrams

Page 56: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Transformers

• Transformers can change the voltage to a

higher or lower value.

• They consist of an iron core on which two

separate coils of wire are wound, called the

primary and a secondary “windings”.

• The primary winding is the transformer

input, the secondary winding is the output.

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Transformers

• Current enters the primary winding and it induces a current on the secondary winding.

• The number of turns in the windings determines whether the voltage is stepped up or stepped down.

• More turns in the secondary winding yields a higher voltage than the primary.

• Fewer turns in the secondary winding yields a lower voltage than the primary.

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Transformers

• Transformer wasted energy is in the form of heat which must be dissipated.

• Transformer sizes are defined in volt-amps (VA), or kilovolt-amps (KVA).

• Transformer windings are insulated to protect them from overheating.

• Common types are dry-type and liquid filled.

Page 59: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Transformers

• Failures can be catastrophic.

• Transformers can be noisy.

• Generally speaking, dry-type transformers

located indoors do not require specially

constructed rooms.

• Liquid filled transformers located indoors

may have more complicated requirements

(liquid containment, vault construction).

Page 60: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Single Phase Transformer

Connections

• Two-wire secondary

– One wire grounded, becomes the neutral.

– One single phase voltage available. (e.g. 120V)

• Three wire secondary

– Center tap is grounded

– Two single phase voltages available. (e.g. 120V-1ph and 240V-1ph).

Page 61: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

3-Phase Transformer

Connections

• There are two basic types of connections:

– Delta

– Wye or Star

• When a neutral connection is desired, it

may be taken from the center point of the

Wye or from the midpoint of a Delta leg.

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3-Phase Transformer

Connections

Page 63: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Typical Transformer

Voltages• The delta-wye configuration is commonly used in

commercial work.

• Typical commercial system voltages are 480Y/277V-3ph-4w and 208Y/120V-3ph-4w.

• Typical residential system voltage is 120/240V-1ph-3w.

• Older buildings sometimes had two services, 120/240V-1ph-3w and a 240V-3ph-3w delta for motor loads.

Page 64: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

480Y/277V Transformer

Voltage Relationships• 480Y/277V-3ph-4w means:

– Line to line voltages

VL-L = VA-B = VA-C = VB-C = 480V

– Line to neutral voltages

VL-N = VA-N = VB-N = VC-N = 277V

– Or mathematically,

VL-L= VL-N x 1.732

480 = 277 x 1.732

Page 65: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

208Y/120V Transformer

Voltage Relationships• 208Y/120V-3ph-4w means:

– Line to line voltages

VL-L = VA-B = VA-C = VB-C = 208V

– Line to neutral voltages

VL-N = VA-N = VB-N = VC-N = 120V

– Or mathematically,

VL-L= VL-N x 1.732

208 = 120 x 1.732

Page 66: BUILDING SYSTEMS (PART ONE OF TWO) · 19/02/2013  · Mechanical and Electrical Layout The Problem Complete the partially completed reflected ceiling plan on the work screen by: (1)

Transformer Calculations

• Transformer capacity is measured in kilovolt-amps (KVA)

• KVA = V x I x 1.732 x 1000

• A 75 KVA 3-phase transformer has a primary voltage of 480V-delta. Find the primary amps:

• (75 KVA x 1000) / (480V x 1.732) = 90A

• Size the breaker: 90A x 1.25 = 112.5A

• Round to the next standard size: 125A

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Transformer Calculations

• A 75 KVA 3-phase transformer has a secondary voltage of 208Y/120V. Find the secondary amps:

• (75 KVA x 1000) / (208V x 1.732) = 208A

• Size the breaker: 90A x 1.25 = 260A

• Round to the next standard size: 300A

• Remember, KVA is KVA!

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So which voltage do I

choose?• Considerations

– What is available from the power company?

– What is the building square footage?

– Building use? (industrial, office, retail, warehouse,

multi-tenant, high-rise)

– Standard distribution equipment sizes and how they

affect architecture (NEC workspace clearances,

multiple doors to main electrical room, doors with

panic hardware opening outward from the room)

– What makes sense for the major equipment?

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Other things to

consider:• Higher voltage = lower current = smaller wire

• Higher voltage = better tolerance for voltage drop

• Higher voltage = more load per circuit = fewer circuits used

These concepts are based on the inverse relationship between voltage and current in the power equation. (Simply P/V = I, and P/I = V)

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Electrical Heat

• 100% efficient (electrically speaking) since all electricity is turned into heat.

• Inefficient in the big picture, since heat is used to generate electricity at 30% efficiency, which we are turning around to re-generate heat.

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Electric Lighting

• Switching can be done centrally via circuit breakers in a panel, or locally via wall switches.

• Switch types:

– One switch location –need (1) 2-way switch.

– Two switch locations –need (2) 3-way switches, one at

each location.

– More than two switch locations - need (2) 3-way

switches at each “end” location and a 4-way switch at

every “in between” location.

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Switching Diagrams

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Alternate Switch NamesSPST=2-way, SPDT=3-way, DPDT=4-way

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Motors

Four types in general use:

1. DC motor – small scale applications, elevators for

smooth continuous acceleration

2. Single phase AC motors – typically 3/4HP or less

such as exhaust fans, pumps, etc.

3. 3-phase induction motors – larger motors, constant

rpm such as air handlers, power factors from 0.7 to

0.9

4. Universal motors – DC or AC, variable speed, such as

hand drills, mixers, similar appliances

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Motor Protection

• Thermal overload protection via thermal

relays which shut the motor off to prevent

permanent damage. (example: the reset

button on a garbage disposal)

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Capacitors

• Stores energy for later use.

• Simplest form is two plates separated by a small insulating layer.

• Used for power factor correction, improving efficiency and overall performance. Reduce the utility company KVAR charge.

• Capacitive loads typically “lead”, and can be used to shift the “lagging” current bogged down by inductive motor loads.

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Receptacles

• Also called an outlet. (A plug is what you put into the outlet.)

• All 120V outlets should be 3-prong where the third prong is the ground.

• In a large room, provide more than one circuit for the outlets.

• Receptacle placement in commercial buildings:

– show windows

– HVAC equipment outlets

– No “code” for convenience outlets

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Receptacle Placement

• NEC 210.52 has very specific rules on the placement of receptacles in dwelling units and hotels.

– 12 feet on center on any wall space greater than 2 feet in width for the following rooms: kitchen, family room, dining room, living room, parlor, library, den, sunroom, bedrooom, or similar.

– At least one receptacle for hallways of 10 feet or more in length.

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Receptacle Placement

– One in each basement

– One for laundry

– One for garage

– One for each bathroom basin

– One at the front and back exterior of each dwelling unit

– One at each porch/balcony/deck

– Kitchen countertops and islands

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Receptacle Types

• Required types for the above listed locations also are covered in NEC 210

– Ground-fault circuit interrupter

– Weather-resistant

– Arc-fault

– Tamper resistant (pediatric)

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Panelboards

• Types: Fuse box, Circuit breaker panelboard, Frankenstein panel (really old)

• Provides a central distribution point for branch circuit wiring.

• Each breaker serves a single circuit.

• Overcurrent protection is designed to protect the wiring to the load.

• Power riser diagram includes all of the panelboards and their interconnections (conduit and wire) all of the way to the service entrance.

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Panelboard Clearances

(NEC 110.26 simplified)• The panel feeder connects to the panels

main circuit breaker (MCB) or main lugs (MLO).

• Panels require a minimum 3 feet of clearance in front.

• Panels require a 30” wide working space.

• Panels require 6 feet of headroom.• Good link: see http://ecmweb.com/code-

basics/determining-working-clearances

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Wiring

• Electrical wiring sizes are standardized

using American Wire Gage (AWG).

• Power wiring is 12 AWG minimum.

• Power wiring is copper.

• Wiring is sized at 125% of a motor load,

and for any load expected to operate for

three hours or more (continuous load).

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Aluminum Wiring

• Aluminum wiring has been blamed for

causing fires.

• Questions about oxidation, connection

deterioration, metal fatigue over a long

period of time.

• For feeders 200 amps and above,

aluminum alloy wiring as made by Stabiloy

has been an acceptable substitute for cost

savings.

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Conduit

• Wires must be physically protected in

addition to being insulated. Conduit

provides this protection.

• Conduit size is designated by its interior

diameter.

• NEC specifies the number of wires allowed

in a conduit.

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Conduit Types

• Rigid – steel, safest, same wall thickness as

Sch 40 PVC, connections are threaded,

galvanized for exterior use.

• Intermediate Metallic Conduit (IMC) –

steel, slightly thinner and less expensive

than rigid.

• Electrical Metallic Tubing (EMT) – thinnest

of the simple metal conduits, compression

fittings, galvanized, called “thin wall”.

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More Conduit Types

• Flexible Metallic Conduit – available with or

without a waterproof (liquidtight) jacket,

called “flex” or “Greenfield”, can be used

everywhere except underground.

• Interlocked Armored Cable (BX) – factory

assembled wires encased in an interlocking

spiral metal armor, (cannot add wires to it),

cannot be used underground or embedded

in concrete.

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NEC Table 310-16

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NEC Ground Sizing Table

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NEC Ground Sizing Table

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NEC Conduit Fill Table

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Sizing a Transformer

Primary Feeder…

• Given: Transformer is 45KVA, 480V delta

• (45KVA x 1000) / (480 x 1.732) = 54A

• 54A x 1.25 = 67.5A

• 70A is next standard size breaker

• Table 310.16: 3#4

• Table 250.66: 1#8(G)

• Table 9: 1-1/4”C.

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Sizing the Transformer

Secondary Feeder…

• Given: 45KVA, 208Y/120V

• (45KVA x 1000) / (208 x 1.732) = 125A

• 125A x 1.25 = 156A

• 175A is next standard size breaker

• Table 310.16: 4#2/0

• Table 250.66: 1#4(G)

• Table 9: 2”C.

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Calculations

• Load Estimation

– Use watts per square foot load factors based on general experience for various building functions.

– Use watts per square foot allowances as listed in the applicable energy code.

– Use RS Means watts per square foot factors for various building types.

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Calculation Tool

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Safety Considerations

• Short Circuits

– Short circuits occur when two conductors lose enough insulation to allow current to flow between them (arc).

– Little to no resistance results in a high current which can cause the conductors to get hot enough to start a fire.

– A fire within a wall can go undetected for some time, allowing it to spread.

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Short Circuit Protection

• Fuses

– Soft metal link in a glass plug or fiber cartridge

– One and done, can’t be reused.

– Largest glass plug fuse is 30 amps

• Circuit breakers

– Automatically disconnects under fault condition

– Resettable

– More expensive than fuses, but low maintenance cost

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Short Circuit Protection

• Ground fault circuit interrupter

– It is a form of protection against short circuits.

– It may actuate even without anything connected to it.

– It can completely disconnect a circuit.

– On a GFCI receptacle it can be reset.

– GFCI receptacles can be daisy chained.

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Ground fault circuit

interrupters– Detects continual current lost to ground and

breaks the circuit.

– Required locations are covered by NEC.

– Also required for electrical service equipment that is 480Y/277V-3ph-4w services and 1,000

amps.

– Sample of required locations: EWC, vending machines, commercial kitchens, within 6 feet of water.

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Grounding

• Ground wire provides a low resistance path

for current to the ground in the event of a

short circuit.

• Ground wires are usually bare or covered

with green insulation.

• All ground wires are ultimately connected

to a grounding electrode, such as building

steel, a metal water pipe, etc. per NEC

Article 250.

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Electrical Service

(Note: These are the “ARE 4.0 book definitions”, not

necessarily NEC definitions.)

• Service drop – consists of the wires from

the main line, a transformer, a meter, and a

disconnect switch.

• The distance from the meter to the

transformer should not exceed 150 feet.

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Electrical Service

• The minimum service size for a residence is

100 amps.

• In residences the panel and disconnect are

usually located outside of the building

where they are accessible to firefighters.

• In commercial construction they may be

located inside the building but must be

directly accessible from an exterior door.

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Riser Diagram Components

• Service entrance.

• Main service disconnect.

• Main switchboard.

• Feeders

• Distribution boards

• Feeders

• Branch panelboards

• Branch circuits

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Meters

• Residential meters measure total

consumption (kwh). Cost per kwh range

from 8 to 18 cents.

• Commercial meters often measure

consumption and peak demand.

• Think of load as the energy being used at

any given time.

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Peak Demand

• Think of “peak demand” as the time when

demand for energy is at its highest.

• Utility companies must be capable of

providing enough power to meet the peak

demand. This power sits there, idle, during

non-peak times.

• The charge associated with the peak

demand is called a demand surcharge.

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Emergency Power Sources

• Emergency power is required for

emergency egress lighting and exit signs.

• IBC Section 1006 covers Means of Egress

lighting levels and power sources.

• Hospitals have special emergency power

requirements.

• Stand-by power may be required for

elevators.

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Emergency Power Sources

• Lighting backup power is often provided

through emergency battery ballasts that

are continuously recharged while power is

on, and operate one or more lamps when

normal power is lost.

• Emergency lighting may be provided

through individual emergency lighting

units that contain a rechargeable battery.

These operate only when normal power is

lost.

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Generators

• Generator fuel supply should be enough for

two hours of runtime under load.

• Common fuel supplies are diesel, natural

gas, and propane.

• Separate automatic transfer switches are

provided for life safety, stand-by, and

optional equipment loads.

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Building Automation

• Lighting can be controlled by:

– Photocells

– Time clocks

– Smart breakers or smart relays

– Occupancy or vacancy sensors

– Daylighting controls

• Lighting controls are mandated by the

energy code.

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3 Second

Break

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LIGHTING REVIEW

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What is Light?

• Light is defined as that part of the

electromagnetic radiation spectrum that

can be perceived by the human eye.

• It ranges from:

– Blue light (450 to 475 nanometers)

– Through green and yellow light (at 525nm and 575nm)

– To red light (at 650nm)

• White light is the combination of all

wavelengths.

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The Eye

• Lens focuses

• Iris controls the amount of light entering

• Retina senses light

– Rods sense black and white – low light level

– Cones sense color – require more light

• Can adjust from levels below 1fc to over

10,000fc in moments.

• Extreme contrasts are know as glare.

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Perception and the Mind

• Eyes sense 2-D info, brain processes a 3-D

image interpreting color, shadows.

• Misinterpretations are common in optical

illusions.

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Lighting Terms

• Transmission

– Transmitted light passes through a material.

– Transparent means the image is transmitted.

– Refraction changes the transmitted image.

– Translucent means light is transmitted with no image.

– Opaque means no light passes through.

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More Lighting Terms

• Reflective materials bounce light.

– Specular means the image is maintained.

– Diffusing means the image is not maintained.

• Diffuse or Ambient Light

– No sharp shadows like on a cloudy day.

– Light from all directions.

– Referred to area lighting.

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More Lighting Terms

• Direct Light

– Very sharp shadows.

– Distinct reflections from shiny objects.

– Examples: sunlight, projector, desk lamp.

– Most useful when aimed at a task.

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Color Rendition Index

(CRI)

• CRI is the measure of how well light shows

true color.

• Highest rating is 100, meaning there are no

colors missing.

• CRI is calculated by comparing the color

rendering of a test source to a perfect

source.

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More CRI

• The perfect source is an incandescent lamp

for lower CCTs and daylight for higher

CCTs.

• Typical CRIs

– Incandescent: 100; T8 linear fluorescent: 75-85; Cool white linear fluorescent: 62; Compact fluorescent: 82; Standard metal halide: 65; Standard high pressure sodium: 22; Daylight: 100.

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So What Is CCT?

• CCT, or correlated color temperature, is a

specification of the color appearance of the

light emitted by a lamp, relating its color to

the color of a light from a reference source

when heated to a particular temperature….

• …which happens to be an incandescent

source.

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More CCT

• The color temperature is actually the

temperature of the filament of the

incandescent lamp in degrees Kelvin,

ranging from 3000 K to 3200 K.

• For other sources, the color temperature is

correlated based on appearance.

• Remember, CCTs below 3200 K are warm,

and CCTs above 4000 K are cool.

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Basic Lighting Physics

• Power or Intensity

– Represented by the letter “I” for Intensity

– Measured in candlepower (cp) (archaic)

– New unit of measurement is candela (cd)

– Definition: Light emitted in a particular direction.

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Basic Lighting Physics

• Flux

– Represented by the letter “F”

– Measured in lumens (l) or (lm)

– Definition: Time rate flow of light.

• Luminance

– Represented by the letter “L”

– Measured in foot-Lamberts (fL)

– Definition: Perception of brightness.

– Proportional to Illumination

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Basic Lighting Physics

• Illumination

– Represented by the letter “E”

– Measured in foot-candles (fc)

– Measured in lux (lx)

– Definition: Lumens per unit area, where:

• Foot-candles (fc) = lm / ft2

• Lux (lx) = lm / m2

• 1 fc = 10.76 lx

– E (fc) = F (lm) / Area

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Inverse Square Law

• If the light source may be approximated as

a point source, then the flux (F), and

resultant illumination (E), is inversely

proportional to the square of the distance

from the surface.

• Formula: E (fc) = I (cp) / d2

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Sample Illumination

Calculation

• Given a lamp with an intensity of 1,600 cp

and a perpendicular surface 10 feet away,

calculate E.

• E1 (fc) = I (cp) / d12

• E1 = 1,600 cp / (10 x 10) ft2

• E1 = 16 fc

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Sample Illumination

Calculation

• Given a lamp with an intensity of 1,600 cp

and a perpendicular surface 20 feet away,

calculate E.

• E2 (fc) = I (cp) / d22

• E2 = 1,600 cp / (20 x 20) ft2

• E2 = 4 fc

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Relationship between

Illumination and Distance

• In both examples, the intensity “I” was the same, 1,600 fc. So there is a relationship between the illuminations “E’s” and distances “d’s”.

• E1 x d12 = I = E2 x d2

2

• This equation can be rearranged as follows:

• E1 x (d1 / d2)2 = E2

• 16 fc x (10 / 20)2 = E2 = 4 fc

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Luminance Calculation #1

• Given the same lamp with an intensity of

1,600 cp placed 10 feet from a white wall

with surface reflectance of 75%, calculate

the luminance “L”.

• L (fL) = E (fc) x % reflectance

• We previously solved for E at 10 feet which

was 16 fc.

• L = 16 fc x 0.75 = 12 fL

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Luminance Calculation #2

• Given the same lamp with an intensity of

1,600 cp placed 10 feet from a piece of

frosted glass with a transmittance of 65%,

calculate the luminance “L”.

• L (fL) = E (fc) x % reflectance

• From previously, E at 10 feet was 16 fc.

• L = 16 fc x 0.65 = 10.4 fL

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Lighting Systems

• Incandescent

– Contains a tungsten alloy filament

– Filament is heated by a current flowing through

– Filament glows and produces light and heat

– Lamp is filled with an inert gas (nitrogen or argon)

– Typically warmer than sunlight/daylight, rich in yellows and reds, weak in greens and blues.

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Incandescent

• Least efficient type of artificial light

• Output of 15 to 18 lumens per watt

• Lifetime of 2,000 hours

• Dimming increases life span of the lamp

• Various shapes, sized in terms of wattage,

and in multiples of 1/8”. (Ex. A-19 lamp is

19 x 1/8” or 2.375” in diameter)

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Incandescent

• R lamps and PAR lamps include an internal

reflector so that all of the light comes out

of the front of the lamp.

• There are low voltage lamps that operate

at 12 or 24 volts, which allows a smaller

filament and results in better focus of the

beam. Example MR16 lamps.

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Tungsten Halogen

• Lamp is housed in an inner quartz envelope.

• Lamps contain halogen gas which prevents

evaporated metal from the filament from

depositing on the inner surface of the quartz,

allowing the filament to run at much higher

temperatures.

• Lamps produce more light with better color.

• Lamp life extended slightly by redeposition of

the metal back onto the filament.

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Fluorescent

• Fluorescent lighting is based on passing a

current through gases inside a glass tube.

• Energy is released in the form of gas ions

and free electrons.

• The glass tube is lined with phosphors

which are excited by the ions, and in turn

glow in characteristic colors.

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Fluorescent

• Good color rendition is achieved by

combining the phosphors correctly.

• Ballasts are required to get the current to

arc through the lamp.

• Ballasts are sometimes noisy, and are

assigned Sound Ratings from A to E, A

being the quietest.

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Fluorescent

• Linear fluorescent lamp technology has

evolved from the use of T12, to T8, to T5

lamps.

• Cool white produces the most lumens per

watt, but the color is unflattering.

• Warm white looks better.

• The deluxe series and SP series lamps

provide better CRI at a higher cost.

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Fluorescent

• Output of 60 to 80 lumens per watt

• Average lifetime of 10,000 hours, based on

a three hour burning period each time the

lamp is turned on.

• Switching decreases the life span of the

lamp.

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High Intensity Discharge

(HID)

• HID lamps consist of a lamp within a lamp

which is run at very high voltage.

• General types:

– Mercury vapor – very bright bluish light

– Mercury vapor deluxe – phosphors added for better color.

– Mercury vapor lifetime is 24,000 hours and its

output is 50 lumens per watt.

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HID Types

• Metal halide

– Typically iodine gas in the inner envelope.

– Better color than mercury vapor.

– Improved output to 80 lumens per watt.

– Lifetime dropped to 10,000 hours.

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HID Types

• High Pressure Sodium

– Most efficient with an output of 110 lumens per watt, but color rendering is not so good.

– Lifetime expectancy of 24,000 hours.

• Low Pressure Sodium

– Highest ratings in lifetime and lumens per watt.

– Worst CRI. Good for security lighting only.

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Lamp Output Rankings

• Normal incandescent is the least efficient.

• Tungsten halogen is a slight improvement.

• Mercury vapor

• Fluorescent and metal halide

• High pressure sodium is the most efficient.

• Low pressure sodium (security lighting

only)

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Lighting Calculations

• Point Grid Method

– Works best for a single or small quantity of fixtures.

– Ignores surrounding reflection

• Zonal Cavity Method

– Based on a uniform distribution of a large number of fixtures.

– Accounts for reflectivity and volume of space.

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Point Grid Method

• Based on the formula:

E = I cos G / d2

• Where:

– E = illumination (fc)

– I = intensity (cp) at the source at a given angle

– G = the angle between perpendicular to the receiving surface and a line from the source to the service

– D = the distance from the source to the surface

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Abney’s Law

• States that the light arriving at a surface is

the sum of the light arriving from all of the

sources, and can be expressed by repeating

the point grid formula for each source:

E = I1 cos G1/ d12 + I2 cos G2/ d2

2 + …

…+ In cos Gn/ dn2

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Candlepower Distribution

Curves

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Zonal Cavity Method

• Most commonly used for office,

commercial, factory space calculations.

• Based on a coefficient of utilization (CU)

for each fixture type.

• Accounts for fixture performance, room

shape, reflectances, maintenance factors,

and dirt.

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Zonal Cavity Method

EquationE (fc) = (N x n x LL x LLD x DDF x CU) / A

• Where:

– E = illumination in fc

– N = number of fixtures

– n = number of lamps

– LL = lumens per lamp

– LLD = lamp lumen depreciation factor

– DDF = dirt depreciation factor

– CU = coefficient of utilization

– A = area of the work plane

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Zonal Cavity Method

Equation

• Rearrange the equation to solve for the

number of fixtures (N), given a desired

foot-candle level:

N = (E x A) / (n x LL x LLD x DDF x CU)

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Recommended Illumination

• The optimum lighting is called the

equivalent spherical illumination (ESI).

• It is based on a theoretical sphere

surrounding the object being illuminated

with the light cast evenly from all

directions, eliminating any shadows and

any reflected bright spots.

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Daylighting

• Turning off lights reduces the electrical

load and the heat generated within the

space.

• Daylight itself is a diffuse light source with

perfect CRI.

• Direct sunlight should be shielded, diffused

by diffusing glass, or bounced off diffusing

surfaces.

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Daylighting

• Provide artificial lighting in daylit areas.

• Zone lighting for daylight sensor control

via switching fixtures off or dimming them.

• Dimming can be accomplished using

continuous dimming ballasts or stepped

diming ballasts.

• The daylight zone extends 15 feet from the

window, skylight, etc. into the room.

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Daylighting Strategies

• Light shelf is an overhang with glass above

which reflects light into the room and up

on the ceiling.

• Glass transoms le light pass from one room

to another.

• A sawtooth roof facing north to let in

diffuse light.

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Lighting and Sustainable

Design

• Goal is to balance the use of natural and

artificial lighting sources.

– Incorporate daylighting in the architecture.

– Use efficient lighting sources that reduce energy cost and increase visual comfort.

– Install lighting controls (code mandated).

– Computer modeling for an optimum design.

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Daylight Calculating

Methods

• Lumen Method

– Developed in the USA.

– Amount of daylight is calculated in three places: 5 feet from the window, the middle of the room, and 5 feet from the back of the room.

– Can be used to calculate the amount of daylight from one window wall, or two opposing window walls, but not from a corner window.

– Works for clear and cloudy skies.

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Daylight Calculating

Methods

• Daylight Factor Method

– Developed in Europe.

– Assumes overcast or diffuse sky conditions.

– Can be used to calculate the amount of daylight anywhere in a room, including the effect of corner windows.

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Daylight Factor

Calculation Results

• The calculation result is a number that

expresses the amount of light at a

particular location inside as a percentage of

the light available on the exterior surface.

• A daylight factor of 3 means that 3% of the

available light outside would arrive on the

workplane in that area.

• For 2,000 fc outside, 3% would yield 60 fc

inside.

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Emergency and Exit

Lighting

• IBC Section 1006 covers emergency egress

lighting requirements.

• IBC Section 1016 covers exit sign

requirements.

• Emergency lighting may be powered by

generators, battery packs, or emergency

ballasts.

• NiCad rechargeable batteries are common.

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Emergency and Exit

Lighting

• Exit signs are typically LED due to energy

code constraints.

• Exit signs should be visible even with the

lights out.

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General Tips…

� Study comprehensively…

� Save practice tests until the end.

� Don’t get stuck.

� Answer every question!

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Suggested Resources� “References Available During the Test” document

� Mechanical and Electrical Equipment for Buildings

� Heating, Cooling, Lighting (Lechner)

� Dustin Goffron’s “Helpful Links for the ARE 4.0”� http://www.dustingoffron.com/ARE/

� Alkikat’s Study Guide

� Jenny’s Notes (AREndurance)� www.arendurance.files.wordpress.com

� YouTube’s “WikiEngineering” channel

� MEEB’s Student Companion Site� http://bcs.wiley.com/he-

bcs/Books?action=contents&itemId=0471465917&bcsId=2879

� CED Engineering.com’s “Design Options for HVAC Distribution Systems”� http://www.cedengineering.com/upload/Design%20Options%20for%20HVA

C%20Distribution%20Systems.pdf

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QUESTIONS…?

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THANKS!