HVAC Design: Level 1 - Essentials

HVAC Design:HVAC Design:Level I Essentials

Joel Primeau, PEng, HBDP&&

Donald Brandt, CEM

Copyright MaterialsCopyright Materials

Copyright 2014 by ASHRAE All rights reservedCopyright 2014 by ASHRAE. All rights reserved. No part of this presentation may be reproduced

without written permission from ASHRAE, nor may any part of this presentation be reproduced, stored in a retrieval system or transmitted in any form or by any means (electronic, photocopying, recording or other) without written permission from ASHRAE.

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AIA/CES Registered ProviderAIA/CES Registered Provider ASHRAE is a Registered Provider with The American Institute of

Architects Continuing Education Systems. Credit earned on completion of this program will be reported to CES Records for AIA members. Certificates of Completion for non-AIA members are pavailable on request.

This program is registered with the AIA/CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

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USGBC Education ProviderUSGBC Education Provider

ASHRAE

HVAC D i L l I E ti l [ID# 90009904]

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HVAC Design: Level I Essentials [ID# 90009904]

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Learning ObjectivesLearning Objectives

Calculate heating and cooling loadsg g Explain the basics of psychrometrics, hydronic system

design and air system design Discuss system selection Describe HVAC equipment and systemsDescribe HVAC equipment and systems Explain controls and building automation systems Discuss codes and standards Discuss codes and standards Describe building commissioning

E l i t h i l l d j t t kill Explain technical sales and project management skills5

Course OutlineCourse Outline

Fundamentals Design Process

Load Calculations

Psychrometrics

g

HVAC Systems II

BAS/ControlsPsychrometrics

System Selection

Air Systems

BAS/Controls

Codes & Standards

Commissioning & Air Systems

Hydronic Systems

HVAC Eq ipment

Commissioning & Standard 180

Technical Sales HVAC Equipment

HVAC Systems I Project Management

ConclusionConclusion

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FUNDAMENTALSFUNDAMENTALS

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Lesson ContentLesson Content

Components of HVACComponents of HVAC Heat transfer

/ l Fan/pump laws Refrigeration Cycle

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COMPONENTS OF HVACCOMPONENTS OF HVAC

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What is Air Conditioning?

5 Functions:

Heating

Cooling

h d f Dehumidification

HumidificationHumidification

Filtration/Ventilation

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Why HVAC?

To create a comfortable environment inTo create a comfortable environment in which humans can work, live, sleep, playplay,

(Thi f tl h f t(This course focuses mostly on human comfort solutions, although the lessons learned can certainly be applied to process applications )be applied to process applications.)

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UNDERSTANDING HEAT TRANSFERUNDERSTANDING HEAT TRANSFER

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Processes of Heat TransferProcesses of Heat Transfer

convectionconvectionwarm airwarm air

radiationradiation

hotwater

hotwater conductionconduction

cool aircool airwaterwater

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Transferring HeatTransferring Heat

1 lbwater

60F 61F1 Btu

1 kcal

1 kgwater

15C 16C1 kcal

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Transferring Heat (cont )Transferring Heat (cont.)

+ 152 Btu =1 lb

water

60F 212F

1 kg + 85 kcal =1 kgwater

15C 100C15

Sensible HeatSensible Heat

1 lb

60F 61F1 Btu

water

1 kcal

1 kgwater

15C 16C1 kcal

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+ 970.3 Btu =1 lbwater

212F 1 lbsteam

212F

+ 244 5 kcal =1 kg + 244.5 kcal =

1 k

1 kgwater

100C 1 kgsteam

100C17


1 lbsteam - 970.3 Btu =

212F1 lb

water212F

1 kg

1 kg

1 kgsteam - 244.5 kcal =

100Cg

water100C18

Latent HeatLatent Heat

1 lbsteam1 lb

t

212F970.3 Btu

water

212F

1 kg

244 5 kcal

1 kgsteam1 kgwater

100C244.5 kcal

100C19

Specific Heatp

140F 200F140F 200F

AA BB

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Heat Transfer FormulaHeat Transfer Formula

Q = Mass Flow x Specific Heat x T

This formula is used to create f

the formulas for water & air

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Heat Transfer with WaterHeat Transfer with Water

Q = (gal/min x 60 min/hr x 8.33 lb/gal) /lb ( )x 1 Btu/lb- oF x T (oF)

Q = 500 x GPM x Change in Temperature (T2 T1)

Q = Btu/hr

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Airflow Heat TransferAirflow Heat Transfer

SensibleSensible

Q = (ft3/min x 60 min/hr x 1 lb/13.33 ft3) x 0.24 Btu/lb- oF x T oF

Q = 1 10 x CFM x Change in Temperature (oF)Q = 1.10 x CFM x Change in Temperature ( F)

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Airflow Heat Transfer (cont )Airflow Heat Transfer (cont.)

LatentLatent

Q (f 3/ i 60 i /h 1 lb/13 33 f 3) (Q = (ft3/min x 60 min/hr x 1 lb/13.33 ft3) x (Humidity Ratio in lbs of H20/lbs of dry air)

1061 B /lb (l h f i i )x 1061 Btu/lb (latent heat of vaporization)

Q = 4840 x CFM x (Wo-Wc)

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Air Total HeatAir Total Heat

EnthalpyEnthalpy

Q (f 3/ i 60 i /h lb/ 3 33 f 3)Q = (ft3/min x 60 min/hr x 1 lb/13.33 ft3) x Enthalpy (Btu/lb)

Q = 4 5 x CFM x Change in EnthalpyQ 4.5 x CFM x Change in Enthalpy

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Whats a Ton of Refrigeration?What s a Ton of Refrigeration?

One ton of refrigerationOne ton of refrigeration produces the same cooling effect as the melting of 2000 lb gof ice over a 24-hour period.

When 1 lb of ice melts, it absorbs 144 Btu. Therefore, when 1 ton (2000 lb) of ice melts, it absorbs 288,000 Btu (2000 x 144). Consequently, 1 ton of refrigeration absorbs 288,000 Btu within a 24-hour period or 12,000 B /h (288 000/24)Btu/hr (288,000/24).

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HEAT TRANSFERHEAT TRANSFERTHROUGH A SURFACE

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Heat Conduction through SurfacesHeat Conduction through Surfaces

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Conduction through a Shaded WallConduction through a Shaded Wall

SimplestSimplest application

Q = U A TQ = U A T

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Heat Conductance vs. ResistanceHeat Conductance vs. Resistance

Conductance = U (Btu/hrft2F)Conductance = U (Btu/hrft F) Resistance = R (ft2F/Btu/hr)

/ U = 1/R R1 + R2 + R3 + = Rtotal Utotal = 1/Rtotal

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U-factor for Learning Center WallU factor for Learning Center Wall

U =11

U = RtotalRtotal

U = 0.059 Btu/hrft2F

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Heat Transfer through a WindowHeat Transfer through a WindowThe process is the same as a wall

Except:Wi d f ill ll id h U Window manufacturers will usually provide the U-factor for the glass (measured at the center of the glass)glass).

The designer has to determine the U-factor for the Window Assembly, to consider the losses throughWindow Assembly, to consider the losses through the window frame.

http://windows.lbl.gov/software/window/6/index.p // g / / / /html

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FAN & PUMP LAWSFAN & PUMP LAWS

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The LawsThe Laws

Pump Affinity Laws

Fan Laws

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Law #1Law #1

Speed is directly related to flowSpeed s d ect y e ated to oand is directly related to diameter

CFM2 = CFM1 X (RPM2 / RPM1)CFM2 = CFM1 X (DIA2 / DIA1)GPM GPM X (RPM / RPM )GPM2 = GPM1 X (RPM2 / RPM1)GPM = GPM X (DIA / DIA )GPM2 = GPM1 X (DIA2 / DIA1)

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Law #1: ExampleLaw #1: Example

TAB contractor Bill wants to increase the air flow in a vaneaxial fan from 1200 CFM to 1300 CFM. If the fan is currently turning at 1050 rpm, how fast does the fan need to go?the fan need to go?

RPM = RPM X (CFM / CFM )RPM2 = RPM1 X (CFM2 / CFM1)= 1050 rpm x (1300/1200)

= 1137 5 rpm= 1137.5 rpm

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What size of sheave would be required to makeWhat size of sheave would be required to make this happen (assuming the fan is not on a speed drive) if the diameter of the sheave wasdrive) if the diameter of the sheave was originally 4 inches?

Dia = Dia X (CFM / CFM )Dia2 = Dia1 X (CFM2 / CFM1)= 4 in. x (1300/1200)

= 4 1/3 in= 4-1/3 in.

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Law #2Law #2

Pressure changes as the square of the flow essu e c a ges as t e squa e o t e o(or speed)

P2 = P1 X (CFM2 / CFM1)2P = P X (RPM / RPM )2P2 = P1 X (RPM2 / RPM1)2

CFM2 = CFM1 X NOTE: GPM can substituted for CFMNOTE: GPM can substituted for CFM

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Law #3Law #3

Horsepower varies as the cube of the speed (or flow)(or flow)

BHP2 = BHP1 X (RPM2 / RPM1)32 1 ( 2 1)BHP2 = BHP1 X (CFM2 / CFM1)3

NOTE: GPM can be substituted for CFM

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Bill is now being asked what will the increase in speedBill is now being asked what will the increase in speed cost in energy? Lets assume that for every additional HP, the fan will cost $2.75 more per month to operate. The HP draw on the fan was originally 5 HP.

BHP2 = BHP1 X (CFM2 / CFM1)3

= 5 HP x (1300/1200)3

= 6.36 HPOr $17.48/month

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THE REFRIGERATION CYCLETHE REFRIGERATION CYCLE(A BRIEF INTRODUCTION)

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Refrigeration CycleRefrigeration Cycle

CCondenserondenser

EExpansionxpansionDeviceDevice CCompressorompressor

EEvaporatorvaporator42

Pressure-Enthalpy Diagrampy g

d

expansion

condenser

s

u

r

e

pdevice

p

r

e

s

s

compressor

evaporator

enthalpy43

Questions?Questions?

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HVAC Design: Level 1 - Essentials