HVAC Simplified Solution Manual

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Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh Not for Resale, 09/09/2009 18:54:17 MDT

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Solutions Manual

© 2006, American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). HVAC Simplified Solutions Manual. For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE's prior written permission.



About the Author Stephen P. Kavanaugh, PhD, Fellow ASHRAE, has been a professor of mechanical engineering at The University of Alabama since 1985, where he teaches HVAC and is faculty advisor for the ASHRAE Student Chapter as well as a Habitat for Humanity Student Affiliate. Kavanaugh is co-author of Ground-Source Heat Pumps—Design of Geothermal Systems for Commercial and Institutional Buildings, published by ASHRAE in 1997. He has presented over 100 engineering seminars for more than 2,500 designers on the topics of energy efficiency, ground-source heat pumps, and HVAC. He maintains the Web site www.geokiss.com, where there is more information about HVAC and ground-source heat pump design tools. He is past chair and current handbook subcommittee chair of ASHRAE Technical Committee 6.8, Geothermal Energy, as well as past chair of ASHRAE Technical Committee 9.4, Applied Heat Pumps and Heat Recovery. Kavanaugh is also a Fellow of the American Society of Mechanical Engineers and a board member and past president of Habitat for Humanity—Tuscaloosa.

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Any updates/errata to this publication will be posted on the ASHRAE Web site at www.ashrae.org/publicationupdates.



HVAC Simplified Solutions Manual

Stephen P. Kavanaugh

American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc.

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ISBN 978-1-933742-09-0 ©2006 American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. 1791 Tullie Circle, NE Atlanta, GA 30329 www.ashrae.org All rights reserved. Printed in the United States of America

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ASHRAE has compiled this publication with care, but ASHRAE has not investigated, and ASHRAE expressly disclaims any duty to investigate, any product, service, process, procedure, design, or the like that may be described herein. The appearance of any technical data or editorial material in this publication does not constitute endorsement, warranty, or guaranty by ASHRAE of any product, service, process, procedure, design, or the like. ASHRAE does not warrant that the information in the publication is free of errors, and ASHRAE does not necessarily agree with any statement or opinion in this publication. The entire risk of the use of any information in this publication is assumed by the user. No part of this book may be reproduced without permission in writing from ASHRAE, except by a reviewer who may quote brief passages or reproduce illustrations in a review with appropriate credit; nor may any part of this book be reproduced, stored in a retrieval system, or transmitted in any way or by any means—electronic, photocopying, recording, or other—without permission in writing from ASHRAE.

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Contents Author’s Note to Users. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Nomenclature—HVAC Terms, Abbreviations, and Subscripts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ix Solutions to Chapter 2—HVAC Fundamentals: Refrigeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Solutions to Chapter 3—HVAC Fundamentals: Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Solutions to Chapter 4—HVAC Fundamentals: Psychrometrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Solutions to Chapter 5—HVAC Equipment, Systems, and Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Solutions to Chapter 6—Comfort, Air Quality, and Climatic Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Solutions to Chapter 7—Heat and Moisture Flow in Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Solutions to Chapter 8—Cooling Load and Heating Loss Calculations and Analysis . . . . . . . . . . . . . . . . . . . . . . . . 35 Solutions to Chapter 9—Air Distribution System Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Solutions to Chapter 10—Water Distribution System Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Solutions to Chapter 11—Motors, Lighting, and Controls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Solutions to Chapter 12—Energy, Costs, and Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

v Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS


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Author’s Note to Users Several of the solutions in this manual incorporate the use of the spreadsheet programs that are provided with HVAC Simplified, such as E-Pipelator.xls, E-Ductulator.xls, HVACSysEff.xls, PsychProcess.xls, or TideLoad.xls. These programs are updated periodically; the most current version can be obtained for free from the ASHRAE Web site at www.ashrae.org/publicationupdates. The solutions in this text correspond to the 2006 versions of these programs.

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vii Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS


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AC adp ADPI ASD bhp, BHP Btu/h c C Cv CF CFC cfm CLF CLTD COP Δ Δh, ΔH Δp, ΔP D dB db, DB dp E EER ESP f FCU FPVAV ft

alternating current, air cooled, or air-conditioning apparatus dew point Air Diffusion Performance Index adjustable-speed drive (a.k.a. variablespeed drive, VSD) brake horsepower heat rate unit (British thermal units per hour) cooling loss coefficient (duct fittings) flow coefficient (flow in gpm that results in Δp = 1.0 psi) correction factor chlorofluorocarbon (refrigerants) cubic feet per minute (airflow rate) cooling load factor cooling load temperature difference (°F) coefficient of performance (watts/watt) delta (difference) differential head differential pressure diameter decibel (sound power or pressure) or dry bulb (temperature) dry bulb (temperature) dew point or differential pressure energy (electrical ≡ kWh or thermal ≡ Btu) energy efficiency ratio (Btu/W·h or MBtu/kWh) external static pressure (in. of water) frequency (Hz, cycles per second) fan-coil unit fan-powered variable air volume feet (distance or unit of head [ft of water])

gpm η h

H HDPE HRU, hru hp, HP HVAC Hz IAT (ti) K kW kWh kW/ton L, l Lp Lw LMTD MBtu/h MERV μ NC OA OAT (to) ODP psi psia psig

gallons per minute efficiency heating (Btu/h, kW), head of liquid (ft), specific enthalpy (Btu/lb), heat transfer coefficient (Btu/h·ft2⋅°F) heat (Btu, J); enthalpy (Btu) high-density polyethylene (piping material) heat recovery unit unit of power (horsepower = 0.746 kW) or heat pump heating, ventilating, air-conditioning frequency unit (cycles per second) indoor air temperature turbine flow meter constant (cycles per gallon) kilowatt (unit of power or heat rate) kilowatt-hour (unit of electrical energy) electrical demand per unit cooling capacity (kWrefrig./kWelect.) latent heat or liter sound pressure level (dB) sound power level (dB) log-mean temperature difference (°F) heat rate unit (British thermal units per hour × 1000) minimum efficiency reporting values (for air filters) fluid viscosity (lb/ft·s) noise criteria outside air (a.k.a. ventilation air) outdoor air temperature ozone depletion potential pounds per square inch (unit of pressure) pounds per square inch, absolute pounds per square inch, gage

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Nomenclature—HVAC Terms, Abbreviations, and Subscripts


q Q R Ra Re RH RTU rpm, RPM ρ s S SC SCL SHR t, T TC

heat rate (Btu/h or kW) volumetric flow rate (gpm, cfm, Lps, m3/s) thermal resistance (a.k.a. R-value ≡ h·°F·ft2/Btu, °C·m2/W) gas constant for air (ft·lbf/lbm·°F) Reynolds number (Re = ρDV/μ) relative humidity (%) rooftop unit revolutions per minute density (lb/ft3) specific entropy (Btu/lb·°F) entropy (Btu/°F) shade coefficient solar cooling load factor (Btu/h·ft2) sensible heat ratio temperature (°F, °C) total cooling (capacity)

TH ton

TP TSP u U V

VAV VSD w, W wb, WB w.c. x

total heating (capacity) cooling capacity (12,000 Btu/h, rate required to freeze 2000 lb of water (32°F) in 24 hours) total pressure (also p) total static pressure (also ps) specific internal energy (Btu/lb) internal energy (Btu) velocity (fps, fpm, m/s) and in some cases volumetric airflow (ASHRAE Standard 62.1) variable air volume (airflow rate) variable-speed drive (a.k.a. ASD) power (kW) wet bulb (temperature) water column (inches of water head) mole fraction

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Solutions to Chapter 2— HVAC Fundamentals: Refrigeration Problem 2.1

Solution

Find the Carnot COP and the ideal COP for a system that uses R-134a refrigerant at an evaporating temperature of 45°F and a condensing temperature of 120°F. Find the suction pressure and discharge pressures in psia and psig and the temperature of the refrigerant leaving the compressor (assuming the ideal cycle conditions). ( 460° + 45°F )°R 500 Carnot COP c = ------------------------------------------ = --------- = 6.73 120°F – 45°F

75

h1 – h4 110 – 52.5 - = ------------------------- = 5.2 Ideal COP c = ---------------h2 – h1


121 – 110


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Ideal COP (Using Figure B.1 in Appendix) Point 1: Saturated vapor @ 45°F, h1 = 110 Btu/lb (s1 = 0.222 Btu/lb·°F) Point 2: Superheated vapor @ ~190 psia (the saturation pressure for 120°F) and s2 = s1 = 0.222 Btu/lb·°F, h2 = 121 Btu/lb Point 3: Saturated liquid @ 120°F, h3 = 52.5 Btu/lb Point 4: Mixture @ t = 45°F and h4 = h3 = 52.5 Btu/lb


Solution

Problem 2.3

Solution

A scroll compressor (Table 2.3) with R-134a refrigerant operates with a 45°F evaporating temperature and a 120°F discharge temperature. Find the cooling capacity (20°F suction superheat and 15°F liquid subcooling), compressor input power, EER, suction pressure, and discharge pressure (psig). @ te = 45°F, tc = 120°F, SH = 20°F, and SC = 15°F qr = 28.9 MBtu/h (25,900 Btu/h), wc = 2.25 kW (2,250 W) EER = qc/wc = 28.9/2.25 = 12.8 MBtu/kWh (= 28,900/2250 = 12.8 Btu/Wh) Interpolating between P = 50 psia @ 40.3°F and P = 75 psia @ 62.2°F to P @ 45°F P (suction) ≈ 55.4 psia = 40.7 psig Interpolating between P = 175 psia @ 115.8°F and P = 200 psia @ 125.3°F to P @ 120°F P (discharge) ≈ 186 psia = 171.3 psig What increase in capacity and EER can be expected if the superheat is lowered to 10°F and the condensing temperature is lowered to 100°F? What is the disadvantage of doing this? @ te = 45°F, tc = 100°F, SH = 20°F, and SC = 15°F qr = 32.3 MBtu/h, wc = 1.77 kW (@ SH = 20°F) qr (@ SH = 10°F) = 32.3 MBtu/h × (ρ @ SH = 10°F/ρ @ SH = 20°F) = 32.3 MBtu/h × (ρ @ p ≈ 55 psia and t = 55°F/ρ @ p ≈ 55 psia and t = 65°F) = 32.3 MBtu/h × (1.11 lb/ft3 ÷ 1.09 lb/ft3) = 32.8 MBtu/h EER = 32.8 ÷ 1.77 = 18.5 Btu/Wh This represents a 13% increase in capacity and a 45% increase in efficiency. The disadvantage of doing this is that the condenser will most likely have to be cooled with water to lower the temperature to 100°F, and the 10°F lower superheat provides a smaller margin of error to prevent liquid refrigerant from entering the compressor.

Problem 2.4 Solution

Sketch the atomic makeup of R-22, R-12, and R-123. Refrigerant numbering system = R[Carbons–1] [Hydrogens+1][Fluorine] For R-22 (which is really 022) Number of carbon atoms – 1 = 0, thus number of carbon atoms = 1 Number of hydrogen atoms + 1 = 2, thus number of hydrogen atoms = 1 Number of fluorine atoms = 2 Since the structure of the single carbon atom permits four atoms and there are two fluorine atoms and only one hydrogen, the remaining bond is filled with a chlorine atom. For R-12: Carbon = 1, Hydrogen = 0, Fluorine = 2, Chlorine = 4 – 0 –2 = 2 For R-123: Carbon = 2 (6 bonds now available), Hydrogen = 1, Fluorine = 3, Chlorine = 6 – 1 –3 = 2

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Problem 2.2

Chapter 2—HVAC Fundamentals: Refrigeration

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Problem 2.5

How can you determine if a refrigerant has chlorine in its structure from the R-xxx designation?

Solution

If the R number of the refrigerant has only two digits (which means the first digit of the three-digit designation is 0), the sum of the remaining two numbers [(H + 1) and (F)] must be 5 to ensure chlorine is not present. If the first digit of the three-digit designation is 1, the sum of the remaining two numbers [(H + 1) and (F)] must be 7 to ensure chlorine is not present.

Problem 2.6

Compare the ideal COP of R-134a and R-22 at an evaporating temperature of 40°F with 20°F superheat and a condensing temperature of 120°F with 15°F subcooling with the actual compressor COPs calculated from the manufacturer’s performance tables.

Solution

For R-134a using the P-h diagram (Figure B.1): @ te = 40°F (~50 psia) and SH = 20°F, t1 = 60°F and h1 = 113 Btu/lb To find point 2, follow a line of constant entropy (s) to p =190 psia (saturated pressure for tc = 120°F), h2 = 126 Btu/lb. To find point 3, follow a line of constant pressure (p = 190 psia) to the left, cross the saturated liquid line, and go to a point 15°F below the saturated temperature (120°F), or t3 = 105°F, h3 = 47 Btu/lb. To find point 4, follow a line of constant enthalpy (h) downward to te = 40°F (~50 psia), h4 = h3 = 47 Btu/lb. h1 – h4 113 – 47 - = ------------------------ = 5.1 Ideal COP c = ---------------h2 – h1

126 – 113

From Table 2.3 @ te = 40°F and tc = 120°F, qr = 25.9 MBtu/h and wc = 2.27 kW. Thus, EER = 25.9 ÷ 2.27 = 11.4 MBtu/kWh = 11.4 Btu/Wh and COP = EER ÷ 3.412 Btu/Wh = 11.4 Btu/Wh ÷ 3.412 Btu/Wh = 3.34. For R-22, using the P-h diagram (Figure B.2): Point 1: (te = 40°F), p1 ≈ 83 psia, t1 = 60°F, and h1 = 111 Btu/lb. Point 2: (tc = 120°F), p2 ≈ 275 psia, t2 ≈ 160°F, and h2 = 124 Btu/lb. Point 3: (tc = 120°F), p3 ≈ 275 psia, t3 = 105°F, and h3 = 42 Btu/lb. Point 4: t4 = te = 40°F, p4 = p1 ≈ 83 psia, h4 = h3 = 42 Btu/lb h1 – h4 111 – 42 - = ------------------------ = 5.3 Ideal COP c = ---------------h2 – h1

124 – 111

From Table 2.4 @ te = 40°F and tc = 120°F, qr = 32.4 MBtu/h and wc = 2.74 kW. Thus, EER = 32.4 ÷ 2.74 = 11.8 MBtu/kWh = 11.8 /Wh and COP = EER ÷ 3.412 Btu/Wh = 11.8 Btu/Wh ÷ 3.412 Btu/Wh = 3.47. Problem 2.7

Solution

A set of pressure gauges on a manifold (see figure in “Refrigerant Charging” insert above) read 35 psig and a thermometer placed in close contact with the compressor inlet reads 67°F. The discharge pressure is 200 psig with an outdoor temperature of 95°F, and the refrigerant is R-134a. Is this system properly charged? If not, what range of temperature should be expected for these pressures? @ 35 psig, te = 40°F for R-134a Check @ p = 14.7 + 35 = 49.7 psia, te ≈ 40°F (as shown in Table 2.1) Superheat = t1 – te = 67°F – 40°F = 27°F The unit appears to be undercharged since proper operation typically dictates that the superheat be in the 10°F to 20°F range when nearly fully loaded, as indicated with the 95°F outdoor air temperature.




Solution

A manufacturer recommends that their R-22 equipment operate with a suction pressure of 72 psig and a return gas temperature of 53°F with a specified air temperature (75°F) and flow rate (400 cfm/ton). What are the corresponding evaporating temperature and superheat? P1 = 72 psig = 84.7 psia The pressure gauge shown in Figure 2.12 indicates te @ 72 psig ≈ 43°F. Thus, SH = t1 – te = 53°F – 43°F = 10°F.

Problem 2.9

With regard to the use of refrigerant mixtures as substitutes for CFCs, explain the difference between azeotropes and zeotropes. What is “glide”?

Solution

Azeotropes are refrigerant mixtures that behave as pure substances. When the refrigerant exists in a mixture of vapor and liquid, the lines of constant temperature are parallel with the lines of constant pressure with changing vapor-liquid fraction on a P-h diagram. Both lines are horizontal in the dome-shaped region of the chart bounded by the saturated liquid and saturated vapor lines. Zeotropes are refrigerant mixtures whose components evaporate and condense at a “gliding” temperature that depends on both the pressure and vapor-liquid fractions. The lines of constant temperature within the “vapor dome” region of a P-h diagram are not perfectly horizontal.

Problem 2.10

A refrigerant has an ASHRAE Standard 34 designation of A2 and B2. What does this mean? It also has an ODP of 0.75. Is this good, acceptable, or unacceptable?

Solution

The A2 designation indicates a low level of toxicity (A being nontoxic and B being toxic). The value of 2 indicates a low lower flammability limit (LFL) with 1 being no propagation in air and 3 having a high LFL. An ODP (ozone depletion potential) of 0.75 is unacceptable since many of the CFCs that have been banned have ODPs around 1.0.



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Problem 2.8

Solutions to Chapter 3— HVAC Fundamentals: Heat Transfer Problem 3.1

A stream of water flowing at 25 gpm must be cooled from 80°F to 70°F with chilled water at 50°F flowing at 20 gpm in a coaxial counterflow heat exchanger with an overall U-factor of 450 Btu/h⋅ft2⋅°F and 1.25 in. diameter inner tube. Calculate the required length of heat exchanger tubing.

q = mcp (two – twi) = ρQcp (two – twi) For hot fluid side, water at 75°F, ρ = 62.3 lb/ft3, cp = 1.0 Btu/lb·°F: q (Btu/h) = (62.3 lb/ft3 × 1.0 Btu/lb·°F × 60 min/h ÷ 7.48 gal/ft3) Q (gal/min) × (thwo – thwi)°F = 500 × Q (gal/min) × (thwo – thwi)°F = 500 × 25 (gal/min) × (70 – 80)°F = –125,000 Btu/h Rearrange the equation to find the cold fluid out temperature: tcwo = tcwi + q ÷ [500 × Q (gpm)] = 50°F + {125,000* ÷ [500 × 20 (gpm)]} = 62.5°F, Δt2 = 80 – 62.5 = 17.5°F Note the sign of q is changed from – to + since the energy balance convention has changed to the cold side and the addition of heat to the cold stream will result in an increase in temperature.



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Solution


Rearrange Equation 3.18: 2 q 125,000 Btu/h q 125, 000 A o = ------------------------- = --------------------------------------- = --------------------------------------------------------------------- = ------------------------- = 14.8 ft Btu ⎞ ( 20 – 17.5 )°F Δt 1 – Δt 2 450 × 18.7 U o LMTD ⎛ 450 ------------------- ---------------------------------U o ------------------------------⎝ ⎠ 2 h·ft ·°F ln ( 20 ⁄ 17.5 ) ln ( Δt 1 ⁄ Δt 2 )

Ao = πDoL, thus: L = Ao ÷ πDo = 14.8 ft2 ÷ π(1.25 in. ÷ 12 in./ft) = 45.2 ft Problem 3.2

Solution

Find the overall heat transfer coefficient for a schedule 40 steel pipe (do = 1.9 in., di = 1.61 in., k = 41 Btu/h⋅ft⋅°F) with an internal heat transfer coefficient of 48 Btu/h⋅ft2⋅°F and an external coefficient of 20 Btu/h⋅ft2⋅°F. 1.61 /2 in. 1.9 /2 in. 1 U o = ---------------------------------------------- : r o = --------------------- = 0.0792 ft : r i = ----------------------- = 0.0671 ft ro 12 in./ft 12 in./ft r o ln ----ro ri 1--------- + ---------------- + ----ri hi k ho

1 U o = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------0.0792 ft 0.0792 ft ln ⎛ ----------------------⎞ ⎝ 0.0671 ft⎠ 1 0.0792 ft ----------------------------------------------------------------- + -------------------------------------------------------- + ------------------------------------2 2 41 Btu/h·ft·°F 20 Btu/h·ft ·°F 0.0671 ft × 48 Btu/h·ft ·°F 2

U o = 13.3 Btu/h·ft ·°F

Problem 3.3

Solution

A wall is made of a 4 in. thick layer of masonry (0.9 Btu/h⋅ft⋅°F) and a 1 in. layer of insulation (k = 0.03 Btu/h⋅ft⋅°F). Find the overall thermal resistance if the inner and outer surfaces have heat transfer coefficients of 5.0 Btu/h⋅ft2⋅°F. 1 Δx mas'ry Δx ins 1 R ov = R i + R mas'ry + R ins + R o = ---- + ----------------------- + -------------- + -----k ins h o h i k mas'ry

R ov

1 in. 4 in. ------------------------------------1 12 in./ft 12 in./ft 1 - + ------------------------------------ + --------------------------------------- + ---------------------------------= --------------------------------2 2 0.03 Btu/h·ft·°F 0.9 Btu/h·ft·°F 5 Btu/h·ft ·°F 5 Btu/h·ft ·°F 2

R ov = 0.2 + 0.37 + 2.78 + 0.2 = 3.55 h·ft ·°F/Btu

6

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Chapter 3—HVAC Fundamentals: Heat Transfer

Problem 3.4

Solution

Repeat problem 3.3 if an added layer of ½ in. plywood (0.2 Btu/h⋅ft⋅°F) covers 50% of the wall and the remaining 50% is covered by ½ in. thick additional insulation. Based on 1 ft2 (Awall = 1.0 ft2) and rearranging Equation 3.16 to solve for Rov : Δx mas'ry Δx ins Δx ply&ins ⎛ 1 1 ⎞ R ov = A wall ⎜ ----------------- + --------------------------------- + ----------------------- + ------------------------------------------------------------------- + ------------------⎟ ⎝ h i A wall k mas'ry A wall k ins A wall 0.5k ply A wall + 0.5k ins A wall h o A wall⎠

R ov

4 in. 1 in. ------------------------------------⎛ 12 in./ft 12 in./ft 1 ⎜ ----------------------------------------------------------------------------------------------------------------------------------------------------------= 1 ft × + + 2 2 2 2 ⎜ 5 Btu/h·ft ·°F × 1 ft 0.9 Btu/h·ft·°F × 1 ft 0.03 Btu/h·ft·°F × 1 ft ⎝ 2

0.5 in. ------------------⎞ 1 12 in./ft - + --------------------------------------------------- ⎟ + ----------------------------------------------------------------------------------------------------------------------------------2 2 2 2⎟ 5 Btu/h·ft ·°F × 1 ft ⎠ 0.5 × 0.2 Btu/h·ft·°F1 ft + 0.5 × 0.03 Btu/h·ft·°F1 ft 2

Problem 3.5

Solution

A condenser is to be fabricated from the heat exchanger tubing described in Problem 3.1 for a compressor that flows 950 lb/h of R-134a refrigerant. Find the total required heat transfer rate, the heat required to desuperheat the gas, and the required length of tubing if the overall U-factor is 500 Btu/h⋅ft2⋅°F, the temperature leaving the compressor is 200°F, and the pressure is 185 psig. The condenser exit is saturated liquid at 185 psig and the water temperatures entering and leaving the condenser are 70°F and 80°F, respectively. Find refrigerant enthalpy at inlet (h2), saturated vapor (hsat), and outlet (h3). h2 is a superheated vapor @ 185 psig (~200 psia), h2 = 139 Btu/lb. For a saturated vapor @ 200 psia (125°F), hsat = 119 Btu/lb. For a saturated liquid @ 200 psia, hsat = h3 = 54 Btu/lb. qr = mr(h2 – h3) = 950 lb/h × (139 – 54) Btu/lb = 80,750 Btu/h Heat required to desuperheat: qr(ds) = mr(h2 – hsat) = 950 lb/h × (139 – 119) Btu/lb = 19,000 Btu/h Heat required to condense from saturated vapor to saturated liquid: qr(cond) = mr(hsat – h3) = 950 lb/h × (119 – 54) Btu/lb = 61,750 Btu/h

To size condenser, break into two sections so that LMTD can be calculated for both sections.



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R ov = 0.2 + 0.37 + 2.78 + 0.36 + 0.2 = 3.91 h·ft ·°F/Btu


Condensing section: qr(cond) = 61,750 Btu/h Recall that for water q (Btu/h) = mcp(two – twi) ≈ 500 Q (gpm) (two – twi) (°F) Thus: two = twi + qr(cond) ÷ 500 gpm = 70°F + 61,750 ÷ (500 × 16 gpm) = 77.7°F Δt 1 – Δt 2 ( 125 – 70 ) – ( 125 – 77.7 ) LMTD cond = --------------------- = -------------------------------------------------------------- = 51.1°F Δt 1 ( 125 – 70 ) ln -----------------------------ln ------( 125 – 77.7 ) Δt 2 q cond 2 61,750 Btu/h - = 2.42 ft A cond = ------------------------ = -------------------------------------------------------------2 U o LMTD 500 Btu/h·ft ·°F × 51.1°F

For desuperheating section: Δt 2 – Δt 3 ( 125 – 77.7 ) – ( 200 – 80 ) LMTD DS = ------------------------ = -------------------------------------------------------------- = 78.1°F ln Δt 2 ⁄ Δt 3 ( 125 – 77.7 ) ln -----------------------------( 200 – 80 ) q r ( ds ) 19, 000 Btu/h 2 - = 0.49 ft A DS = ------------------------ = -------------------------------------------------------------2 U o LMTD 500 Btu/h·ft ·°F × 78.1°F

L = Ao ÷ πDo = Acond + ADS ÷ πDo = (2.42 + 0.49) ft2 ÷ π(1.25 in. ÷ 12in./ft) = 8.9 ft

Problem 3.6

Hot waste water flowing at 20 gpm at 200°F is used to heat 15 gpm of incoming water at 85°F to 125°F in a coaxial-counterflow heat exchanger. The copper (k = 220 Btu/h⋅ft⋅°F) inside tube has an outer diameter of 1.125 in. and inside diameter of 1.00 in. Compute the required length of tube for an internal heat transfer coefficient of 750 Btu/h⋅ft2⋅°F and an outer heat transfer coefficient of 900 Btu/h⋅ft2⋅°F.

Solution

q = mcp (two – twi) = ρQcp(two – twi) For cold fluid side, 15 gpm water at 85°F heated to 125°F: q (Btu/h) ≈ 500 × Q (gal/min) × (thwo – thwi)°F = 500 × 15 (gal/min) × (125 – 85)°F = 300,000 Btu/h Rearrange the equation to find the hot fluid outlet temperature: tcwo = tcwi + q ÷ [500 × Q (gpm)] = 200°F – {300,000* ÷ [500 × 20 (gpm)]} = 170°F, for counterflow: Δt1 = 200 – 125 = 75°F and Δt2 = 170 – 85 = 85°F Find Uo: ⎛ 1.125 ⎛ 1.0 -------------⎞ in. -------⎞ in. ⎝ 2 ⎠ ⎝ 2⎠ 1 U o = ------------------------------------------- : r o = --------------------------- = 0.0469 ft : r i = --------------------- = 0.0417 ft ro 12 in./ft 12 in./ft r o ln ---ro ri 1 -------- + --------------- + ----ri hi k ho 1 U o = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------0.0469 ft 0.0469 ft ln ⎛ ----------------------⎞ ⎝ 0.0417 ft⎠ 1 0.0469 ft ⎛ ------------------------------------------------------------------⎞ - + -------------------------------------------------------- + ---------------------------------------2 2 ⎝ ⎠ 220 Btu/h·ft·°F 900 Btu/h·ft ·°F 0.0417 ft × 750 Btu/h·ft ·°F 2

U o = 379 Btu/h·ft ·°F


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Chapter 3—HVAC Fundamentals: Heat Transfer

Rearrange Equation 3.18: 300, 000 q q 2 300, 000 Btu/h A o = ------------------------ = ----------------------------- = ---------------------------------------------------------------------- = ------------------------- = 9.96 ft 377 × 79.9 U o LMTD ( 75 – 85 )°F Δt 1 – Δt 2 2 377 Btu/h·ft ·°F ----------------------------U o --------------------75 Δt 1⎞ ln ⎛ ------⎞ ⎛ ln ------⎝ 85⎠ ⎝ Δt 2⎠

Ao = πDoL, thus: L = Ao ÷ πDo = 9.96 ft2 ÷ π(1.125 in. ÷ 12 in./ft) = 33.8 ft

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Solutions to Chapter 4— HVAC Fundamentals: Psychrometrics

Solution

A sling psychrometer measures the air temperatures to be 85°F dry bulb and 72°F wet bulb. Find: relative humidity, dew-point temperature, humidity ratio (in lbmv/lbma and grains), specific volume, and enthalpy. Show results on a chart and verify with the program PsychProcess.xls (on the accompanying CD). Assume sea level elevation. Tdb1 Twb1 Elevation AtmPress APinHg HRatio1 RelHum1 SpHt1 Enal1 SpVol1 DewPt1 lbpCuFt

Problem 4.2

Solution

85 72 0 14.70 29.92 0.0139 97.0 53.7 0.246 35.6 14.04 66.4 0.000988

°F °F ft. psia in Hg

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Problem 4.1

lbw/lba Grains % Btu/lb-°F Btu/lb cu.ft./lb °F lbw/ft3

Air flowing at 4000 cfm is heated from 70°F (RH = 40%) at rate of 95,000 Btu/h. Find the outlet air conditions (db, RH, wb, υ). Sketch the process on a psychrometric chart. Q × 60 (min/h) q = ------------------------------------ × c p × ( t 2 – t 1 ) υ qυ 60Qc p

2

95, 000 (Btu/h) × 13.47 (ft /lb)

- = 92.2°F Thus, t 2 = t 1 + ---------------- = 70 + --------------------------------------------------------------------------------------------------------------3


60 (min/h) × 4000 (ft /min) × 0.24 (Btu/lb·°F)



Problem 4.3

Outside air (100°F/75°F) flowing at 1000 cfm is mixed with return air (75°F/ 63°F) at 5000 cfm. Find the mixed air conditions (db, RH, wb, υ, and h). Sketch on the psychrometric chart.

Solution 5000 cfm 4871 scfm

Qair2 QSair2

1000 cfm 926 scfm

mflow1

Stream 1 21921 lb/hr

mflow2

Stream 2 4166 lb/hr

Stream 3 mflow3 26087 Qair3 6000 Tdb3 79.0 HRatio3 0.0101 70.5 Twb3 65.2 RelHum3 47.9 SpHt3 0.244 Enal3 30.0 SpVol3 13.80 DewPt3 57.5

Problem 4.4

Solution

(mixed) lb/hr cfm °F lbw/lba Grains °F % Btu/lb-°F Btu/lb cu.ft./lb °F

minute. For additional information purposes, these values are corrected to air at standard conditions of ρ=0.075 lb/cu.ft. (QSair1 and QSair2).

Stream 1 @ Qair1(cfm), Tdb1(°F), & Twb1(°F)


Stream 2 @ Qair2(cfm), Tdb2(°F), & Twb2(°F) 2 1

3

A gas furnace produces 60,000 Btu/h with an airflow of 1400 cfm heated air with an inlet condition of 65°F (RH = 45%). Find the outlet air conditions (db, RH, wb, υ). Sketch the process on a psychrometric chart. 3

60, 000 (Btu/h) × 13.3 (ft /lb) qυ - = 104.6°F t 2 = t 1 + ---------------- = 65 + --------------------------------------------------------------------------------------------------------------3 60Qc p 60 (min/h) × 1400 (ft /min) × 0.24 (Btu/lb·°F)

From psychrometric chart, RH2 = 13%, t2wb =67.5°F, υ2 = 14.3 ft3/lb.



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Qair1 QSair1

Chapter 4—HVAC Fundamentals: Psychrometrics

Problem 4.5

Solution

Outside air (95°F/75°F) flowing at 2500 cfm is mixed with return air (75°F/63°F) at 7500 cfm. Find the mixed air conditions (db, RH, wb, υ, and h). Sketch on the psychrometric chart. Q2 2500 cfm 1 – 3 = 1 – 2 -------------------- = 2.05 in. ---------------------------------------------------- = 0.51 in. 7500 cfm + 2500 cfm Q1 + Q2

Point 3 is on a line drawn from point 1 to point 2 at a distance of 0.51 in. from point 1. Note that point 3 will be closer to the condition (point 1) with the larger flow rate. From psychrometric chart, t3 = 79.8°F, t3wb = 66.5°F, RH3 = 49%, υ3 = 13.8 ft3/lb, h3 = 30.9 Btu/lb

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Problem 4.6

Solution

A quantity of 1600 cfm of air at 80°F/67°F enters an evaporator coil with a 0.12 bypass factor and a 45°F apparatus dew point. Find the outlet air conditions (db, wb, RH, h), the sensible cooling capacity, the latent cooling capacity, total cooling capacity, and the SHR of the coil. Sketch on the psychrometric chart. Q = 1600 cfm, t1 = 80°F, t1wb = 67°F, tadp = 45°F, BF = 0.12 t2 = BF(t1 – tadp) + tadp = 0.12(80 – 45) + 45 = 49.2°F A line is drawn on the psychrometric chart from point 1 [80°F (db)/67°F (wb)] to tadp = 45°F, which is located on the saturation (RH = 100%) line. Point 2 is located on the intersection of this line and the line for t2 = 49.2°F. From psychrometric chart, t2wb = 48°F, RH2 = 92%, h2 = 19.3 Btu/lb. qs (Btu/h) ≈ 1.08 · Q (cfm) · (t2 – t1)°F = 1.08 · 1600 cfm · (80 – 49.2) = 53,200 Btu/h qL (Btu/h) ≈ 4680 · Q (cfm) · (W2 – W1) lbw/lba = 1.08 · 1600 cfm · (0.0110 – 0.007) ≈ 30,200 Btu/h q = qs + qL = 53,200 + 30,000 = 83,200 Btu/h SHRcoil = qs ÷ q = 53,200 ÷ 83,200 = 0.64




Problem 4.7

Solution

A 500 cfm outdoor air heat recovery unit (HRU) has a total effectiveness of 75% (both sensible and latent are equal). If the exhaust and makeup airflow rates are equal, find the conditions of the air (db, wb, h) leaving the HRU and entering the room when outdoor conditions are 94°F/77°F and the room air entering the HRU is 75°F/63°F. What is the capacity of this unit?

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Since εs = εL, εT = εs = εL = 0.75 Since exhaust and inlet flows are equal, mmin/ms = 1.0 hhru = ho – εT · (mmin/ms) · (ho – hr) @ to = 94°F (db) and 77°F wet bulb, ho = 40.3 Btu/lb @ tr = 75°F (db) and 63°F wet bulb, hr = 28.4 Btu/lb hhru = ho – εT · (mmin/ms) · (ho – hr) = 40.3 – 0.75 · 1.0 · (40.3 – 28.4) = 31.4 Btu/lb thru = to – εT · (mmin/ms) · (to – tr) = 94 – 0.75 · 1.0 · (94 – 75) = 79.8°F thru-wb = 67°F from psychrometric chart 3

500 (ft /min) × 60 (min/h) Q × 60 (min/h) - × ( 40.3 – 31.4 ) Btu/lb q hru = ------------------------------------ × ( h o – h hru ) = --------------------------------------------------------------3 υ 14.3 (ft /lb)

qhru = 18,700 Btu/h or qhru (Btu/h) ≈ 4.4 · Q (cfm) · (ho – hhru) (Btu/lb) = 4.4 · 500 · (40.3 – 31.4) ≈ 19,600 Btu/h (Discrepancy results since 4.4 in the above equation assumes room air conditions) Results using PsychProcess06.xls HRU Effectiveness 75 % 75 % Outdoor QairOut 500 cfm QSairOut 465 scfm mflowOut 2094 lb/hr SupFankW 0 kW ExFankW 0 kW Exhaust (not req'd) QairEx 500 cfm QSairEx 463 scfm Room Air HRatioRm 0.0095 lbw/lba 66.8 Grains RelHumRm 51.6 % SpHtRm 0.244 Btu/lb-°F EnalRm 28.4 Btu/lb SpVolRm 13.69 cu.ft./lb DewPtRm 56.0 °F

Room Air TdbRm TwbRm Outdoor Air TdbOut TwbOut Elev.

SenEff LatEff

Problem 4.8

Solution

Capacities 18.7 7.4 0.40 Outdoor Air HRatioOut 0.0161 112.6 RelHumOut 46.8 SpHtOut 0.247 EnalOut 40.3 SpVolOut 14.32 DewPtOut 70.6 qTotalHru qSenHru SHRHru

75 63 °F °F 94 77 °F 0 ft. HRU Outlet

Exhaust Air

Outdoor Air

Temps. MBtu/h MBtu/h

TdbHru TwbHru

lbw/lba Grains % Btu/lb-°F Btu/lb cu.ft./lb °F

HRU Outlet 0.0112 78.2 RelHumHru 51.6 SpHtHru 0.245 EnalHru 31.4 SpVolHru 13.85 DewPtHru 60.4 HRatioHru

lbw/lba Grains % Btu/lb-°F Btu/lb cu.ft./lb °F

A sensible heat recovery unit (HRU) with 80% efficiency draws in 1000 cfm of outside air at –10°F and exhausts an equal amount of room air at 70°F. Calculate the air temperature leaving the HRU and entering the room. What is the capacity of this unit? What is the capacity for 40°F outside air? Calculate the EER (= capacity in Btu/h ÷ power input in W) for both conditions if two fans that draw 700 W each are used. @ to = –10°F thru = to – εT · (mmin/ms) · (to – tr) = –10 – 0.80 · 1.0 · (–10 – 70) = 54°F @ to = 40°F thru = to – εT · (mmin/ms) · (to – tr) = 40 – 0.80 · 1.0 · (40 – 70) = 64°F @ to = –10°F, qs ≈ qhru ≈ 1.08 · Q (cfm) · (to – tr)°F = 1.08 · 1000 · (–10 – 54) = 69,100 Btu/h EER = q ÷ W = 69,100 Btu/h ÷ (2 · 700 W) = 49.4 Btu/Wh @ to = 40°F qhru ≈ 1.08 · 1000 · (40 – 64) = 25,900 Btu/h EER = q ÷ W = 25,900 Btu/h ÷ (2 · 700 W) = 18.5 Btu/Wh


79.8 °F 67.0 °F


Chapter 4—HVAC Fundamentals: Psychrometrics

Problem 4.9

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Solution

A quantity of 2500 cfm of air at 82°F/70°F enters an evaporator coil with a 0.08 bypass factor and a 45°F apparatus dew point. Find the outlet air conditions (db, wb, RH, h), the sensible cooling capacity, the latent cooling capacity, total cooling capacity, and the SHR of the coil. Sketch on the psychrometric chart. Q = 2500 cfm, t1 = 82°F, t1wb = 70°F, tadp = 45°F, BF = 0.08 t2 = BF(t1 – tadp) + tadp = 0.08(82 – 45) + 45 = 48°F A line is drawn on the psychrometric chart from point 1 [82°F (db)/70°F (wb)] to tadp = 45°F, which is located on the saturation (RH = 100%) line. Point 2 is located on the intersection of this line and the line for t2 = 48°F. From psychrometric chart, t2wb = 48°F, RH2 = 98%, h2 = 19.2 Btu/lb qs (Btu/h) ≈ 1.08 · Q (cfm) · (t2 – t1)°F = 1.08 · 2500 cfm · (82 – 48) = 91,800 Btu/h qL (Btu/h) ≈ 4680 · Q (cfm) · (W2 – W1) lbw/lba = 1.08 · 2500 cfm · (0.013 – 0.007) ≈ 69,600 Btu/h q = qs + qL = 91,800 + 69,600 = 161,400 Btu/h SHRcoil = qs ÷ q = 91,800 ÷ 161,400 = 0.57

Problem 4.10

Solution

A room at 75°F/63°F has a 36,000 Btu/h total capacity with a room SHR of 0.90 and an outdoor air (95°F/75°F) requirement of 400 cfm. Find the required sensible capacity and total cooling capacity of a unit to handle the building and outdoor air loads. t1 = 75°F (db) and 63°F (wb), h1 = 28.4 Btu/lb qroom = 36,000 Btu/h, SHRroom = 0.9 qs(room) = SHRroom · qroom = 0.9 · 36,000 = 32,400 Btu/h qL(room) = qroom – qs(room) = 36,000 – 32,400 = 3,600 Btu/h qs(OA) ≈ 1.08 · QOA (cfm) · (to – ti)°F = 1.08 · 400 cfm · (95 – 75) = 8,600 Btu/h qL(OA) ≈ 4680 · QOA (cfm) · (Wo – Wi) Btu/lb = 4680 · 400 cfm · (0.0142 – 0.0096) ≈ 8,600 Btu/h Required equipment size to handle the room load and the outdoor air load: qs = qs(room) + qs(OA) = 32,400 + 8,600 = 41,000 Btu/h qL = qL(room) + qL(OA) = 3,600 + 8,600 = 12,200 Btu/h q = q(room) + q(OA) = 41,000 + 12,200 = 53,200 Btu/h




Solution

Air flowing at 1500 cfm is heated from 65°F (RH = 35%) at a rate of 50,000 Btu/h. Find the outlet air conditions (db, RH, wb, υ). Sketch the process on a psychrometric chart. Q × 60 (min/h) q = ------------------------------------ × c p × ( t 2 – t 1 ) υ qυ 60Qc p

3

50, 000 (Btu/h) × 13.3 (ft /lb)

- = 95.8°F Thus, t 2 = t 1 + ---------------- = 65 + --------------------------------------------------------------------------------------------------------------3

Problem 4.12

Solution

60 (min/h) × 1500 (ft /min) × 0.24 (Btu/lb·°F)

Air flowing at a rate of 2000 cfm at 78°F/65°F enters a cooling unit with a total capacity (TC) of 60,000 Btu/h and a sensible heat ratio (SHR) of 0.75. Calculate the dry bulb, wet bulb, and relative humidity of the air leaving the coil. Determine the apparatus dew point and the bypass factor. Q = 2000 cfm, t1 = 78°F, t1wb = 65°F, qcoil = 60,000 Btu/h h1 = 30.0 Btu/lb, W1 = 0.0103 lbw/lba qs-coil = SHRcoil · qcoil = 0.75 · 60,000 = 45,000 Btu/h q s-coil (Btu/h) 45, 000 Btu/h t 2 ( °F ) ≈ t 1 – ------------------------------------ = 78°F – ---------------------------------------- = 57.1°F 1.08 × 2000 cfm 1.08 × Q (cfm) q coil (Btu/h) 60, 000 Btu/h h 2 (Btu/lb) ≈ h 1 – --------------------------------- = 30.0 Btu/lb – ------------------------------------- = 23.2 Btu/lb 4.4 × 2000 cfm 4.4 × Q (cfm)

Find point 2 on the psychrometric chart using t2 = 57.1°F and h2 = 23.2 Btu/lb

Then read properties at point 2 from chart:

t 2 – t adp 57.1 – 51 = ---------------------- = 0.23 t2wb = 55°F, RH2 = 85%, tadp = 51°F, BF = ------------------t 1 – t adp

78 – 51



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Problem 4.11

Solutions to Chapter 5— HVAC Equipment, Systems, and Selection Problem 5.1

Find the total cooling capacity (gross), sensible cooling capacity (gross), and input kW for a Model 150 rooftop unit when the outdoor temperature is 95°F, indoor temperature is 80°F/67°F, and airflow is 5000 cfm.

Solution

For a Model 150 RTU with return air at 80°F (db)/67°F (wb), outdoor air of 95°F, and an indoor airflow of 5000 cfm from Table 5.2, TC = 138 MBtu/h (138,000 Btu/h), SC = 99 MBtu/h (99,000 Btu/h), and kW = 10.2 kW

Problem 5.2

Find the required fan power to deliver 5000 cfm at an ESP of 1.2 in. w.g. for the unit selected in Problem 5.1.

Solution

For a Model 150 RTU with an air flow of 5000 cfm requiring an ESP of 1.2 in. of water, Interpolate between values for ESP = 1.0 and 1.5 in BHP1.0 = 2.95 hp and BHP1.5 = 3.67 hp → BHP1.2 = 3.24 hp kW1.0 = 2.57 and kW1.5 = 3.19 → kW1.2 = 2.82 RPM1.0 = 1234 and RPM1.5 = 1420 → RPM1.2 = 1308

Problem 5.3

Repeat Problem 5.1 for an indoor temperature of 74°F/62°F.

Solution

For a Model 150 RTU with return air at 74°F (db)/62°F (wb), outdoor air of 95°F, and an indoor airflow of 5000 cfm from Table 5.2, TC = 131 MBtu/h @ wb = 62°F, SC = 122 MBtu/h @ db = 80°F and wb = 62°F SC74/62 = SC80/62 + 1.1 · (1 – BF) · (cfm/1000) · (EAT – 80) = 122 + 1.1 · (1 – 0.5) · (5000/1000) · (74 – 80) = 90.7 MBtu/h kW62 = 10.1 kW

Problem 5.4

Correct the results of Problem 5.3 for fan heat to obtain total capacity (net), sensible cooling capacity (net), and resulting sensible heat ratio (SHR).

Solution

TC62 (net) = TC62 (gross) – 3.41 kWfan From Problem 5.2, kWfan = 2.82 TC62 (net) = 131 MBtu/h – 3.41 MBtu/kWh · 2.82 kW = 121.4 MBtu/h SC74/62 (net) = 90.7 MBtu/h – 3.41 MBtu/kWh · 2.82 kW = 81.1 MBtu/h SHR = SC74/62 (net) ÷ TC62 (net) = 81.1 ÷ 121.4 = 0.67 --`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---




Problem 5.5

Solution

A building in St. Louis, Missouri, has a sensible heat gain of 23,000 Btu/h and a total load of 33,000 when outdoor conditions are 97°F/76°F and mixed indoor air conditions entering the cooling coil are 80°F/67°F. Select a cooling unit from Table 5.3 to meet the load and SHR requirement. Specify the required cfm and resulting EER. qs = 23,000 Btu/h (23 MBtu/h), q = 33,000 Btu/h (33 MBtu/h) @ to = 97°F/76°F and ti = 80°F/67°F, Q = 1200 cfm SHRLoad = qs ÷ q = 23 MBtu/h / 33 MBtu/h = 0.70 For a Model 036, interpolate between values for OAT (aka to) = 95°F and 105°F @ 95°F OAT and EAT (aka ti) 80°F/67°F, TC = 35.8 MBtu/h, SC = 26.4 MBtu/h, and kW = 2.97 @ 105°F OAT and EAT = 80°F/67°F, TC = 34.5 MBtu/h, SC = 25.9 MBtu/h, and kW = 3.31 Via interpolation, TC97 = 35.5 MBtu/h, SC97 = 26.3 MBtu/h, and kW97 = 3.04 Since this manufacturer reports net capacity values, no fan heat deduction is required. SHRUnit = SC97 ÷ TC97 = 26.3 ÷ 35.3 = 0.74 Since SHRUnit ≥ SHRLoad, the unit will not meet the SHR (dehumidification) requirement at the rated 1200 cfm airflow. Lower cfm to reduce SHR and improve dehumidification. Try lowering flow to 80% of rated flow = 0.80 · 1200 cfm = 960 cfm CFTC = 0.97, TC97/960 = 0.97 · 35.5 = 34.4 CFSC = 0.90, SC97/960 = 0.90 · 26.3 = 23.7, SHRUnit = 23.7 ÷ 34.4 = 0.69 → OK CFkWc = 0.975, kW97/960 = 0.975 · 3.04 = 3.04 EER = TC (net) ÷ kW (total) = 34.4 MBtu/h ÷ 3.04 kW = 11.3 MBtu/kWh (Btu/Wh)


Repeat Problem 5.5 for an indoor condition of 75°F/63°F. qs = 23,000 Btu/h (23 MBtu/h), q = 33,000 Btu/h (33 MBtu/h) @ to = 97°F/76°F and ti = 75°F/63°F SHRLoad = qs ÷ q = 23 MBtu/h / 33 MBtu/h = 0.70 For a Model 036, interpolate between values for OAT (aka to) = 95°F and 105°F @ 95°F OAT and EAT (aka ti) 75°F/63°F, TC = 33.4 MBtu/h, SC = 25.7 MBtu/h, and kW =2.94 @ 105°F OAT and EAT = 75°F/63°F, TC = 32.2 MBtu/h, SC = 25.3 MBtu/h, and kW = 3.28 Via interpolation, TC97 = 33.2 MBtu/h, SC97 = 25.6 MBtu/h, and kW97 = 3.01 Since this manufacturer reports net capacity values, no fan heat deduction is required. SHRUnit = SC97 ÷ TC97 = 25.6 ÷ 33.2 = 0.77 → too high Reduce airflow and correct performance to see if the unit can meet requirements. Try lowering flow to 80% of rated value = 0.80 · 1200 cfm = 960 cfm CFTC = 0.97, TC97/960 = 0.97 · 33.2 = 32.2 → TC too low CFSC = 0.90, SC97/1080 = 0.90 · 25.6 = 23.0, SHRUnit = 23.0 ÷ 32.2 = 0.72 → too high Unit will not meet requirements since increasing flow to meet TC requirement will raise SHR, which is already too high. Try a larger unit at the lowest possible airflow rate (this may create more problems since oversized units cycle more frequently and exacerbate humidity problems).


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Chapter 5—HVAC Equipment, Systems, and Selection

For a Model 042 @ 1300 cfm, interpolate between OAT (aka to) = 95°F and 105°F @ 95°F OAT and EAT (aka ti) 75°F/63°F, TC = 40.7 MBtu/h, SC = 31.1 MBtu/h, and kW = 3.47 @ 105°F OAT and EAT = 75°F/63°F, TC = 39.1 MBtu/h, SC = 30.4 MBtu/h, and kW = 3.88 Via interpolation, TC97 = 40.4 MBtu/h, SC97 = 31.0 MBtu/h, and kW97 = 3.55 @ 80% airflow = 1040 cfm CFTC = 0.97, TC97/1040 = 0.97 · 40.4 = 39.2 → OK CFSC = 0.90, SC97/1040 = 0.90 · 31.0 = 27.9 SHRUnit = 27.9 ÷ 39.2 = 0.71 → still too high This result is typical for high-efficiency equipment that frequently cannot meet latent requirements. Either use a smaller indoor coil with lower airflow and lower efficiency to meet latent requirements or use the Model 036 since it is not oversized. EER = TC (net) ÷ kW (total) = 40.4 MBtu/h ÷ 3.55 kW = 11.4 MBtu/kWh (Btu/Wh) Problem 5.7

Solution

Problem 5.8

Solution

Determine if the unit selected in Problem 5.6 can meet an SHR of 0.68 for an outdoor condition of 85°F and indoor condition of 75°F/63°F at the design cfm. Can it meet the SHR at a lower cfm? SHRLoad = 0.68 @ to = 85°F and ti = 75°F/63°F SHRLoad = qs ÷ q = 23 MBtu/h / 33 MBtu/h = 0.70 For Model 036 at 1200 cfm @ 85°F OAT and EAT 75°F/63°F TC = 34.6 MBtu/h, SC = 26.3 MBtu/h, and kW =2.66 SHRUnit = 26.3 ÷ 34.6 = 0.76 → too high Correct to 80% airflow (960 cfm) SHRUnit = 23.7 ÷ 33.6 = 0.71 → too high The building heat loss is 37,000 Btu/h when the indoor temperature is 70°F and the outdoor temperature is 20°F. Use the heating data of the Problem 5.6 heat pump to determine the unit’s capacity (with a 10% defrost cycle deduct) and size the electric resistance supplementary backup if necessary. Find the system COP. qh = 37 MBtu/h @ 20°F OAT @ 27°F OAT and EAT 70°F, TH = 24.3 MBtu/h, kW = 2.80 @ 17°F OAT and EAT 70°F, TH = 22.6 MBtu/h, kW = 2.76 Interpolated to 20°F OAT and EAT 70°F, TH = 23.1 MBtu/h, kW = 2.77 @ 80% rated flow (960 cfm) CFTH = 0.98, TH20/960 = 0.98 · 23.1 = 22.6 MBtu/h CFkW = 1.05, kW20/960 = 1.05 · 2.77 = 2.91 Deduct 10% for defrost: TH20/960 (with defrost penalty) = 0.9 · 22.6 = 20.4 MBtu/h Auxiliary heating requirement: qAux = qh – TH = 37 – 20.4 = 16.6 MBtu/h kWAux = qAux ÷ 3.41 = 16.6 ÷ 3.41 = 4.9 kW ( TH + q Aux ) ⁄ 3.41 ( 20.4 + 16.6 ) ⁄ 3.41 COP = -------------------------------------------- = ---------------------------------------------- = 1.39 2.91 + 4.9 kW + kW Aux

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19



Meet the requirements of Problem 5.8 by selecting a natural gas furnace for an indoor temperature of 70°F. qh = 37 MBtu/h @ 20°F OAT Find a furnace with TH > 37 MBtu/h and Q ≤ 960 cfm (the cooling mode airflow from previous problems). Option 1 for noncondensing furnace, use Model 060 (Table 5.4) that has a TH of 47 MBtu/h and set the fan speed tap on low (if ESP ≈ 0.4 in. for air distribution system) or med/low (if ESP ≈ 0.8 in.). Option 2 for noncondensing furnace, use Model 060 (Table 5.5) that has a TH of 47 MBtu/h and set the fan speed tap on low (if ESP ≈ 0.4 in. for air distribution system) or med/low (if ESP ≈ 0.8 in.). Option 3 (with very little cushion for extremely cold days), use Model 040 (Table 5.5) that has a TH of 38 MBtu/h and set the fan speed on high and specify that the air distribution system required ESP does not exceed 0.4 in.

Problem 5.10

Solution

Repeat Problems 5.6 and 5.8 using a water-to-air heat pump with a 90°F entering water temperature in cooling and a 45°F entering water temperature in heating. Assume a pump power requirement of 160 W (this replaces the outdoor fan of an air unit) and indoor fan is included in the total kW. qs = 23,000 Btu/h (23 MBtu/h), q = 33,000 Btu/h (33 MBtu/h) @ EWT = 90°F and ti = 75°F/63°F, wpump = 160 W = 0.16 kW SHRLoad = qs ÷ q = 23 MBtu/h / 33 MBtu/h = 0.70 From Table 5.6, try a Model 036 using Qwater = 9 gpm (analysis could also use 7 gpm as starting point) and 80°F/67°F EAT and EWT = 90°F, TC67 = 34.5 MBtu/h Correct TC to twb =63°F, TC63 = CF63 · TC67 = 0.93 · 34.5 = 32.1 MBtu/h → too small Try a Model 042 using 8 gpm (analysis could also use 11 gpm as starting point) and 80°F/67°F EAT, and EWT = 90°F, TC67 = 42.2 MBtu/h, SC = 30.8 MBtu/h, and kWc = 3.22 Correct TC to twb =63°F, TC63 = CF63 · TC67 = 0.93 · 42.2 = 39.2 MBtu/h → OK Correct SC to tdb =75°F, twb = 63°F, SC75/63 = CF75/63 · SC80 = 0.96 · 30.8 = 29.6 MBtu/h SHRUnit = SC75/63 ÷ TC63 = 29.6 ÷ 39.2 = 0.75 → too high, since SHRLoad = 0.70 Lower cfm to 80% of rated Qair = 0.80 · 1400 = 1120 cfm TC80% = CF80% · TC100% = 0.97 · 39.2 = 38.0 MBtu/h SC80% = CF80% · SC100% = 0.90 · 29.6 = 26.6 MBtu/h SHRUnit = SC80% ÷ TC80% = 26.6 ÷ 38.0 = 0.70 → OK Correct kW to tdb = 75°F, twb = 63°F, and Qair = 1120 cfm kWc = CF63 · CF80% · kWc @ 67 wb,1400 cfm = 0.98 · 0.975 · 3.22 = 3.08 kW For Model 042 @ 1120 cfm, 8 gpm: EER = TC ÷ (kWc + kWpump) = 38.0 MBtu/h ÷ (3.08 + 0.16) = 11.7 MBtu/kWh Heating @ EAT = 70°F, EWT = 45°F, Qair = 1120 cfm, Qwater = 8 gpm @ EWT = 50, TH = 38.1 MBtu/h, @ EWT = 40°F, TH = 33.1 MBtu/h Via interpolation, TH = 35.6 MBtu/h TH = CF80% · TH100% = 0.98 · 35.6 = 34.9 MBtu/h kWh = 2.64 @ 45°F, kWh@80% = CF80% · kWh@100% = 1.05 · 2.64 = 2.77 kW qAux = qh – TH = 37 – 34.9 = 2.1 MBtu/h, kWaux = 2.1 ÷ 3.412 = 0.62 TH + q aux 34.9 + 2.1 COP = ---------------------------------------------------------------------------------- = ------------------------------------------------------------------- = 3.05 3.412 ⋅ ( 2.77 + 0.62 + 0.16 ) 3.412 ⋅ ( kW h + kW aux + kW pump )

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Problem 5.11

A building has a sensible heat gain of 140 MBtu/h and a total load of 190 MBtu/h when outdoor conditions are 95°F/75°F and mixed indoor air conditions entering the cooling coil are 78°F/64.5°F. Select a rooftop cooling unit from Table 5.2 to meet the load. Specify the required cfm, SHRunit, and fan motor size to deliver 1.2 in. of water external static pressure (ESP) and the resulting EER.

Solution

q = 190 MBtu/h, qs = 140 MBtu/h @ 95°F OAT SHRLoad = 140 ÷ 190 = 0.74 Try Model 240 @ mid-range flow rate (8000 cfm) from Table 5.2 since SHR is normal @ 95°F OAT, twb = 67°F, TC* = 252 MBtu/h, SC* = 183 MBtu/h, kW = 19.3 @ 95°F OAT, twb = 62°F, TC* = 232 MBtu/h, SC* = 218 MBtu/h, kW = 18.7 Via interpolation, @ 95°F OAT, twb = 64.5°F, TC* = 242 MBtu/h, SC* = 201 MBtu/h, and kW = 19.0 * gross capacities, must be corrected for fan heat

TCnet = TCgross – 3.41 kWfan and SCnet = SCgross – 3.41 kWfan @ 8000 cfm and 1.2 in. ESP, kWfan = 5.72 kW (via interpolation between ESP = 1.0 and 1.5) Note fan BHP = 6.8 hp and increased to next standard size = 7.5 hp (see Chapter 11) TCnet = 242 – 3.41 · 5.72 = 222 MBtu/h → OK since requirement is 190 MBtu/h SCnet @ 80 EAT = SCgross @ 80 EAT – 3.41 kWfan = 201 – 3.412 · 5.72 = 181.5 Correct for EAT = 78°F SCnet @ 78 EAT = SCnet @ 80 EAT + 1.1 × (1 – BF) × (cfm/1000) × (EAT – 80) SCnet @ 78 EAT = 181.5 + 1.1 × (1 – 0.06) × (8000/1000) × (78 – 80) = 165 SHRUnit = SCnet @ 78 EAT ÷ TCnet = 165 ÷ 222 = 0.74 → OK since SHRLoad = 0.74 However, consider running a slightly lower airflow since extra capacity is available and SHRUnit is so close to SHRLoad 222 TC EER = ---------------------------------- = -------------------------------- = 9.0 ( 19.0 + 5.72 ) ( kW + kW fan )

Summary: Use Model 240, Qair = 8000 cfm, BHPfan = 7.5 hp (6.8 hp), and SHRUnit = 0.74. Problem 5.12

A building zone has a total sensible heat gain of 105,000 Btu/h (walls, roof, windows, internal, people) and a latent gain of 20,000 Btu/h. The required outdoor air ventilation rate is 800 cfm. Indoor conditions are 75°F/63°F and outdoor conditions are 95°F/75°F, and outside air is mixed with the return air before entering the unit. Select a rooftop unit to cool this zone. The fan must deliver 1.0 in. water of external static pressure (ESP). Recall the capacities given are gross. You must convert them to total net capacities by deducting the fan heat.

Solution

Loads Room: qRS = 105 MBtu/h, qRL = 20 MBtu/h, qR = qRS + qRL = 105 + 20 = 125 MBtu/h Outdoor air: @ to = 95°F/75°F, ho = 38.3 Btu/lb, @ ti = 75°F/63°F, hi = 28.4 Btu/lb qOAS ≈ 1.08 · Qo · (to – ti) ≈ 1.08 · 800 cfm · (95 –75) ≈ 17,300 Btu/h ≈ 17.3 MBtu/h qOA ≈ 4.44 · Qo · (ho – hi) ≈ 4.44 · 800 cfm · (38.3 – 28.4) ≈ 35,200 Btu/h ≈ 35.2 MBtu/h Totals: qs = qRS + qOAS = 105 + 17.3 = 122 MBtu/h q = qR + qOA = 125 + 35.2 = 160 MBtu/h SHRLoad = 122 ÷ 160 = 0.76 Must now find mixed air conditions, which means the supply air or recirculated air quantity must be known (or assumed) to compute the mixed air conditions when mixed with the 800 cfm outdoor air. Note the Model 180 rooftop unit is rated at

21

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6000 cfm for the mid-range value. Since the SHRLoad is also a mid-range value, use this flow rate for first computation; 5200 cfm of recirculated air at is mixed with 800 cfm of outside air 95°F/75°F to provide 6000 cfm of supply air. Using PsychProcess.xls (mixing) provides a mixed air condition ≈ 78°F/65°F entering the rooftop unit. Qair1 QSair1

5200 cfm 5064 scfm

Qair2 QSair2

800 cfm 740 scfm

mflow1

Stream 1 22789 lb/hr

mflow2

Stream 2 3332 lb/hr

Stream 3 mflow3 26120 Qair3 6000 Tdb3 78.2 HRatio3 0.0100 69.8 Twb3 64.8 RelHum3 48.7 SpHt3 0.244 Enal3 29.7 SpVol3 13.78 DewPt3 57.2

(mixed) lb/hr cfm °F lbw/lba Grains °F % Btu/lb-°F Btu/lb cu.ft./lb °F

Note: Input values (Qair1 and Qair2) are in cubic feet per minute. For additional information purposes, these values are corrected to air at standard conditions of ρ=0.075 lb/cu.ft. (QSair1 and QSair2).



Stream 2 @ Qair2(cfm), Tdb2(°F), & Twb2(°F) 2 1

3

Via interpolation, @ 95°F OAT, twb = 65°F for a Model 180 rooftop unit TC = 184 MBtu/h, SC = 147 MBtu/h, kW = 14.0 (gross capacities) To correct for fan heat, go to fan data at 6000 cfm and 1.0 in. ESP: kWfan = 3.32, BHP = 3.89 hp (need 5 hp motor) TCnet = TCgross – 3.41 kWfan and SCnet = SCgross – 3.41 kWfan TCnet = 184 – 3.41 · 3.32 = 173 MBtu/h → OK since requirement is 160 MBtu/h SCnet @ 80 EAT = SCgross @ 80 EAT – 3.41 kWfan = 147 – 3.412 · 3.32 = 136 Correct for EAT = 78°F SCnet @ 78 EAT = SCnet @ 80 EAT + 1.1 · (1 – BF) · (cfm/1000) · (EAT – 80) SCnet @ 78 EAT = 136 + 1.1 · (1 – 0.04) · (6000/1000) · (78 – 80) = 123 SHRUnit = SCnet @ 78 EAT ÷ TCnet = 123 ÷ 173 = 0.71 → Excellent, since SHRLoad = 0.76 (flow can be increased since SHRUnit is lower, but this is not necessary since unit is slightly oversized). TC 173 EER = ---------------------------------- = -------------------------------- = 10.0 ( kW + kW fan ) ( 14.0 + 3.32 )

Model 180 operating at 6000 cfm with 5 hp fan motor is suitable.

22

Problem 5.13

A water-cooled chiller must provide water at 45°F to ten fan coil units that require 45 MBtu/h (net) each with fans that draw 600 W each. The condenser water is cooled with a cooling tower that can provide 85°F. a. Select a chiller to meet this load. b. Calculate the required chilled water flow in gpm for a 55°F chiller entering temperature (base answer on chiller capacity). c. Calculate the required condenser water flow based on 3.0 gpm per ton of chiller capacity. d. Determine the head loss in feet of water across the evaporator and condenser. e. Determine the chiller gross kW/ton (gross) and EER (Btu/W·h). f. Determine system net kW/ton and EER if two pumps (chilled water and condenser water) draw 2.0 kW and 2.25 kW, respectively.

Solution

qLoad = 10 FCUs · (45,000 Btu/h + 3.41 · 600 W) = 470,500 Btu/h = 39.2 tons a. A Model 040 (Table 5.10 ) water-cooled scroll compressor chiller will deliver:

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b. c. d. e. f.

TC = 40.1 tons = 40.1 · 12,000 Btu/ton-h = 481,200 Btu/h @ 45°F LWT and 85°F Condenser EWT Compressor demand will be 30.4 kW (30,400 W) gpm(Evap.) = 481,200 Btu/h ÷ [500 · (55°F – 45°F)] = 96 gpm gpm(Cond.) = 3 gpm/ton · 40.1 tons = 120 gpm From Figure 5.13 for 040 Chiller @ 96 gpm: hEvap = 13 ft of water From Figure 5.14 for 040 Chiller @ 120 gpm: hCond = 15.3 ft of water kW/ton (gross) = 30.4 kW ÷ 40.1 tons = 0.76 kW/ton EER (gross) = 481,200 Btu/h ÷ 30,400 W = 15.8 Btu/Wh q Gross – q Fan EER Net = --------------------------------------------------------------------------------------------------------------------W Chiller + W FcuFans + W CWPump + W CHWPump 481, 200 Btu/h – 10 ⋅ ( 3.412 ⋅ 600 W ) = -------------------------------------------------------------------------------------------------------------30, 400 W + 10 ⋅ 600 W + 2250 W + 2000 W = 11.3 Btu/Wh

Problem 5.14.

A four-zone building has the loads shown below. The room air entering the coils is 80°F/67°F and chilled water at 45°F is supplied. Select fan coil units (assuming a 10% deduction for fan heat) and specify airflow and water flow while attempting to maintain a coil outlet temperature of 55°F ±2.0°F.

Zone 1 Zone 2 Zone 3 Zone 4 Total

Solution

10 a.m. Cooling Loads (MBtu/h) Sensible Total 30 40 45 60 25 35 30 38

3 p.m. Cooling Loads (MBtu/h) Sensible Total 42 60 35 45 38 54 40 55

Increase loads given in table by 10% to account for fan heat (this will be added to both the sensible and totals loads). 10 a.m. Zone 1 Zone 2 Zone 3 Zone 4 Totals

Sensible 34 51 28.5 33.8 147.3

3 p.m. Total 44 66 38.5 41.8 190.3

Sensible 48 39.5 43.4 45.5 176.4

Total 66 49.5 59.4 60.5 235.4

Zone 1: Peak load occurs at 3 p.m.: q1-Load = 66 MBtu/h and SHR1-Load = 48 ÷ 66 = 0.73 A model 60-HW-4 coil at 2000 cfm, 80°F/67°F air and 13 gpm: TC = 65 MBtu/h (Too low) at 21 gpm: TC = 75.2 MBtu/h (High) Reduce flow to 17 gpm and by interpolation: TC = 70.1 MBtu/h, SC = 46.7 MBtu/h Check SHRFCU = 46.7 ÷ 70.1 = 0.67 (OK) Check outlet water temperature: 70, 100 Btu/h TC ( Btu/h ) t o = t i ( °F ) + ----------------------------------- = 45°F + --------------------------------- = 53.2°F 500 × 17 gpm 500 × Q ( gpm )

Flow of 17 gpm is acceptable, but try a lower flow. Reduce flow to 15 gpm and by interpolation: TC = 67.6 MBtu/h, SC = 45.8 MBtu/h Check SHRFCU = 45.8 ÷ 67.6 = 0.68 (OK) --`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---




Check outlet water temperature: 67, 600 Btu/h TC ( Btu/h ) t o = t i ( °F ) + ----------------------------------- = 45°F + --------------------------------- = 54.0°F 500 × 15 gpm 500 × Q ( gpm )

Zone 2: Use same coil and water flow as zone 1 to meet 10 a.m. load, which is also 66 MBtu/h with slightly higher SHR. So Model 60-HW-4 coil at 2000 cfm and 15 gpm will work. Zone 3: Use same coil as zones 1 and 2, but water flow can be lowered to 13 gpm: TC = 65 MBtu/h, SC = 44.9, SHRLoad = 43.4 ÷ 59.4 = 0.73; SHRFCU = 44.9 ÷ 65.0 = 0.69 (OK) Check outlet water temperature: 65, 000 Btu/h TC ( Btu/h ) t o = t i ( °F ) + ----------------------------------- = 45°F + --------------------------------- = 55.0°F 500 × 13 gpm 500 × Q ( gpm )

So Model 60-HW-4 coil at 2000 cfm and 13 gpm will work. Zone 4: Use same coil and water flow as zone 3 to meet 3 p.m. load, which is 60.5 MBtu/h. TC = 65 MBtu/h, SC = 44.9, SHRLoad = 45.5 ÷ 60.5 = 0.75; SHRFCU = 44.9 ÷ 65.0 = 0.69 (OK) So Model 60-HW-4 coil at 2000 cfm and 13 gpm will work. Problem 5.15

Solution

Select a chiller (or chillers) to meet the combined loads of the coils in Problem 5.14. Specify unit model number, required water flow, and gross kW/ ton and EER. The peak block load occurs at 3 p.m. (although load in zone 2 peaks at 10 a.m.). At 3 p.m., qLoads = 235.4 MBtu/h = 19.6 tons A Model 020 (Table 5.10) water-cooled scroll compressor chiller will deliver: TC = 20.4 tons = 20.4 · 12,000 Btu/ton-h = 244,800 Btu/h @ 45°F LWT and 85°F Cond. EWT Compressor demand will be 15.4 kW (15,400 W). gpm(Evap.) = 244,800 Btu/h ÷ [500 · (55°F – 45°F)] = 49 gpm kW/ton (gross) = 15.4 kW ÷ 20.4 tons = 0.75 kW/ton EER (gross) = 244,800 Btu/h ÷ 15,400 W = 15.9 Btu/Wh

24 --`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---



Solutions to Chapter 6— Comfort, Air Quality, and Climatic Data Problem 6.1 Solution


Compute the heat rate of a 5 ft, 10 in., 160 lb male machinist at work. AD = 0.108 · m0.425 · l 0.725 = 0.108 · 160 lb.0.425 · 70 in.0.725 = 20.3 ft2 Doing light machine work generates 37 to 44 Btu/h·ft2 (mid-range = 40.5 Btu/h·ft2) Thus, Qmachinist = 40.5 Btu/h·ft2 · 20.3 ft2 = 820 Btu/h Repeat Problem 6.1 for a 5 ft, 6 in., 120 lb performing ballerina. AD = 0.108 · m0.425 · l 0.725 = 0.108 · 120 lb.0.425 · 66 in.0.725 = 19.1 ft2 A ballerina generates 44 to 81 Btu/h·ft2 (mid-range = 62.5 Btu/h·ft2) Thus, qballerina = 62.5 Btu/h·ft2 · 19.1 ft2 = 1190 Btu/h

Problem 6.3

What range of indoor temperature and humidity is best to satisfy occupants in the summer? In the winter? Why is there a difference?

Solution

At the upper relative humidity level of 60%, the indoor temperature should be in the 73°F to 79ºF range to satisfy the most individuals in the summer. At the lower relative humidity level of 30%, the indoor temperature should be in the 74°F to 81°F range to satisfy the most individuals in the summer. At the upper relative humidity level of 60%, the indoor temperature should be in the 68°F to 74ºF range to satisfy the most individuals in the winter (this condition is difficult to maintain in the winter because the outside air is drier). At the lower relative humidity level of 30%, the indoor temperature should be in the 69°F to 76°F range to satisfy the most individuals in the winter. The temperatures are lower in the winter because occupants are typically dressed with heavier clothing in the winter because of the lower outdoor temperature.

Problem 6.4

Why are people more comfortable in the winter with a lower thermostat setting?

Solution

Occupants are comfortable with a lower setting in the winter because they are typically dressed with heavier clothing because of the lower outdoor temperature.

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Problem 6.5

Solution

Find the required ventilation air for a 1500 ft2 college classroom with 40 students. The ventilation air is delivered through ceiling vents and returned though a grille near the floor. Vbz = RPPZ + RaAZ = 10 cfm/person · 40 people + 0.12 cfm/ft2 · 1500 ft2 = 400 + 180 = 580 cfm Since the ventilation air is delivered at the ceiling and exhausted near the floor, the air will be delivered to the breathing zone and the zone air distribution effectiveness (EZ) is 1.0. Thus, Voz = Vbz ÷ EZ = 580 ÷ 1.0 = 580 cfm


Repeat Problem 6.5 if the return is in the ceiling and the HVAC unit is in cooling. Vbz = RPPZ + RaAZ = 10 cfm/person · 40 people + 0.12 cfm/ft2 · 1500 ft2 = 400 + 180 = 580 cfm When cold air is delivered at the ceiling, it will tend to fall into the breathing zone since it is more dense than the room air. It will mix with the room air and fulfill its intended effect before it is exhausted. The zone air distribution effectiveness (EZ) is 1.0. Thus, Voz = Vbz ÷ EZ = 580 ÷ 1.0 = 580 cfm


Repeat Problem 6.6 if the unit is in heating and the delivery temperature is 100°F. Vbz = RPPZ + RaAZ = 10 cfm/person · 40 people + 0.12 cfm/ft2 · 1500 ft2 = 400 + 180 = 580 cfm Since the warm ventilation air is delivered at the ceiling, it will tend to stay near the ceiling and not completely mix with the room air before it is exhausted near the ceiling. The zone air distribution effectiveness (EZ) is 0.8. Thus, Voz = Vbz ÷ EZ = 580 ÷ 0.8 = 725 cfm

Problem 6.8

You are required to design a ventilation air system for a 3000 ft2 library with supply and return in the ceiling, but no occupancy is provided. Specify the required ventilation airflow rate.

Solution

If the building owner or building owner’s representative does not provide occupancy, use the default values in Table 6.2. In the case of the library, the value is 10 people per 1000 ft2. It is advisable that this be noted in the design documentation or provided directly to the owner or owner’s representative in writing. Thus, PZ = (10 people/1000 ft2) · 3000 ft2 = 30 people Vbz = RPPZ + RaAZ = 5 cfm/person · 30 people + 0.12 cfm/ft2 · 3000 ft2 = 150 + 360 = 510 cfm In cooling EZ = 1.0 (cold air supply in ceiling, return in ceiling) Voz = Vbz ÷ EZ = 510 ÷ 1.0 = 510 cfm In heating EZ = 0.8 (warm air supply in ceiling, return in ceiling) Voz = Vbz ÷ EZ = 510 ÷ 0.8 = 640 cfm


Determine the required ventilation air rate for a 3000 ft2, five-bedroom, threebathroom home. Qfan = 0.01 · A (ft2) + 7.5 · (Nbedrooms + 1) = 0.01 · 3000 ft2 + 7.5 · (5 +1) = 75 cfm

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Chapter 6—Comfort, Air Quality, and Climatic Data

Problem 6.10

A building with four zones has the airflow requirements below. Determine the required ventilation air rate for a multi-zone ventilation air system. Supply air: Zone 1 = 800 cfm, Zone 2 = 1200 cfm, Zone 3 = 700 cfm, Zone 4 = 1500 cfm Ventilation air: Zone 1 = 200 cfm, Zone 2 = 275 cfm, Zone 3 = 150 cfm, Zone 4 = 500 cfm

Solution

Zone 1 2 3 4

Vz 200 275 150 500

Vp 800 1200 700 1500

Zp (Vz/Vp) 0.25 0.23 0.21 0.33 ←Zpmax

Assume EZ = 1.0, Vou = ΣVbz ÷ EZ = (200 + 275 + 150 + 500) ÷ 1.0 = 1125 cfm Ev = 0.8 since Zp (max) ≤ 0.35 Vot = Vou ÷ EV = 1125 ÷ 0.8 = 1406 cfm Problem 6.11

An office with six zones is served with a single rooftop unit that provides 1.0 cfm/ft2 of supply air through the ceiling. The return is also in the ceiling. Ventilation air is supplied at the rooftop unit return. Compute the required ventilation air rate in the summer and winter given the following table. Zone 1 2 3 4 5 6

Solution

Use Reception Office Office Office Conference Office

Area (ft2) 700 400 800 700 500 400

People 5 2 8 4 10 1

For summer [EZ = 1.0 (cold air supply in ceiling, return in ceiling)] and assuming that occupants move from their office to occupy the conference room so that the normal number of people in the building is 20 (not 30). Multi-Zone Systems Only Zone Description Reception Office Office Office Conference Office

Zone Number 1 2 3 4 5 6 7 8 9 10

Rp

20 0.67 1.00 205 1.00 0.8 256

people

5 5 5 5 5 5

Totals Max Building occupants Diversity Ez Vou Estimated Vp/A Ev Vo

No. of People 5 2 8 4 10 1

Rp*people 25 10 40 20 50 5 0

30

150

Ra 0.06 0.06 0.06 0.06 0.06 0.06

A (ft2) 200 300 300 400 250 300

1750

Ra*Area 12 18 18 24 15 18 0

Vbz 37 28 58 44 65 23 0 0 0 0

Vp 200 300 300 400 250 300 0

105

Vp

Zpmax

Zp 0.19 0.09 0.19 0.11 0.26 0.08 0.00 0.00 0.00 0.00 0.26

cfm cfm/ft2 cfm

For winter, EZ = 0.8 (warm air supply in ceiling, return in ceiling) Multi-Zone Systems Only Zone Description Reception Office Office Office Conference Office

Zone Number 1 2 3 4 5 6 7 8 9 10

Rp

20 0.67 0.80 256 1.00 0.8 320

people

5 5 5 5 5 5

Totals Max Building occupants Diversity Ez Vou Estimated Vp/A Ev Vo

No. of People 5 2 8 4 10 1

Rp*people 25 10 40 20 50 5 0

30

150

Ra 0.06 0.06 0.06 0.06 0.06 0.06

A (ft2) 200 300 300 400 250 300

1750

Ra*Area 12 18 18 24 15 18 0

Vbz 37 28 58 44 65 23 0 0 0 0

105

Vp

Zpmax

Zp 0.19 0.09 0.19 0.11 0.26 0.08 0.00 0.00 0.00 0.00 0.26

cfm cfm/ft2 cfm

27

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Vp 200 300 300 400 250 300 0



Problem 6.12

Solution




Find the following for Chicago: Elevation Dry-bulb temperature at 0.4% cooling design Mean wet-bulb temperature at 0.4% cooling design Temperature at 99.6% heating design condition 0.4% design wet-bulb Dry-bulb temperature at 0.4% WB condition Daily range on cooling design day For Chicago from Table 6.5: Elevation = 673 ft above sea level tdb @ 0.4% = 91°F tMWB @ 0.4% tdb = 74°F tdb @ 99.6% = –6°F twb @ 0.4% = 77°F tMDB @ 0.4% twb = 88°F DR = 20°F Repeat Problem 6.12 for Tuscaloosa, Alabama. For Tuscaloosa from Table 6.5: Elevation = 171 ft above sea level tdb @ 0.4% = 95°F tMWB @ 0.4% tdb = 77°F tdb @ 99.6% = 20°F twb @ 0.4% = 80°F tMDB @ 0.4% twb = 90°F DR = 20°F Explain the meaning of 99.6% and 0.4% design conditions in Table 6.3. This notation means that the value of the temperature in the table is exceeded during 99.6% of the hours during a normal year. Likewise, the 0.4% temperature is exceeded during only 0.4% of the hours during a normal year.

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Solutions to Chapter 7— Heat and Moisture Flow in Buildings Problem 7.1 --`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Find the overall U-factor and R-value for a 2 × 4 in. wall with R-13 fiberglass batts (approximately 20% of the wall is framing). The exterior wall is 5/8 in. hardboard (standard tempered) over 1/2 in. vegetable fiberboard with no air gap. The interior finish is 1/2 in. gypsum board.

Solution R-Insul. Path = R-Wood Path = U (Overall)= R(Total)=

Problem 7.2

R-Insul. Path = R-Wood Path = U (Overall)= R(Total)=

Solution

0.074 13.59

Insulation 13

Wood

Sheath 1.32 1.32

4.375

Other

=0.20*7.925+0.8*16.55

Btu/hr-ft2-F

Int. Finish In. Surface 0.45 0.68 0.45 0.68

R-Path Total 16.55 7.925

=1/0.074

hr-ft2-F/Btu

Repeat Problem 7.1 if the exterior finish is 110 lb/ft3 4 in. face brick.

Solution

Problem 7.3

% Total Area Out. Surface Out. Finish 80 0.25 0.85 20 0.25 0.85

% Total Area Out. Surface Out. Finish 80 0.25 0.45 20 0.25 0.45 0.076 13.14

Insulation 13

Wood

Sheath 1.32 1.32

4.375

Other

=0.20*7.525+0.8*16.15

Btu/hr-ft2-F

Int. Finish In. Surface 0.45 0.68 0.45 0.68


=1/0.076

hr-ft2-F/Btu

Find the overall U-factors (Btu/h⋅°F⋅ft2) and R-values (h⋅°F⋅ft2/Btu) of a 2.25 in. thick solid wood door with and without a metal storm door. From Table 7.1 for 2.25 in. solid wood door U = 0.27 Btu/h·°F·ft2, R = 3.7 h·°F·ft2/Btu With storm door U = 0.20 Btu/h·°F·ft2, R = 5.0 h·°F·ft2/Btu


Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of the wall in the building shown below. Assuming the 1 in. insulation is expanded polystyrene (aka beadboard): R(Total)= R(Total)= U (Overall)=

Out. Surface FaceBrick 0.25 0.45 7.28 hr-ft2-F/Btu 0.137 Btu/hr-ft2-F

Insulation 3.5

Other

Other 8"LW block In. Surf. 2.4 0.68

7.28

=1/7.28

Assuming the 1 in. insulation is extruded polystyrene (aka pink or blue board): R(Total)= R(Total)= U (Overall)=


Out. Surface FaceBrick Insulation Other 0.25 0.45 5 8.78 hr-ft2-F/Btu =1/8.78 0.114 Btu/hr-ft2-F

Other

8"LW block In. Surf. 2.4 0.68


8.78


Problem 7.5

Solution

Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of the roof/ ceiling in the building shown below if the insulation is polyisocyanuarate.

R(Total)= R(Total)= U (Overall)=

Out. Surface 0.25 17.25 0.058

1/2"slag 0.05

2" Polyiso 12

2

hr-ft -F/Btu

2" concrete Air gap 0.28 2.2

Acos.tile 1.79

In. Surf. 0.68

ΣR 17.25

=1/17.25

Btu/hr-ft2-F

Problem 7.6

A roof-ceiling system consists of a metal roof, 6 in. fiberglass insulation, 24 in. attic space, and ½ in. acoustical ceiling tile. Find the R-value and the most appropriate roof number.

Solution

Assuming the attic is not ventilated, with no reflective barrier, 100°F ventilation air temperature (see Table 7.2): R(Total)= R(Total)= U (Overall)=

Out. Surface 0.25 25.62 0.039

Metal 0

NoVentAttic 6" fiber ins Air gap 1.9 21

2

hr-ft -F/Btu

Acos.tile 1.79

In. Surf. 0.68

ΣR 25.62

=1/25.62

Btu/hr-ft2-F

If attic is naturally ventilated, with no reflective barrier, 100°F ventilation air temperature (see Table 7.2): R(Total)= R(Total)= U (Overall)=

Out. Surface 0.25 26.42 0.038

Metal 0

NatVentAttic 6" Fiber.ins. AcosTile 2.7 21 1.79

2

hr-ft -F/Btu

other

In. Surf. 0.68

ΣR 26.42

=1/26.42

2

Btu/hr-ft -F

In both cases, the most appropriate roof is #4 since the insulation is greater than 20 h·°F·ft2/Btu. Problem 7.7

Find the overall U-factor and R-value for a wall with 2 × 4 in. fir studs on 16 in. O.C. with R13 fiberglass batt insulation (approximately 15% of the wall is framing). The wall exterior is 4 in. face brick (110 lb/ft3) and 1/2 in. extruded polystyrene with a 1/2 in. air gap (no reflective foil). The interior finish is 1/2 in. gypsum board.

Solution R-Insul. Path = R-Wood Path = U (Overall)= R(Total)=

Problem 7.8

R-Insul. Path = R-Wood Path = U (Overall)= R(Total)=


0.061 16.39

Btu/hr-ft2-F

Insulation 3½"wood½" extr,poly Air gap 13 2.5 1.2 4.375 2.5 1.2

=0.15*9.905+0.85*18.53

1/2" Gyp. In. Surface 0.45 0.68 0.45 0.68


=1/0.061

hr-ft2-F/Btu

Repeat problem 7.7 for wall with 2 × 6 in. studs on 24 in. O.C. with R19 batts.

Solution

30

% Total Area Out. Surface Face Brick 85 0.25 0.45 15 0.25 0.45

% Total Area Out. Surface Face Brick 90 0.25 0.45 10 0.25 0.45 0.045 22.35

Btu/hr-ft2-F hr-ft2-F/Btu

Insulation 5½"wood½" extr,poly Air gap 19 2.5 1.2 6.875 2.5 1.2

=0.10*12.405+0.80*24.53

1/2" Gyp. In. Surface 0.45 0.68 0.45 0.68

=1/0.045

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---



Chapter 7—Heat and Moisture Flow in Buildings

Problem 7.9

Solution

Find the overall U-factors (Btu/h⋅°F⋅ft2) and R-values (h⋅°F⋅ft2/Btu) of 1.75 in. thick steel door with a urethane core (no thermal break) with and without a metal storm door. From Table 7.1 for 1.75 in. steel door with a urethane core and no thermal break: U = 0.40 Btu/h·°F·ft2, R = 2.5 h·°F·ft2/Btu

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

With storm door: U = 0.26 Btu/h·°F·ft2, R = 3.8 h·°F·ft2/Btu Problem 7.10

Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of a vinyl frame, double-glass window with a ½ in. air gap and no thermal break.

Solution

From Table 7.3 (assuming resistance of a ½ in. air gap ≈ resistance of a ¼ in. air gap): U = 0.55 Btu/h·°F·ft2, R = 1/U = 1.82 h·°F·ft2/Btu

Problem 7.11

Find the R-value of a wall that is 4 in. face brick, 2 in. insulation, and 8 in. heavyweight concrete walls. What wall number most closely matches this wall?

Solution

R(Total)= R(Total)= U (Overall)=

Out. Surface Face brick 0.25 0.45 12.12 hr-ft2-F/Btu 0.083 Btu/hr-ft2-F

2" extr,poly 8" HWblock other 10 0.74

other

In. Surf. 0.68

ΣR 12.12

=1/12.12

Wall #16 Problem 7.12

Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of an aluminum frame, double-glass window with a ½ in. air gap and no thermal break.

Solution

From Table 7.3 (assuming resistance of a ½ in. air gap ≈ resistance of a ¼ in. air gap): U = 0.87 Btu/h·°F·ft2, R = 1/U = 1.15 h·°F·ft2/Btu

Problem 7.13

Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of a 1.75 in. thick wood door with and without a metal storm door.

Solution

From Table 7.1 for 1.75 in. solid wood door: U = 0.40 Btu/h·°F·ft2, R = 2.5 h·°F·ft2/Btu With storm door: U = 0.26 Btu/h·°F·ft2, R = 3.8 h·°F·ft2/Btu



Find the overall U-factor (Btu/h⋅°F⋅ft2) and R-value (h⋅°F⋅ft2/Btu) of a vinyl frame, double-glass window with a ½ in. argon gap. From Table 7.3: U = 0.49 Btu/h·°F·ft2, R = 2.04 h·°F·ft2/Btu Find the CLF and SCL zone type letters for the top floor of a building made with walls like the room of Problems 7.4 and 7.5. Heavy walls with lightweight (LW) block, top floor Thus: CLF zone type = C SCL zone type = B




Problem 7.16

Solution

Find the shade coefficient for the window of Problem 7.12 with (a) no interior shade, (b) closed, medium-colored blinds, and (c) dark-colored drapes with a closed weave fabric. Aluminum frame, double pane (from Tables 7.3 and 7.4): (a) no shade, SC = 0.76 (b) medium blind, closed, SC = 0.63 (c) dark drapes, closed weave, SC = 0.35

Problem 7.17

Find the shade coefficient for the window of Problem 7.10 with (a) no interior shade, (b) dark roller shades, and (c) light-colored drapes with an open weave fabric.

Solution

Vinyl frame, double pane (from Tables 7.3 and 7.4): (a) no shade, SC = 0.63 (b) medium blind (closed), SC ≈ SCTable7.4 · (SCVinyl-NoShade ÷ SCAL-NoShade) ≈ 0.63 · (0.63 ÷ 0.76) ≈ 0.52 (c) dark drapes, closed weave, SC = 0.35 SC ≈ SCTable7.4 · (SCVinyl-NoShade ÷ SCAL-NoShade) ≈ 0.35 · (0.63 ÷ 0.76) ≈ 0.29

Problem 7.18

Ductwork is to be located in an attic of a 5000 ft2 building with a roof of R = 3 h⋅ft2⋅°F/Btu and an R = 20 h⋅ft2⋅°F/Btu ceiling. The ductwork consists of a 100 ft main rectangular metal duct (20 × 28 in.) and 150 linear ft of 10 in. round metal duct. The duct is not sealed but is wrapped with insulation to provide R = 6 (h⋅ft⋅°F/Btu). The average main duct ESP is 1.3 in. of water and the round duct is at 0.75 in. of water. The outdoor temperature is –5°F, the indoor temperature is 68°F, and the indoor fan of a 200 MBtu/h output furnace delivers 5200 cfm. The attic is ventilated naturally at 0.1 cfm/ft2. Compute the duct losses.

Solution

AFloor = 5000 ft2 Duct: 100 ft × 20 in. × 28 in. rectangular: AMainDuct = 2 · (20 + 28) ÷ 12 in./ft · 100 ft = 800 ft2 150 ft × 10 in. round: ATakeOff = π(10 ÷ 12 in./ft) · 150 ft = 393 ft2 AD = 800 + 393 = 1193 ft2 Leakage: (Leakage class (CL) values from Table 7.7) QDL-Main = [(CL · Δps0.65) ÷ 100] · AD = [(48 · 1.30.65 ÷ 100] · 800 = 455 cfm QDL-Round = [(30 · 0.750.65) ÷ 100] · 393 = 98 cfm 455 + 98 = 553 cfm Attic ventilation: QAV = 0.1 cfm/ft2 · 5000 ft2 = 500 cfm Supply air temperature: ts = ti + HC ÷ (1.08 · Qs) = 68°F + 200,000 Btu/h ÷ (1.08 · 5000 cfm) = 103.6°F Energy balance to find attic temperature:

tA

32

AD AC A ------- × t i + ⎛ -----R- + 1.08 × Q AV⎞ × t o + ⎛ ------ + 1.08 × Q DL⎞ × t s ⎝ RR ⎠ ⎝ RD ⎠ RC = ----------------------------------------------------------------------------------------------------------------------------------------------A Floor A R A Duct --------------+ ------ + -------------- + 1.08 × ( Q CSV + Q DL ) R Floor R R R Duct

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---



Chapter 7—Heat and Moisture Flow in Buildings

Compute duct heat loss: qDuct = qDuctCond + qDuctLeak = (AD ÷ RD) · (ts – tA) + 1.08 · QDL · (ts – tA) = (1193 ÷ 6) · (103.6 – 27.2) + 1.08 · 553 · (103.6 – 27.2) = 15,190 + 45,630 = 60,800 Btu/h Problem 7.19 Solution

Compute the losses through the roof and the attic ventilation air in Problem 7.18. qR = (AR ÷ RR) · (tA – to) and qAV = 1.08 · QAV · (tA – to) from Problem 7.18 qR = (5000 ft2 ÷ 3 h·°F·ft2/Btu) · [27.2°F – (–5°F)] = 53,700 Btu/h and qAV = 1.08 · 533 cfm · [27.2°F – (–5°F)] = 18,500 Btu/h

Problem 7.20

Repeat Problem 7.18 if the duct is located in a 5 ft high crawlspace that has a 1 ft exterior exposure with no insulation. The building is 50 × 100 ft and has an R = 5 h⋅ft2⋅°F/Btu floor. Crawlspace ventilation is 0.05 cfm/ft2.

Solution

AFloor = 5000 ft2, PBldg = 2 · 50 + 2 · 100 = 300 ft Duct: 100 ft × 20 in. × 28 in. rectangular: AMainDuct = 2 · (20 + 28) ÷ 12 in./ft · 100 ft = 800 ft2 150 ft × 10 in. round: ATakeOff = π(10 ÷ 12 in./ft) · 150 ft = 393 ft2 Duct leakage: QDL-Main = [(CL · Δps0.65) ÷ 100] · AD = [(48 · 1.30.65 ÷ 100] · 800 = 455 cfm QDL-Round = [(30 · 0.750.65) ÷ 100] · 393 = 98 cfm 455 + 98 = 553 cfm Crawlspace ventilation: QCS = 0.05 cfm/ft2 · 5000 ft2 = 250 cfm Crawlspace wall HT coefficient (Table 7.6): For Lag = 1 ft and Lcsb = 5 ft and no insulation: Fscb = 4.42 Btu/h·ft2·°F Supply air temperature: ts = ti + HC ÷ (1.08 · Qs) = 68°F + 200,000 Btu/h ÷ (1.08 · 5000 cfm) = 103.6°F Energy balance to find crawlspace temperature:

t CS

A Floor AD --------------- × t i + ( F csb × P Bldg + 1.08 × Q CSV ) × t o + ⎛ ------ + 1.08 × Q DL⎞ × t s ⎝ ⎠ R Floor RD = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------A Floor AD --------------- + P Bldg × Q CSV + ------ + 1.08 × ( Q CSV + Q DL ) R Floor RD

1193 ⎛ 5000 ------------⎞ × 68°F + ( 4.42 × 300 + 1.08 × 250 ) × ( – 5 ) + ⎛ ------------ + 1.08 × 553⎞ × 103.6 ⎝ 5 ⎠ ⎝ 6 ⎠ t CS = ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- = 42°F 5000 1193 ------------ + 4.42 × 300 + ------------ + 1.08 × ( 250 + 553 ) 5 6

Compute duct heat loss: qDuct = qDuctCond + qDuctLeak = (AD ÷ RD) · (ts – tcs) + 1.08 · QDL · (ts – tcs) = (1193 ÷ 6) · (103.6 – 42) + 1.08 · 553 · (103.6 – 42) = 12,250 + 36,800 = 49,050 Btu/h



--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

tA

5000 5000 1193 ------------ × 68°F + ⎛ ------------ + 1.08 × 500⎞ × – 5 + ⎛ ------------ + 1.08 × 553⎞ × 103.6 ⎝ 3 ⎠ ⎝ 6 ⎠ 20 = ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- = 27.2°F 1193 5000 5000 ------------ + ------------ + ------------ + 1.08 × ( 500 + 553 ) 6 3 20

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---



Solutions to Chapter 8— Cooling Load and Heating Loss Calculations and Analysis Problem 8.2

Compute the 3 p.m. heat gain of the roof/ceiling of the single office described in Figure C.4* in Appendix C if it is located in St. Louis, Missouri, and room conditions are 75°F/63°F (db/wb).

Solution

For St. Louis, to = 95°F, DR = 18°F, lat = 39°N (Table 6.5 or 8.2) From problem statement, ti = 75°F Use roof #4 (see Figure 7.5 or 7.6)—metal roof, Rins > 20, vented attic From Table 8.4: CLTDTable = 66°F (lat = 36°N), CLTDTable = 64°F (lat = 42°N), via interpolation CLTDTable = 65°F (lat = 39°N) CLTDCor = CLTDTable + (78 – ti) + [(to – DR/2) – 85] = 65 + (78 – 75) + [(95 – 18/2)–85] = 69°F R(Total)= R(Total)= U (Overall)=

Out. Surface 0.25 30.62 0.033

metal 0

hr-ft2-F/Btu Btu/hr-ft2-F

Attic Space 2.9

Insulation Asc tile 25 1.79

other

In. Surf. 0.68

ΣR 30.62

=1/30.62

qRoof = U · A · CLTDCor = 0.033 Btu/h·ft2·°F · (42 ft · 24 ft) · 69°F = 2270 Btu/h Problem 8.2

Solution

Compute the 3 p.m. heat gain of the walls of the single office described in Figure C.4* in Appendix C if it is located in St. Louis. For St. Louis, to = 95°F, DR = 18°F, lat = 39°N (Table 6.5 or 8.2) From problem statement, ti = 75°F Use wall #16 (see Figure 7.4) R(Total)= R(Total)= U (Overall)=

Out. Surface Face brick 0.25 0.45 13.38 hr-ft2-F/Btu 0.075 Btu/hr-ft2-F

2" extr,poly 8" LWblock 10 2

other

other

In. Surf. 0.68

ΣR 13.38

=1/13.38

From Table 8.3 for east wall: CLTDTable = 28°F (lat = 36°N), CLTDTable = 29°F (lat = 42°N) → CLTDTable = 28.5°F (lat = 39°N) CLTDCor = CLTDTable + (78 – ti + [(to – DR/2) – 85] = 28.5 + (78 – 75) + [(95 – 18/2) – 85] = 32.5°F qEWall = U · A · CLTDCor = 0.075 · [10 ft · 42 ft – (3 ft · 6 ft · 5 windows)] · 32.5°F = 800 Btu/h *See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.4. --`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---




From Table 8.3 for south wall: CLTDTable = 12°F (lat = 36°N), CLTDTable = 15°F (lat = 42°N) → CLTDTable = 13.5°F (lat = 39°N) CLTDCor = CLTDTable + (78 – ti + [(to – DR/2) – 85] = 13.5 + (78 – 75) + [(95 – 18/2) – 85] = 17.5°F qSWall = U · A · CLTDCor = 0.075 · [10 ft · 24 ft – (3 ft · 6 ft · 4 windows)] · 17.5°F = 220 Btu/h qNWall = qWWall = 0 (walls face conditioned space, Δt = 0) Problem 8.3 Solution

Compute the 3 p.m. heat gain of the windows of the single office described in Figure C.4* in Appendix C if it is located in St. Louis. Double-pane, aluminum frame windows: U = 0.87 Btu/h·ft2·°F (Table 7.3) Light blinds: SC = 0.58 (assuming a 45° position—Table 7.4) For conduction see Table 8.4; all walls, h = 15 CLTDTable = 14°F, CLTDCor = 14 + (78 – 75) + [(95 – 18/2) – 85] = 18°F To compute solar, SCL zone types based on wall types, see Figure 7.4 Heavy wall, top floor SCL zone type = B From Table 8.6, east wall, h = 15 SCLE = 48 @ lat = 36°N, CLTDTable = 48 @ lat = 42°N, thus SCLE = 48 Btu/h·ft2 @ lat = 39°N From Table 8.6, south wall, h = 15 SCLS = 52 @ lat = 36°N, CLTDTable = 81 @ lat = 42°N, thus SCLS = 67 Btu/h·ft2 @ lat = 39°N qEWin = qcond + qsolar = U · A · CLTDCor + SC · A · SCL = 0.87 · (3 ft · 6 ft · 5 windows) · 18°F + 0.58 · (3 ft · 6 ft · 5 windows) · 48 Btu/h·ft2 = 1410 + 2510 = 3920 Btu/h qSWin = qcond + qsolar = U · A · CLTDCor + SC · A · SCL = 0.87 · (3 ft · 6 ft · 4 windows) · 18°F + 0.58 · (3 ft · 6 ft · 4 windows) · 48 Btu/h·ft2 = 1130 + 2800 = 3930 Btu/h

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Problem 8.4

Solution

Compute the 3 p.m. internal heat gain due to lighting and office equipment of the single office described in Figure C.4* in Appendix C if it is located in St. Louis and is occupied for 12 hours per day. Lights: qSLights = 3.412 Btu/Wh · CLF15 · wS/bulb · Nbulbs For a heavy wall, LW block, CLF zone type = C (Figure 7.4), CLF15 = 0.92 (12 h day—Table 8.13) qSLights = 3.412 Btu/Wh · 0.92 · 31 wS/bulb (E-ballast) · 20 · 2 bulbs = 3890 Btu/h Unhooded equipment: qSEquip = 3.412 Btu/Wh · CLF15 · ΣwS = 3.412 Btu/Wh · CLF15 · wSComp + wSPrinters For a heavy wall, LW block, CLF zone type = C (Figure 7.4), CLF15 = 0.89 (12 h day—Table 8.8) qSEquip= 3.412 Btu/Wh · 0.89 · (15 · 125 + 4 · 160) = 7640 Btu/h

Problem 8.5

Solution

Compute the 3 p.m. heat gain (sensible and latent) due to ventilation air of the single office described in Figure C.4* in Appendix C if it is located in St. Louis. Assume it is a single zone. Find outdoor ventilation rate for 15 people in a 1008 ft2 office (single zone). Vbz = RP · P + Ra · A = 5 cfm/person · 15 people + 0.06 cfm/ft2 · 1008 ft2 = 135 cfm

*See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.4.



Chapter 8—Cooling Load and Heating Loss Calculations and Analysis

For cool air delivered through ceiling, EZ = 1.0, for single zone EV = 1.0 Thus, Vot = Vbz ÷ (EZ · EV) = 135 cfm ÷ (1.0 · 1.0) =135 cfm For St. Louis: = 95°F MWB = 76°F, Wo = 0.0142 lb/lb and for building ti = 75°F, twb = 63°F, Wi = 0.0094 lb/lb qSOA ≈ 1.08 · Q (cfm) · (to – ti) = 1.08 · 135 cfm · (95°F – 75°F) = 2920 Btu/h qLOA ≈ 4680 · Q (cfm) · (Wo – Wi) = 4680 · 135 cfm · (0.0142 – 0.0094) = 3030 Btu/h Problem 8.6

Solution

Compute the total cooling load, sensible cooling load, latent cooling load, and sensible heat ratio of the single office described in Figure C.4* in Appendix C if it is located in St. Louis. Provide results for all three design conditions and compare with estimates given in Table 8.15. Use TideLoad06b.xls or later and enter values as shown below. The main screen input for the maximum dry-bulb (95°F) mean wet-bulb (76°F) condition is shown below.

The main screen input for the maximum wet-bulb (79°F) mean dry-bulb (90°F) condition is shown below.


Compute the heat loss of the single office described in Figure C.4* in Appendix C if it is located in St. Louis and room temperature is 70°F (db). qh = 30.6 MBtu/h (total loss) qh = 18.0 MBtu/h (net loss = total loss – internal heat gain) Note: Problem 8.7 results are also shown in the tables that follow.




--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

The main screen input for the maximum humidity ratio (dry bulb = 85°F, wet bulb = 78°F) condition is shown below.


Zone input values and output results for design dry-bulb and MWB conditions. Zone 1 Morning Afternoon Morning Area 1008 Cooling Cooling Heating 2 Area-ft qc (am) qc(pm) qh 0.00 0.00 90 7.46 2.51 72 2.30 2.80 0.00 0.00 0.00 0.00

For Printout of 8 zones - Use Legal Paper See other sheets for CLTD, CLF, SCL & U-values.

Solar Windows (N) Windows (E) Windows (S) Windows (W) Other

Shade Coeff. SCL (am) SCL (pm) 35 35 143 48 0.58 55 67 0.58 32 140

Enter CLTDs directly from Tables. Program will correct for temperatures.

Conduction Windows (N) Windows (E) Windows (S) Windows (W) Other Other Conduction Walls (N) Walls (E) Walls (S) Walls (W) Other Other Conduction Doors (N) Doors (E) Doors (S) Doors (W) Other Other Roof/Ceiling Type A Type B Floor

CLTD(pm) 14 14 14 14

ΗT

4 4 4 4

ΗT

7 12 8 13

CLTD(pm) 10 28 13 12

CLTD(pm)

ΗT

U (Btu/hr-ft2-F) CLTD(am) 0.87 0.87

U(Btu/hr-ft2-F) 0.075 0.075

U(Btu/hr-ft2-F)

CLTD(am)

CLTD(am)

U(Btu/hr-ft2-F) CLTD(am) CLTD(pm) 7 65 0.033

Latent Lighting Net Sen.w/o duct Ductwork Sensible Latent Total Vent Air Total Sensible Total Latent Total Gain Total Loss


TSP =

0.00 1.33 1.06 0.00 0.00 0.00

0.00 5.32 4.26 0.00 0.00 0.00

0.00 0.37 0.14 0.00 0.00 0.00

0.00 0.77 0.20 0.00 0.00 0.00

0.00 1.68 0.86 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.33 0.00

2.26 0.00

2.26 0.00

2

330 168

2

68 68

Area-ft 1008

Area-ft

2

68 ΗT(slab) Perim.-ft 68 66 ΗT cfm 68 135

2.42

2.82

135

3.20

3.20

29

0.10

0.10

2.74 3.00

3.34 3.00

6.26

7.64

0.00

0.00

3.55 26.66

3.89 28.72

3.46 0.85

3.73 0.78

3.52

135 cfm 30.1 32.4 7.0 7.0 37.2 39.4 0.81 0.82 30.6 NetLoss

10%

1 in. wtr.

Btu/person CLF(am) CLF(pm) 0.73 0.89 250 200 1 1 CLF(am) CLF(pm) 0.73 0.89 3.412

People 15 15 Watts 2515 Btu/h

1 1 CLF(am) CLF(pm) F(ballast) 0.84 0.92 3.412 Insulation Leakage R4 Wrap-Unsealed

(MBtu/h) (MBtu/h) (MBtu/h) SHR (MBtu/h)

0.00 0.55 0.44 0.00 0.00 0.00

Area-ft2

ΗT

ΗT(flr)

U(Btu/hr-ft2-F)

Ins. Position Insulation UP Vertical R5 x 24 in 0.58 Ventilation ΗT(am) ΗT(pm) Sensible 1.1 16.3 19 ΗW ΗW HRU Eff. (sen.) = Latent 4840 0.0049 0.0049 HRU Eff. (lat.) =

People Sensible Latent Internal Sensible

90 72

Area-ft 68 68 68 68

2

68 68 68 68

Slab/Basemt

Vent Air Fan

Area-ft 68 68 68 68

Watts 1 1240 Location S-Ext

Entire Building Totals 135 cfm 10% 30.1 32.4 Sens. 7.0 7.0 Lat. 37.2 39.4 TotGain 0.81 0.82 SHR 30.6 NetLoss 18.0 TotLoss

0.00 2.60

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---


10.10

27.09

18.0

Chapter 8—Cooling Load and Heating Loss Calculations and Analysis

Zone input values and output results for maximum wet-bulb and MDB conditions. Zone 1 Morning Afternoon Morning Area 1008 Cooling Cooling Heating 2 Area-ft qc (am) qc(pm) qh 0.00 0.00 90 7.46 2.51 72 2.30 2.80 0.00 0.00 0.00 0.00






U (Btu/hr-ft2-F) CLTD(am)

CLTD(pm) 14 14 14 14

T

4 4 4 4

T

7 12 8 13

CLTD(pm) 10 28 13 12

CLTD(pm)

T

0.87 0.87

U(Btu/hr-ft2-F) 0.075 0.075

U(Btu/hr-ft2-F)

CLTD(am)

CLTD(am)


U(Btu/hr-ft2-F)

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---


TSP =

0.00 0.70 0.56 0.00 0.00 0.00

0.00 5.32 4.26 0.00 0.00 0.00

0.00 0.17 0.04 0.00 0.00 0.00

0.00 0.57 0.10 0.00 0.00 0.00

0.00 1.68 0.86 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.07 0.00

2.00 0.00

2.26 0.00

2

330 168

2

68 68

Area-ft 1008

Area-ft

2

68 T(slab) Perim.-ft 68 66 T cfm 68 135

1.23

1.63

135

4.41

4.41

29

0.10

0.10

2.74 3.00

3.34 3.00

6.26

7.64

0.00

0.00

3.55 23.78

3.89 25.84

3.09 2.04

3.36 1.67

3.52

135 cfm 26.9 29.2 9.5 9.1 36.3 38.3 0.74 0.76 30.6 NetLoss

11%

1 in. wtr.

Btu/person CLF(am) CLF(pm) 0.73 0.89 250 200 1 1 CLF(am) CLF(pm) 0.73 0.89 3.412 1 1 CLF(am) CLF(pm) F(ballast) 0.84 0.92 3.412 Insulation Leakage R4 Wrap-Unsealed


0.00 -0.08 -0.06 0.00 0.00 0.00

Area-ft2

T

T(flr)

Ins. Position Insulation UP Vertical R5 x 24 in 0.58 Ventilation T(pm) T(am) Sensible 1.1 8.3 11 HRU Eff. (sen.) = W W Latent 4840 0.0068 0.0068 HRU Eff. (lat.) =


90 72

Area-ft 68 68 68 68

2

68 68 68 68

Slab/Basemt

Vent Air Fan

Area-ft 68 68 68 68

People 15 15 Watts 2515 Btu/h Watts 1 1240 Location S-Ext


0.00 2.60 10.10

27.09

18.0




Zone input values and output results for maximum humidity ratio conditions. Zone 1 Morning Afternoon Morning Area 1008 Cooling Cooling Heating 2 Area-ft qc (am) qc(pm) qh 0.00 0.00 90 7.46 2.51 72 2.30 2.80 0.00 0.00 0.00 0.00

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---






CLTD(pm) 14 14 14 14

°T

4 4 4 4

°T

7 12 8 13

CLTD(pm) 10 28 13 12

CLTD(pm)

°T

U (Btu/hr-ft2-F) CLTD(am) 0.87 0.87

U(Btu/hr-ft2-F) 0.075 0.075

U(Btu/hr-ft2-F)

CLTD(am)

CLTD(am)



TSP =

0.00 0.39 0.31 0.00 0.00 0.00

0.00 5.32 4.26 0.00 0.00 0.00

0.00 0.07 -0.01 0.00 0.00 0.00

0.00 0.47 0.05 0.00 0.00 0.00

0.00 1.68 0.86 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

-0.07 0.00

1.86 0.00

2.26 0.00

2

330 168

2

68 68

Area-ft 1008

Area-ft

2

68 ° T(slab) Perim.-ft 68 66 °T cfm 68 135

0.64

1.04

135

5.03

5.03

29

0.10

0.10

2.74 3.00

3.34 3.00

6.26

7.64

0.00

0.00

3.55 22.34

3.89 24.40

2.90 4.22

3.17 2.83

3.52

135 cfm 25.2 27.6 12.2 10.9 37.5 38.4 0.67 0.72 30.6 NetLoss

11%

1 in. wtr.

Btu/person CLF(am) CLF(pm) 0.73 0.89 250 200 1 1 CLF(am) CLF(pm) 0.73 0.89 3.412 1 1 CLF(am) CLF(pm) F(ballast) 0.84 0.92 3.412 Insulation Leakage R4 Wrap-Unsealed


0.00 -0.39 -0.31 0.00 0.00 0.00

Area-ft2

°T

° T(flr)

U(Btu/hr-ft2-F)

Ins. Position Insulation UP Vertical R5 x 24 in 0.58 Ventilation ° T(am) ° T(pm) Sensible 1.1 4.3 7 °W °W HRU Eff. (sen.) = Latent 4840 0.0077 0.0077 HRU Eff. (lat.) =


90 72

Area-ft 68 68 68 68

2

68 68 68 68

Slab/Basemt

Vent Air Fan

Area-ft 68 68 68 68

People 15 15 Watts 2515 Btu/h Watts 1 1240 Location S-Ext


0.00 2.60



10.10

27.09

18.0

Solutions to Chapter 9— Air Distribution System Design Problem 9.1

Room 255 (Appendix C, Figure C.7*) has a 9 ft ceiling and an 18 × 20 ft floor. Select square ceiling diffusers to distribute 1400 cfm of air and provide an ADPI > 90%, an NC < 30, and a TP < 0.1 in. of water. The solution should include a sketch.

Solution

Q = 1400 cfm: The shape of the room lends itself to two diffusers, 700 cfm each, that have to throw air in near equal distance in four directions. A four-way diffuser is appropriate. To achieve an ADPI > 90%, T50/L = 1.4 to 2.7 with louvered ceiling diffusers (Table 9.2) For L = 9 ft: T50 = (T50/L) · L = 1.4 · 9 = 12.6 ft to L = 2.7 · 9 = 24.3 ft For L = 10 ft: T50 = 1.4 · 10 = 14.0 ft to L = 2.7 · 10 = 27.0 ft So the diffuser must have a T50 between 14.0 and 24.3 ft in four directions

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Try a 16 × 16 in. (Table 9.4): at 610 cfm T50 = 20 ft and at 765 cfm T50 = 25 ft Via interpolation: at 700 cfm T50 ≈ 23 ft → OK, and NC = 27 → OK TP = TPRated · (Q/QRated)2 = 0.094 · (700/765)2 = 0.079 in. of water → OK Problem 9.2

Repeat Problem 9.1 for room 256 (Appendix C, Figure C.7*), which has a 40 × 40 ft floor with 2200 cfm.

Solution

*See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.7. Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS



To achieve an ADPI > 90%, T50/L = 1.4 to 2.7 with louvered ceiling diffusers (Table 9.2) For L = 10 ft: T50 = 1.4 · 10 = 14.0 ft to L = 2.7 · 10 = 27.0 ft So the diffuser must have a T50 between 14.0 and 27 ft in four directions Try a 12 × 12 in. (Table 9.4): at 520 cfm T50 = 26 ft and at 610 cfm T50 = 30 ft Via interpolation: at 550 cfm T50 ≈ 27 ft and NC = 32 → too high, but may be OK for VAV Try a 14 × 14 in. (Table 9.4): at 500 cfm T50 = 19 ft and at 625 cfm T50 = 24 ft Via interpolation: at 550 cfm T50 ≈ 21 ft → OK, and NC = 25 → OK TP = TPRated · (Q/QRated)2 = 0.058 · (550/500)2 = 0.07 in. of water → OK Problem 9.3

Solution

Size a MERV 6 filter and a matching filter-grille for room 255 (Appendix C, Figure C.7*) that will limit the final resistance to 0.5 in. of water. One dimension should be a multiple of 24 in. if possible. ΔhFinal ≤ 0.5 in. w.c. and recommendation is ΔhFinal ≈ 4 · ΔhInitial Thus, Δhinitial = Δhfinal/4 = 0.5/4 = 0.125 For a MERV 6, 2 in. thick, pleated filter at 300 fpm, ΔhInitial = 0.13 in. w.c. (Table 9.7) V = VRated · (ΔhInitial/ΔhRated)0.5 = 300 · (0.125/0.13)0.5 = 294 fpm A = Q ÷ V = 1400 cfm ÷ 294 fpm = 4.76 ft2 W = 24 in. = 2.0 ft: H = 4.76 ft2 ÷ 2 ft = 2.38 ft = 29 in. Use 24 in. · 30 in. grille (Table 9.6): TP = 0.024 in. w.c. @ 1431 cfm TPGrille = TPRated · (Q/QRated)2 = 0.024 · (1400/1431)2 = 0.023 in. w.c


Repeat Problem 9.3 for room 256 (Appendix C, Figure C.7*). ΔhFinal ≤ 0.5 in. w.c. and recommendation is ΔhFinal ≈ 4 · ΔhInitial, thus ΔhInitial ≈ ΔhFinal ÷ 4 ≈ 0.5 ÷ 4 ≈ 0.125 in. w.c. For a MERV 6, 2 in. thick, pleated filter at 300 fpm, ΔhInitial = 0.13 in. w.c. (Table 9.7) V = VRated · (ΔhInitial/ΔhRated)0.5 = 300 · (0.125/0.13)0.5 = 294 fpm A = Q ÷ V = 2200 cfm ÷ 294 fpm = 7.5 ft2 W = 24 in. = 2.0 ft: H = 7.5 ft2 ÷ 2 ft = 3.75 ft = 45 in. Use 24 in. · 48 in. grille (Table 9.6): TP = 0.024 in. w.c. @ 2307 cfm TPGrille = TPRated · (Q/QRated)2 = 0.024 · (2200/2307)2 = 0.022 in. w.c.

Problem 9.5

Select a unit to either fit in the hallway outside room 255 (Appendix C, Figure C.7*) or above the ceiling (42 in. vertical space) and route metal supply ductwork with round take-offs and metal return ductwork.

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Solution

42




Chapter 9—Air Distribution System Design

Supply duct: 1400 cfm section use 16 in. Ø duct Δh/100 ft = 0.096 in. /100 ft Δh = Δh/100 ft · [L + Leqv] = 0.096/100 ft · [5 ft + 12 ft + 32 ft + 21 ft] = 0.07 ft Straight, plenum, 90°L

700 cfm section use 12 in. Ø duct Δh/100 ft = 0.10 in. /100 ft Δh = 0.10/100 ft · [20 ft + 5 ft + 5 ft + 15 ft + 35 ft] = 0.08 in. Straight, reducer, 45°L, 90°L, ceiling box

Return: 1400 cfm through a short 30 × 24 in. section Δh is negligible Problem 9.6 Solution

Repeat Problem 9.5 for room 256 (Appendix C, Figure C.7*). The solution is demonstrated in the figure below.

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Problem 9.7

Use the E-Ductulator program on the CD accompanying this book to design an air distribution system to deliver 6000 cfm evenly in the building below. Provide a MERV 6 filter and limit total losses to less than 1.2 in. of water.

Solution

The solution is demonstrated in the figure below and in the spreadsheet sample on the following page.

*See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.7. Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS


43


--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Summary of design calculations using E-Ductalator.xls: E-Ductulator - Equal Friction - Equivalent Length Method - See Eqv. Len. Worksheet for Equivalent Lengths of Common Fittings Duct Width or Height Supply Air Flow Dia. if round 0 if round cfm Inches Inches 2200 18

Velocity Hyd. Dia. Roughness fpm Inches Feet 1245 18.0 Galv Steel (0.00025)

1100

14

1029

14.0 Galv Steel (0.00025)

550

12

700

12.0 Galv Steel (0.00025)

1

16.0 Galv Steel (0.00025)

2

10.0 Galv Steel (0.00025)

1

16 For Sizing

1

Par. Path 10

For Sizing

Par. Path

1

1

183

1.0 Galv Steel (0.00025)

1

1

183

1.0 Galv Steel (0.00025)

Diffuser cfm Rated cfm 550 500 Duct Width or Height Return cfm Dia. if round 0 if round 2200 28 20 1

1

Δh/100' "Wtr./100' 0.103156

Length Leqv A # As Leqv B # Bs Leqv C # Cs Leqv D # Ds L(total) Δh ft. ft. ft. ft. ft. (in.wtr.) 17 37 1 24 54 0.056 Notes--> Plenum 90 L 0.098927 12 10 1 22 0.022 Notes--> Red 0.059537 18 13 1 15 1 35 1 81 0.048 Notes--> Wye 90 L topbox 2.11E-07 0 0.000 Notes--> 1.98E-06 0 0.000 Notes--> 0.114782 0 0.000 Notes--> 0.114782 0 0.000 Notes-->

Rated Δh 0.058

566 183

25.8 Galv Steel (0.00025) 1.0 PVC (0.0001)

0.070

Δh/100' 0.01605

Length Leqv A # As Leqv B # Bs Leqv C # Cs Leqv D # Ds L(total) 15 18 1 24 2 50 1 131 Notes--> Gr,trans Rad. L Plen Ret. 0.114484 0 Notes-->

0.021 0.000

Grill cfm

Notes: Rated cfm Rated Δh 2200 2307 0.024 48" x 24" Rated Vel Face Vel Rated Δh Notes: Δh final Filter cfm Area (sq. ft.) fpm fpm in. water 2" MERV 6 Pleated Δh clean 2200 8 300 275 0.13 0.13" @300 fpm (clean) 3

0.022

Total

0.328 0.566

Problem 9.8

Select a fan with a direct-drive motor to provide 1200 cfm at a TSP of 0.8 in. of water. Specify the required speed tap setting, resulting motor power output, and estimated required power demand.

Solution

Option 1 (using Table 9.12): A 1/3 hp motor at high speed (1075 room) will deliver 1540 cfm @ 0.7 in. w.c. and 1080 cfm @ 0.9 in. w.c. Via interpolation for TSP = 0.8 in. w.c. → Q = 1310 cfm (at medium-high flow rate would be 1070 cfm @ 0.8 in. w.c.) Wfan = 0.33 hp and from Table 11.5, ηMotor = 63% ≈ 63% Thus, WM,In = 0.746 · Wreq’d (hp) ÷ (ηMotor · ηVSD) = 0.746 kW/hp · 0.33 hp ÷ (0.63 · 1.0) = 0.39 kW Option 2 (using Table 9.12): A 1/2 hp motor at medium-low speed will deliver 1240 cfm @ 0.7 in. w.c. and 1080 cfm @ 0.9 in. w.c. Via interpolation for TSP = 0.8 in. w.c. → Q = 1205 cfm At medium-low speed Wfan = 0.25 hp; no efficiency data provided but Table 11.3 indicates the part-load factor at 50% load (0.25 hp/0.5 hp) of a small motor is 0.86. From Table 11.5, ηMotor ≈ 70% at full load WM,In = 0.746 · Wreq’d (hp) ÷ (ηMotor · fPL) = 0.746 kW/hp · 0.25 hp ÷ 0.70 · 0.86 = 0.31 kW



Chapter 9—Air Distribution System Design

Solution

Select a belt-drive fan and motor to provide 2000 cfm at a TSP of 1.2 in. of water. Specify the required fan pulley diameter for a 1750 rpm motor with a 6 in. diameter drive pulley, resulting motor power output, and estimated required power demand. A 1.5 hp fan will deliver 1930 cfm @ 1.2 in. w.c. (too small) A 2.0 hp fan will deliver 2050 cfm @ 1.2 in. w.c. for: A fan speed of 1150 rpm Required 1.5 BHP (which can be delivered by a 2.0 hp motor @ 75% load) From Table 11.2 ηMotor = 84% WM,In = 0.746 · Wreq’d (hp) ÷ (ηMotor · fPL) = 0.746 kW/hp · 1.5 hp ÷ (0.84 · 1.0) = 1.33 kW DFanPulley = DMotorPulley · (rpmM ÷ rpmFan) = 6 in. · (1750 ÷ 1150) = 9.13 in. ≈ 9 1/8in.

Problem 9.10

Select a belt-drive fan and motor to provide 7500 cfm at a TSP of 3.25 in. of water. Specify the required fan pulley diameter for a 1750 rpm motor with an 8 in. diameter drive pulley, resulting motor power output, and estimated required power demand.

Solution

Interpolating between 3.0 in. w.c and 3.5 in. w.c. for 7500 cfm for a 17R (24.5 in.) fan rpm = 1955, BHP = 5.9 hp @ 3.25 in. w.c. Use a 7.5 hp motor, ηMotor = 88% WM,In = 0.746 · Wreq’d (hp) ÷ (ηMotor · fPL) = 0.746 kW/hp · 5.9 hp ÷ (0.88 · 1.0) = 5.0 kW DFanPulley = DMotorPulley · (rpmM ÷ rpmFan) = 8 in. · (1750 ÷ 1955) = 7.16 in. ≈ 7-1/8 in.

Problem 9.11

Select a motor and specify the resulting bhp and fan speed to provide 1.2 in. of water external static pressure (ESP) and 6000 cfm for a Model 180 (Table 5.2 of this book) rooftop unit. Provide the fan pulley diameter for a 1750 rpm motor with a 4 in. diameter drive pulley.

Solution

Interpolation at 6000 cfm for 1.2 in. w.c. M180 6000 cfm

ESP = 1.0 in. rpm kW 1100 3.32

bhp 3.89

ESP = 1.5 in. rpm kW 1236 4.09

bhp 4.79

rpm = 1154, kW = 3.63, bhp = 4.25 → need 5 hp motor (4.25 ÷ 5 = 85% load) From Table 11.2 ηMotor = 87.5%, from Table 11.3 fPL = 1.0 WM,In = 0.746 · Wreq’d (hp) ÷ (ηMotor · fPL) = 0.746 kW/hp · 4.25 hp ÷ (0.875 · 1.0) = 3.63 kW (note: same as value in table) DFanPulley = DMotorPulley · (rpmM ÷ rpmFan) = 4 in. · (1750 ÷ 1154) = 6.06 in. ≈ 6-1/16 in. Problem 9.12

A centrifugal fan has the following characteristics at 1000 rpm: Δh = 2.96 in. of water at 2400 cfm Δh = 2.94 in. of water at 4800 cfm Δh = 2.57 in. of water at 7200 cfm Δh = 2.09 in. of water at 9600 cfm Δh = 1.35 in. of water at 12000 cfm Develop a fan curve for 1000 rpm and calculate the required input power (hp) assuming a 75% efficiency at 7200 cfm, 70% at 4800 and 9600 cfm, and 65% at 2400 and 12,000 cfm. Calculate the required motor input (kW) assuming a 93% efficient motor.

Solution



--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

Problem 9.9


Total Static pressure (Inches of Water)

Fan Performance Curve 3.5 3 2.5 2 1.5 1 0.5 0 0

4000

8000

12000

16000

Air Flow Rate (cfm)


Using the fan laws, develop fan curves for 800 and 600 rpm. Prob 9.13 1000 rpm Q - cfm 2400 4800 7200 9600 12000

800 rpm Q=(800/1000)Q1000

Δh "wtr. 2.96 2.94 2.57 2.09 1.35

Δh=(800/1000)2Δh1000

600 rpm Q=(800/1000)Q1000

Δh=(800/1000)2Δh1000

Q at 800 rpm - cfm Δh at 800 rpm - "wtr. Q at 600 rpm - cfm Δh at 600 rpm - "wtr. 1920 1.89 1440 1.07 3840 1.88 2880 1.06 5760 1.64 4320 0.93 7680 1.34 5760 0.75 9600 0.86 7200 0.49

Problem 9.14

The fan described in Problem 9.12 is connected to an air distribution system that has a loss of 1.8 in. of water at 9500 cfm when the dampers are open and 3.0 in. of water at 5000 cfm when the dampers are set at a minimum opening. Develop a system curve for both situations (dampers full open and minimum) and find the resulting flow when the fan is turning 1000, 800, and 600 rpm for both situations (dampers full open and minimum). Estimate the required power at all six points. Note: efficiency will remain nearly constant with varying fan speed when the fan law [Δh2 = Δh1 × (rpm2/rpm1)2] is applied.

Solution

Fan curve lines represent results of Problem 9.13. System curve for minimum damper found using: Δh = 3.0 · (Q/5000)2 = 3.0 · (4000/5000)2 = 1.92 in. and Δh = 3.0 · (3000/5000)2 =1.08 ft System curve for damper open found using: Δh =1.8 · (Q/9500)2 = 1.8 · (8000/9500)2 = 1.28 in. and Δh = 1.8 · (6000/9500)2 = 0.72 in. 70% 75%

Eff. Curves

100 0 rp m

2.0

800 r pm

Da mp er

Δh - in. water.

3.0

Mi mi mu m

--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---

65%

1.0

2000

4000

600 r pm

6000 Q - cfm

70%

System Curves

Fan Curves

65%

en Op per Dam

8000

10000

12000

WOpen, 1000 rpm= (9500 · 2.2)/(0.69 · 6350) = 4.77 hp WOpen, 800 rpm= (8200 · 1.2)/(0.68 · 6350) = 2.28 hp WOpen, 600 rpm= (6000 · 0.8)/(0.69 · 6350) = 1.10 hp WMinimum, 1000 rpm= (5000 · 2.9)/(0.70 · 6350) = 3.26 hp WMinimum, 800 rpm= (3900 · 1.9)/(0.70 · 6350) = 1.67 hp WMinimum, 600 rpm= (3000 · 1.1)/(0.70 · 6350) = 0.74 hp



Problem 10.1

Solution

Size the piping (Schedule 40 steel) and compute the system head loss for the (direct-return) chilled water system shown in Figure 10.1 (and Figure 5.10). The distance between each FCU is 50 ft, and the distance between the last FCU and the headers is 30 ft. The distance between the first FCU and the chiller is 120 ft. The chiller head loss is found using the specifications of a Model 060 scroll compressor design (Table 5.10). The fan-coils units are Model 60-HW-4 (Table 5.12). Use ball valves for all valves 2 in. and smaller and gate valves for larger valves. E-Pipelator.xls results for “old steel” pipe and 50°F water: Input

Piping Loop Head Loss Calculator - System Designer for HVAC Systems Liquid Water Temp Den Vis

Flow gpm 0 0 0 0 0 0 Flow gpm 160 120 80 40 20 0

Coils 50 ºF 62.38 lbm/ft3 8.88E-04 lbm/ft-s

HDPE Piping Nom. Dia. Inches 4 1.5 2 1.5 0.75 0.75 Steel/Brass Nom. Dia. Inches 4 3 3 2 1.5 3

DR I.D. Roughness OD ÷ t in. (for HDPE in ft.) 11 3.68 0.00007 11 1.55 0.00007 11 1.94 0.00007 11 1.55 0.00007 11 0.86 0.00007 11 0.86 0.00007 Schedule I.D. 40 or 80 in. 40 4.03 40 3.07 40 3.07 40 2.07 40 1.61 40 3.07

Pipe Mat'l for Rghness in ft. Steel-Old Steel-Old Steel-Old Steel-Old Steel-Old Steel-Old

Flow gpm 20 160 0

FCU Chiller

1.32 cps

Vel fps 0.0 0.0 0.0 0.0 0.0 0.0

Re

Vel fps 4.0 5.2 3.5 3.8 3.2 0.0

Re

Rated Flow gpm 21 150 0

Rated Δh @ 60ºF ft. water 10 15 0

Inlet Size inches 1.5 4 0

Inlet Vel fps 3.6 4.1 0.0

Optional Input

Re(in) 31892 95677 0

Rated Vel fps 3.8 3.8 0.0

0 0 0 0 0 0

Δh(ft) Length Fitting Selector 100 ft. ft. 0.00 160 Butt90 0.00 390 Butt90 0.00 0 Butt90 0.00 0 Butt90 0.00 0 Butt90 0.00 0 Butt90

Leqv Qty. Fitting Selector Leqv ft ft 35 2 ButtRed 13 12 2 ButtRed 6 17 2 ButtRed 7 12 2 ButtRed 6 5 2 ButtRed 4 5 2 ButtRed 4

95059 93556 62371 46288 29713 0

Δh(ft) Length Fitting Selector 100 ft. ft. 1.87 240 T-Straight 4.34 100 T-Straight 1.98 100 T-Straight 3.97 100 T-Straight 3.79 60 90 L 0.00 0 T-Straight

Leqv Qty. Fitting Selector Leqv ft ft 6.8 4 Gate Valve 5.7 5.4 2 Reducer 1.8 5.4 4 Reducer 1.8 3.4 4 Reducer 1.1 2.1 4 Reducer 0.8 5.4 2 Reducer 1.8

Other Fittings & Valves ball valve Zone

Flow gpm 20 20 0

Cv @ 60ºF gpm 81 27.5 0

Quanity 2 1 0

Inlet Size inches 1.5 1.5 2


Re(in) 31892 31892 0 Open Systems

Rated Vel fps 14.7 5.0 0.0 Only

Output

Re(rated)

Δh Ft. Liquid 38381 9.2 102808 17.2 0 0.0 Coil sub-total 26.4 Qty. Fitting Selector Leqv Qty. Δh ft Ft. Liquid 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 0.0 2 5-LoopHdrLastTO 30 2 5-LoopHdrLastTO 30 2 0.0 HDPE sub-total 0.0 Qty. Fitting Selector Leqv Qty. Δh ft Ft. Liquid 2 T-Straight 6.8 4 5.7 2 90 L 4.5 0 5.0 2 90 L 4.5 0 2.5 2 90 L 2.8 0 4.6 2 90 L 2.1 0 2.7 0 90 L 4.5 0 0.0 Fe/Br sub-total 20.4 Re(rated) 148043 50261 0 Fitting sub-total ? ? Elevation Total Loss

Δh Ft. Liquid 0.3 1.3 0.0 1.6 0 48.4

Graph (Figure 10.8) and calculation (Equations 10.4 and 10.7) method for “new steel” pipe and 60°F water (the abbreviations following the equivalent length values correspond to the fittings listed in the drop-down boxes in the spreadsheet): 160 gpm section use 4 in., schedule 40 pipe → Δh = 1.6 ft/100 ft Δh160 = 1.6 ft/100 ft · [2 · 120 ft + 4 · 5.7 ft (90°Ls) + 2 · 5.7 (gate v.) + 4 · 6.8 (T-str.)] = 4.8 120 gpm section use 3 in., schedule 40 pipe → Δh = 3.5 ft/100 ft 4.5 Δh120 = 3.5 ft/100 ft · [2 · 50 ft + 2 · 3.5 ft (Red.) + 4 · 5.2 ft (T-str.)] = 80 gpm section use 3 in., schedule 40 pipe → Δh = 1.8 ft/100 ft Δh80 = 1.8 ft/100 ft · [2 · 50 ft + 4 · 5.2 (T-str.)] = 2.2 40 gpm section use 2 in., schedule 40 pipe → Δh = 3.3 ft/100 ft Δh40 = 3.3 ft/100 ft · [2 · 50 ft + 2 · 3.5 ft (Red.) + 4 · 3.4 (T-str.)] = 4.0 20 gpm section use 1 1/2 in., schedule 40 pipe → Δh = 3.0 ft/100 ft Δh20 = 3.0 ft/100 ft · [2 · 30 ft + 4 · 3.4 (90°Ls)] = 2.1 Δhvalves = 2 · 2.31 · (20 gpm/81)2 · (1 1/2 in. ball) + 2.31 · (20 gpm/27.5)2 · (1 1/2 in. zone) = 1.5 ΔhFCU = 10 ft (from Table 5.12) · (20 gpm/21 gpm)2 = 9.1 16.8 ΔhChiller @ 160 gpm (from Figure 5.12) = ΔhTotal = 45.0 Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS


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Solutions to Chapter 10— Water Distribution System Design


Problem 10.1 with pipe sizes

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Problem 10.2

Select a chilled water pump and corresponding motor for the system described in Problem 10.1.

Solution

The system requires a pump that will deliver a 160 gpm flow rate and 45 ft of head, assuming the water treatment program will be good to maintain the condition of the pipe. In this case, a Model #2-1/2 AB pump (Figure 10.14) with a 7 in. diameter impeller will provide 46.5 ft of head at 160 gpm. It will require approximately 2.8 bhp, so a 3 hp, 1750 rpm motor will suffice. Note the selected pump will operate near 73% efficiency, which is near the maximum efficiency point (MEP) of 75.5%. If the water treatment program is uncertain, the old steel pipe roughness and head loss (48.4 ft) should be assumed. Although a Model 5x5x9-3/4B pump with a 7-3/4 in, impeller will provide adequate flow and head, it is not a good choice. Note the efficiency at the resulting operating point is only about 45%—well below its MEP of 79.5%. A more extensive set of pump curves should be consulted to select a pump that will operate at a more favorable efficiency.

Problem 10.3

Solution

Size the piping, compute the required head, and select a pump and motor for the condenser water loop shown in Figure 10.3. Use the Model 060 chiller from Problem 10.1 with a flow rate of 200 gpm. Use SDR 11 high-density polyethylene to eliminate corrosion in this open loop. The distance between the basin and upper tray in the cooling tower is 12 ft. Graph (Figure 10.9) and calculation (Equation 10.4) method for HDPE pipe: 200 gpm use 4 in., DR11 pipe → Δh = 3.3 ft/100 ft Δh160 = 3.3 ft/100 ft · [2 · 250 + 12 ft + 6 · 38 ft (90°Ls) + 4 · 5.7 (gate v.)] = Elevation = ΔhCondenser @ 200 gpm (from Figure 5.15) = ΔhTotal =

25.2 12.0 12.5 49.0 ft.

The Model 5x5x9-3/4B pump with a 7-3/4 in. impellor will provide 54 ft at 200 gpm but is not a good choice. Note the efficiency at the resulting operating point is only about 56%, well below its MEP of 79.5%. A more extensive set of pump curves should be consulted to select a pump that will operate at a more favorable efficiency. If the pump is used, a 5 hp/1750 rpm motor is acceptable.



Chapter 10—Water Distribution System Design

Problem 10.4

Solution Problem 10.5

Design the chilled water piping loop for the system shown using Schedule 40 steel pipe and gate valves on the main piping, with two-ball valves and one motorized zone valve (Cv = 18) on the fan-coil loops. See solution beneath Problem 10.5 solution. Compute the required head loss and select a chilled water pump for Problem 10.5.

Solution

Input

Piping Loop Head Loss Calculator - System Designer for HVAC Systems Liquid Water Temp Den Vis

Flow gpm 0 0 0 0 0 0 Flow gpm 70 50 35 20 15 0

Coils 50 ºF 62.38 lbm/ft3 8.88E-04 lbm/ft-s

HDPE Piping Nom. Dia. Inches 1.5 1.5 2 1.5 0.75 0.75 Steel/Brass Nom. Dia. Inches 3 3 2 1.5 1.25 3

DR I.D. Roughness OD ÷ t in. (for HDPE in ft.) 11 1.55 0.00007 11 1.55 0.00007 11 1.94 0.00007 11 1.55 0.00007 11 0.86 0.00007 11 0.86 0.00007 Schedule I.D. 40 or 80 in. 40 3.07 40 3.07 40 2.07 40 1.61 40 1.38 40 3.07

Pipe Mat'l for Rghness in ft. Steel-Old Steel-Old Steel-Old Steel-Old PVC PVC

For sizing pipe for 15 gpm coils - parallel flow no losses added (L= 0)

Flow gpm 20 70 0

60HW4 Chiller

1.32 cps

Vel fps 0.0 0.0 0.0 0.0 0.0 0.0

Re

Vel fps 3.0 2.2 3.3 3.2 3.2 0.0

Re

Rated Flow gpm 21 80 0

Rated Δh @ 60ºF ft. water 10 7.5 0

Inlet Size inches 1.5 3 0


Optional Input

Re(in) 31892 55811 0

Rated Vel fps 3.8 3.6 0.0

0 0 0 0 0 0

Δh(ft) Length Fitting Selector 100 ft. ft. 0.00 160 Butt90 0.00 390 Butt90 0.00 0 Butt90 0.00 0 Butt90 0.00 0 Butt90 0.00 0 Butt90

Leqv Qty. Fitting Selector Leqv ft ft 12 2 ButtRed 6 12 2 ButtRed 6 17 2 ButtRed 7 12 2 ButtRed 6 5 2 ButtRed 4 5 2 ButtRed 4

54574 38982 40502 29713 25999 0

Δh(ft) Length Fitting Selector 100 ft. ft. 1.53 120 90 L 0.81 120 T-Straight 3.07 100 T-Straight 3.79 200 90 L 3.62 0 T-Straight 0.00 0 T-Straight

Leqv Qty. Fitting Selector Leqv ft ft 4.5 2 Gate Valve 4.5 5.4 2 Reducer 1.8 3.4 2 Reducer 1.1 2.1 2 Reducer 0.8 2.2 0 Reducer 0.7 5.4 2 Reducer 1.8

Other Fittings & Valves ball valve Zone valve

Flow gpm 20 20 0

Cv @ 60ºF gpm 81 27.5 0

Quanity 2 1 0

Inlet Size inches 1.5 1.5 2


Re(in) 31892 31892 0 Open Systems

Rated Vel fps 14.7 5.0 0.0 Only

Output

Re(rated)

Δh Ft. Liquid 38381 9.2 73108 5.9 0 0.0 Coil sub-total 15.1 Qty. Fitting Selector Leqv Qty. Δh ft Ft. Liquid 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 2 5-LoopHdrLastTO 30 2 0.0 2 0.0 2 5-LoopHdrLastTO 30 2 0.0 2 5-LoopHdrLastTO 30 HDPE sub-total 0.0 Qty. Fitting Selector Leqv Qty. Δh ft Ft. Liquid 3 T-Straight 5.4 2 2.4 0 90 L 4.5 0 1.1 2 90 L 2.8 0 3.3 2 90 L 2.1 0 7.8 0 90 L 1.8 0 0.0 0 90 L 4.5 0 0.0 Fe/Br sub-total 14.5 Re(rated) 148043 50261 0 Fitting sub-total ? ? Elevation Total Loss

Δh Ft. Liquid 0.3 1.3 0.0 1.6 0 31.2

Pump requirement: 31.2 ft of head and 70 gpm A model 2AC pump with a 6 in. diameter impeller operating at 1750 rpm will provide 34 ft of water at 70 gpm. The pump will require slightly more than 1.0 bhp, so a 1.5 hp, 1750 rpm motor is recommended to avoid overload.

49

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Solutions to Chapter 11— Motors, Lighting, and Controls Problem 11.1

Solution

An air system design requires 6600 cfm with 3.0 in. of water (static pressure). Find the required motor size, drive pulley diameter (if the blower wheel has a 12 in. diameter pulley and the motor is 4-pole), fan efficiency, and motor demand. A 17R fan (Table 9.13) will deliver: 6000 cfm and 3.0 in. TSP when operating at 1730 rpm and requiring 4.33 bhp 7500 cfm and 3.0 in. TSP when operating at 1920 rpm and requiring 5.68 bhp Via linear interpolation to the operating requirement: 6600 cfm and 3.0 in. TSP when operating at 1810 rpm and requiring 4.9 bhp 6600 (cfm) ⋅ 3.0 (in. water) Q (cfm) ⋅ Δp (in. water) η fan = --------------------------------------------------------- = ------------------------------------------------------------------ = 0.636 = 63.6% 6350 ⋅ 4.9 6350 ⋅ bhp

Use 5 hp, 1725 rpm motor (ηMotor = 86.5%) from Table 11.5 Note: rpm ≈ 7200 ÷ no. of poles ≈ 7200 ÷ 4 ≈ 1800 rpm rpm Fan 1810 D Motor = D Fan ⋅ ----------------------- = 12 in. ⋅ ------------ = 12.6 in. = 12 5/8 in. 1725 rpm Motor 0.746 kW/hp ⋅ W Req'd (hp) 0.746 kW/hp ⋅ 4.9 (hp) W MotorIn = ---------------------------------------------------------------- = ------------------------------------------------------- = 4.2 kW 0.865 ⋅ 1.0 η Motor ⋅ f PL

Problem 11.2

Solution

Compute the demand, KVA, and KVAR of a 6-pole, 20 hp motor at 100%, 75%, and 50% load. rpm ≈ 7200 ÷ no. of poles ≈ 7200 ÷ 6 ≈ 1200 rpm, 20 hp From Table 11.2, ηMotor = 91.0% From Table 11.3, fPL-100% = 1.0, fPL-75% = 1.0, fPL-50% = 0.99 From Table 11.4, PF100% = 0.84, PF75% = 0.81, PF50% = 0.74 For 100% load, W MotorIn 0.746 kW/hp ⋅ 20 (hp) 16, 400 W W MotorIn = ------------------------------------------------------ = 16.4 kW, I = ---------------------------- = ----------------------------------- = 49.0 amps 0.91 ⋅ 1.0 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.84

For 75% load, W MotorIn 0.746 kW/hp ⋅ 0.75 ⋅ 20 (hp) 12, 300 W W MotorIn = --------------------------------------------------------------------- = 12.3 kW, I = ---------------------------- = ----------------------------------- = 38.1 amps 0.91 ⋅ 1.0 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.81




For 50% load, W MotorIn 8, 280 W 0.746 kW/hp ⋅ 0.50 ⋅ 20 (hp) W MotorIn = --------------------------------------------------------------------- = 8.28 kW, I = ---------------------------- = ----------------------------------- = 28.1 amps 0.91 ⋅ 0.99 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.74


Compute the demand, KVA, and KVAR of a 2-pole, 15 hp motor at 100%, 75%, and 50% load. rpm ≈ 7200 ÷ no. of poles ≈ 7200 ÷ 2 ≈ 3600 rpm, 15 hp From Table 11.2, ηMotor = 89.5% From Table 11.3, fPL-100% = 1.0, fPL-75% = 1.0, fPL-50% = 0.99 From Table 11.4, PF100% = 0.89, PF75% = 0.88, PF50% = 0.84 For 100% load, W MotorIn 12, 500 W 0.746 kW/hp ⋅ 15 (hp) W MotorIn = ------------------------------------------------------ = 12.5 kW, I = ---------------------------- = ----------------------------------- = 17.6 amps 0.895 ⋅ 1.0 3 ⋅ E ⋅ PF 3 ⋅ 460 ⋅ 0.89

For 75% load, W MotorIn 0.746 kW/hp ⋅ 0.75 ⋅ 15 (hp) 9380 W W MotorIn = --------------------------------------------------------------------- = 9.38 kW, I = ---------------------------- = ----------------------------------- = 13.4 amps 0.895 ⋅ 1.0 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.88

For 50% load, W MotorIn 0.746 kW/hp ⋅ 0.50 ⋅ 15 (hp) 8, 280 W W MotorIn = --------------------------------------------------------------------- = 6.31 kW, I = ---------------------------- = ----------------------------------- = 9.4 amps 0.895 ⋅ 0.99 3 ⋅ E ⋅ PF 3 ⋅ 460 ⋅ 0.84

Problem 11.4

Solution

A 30 DE-25 in. fan is operated at 1195 rpm to deliver 10,000 cfm at 3.5 in. of water. Select an 1800 rpm motor to drive this fan and specify resulting demand (kW), KVA, and KVAR at the design point and at 6,000 cfm. 30 DE-25 in. fan at 1195 rpm, 10,000 cfm, and 3.5 in. TSP bhp = 7.84, use 10 hp motor @ 78% load From Table 11.2, ηMotor = 89.5% From Table 11.3, fPL-78% = 1.0 From Table 11.4, PF78% = 0.83 For 78% load, W MotorIn 0.746 kW/hp ⋅ 0.78 ⋅ 10 (hp) 6500 W W MotorIn = --------------------------------------------------------------------- = 6.5 kW, I = ---------------------------- = ----------------------------------- = 19.7 amps 0.895 ⋅ 1.0 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.83


Repeat Problem 11.4 if a motor one size larger than required is specified. For a 30 DE-25 in. fan at 1195 rpm, 10,000 cfm, and 3.5 in. TSP bhp = 7.84, but use 15 hp motor @ 52% load From Table 11.2, ηMotor = 91.0% From Table 11.3, fPL-52% = 0.99 From Table 11.4, PF52% = 0.78 For 78% load, W MotorIn 0.746 kW/hp ⋅ 0.52 ⋅ 15 (hp) 6500 W W MotorIn = ---------------------------------------------------------------------- = 6.5 kW, I = ---------------------------- = ----------------------------------- = 20.8 amps 0.91 ⋅ 0.99 3 ⋅ E ⋅ PF 3 ⋅ 230 ⋅ 0.78


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Problem 11.6

Solution

A pump that operates 3000 hours per year and requires 400 gpm and 55 ft of head is significantly oversized. The curve of the existing pump is shown in Figure 10.12. The impellor diameter is 9.75 in. and the motor is 20 hp. The required flow rate is attained by throttling the pump discharge valve. a. Estimate the required horsepower and compute the resulting motor efficiency and power demand at the throttled position. b. Estimate the required horsepower and compute the resulting motor efficiency and power demand for the 20 hp motor if the pump impellor was trimmed to a size shown on the curve that provides the needed flow and head. c. Estimate the required horsepower and compute the resulting motor efficiency and power demand for the adequately sized motor if the pump impellor was trimmed. d. Estimate the annual energy consumption for the three above options. For a cost of $500 to trim the impellor and $1000 to purchase and replace the motor, estimate the simple payback for the options in 11.6b and 11.6c. a. For a Model 5x5x9-3/4 pump with a 9-3/4 in. impeller: @ 400 gpm, Δh = 88 ft, ηpump = 73% Wreq’d = 400 · 88 ft (3960 · 0.73) = 12.2 hp For 20 hp motor, %Load = 12.2 hp ÷ 20 hp = 61% ηMotor = 91%, fPL = 1.0 WMotorIn = 0.746 · 12.2 ÷ (0.91 · 1.0) = 10.0 kW b. For a Model 5x5x9-3/4 pump with an 8-1/4 in. impeller: @ 400 gpm, Δh = 55 ft, ηpump = 75% Wreq’d = 400 · 55 ft (3960 · 0.75) = 7.4 hp For 20 hp motor, %Load = 7.4 hp ÷ 20 hp = 37% ηMotor = 91%, fPL = 0.96 WMotorIn = 0.746 · 7.4 ÷ (0.91 · 0.96) = 6.3 kW c. For a Model 5x5x9-3/4 pump with an 8-1/4 in. impeller and either 7.5 or 10 hp motor: @ 400 gpm, Δh = 55 ft, ηpump = 75% Wreq’d = 400 · 55 ft (3960 · 0.75) = 7.4 hp For 7.5 hp motor, ηMotor = 88.5%, fPL = 1.0 WMotorIn = 0.746 · 7.4 ÷ (0.885 · 1.0) = 6.2 kW d. For a., E = 3000 h · 10 kW = 30,000 kWh For b., E = 3000 h · 6.3 kW = 18,900 kWh For c., E = 3000 h · 6.2 kW = 18,600 kWh

Problem 11.7

Solution

Design a lighting system for a 30 × 40 ft classroom using 2-bulb, 48-in. T-8 fluorescent lighting fixtures with electronic ballasts. Classroom illumination is D or E (Table 11.8). Try higher level first (E, illumination = 750 lumens/m2 or 75 lumens/ft2). A 48-in.-long T-8 lamp, 2710 lumens (mean), 31 W (Table 11.11) Each two-bulb fixture provides 5420 lumens and requires 62 W. Luminaries = 75 lumens/ft2 · 1200 ft2 ÷ 5420 lumens/fixture = 16.6 fixtures → Use 17 or 18



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Chapter 11—Motors, Lighting, and Controls


Compare the demand of the resulting design for problem 11.7 with ASHRAE Standard 90.1-2004 limits for this application. For 18 bulbs: W = 18 fixtures · 62 W/fixture = 1116 W LPD = 1116 W ÷ 1200 ft2 = 0.93 W/ ft2 From Table 11.10, LPDClassroom = 1.4 W/ ft2 Design complies with ASHRAE Standard 90.1-2004 since 0.93 W/ft2 < 1.4 W/ft2.

Problem 11.9

A 1000 ft2 storage area is currently lit with standard 100-W incandescent bulbs to an illumination of 30 footcandles for 80 hours per week. Compare the annual operating cost of using existing bulbs and replacing them with equivalent lighting output compact fluorescent bulbs. Include operating cost (at 8¢/kWh), cost of bulbs (at $3.00 each), and installation cost (1 hour at $20/hour labor).

Solution

For 100-W incandescent bulbs that provide 1710 lumens each with an average life of 750 hours: Luminaries = 30 lumens/ft2 · 1000 ft2 ÷ 1710 lumens/bulb = 18 bulbs W = 18 · 100 W = 1800 W 1800 W 18 bulbs ⋅ $0.30 $20/h Annual Cost = 80 h ⋅ 52 weeks ----------------------------- ⋅ 0.08 $/kW + --------------------------------------- + ------------- = $740 yr 1000 W/kW 750 h 750 h

For compact fluorescent bulbs, use 26-W (33-W actual), which provide 1700 lumens each with an average life of 12,000 hours. 33 ⋅ 18 W 18 bulbs ⋅ $3.00 $20/h Annual Cost = 80 h ⋅ 52 weeks ----------------------------- ⋅ 0.08 $/kW + --------------------------------------- + ------------------- = $223 yr 1000 W/kW 12000 h 12000 h



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Solutions to Chapter 12— Energy, Costs, and Economics Problem 12.1

Compute the HVAC system demand (kW/ton) and efficiency (EER, COP) for a split-system heat pump with a medium-efficiency scroll compressor, an indoor fan with a standard AC motor, and an axial condenser fan.

Solution

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Problem 12.2

Repeat Problem 12.1 using a high-efficiency scroll compressor, an electronically commutated motor (ECM), and an axial condenser fan.

Solution

Problem 12.3

Compute the HVAC system demand (kW/ton) and efficiency (EER, COP) for a packaged rooftop unit with a medium-efficiency reciprocating compressor, an axial condenser fan, an indoor fan that delivers 1.5 in. of water, and a return fan that delivers 1.0 in. of water. Fan motors are 85% efficient and fans are 65% efficient.

Solution


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Chapter 12—Energy, Costs, and Economics

Problem 12.4

Compute the HVAC system demand (kW/ton) and efficiency (EER, COP) for a ground-source heat pump using a high-efficiency scroll compressor, a fan with an ECM, and a 50% efficient pump with a 60% electric motor that delivers 25 ft of water head. Assume the entering water temperature (EWT) to the unit is 85°F.

Solution

Problem 12.5

Compute the HVAC system demand (kW/ton) and efficiency (EER, COP) for a chilled water system with a high-efficiency water-cooled centrifugal compressor, 70% efficient chiller pumps with 50 ft of head, 70% efficient loop pumps with 70 ft of head, air-handling units with 75% efficient supply fans that deliver 5.0 in. of water, 75% efficient return air fans that deliver 2.0 in. of water, 70% efficient condenser pumps with 60 ft of head, an axial fan cooling tower, and fan-powered variable air volume (FPVAV) terminals with ECMs. Assume all motors are 92% efficient (except ECMs).

Solution




Problem 12.6

Repeat Problem 12.5 but replace the VAV system (supply fan, return fans, FPVAVs) with fan-coil units (FCUs) that have a nominal 10 ton/4000 cfm capacity and circulate air with 3 hp fans driven by 85% efficient motors.

Solution

Problem 12.7



Solution


A building in Birmingham, Alabama, is occupied five days per week from 8 a.m. to 8 p.m. During the occupied period, it has a cooling load of 120 MBtu/h at 97°F outside air temperature and a cooling load of 0 MBtu/h at 57°F OAT. During the unoccupied period, it has a cooling load of 40 MBtu/h at 97°F outside air temperature and a cooling load of 0 MBtu/h at 57°F OAT. In heating, the load is 80 MBtu/h at 17°F OAT (occupied), 60 MBtu/h at 17°F OAT (unoccupied), and 0 MBtu/h at 47°F OAT (occupied and unoccupied). It is cooled by a unit with a 125 MBtu/h capacity and 14 kW demand at 97°F and a 141 MBtu/h capacity and 11.4 kW demand at 67°F. It is heated by a unit with a 120 MBtu/h heating capacity with an 80% efficiency with a 1.5 hp 82% efficient fan motor. Compute the annual cost of heating and cooling the building based on 8¢/kWh in the summer and 7¢/kWh in the winter. Natural gas cost is $1.80 per therm (ccf). See table on following pages. Find the savings for the system described in Problem 12.7 if the efficiency of the cooling unit was improved by 20% (same capacity with 20% lower demand), the efficiency of the furnace is 95%, and the fan is reduced to 1 hp with a 90% efficient motor. See table on following pages. Repeat Problem 12.7 using a heat pump with the same cooling capacity and a heating capacity of 120 MBtu/h with an input of 11.3 kW at 47°F and 55 MBtu/h with an input of 9.8 kW at 17°F. See table on following pages.

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Solution to Problem 12.7: Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS

Solution

59



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60

Solution to Problem 12.8: Solution


Solution to Problem 12.9: Copyright ASHRAE Provided by IHS under license with ASHRAE No reproduction or networking permitted without license from IHS

Solution

61



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Problem 12.10

Solution

Discuss the economic value of installing a $10,000 energy efficiency package on a $225,000, 30-year, 6.25% APR home mortgage that will lower monthly utility bills by $40. The energy inflation rate is expected to be 8%. Economic Analysis Using 30-Year Fixed Rate Loan with Inflation Added Cost Loan Interest Energy Inflation of ECO Rate (%) Rate (%) 10000 6.25 8 Year One Year One Energy Savings Maint. Cost* 480 0 Year Energy Savings 1 480.00 2 518.40 3 559.87 4 604.66 5 653.03 6 705.28 7 761.70 8 822.64 9 888.45 10 959.52 11 1036.28 12 1119.19 13 1208.72 14 1305.42 15 1409.85 16 1522.64 17 1644.45 18 1776.01 19 1918.09 20 2071.54 21 2237.26 22 2416.24 23 2609.54 24 2818.30 25 3043.77 26 3287.27 27 3550.25 28 3834.27 29 4141.01 30 4472.29

Maint. Cost* 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Main. Inflation Rate (%) 5 Added Annual Payment ($) 746.03

Gen. Inflation Rate -CPI (%) 5 Added Monthly Payment ($) 62.17

Disc. NCF -266.03 -216.79 -168.85 -122.12 -76.51 -31.93 11.69 54.44 96.39 137.62 178.19 218.18 257.64 296.66 335.28 373.56 411.58 449.38 487.02 524.55 562.03 599.51 637.04 674.67 712.45 750.44 788.66 827.18 866.04 905.28

Pres. Worth -266.03 -482.82 -651.67 -773.78 -850.29 -882.22 -870.53 -816.08 -719.69 -582.07 -403.88 -185.70 71.95 368.60 703.88 1077.44 1489.02 1938.40 2425.42 2949.96 3511.99 4111.50 4748.54 5423.22 6135.67 6886.11 7674.77 8501.95 9367.99 10273.27

Net Cash Flow -266.03 -227.63 -186.16 -141.37 -92.99 -40.75 15.67 76.61 142.42 213.49 290.26 373.16 462.69 559.39 663.82 776.61 898.42 1029.98 1172.06 1325.51 1491.23 1670.21 1863.51 2072.27 2297.74 2541.24 2804.22 3088.24 3394.98 3726.26

The added monthly mortgage payment is $62.17, but the savings is only $40 per month. However, the cost of energy is inflating at a higher rate compared to inflation and the monthly note is fixed, so after 12 years the owner begins to receive a positive return if he/she plans on owning the home for this extended period and the life of the energy efficiency package is more than 12 years. Given the frequency of moving to a new home for the typical US family, this would be a marginal investment. Problem 12.11 Solution


Repeat Problem 12.10 for an energy inflation rate of 4%. When problem 12.10 is repeated with an energy inflation rate of 4% rather than 8%, the investment will require 27 years to receive a positive return. Even though the mortgage payment is fixed, the energy inflation rate is low compared to inflation, so the payback period is extended.

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Repeat Problem 12.10 for a 15-year, 5.75% APR loan. Economic Analysis Using 15-Year Fixed Rate Loan with Inflation Added Cost Loan Interest Energy Inflation of ECO Rate (%) Rate (%) 10000 5.75 8 Year One Year One Salvage Value Energy Savings Maint. Cost* in Year 15 480 0 0.00 Year Energy Savings 1 480.00 2 518.40 3 559.87 4 604.66 5 653.03 6 705.28 7 761.70 8 822.64 9 888.45 10 959.52 11 1036.28 12 1119.19 13 1208.72 14 1305.42 15 1409.85

Maint. Cost* 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Net Cash Flow -532.88 -494.48 -453.00 -408.21 -359.84 -307.60 -251.18 -190.24 -124.43 -53.35 23.41 106.31 195.85 292.54 396.98

Main. Inflation Rate (%) 5 Added Annual Payment ($) 1012.88

Gen. Inflation Rate -CPI (%) 5 Added Monthly Payment ($) 84.41

Disc. NCF -532.88 -470.93 -410.89 -352.63 -296.04 -241.01 -187.43 -135.20 -84.22 -34.39 14.37 62.16 109.05 155.14 200.50

Pres. Worth -532.88 -1003.80 -1414.69 -1767.32 -2063.36 -2304.37 -2491.80 -2627.00 -2711.22 -2745.61 -2731.24 -2669.08 -2560.03 -2404.89 -2204.39

This investment would not be prudent. Problem 12.13

Solution

A complete energy retrofit that will cost $200,000 is estimated to provide an annual savings of $30,000. The energy inflation rate is 6%, while the general and maintenance inflation rates are 5%. However, the system will require an additional $3000-per-year service contract. Compute the discounted ten-year present worth of the project. Discounted 10-Year Economic Analysis with Inflation Added Cost Discount Energy Inflation Main. Inflation Gen. Inflation of ECO Rate (%) Rate (%) Rate (%) Rate -CPI (%) 200000 6 6 5 5 Year One Salvage Value Energy Savings Maint. Cost* in Year 10 30000 3000 0.00 Year Energy Savings Maint. Cost* Net Cash Flow 1 30000.00 3000.00 27000.00 2 31800.00 3150.00 28650.00 3 33708.00 3307.50 30400.50 4 35730.48 3472.88 32257.61 5 37874.31 3646.52 34227.79 6 40146.77 3828.84 36317.92 7 42555.57 4020.29 38535.29 8 45108.91 4221.30 40887.61 9 47815.44 4432.37 43383.08 10 50684.37 4653.98 46030.38

Disc. NCF 27000.00 25741.24 24540.90 23396.27 22304.79 21264.01 20271.58 19325.27 18422.94 17562.57

Pres. Worth -173000.00 -147258.76 -122717.87 -99321.59 -77016.80 -55752.79 -35481.21 -16155.94 2267.00 19829.57

PW10 = $19,829.57

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63


Problem 12.14

Calculate the discounted rate of return on the project described in Problem 12.13 for a ten-year evaluation.

Solution

Discounted 10-Year Economic Analysis with Inflation Added Cost Discount Energy Inflation Main. Inflation Gen. Inflation of ECO Rate (%) Rate (%) Rate (%) Rate -CPI (%) 200000 8.5276 6 5 5 Year One Salvage Value Energy Savings Maint. Cost* in Year 10 30000 3000 0.00 Year Energy Savings Maint. Cost* Net Cash Flow 1 30000.00 3000.00 27000.00 2 31800.00 3150.00 28650.00 3 33708.00 3307.50 30400.50 4 35730.48 3472.88 32257.61 5 37874.31 3646.52 34227.79 6 40146.77 3828.84 36317.92 7 42555.57 4020.29 38535.29 8 45108.91 4221.30 40887.61 9 47815.44 4432.37 43383.08 10 50684.37 4653.98 46030.38

Disc. NCF 27000.00 25141.73 23411.10 21799.35 20298.36 18900.51 17598.74 16386.46 15257.53 14206.24

RR10 = 8.5276% (discount rate that results in PW10 = $0)


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Pres. Worth -173000.00 -147858.27 -124447.18 -102647.82 -82349.47 -63448.96 -45850.22 -29463.75 -14206.22 0.02


Problem 12.15

Solution

A ground-source heat pump costs $5000 more than a conventional heating and cooling system. It saves approximately $400 per year in energy costs and $100 per year in maintenance costs. The owner plans to live in this home for 20 years. The energy inflation rate is 7%, the discount rate is 5%, and the general inflation and maintenance rates are 6%. What is the present worth at 20 years and what is the discounted payback? Discounted 20-Year Economic Analysis with Inflation Added Cost Discount Energy Inflation Main. Inflation Gen. Inflation of ECO Rate (%) Rate (%) Rate (%) Rate -CPI (%) 5000 5 7 6 6 Year One Year One Salvage Value Energy Savings Maint. Cost* in Year 20 400 -100 0.00

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Year Energy Savings Maint. Cost* Net Cash Flow 1 400.00 -100.00 500.00 2 428.00 -106.00 534.00 3 457.96 -112.36 570.32 4 490.02 -119.10 609.12 5 524.32 -126.25 650.57 6 561.02 -133.82 694.84 7 600.29 -141.85 742.14 8 642.31 -150.36 792.68 9 687.27 -159.38 846.66 10 735.38 -168.95 904.33 11 786.86 -179.08 965.95 12 841.94 -189.83 1031.77 13 900.88 -201.22 1102.10 14 963.94 -213.29 1177.23 15 1031.41 -226.09 1257.50 16 1103.61 -239.66 1343.27 17 1180.87 -254.04 1434.90 18 1263.53 -269.28 1532.80 19 1351.97 -285.43 1637.41 20 1446.61 -302.56 1749.17

Disc. NCF 500.00 479.78 460.39 441.79 423.95 406.83 390.41 374.65 359.54 345.04 331.13 317.79 304.98 292.70 280.92 269.61 258.76 248.35 238.37 228.78

Pres. Worth -4500.00 -4020.22 -3559.82 -3118.03 -2694.09 -2287.26 -1896.85 -1522.20 -1162.66 -817.62 -486.48 -168.70 136.29 428.99 709.90 979.51 1238.27 1486.63 1724.99 1953.78

PW20 = $1953.78 DPB ≈ 12.6 years (time at which PW = 0)





Repeat Problem 12.15 for ie = 6%, ig = im = 7%, and d = 6% and compare these results with a simple payback analysis. Discounted 20-Year Economic Analysis with Inflation Added Cost Discount Energy Inflation Main. Inflation Gen. Inflation of ECO Rate (%) Rate (%) Rate (%) Rate -CPI (%) 5000 6 6 7 7 Year One Year One Salvage Value Energy Savings Maint. Cost* in Year 20 400 -100 0.00

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Year Energy Savings Maint. Cost* Net Cash Flow 1 400.00 -100.00 500.00 2 424.00 -107.00 531.00 3 449.44 -114.49 563.93 4 476.41 -122.50 598.91 5 504.99 -131.08 636.07 6 535.29 -140.26 675.55 7 567.41 -150.07 717.48 8 601.45 -160.58 762.03 9 637.54 -171.82 809.36 10 675.79 -183.85 859.64 11 716.34 -196.72 913.05 12 759.32 -210.49 969.80 13 804.88 -225.22 1030.10 14 853.17 -240.98 1094.16 15 904.36 -257.85 1162.21 16 958.62 -275.90 1234.53 17 1016.14 -295.22 1311.36 18 1077.11 -315.88 1392.99 19 1141.74 -337.99 1479.73 20 1210.24 -361.65 1571.89

Disc. NCF 500.00 468.17 438.38 410.48 384.37 359.92 337.03 315.61 295.54 276.76 259.18 242.72 227.30 212.87 199.36 186.70 174.86 163.77 153.38 143.65

PW20 = $750.05 DPB ≈ 16.3 years (time at which PW = 0) SPB = $5000 ÷ [$400/yr – (–$100/yr)] = 10 years



Pres. Worth -4500.00 -4031.83 -3593.45 -3182.97 -2798.60 -2438.68 -2101.65 -1786.05 -1490.50 -1213.74 -954.56 -711.84 -484.54 -271.67 -72.31 114.39 289.25 453.02 606.40 750.05

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HVAC Simplified Solution Manual

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