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HVAC
Simplified
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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.
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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.
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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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SPECIAL PUBLICATIONS
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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
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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
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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
ix
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Nomenclature—HVAC Terms,
Abbreviations, and Subscripts
HVAC Simplified Solutions Manual
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
x
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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
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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
HVAC Simplified Solutions Manual
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
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.
3
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HVAC Simplified Solutions Manual
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.
4
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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
HVAC Simplified Solutions Manual
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
HVAC Simplified Solutions Manual
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
8
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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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
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60 (min/h) × 4000 (ft /min) × 0.24 (Btu/lb·°F)
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HVAC Simplified Solutions Manual
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 3 @
Qair3(cfm),
Tdb3(°F),
& Twb3(°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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
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HVAC Simplified Solutions Manual
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?
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
14
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79.8 °F
67.0 °F
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Chapter 4—HVAC Fundamentals: Psychrometrics
Problem 4.9
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
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HVAC Simplified Solutions Manual
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
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HVAC Simplified Solutions Manual
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)
Problem 5.6
Solution
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
HVAC Simplified Solutions Manual
Problem 5.9
Solution
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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Chapter 5—HVAC Equipment, Systems, and Selection
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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HVAC Simplified Solutions Manual
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 1 @
Qair1(cfm),
Tdb1(°F),
& Twb1(°F)
Stream 3 @
Qair3(cfm),
Tdb3(°F),
& Twb3(°F)
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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Chapter 5—HVAC Equipment, Systems, and Selection
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)
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HVAC Simplified Solutions Manual
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
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Solutions to Chapter 6—
Comfort, Air Quality,
and Climatic Data
Problem 6.1
Solution
Problem 6.2
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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HVAC Simplified Solutions Manual
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
Problem 6.6
Solution
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
Problem 6.7
Solution
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
Problem 6.9
Solution
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
26
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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
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HVAC Simplified Solutions Manual
Problem 6.12
Solution
Problem 6.13
Solution
Problem 6.14
Solution
28
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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
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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
R-Path Total
16.15
7.525
=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
Problem 7.4
Solution
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)=
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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
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8.78
HVAC Simplified Solutions Manual
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)=
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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
R-Path Total
18.53
9.905
=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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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R-Path Total
24.53
12.405
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
Problem 7.14
Solution
Problem 7.15
Solution
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
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HVAC Simplified Solutions Manual
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Copyright ASHRAE
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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
33
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--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Copyright ASHRAE
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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.
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HVAC Simplified Solutions Manual
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.
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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.
Problem 8.7
Solution
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.
*See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.4.
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--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
The main screen input for the maximum humidity ratio (dry bulb = 85°F, wet
bulb = 78°F) condition is shown below.
HVAC Simplified Solutions Manual
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
38
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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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
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10.10
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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
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
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) CLTD(am) CLTD(pm)
7
65
0.033
U(Btu/hr-ft2-F)
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Latent
Lighting
Net Sen.w/o duct
Ductwork
Sensible
Latent
Total Vent Air
Total Sensible
Total Latent
Total Gain
Total Loss
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
(MBtu/h)
(MBtu/h)
(MBtu/h)
SHR
(MBtu/h)
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.) =
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
People
15
15
Watts
2515
Btu/h
Watts
1
1240
Location
S-Ext
Entire Building Totals
135 cfm
11%
26.9
29.2
Sens.
9.5
9.1
Lat.
36.3
38.3
TotGain
0.74
0.76
SHR
30.6 NetLoss
18.0 TotLoss
0.00
2.60
10.10
27.09
18.0
39
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HVAC Simplified Solutions Manual
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
°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)
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
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
(MBtu/h)
(MBtu/h)
(MBtu/h)
SHR
(MBtu/h)
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.) =
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
People
15
15
Watts
2515
Btu/h
Watts
1
1240
Location
S-Ext
Entire Building Totals
135 cfm
11%
25.2
27.6
Sens.
12.2
10.9
Lat.
37.5
38.4
TotGain
0.67
0.72
SHR
30.6 NetLoss
18.0 TotLoss
0.00
2.60
40
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10.10
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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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
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HVAC Simplified Solutions Manual
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
Problem 9.4
Solution
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
*See errata for HVAC Simplified posted to www.ashrae.org/publicationupdates for corrected Figure C.7.
Copyright ASHRAE
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Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
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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
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Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
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43
HVAC Simplified Solutions Manual
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
44
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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
45
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--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Problem 9.9
HVAC Simplified Solutions Manual
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)
Problem 9.13
Solution
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
46
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Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
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
Inlet
Vel
fps
3.6
3.6
0.0
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
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--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Solutions to Chapter 10—
Water Distribution System Design
HVAC Simplified Solutions Manual
Problem 10.1 with pipe sizes
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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.
48
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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
Inlet
Vel
fps
3.6
3.2
0.0
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
Inlet
Vel
fps
3.6
3.6
0.0
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
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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
Problem 11.3
Solution
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
Problem 11.5
Solution
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
Problem 11.8
Solution
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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HVAC Simplified Solutions Manual
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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HVAC Simplified Solutions Manual
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
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HVAC Simplified Solutions Manual
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
Problem 12.8
Solution
Problem 12.9
Solution
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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:
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Solution
59
Chapter 12—Energy, Costs, and Economics
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60
Solution to Problem 12.8:
Solution
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Solution to Problem 12.9:
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Solution
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Chapter 12—Energy, Costs, and Economics
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HVAC Simplified Solutions Manual
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 $40 per
month. However, the cost of energy is inflating at a higher rate compared to inflation
and the monthly note is fixed, so after 12 years the owner begins to receive a positive
return if he/she plans on owning the home for this extended period and the life of the
energy efficiency package is more than 12 years. Given the frequency of moving to a
new home for the typical US family, this would be a marginal investment.
Problem 12.11
Solution
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Repeat Problem 12.10 for an energy inflation rate of 4%.
When problem 12.10 is repeated with an energy inflation rate of 4% rather than 8%,
the investment will require 27 years to receive a positive return. Even though the
mortgage payment is fixed, the energy inflation rate is low compared to inflation, so
the payback period is extended.
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Chapter 12—Energy, Costs, and Economics
Problem 12.12
Solution
Repeat Problem 12.10 for a 15-year, 5.75% APR loan.
Economic Analysis Using 15-Year Fixed Rate Loan with Inflation
Added Cost
Loan Interest Energy Inflation
of ECO
Rate (%)
Rate (%)
10000
5.75
8
Year One
Year One
Salvage Value
Energy Savings Maint. Cost*
in Year 15
480
0
0.00
Year Energy Savings
1
480.00
2
518.40
3
559.87
4
604.66
5
653.03
6
705.28
7
761.70
8
822.64
9
888.45
10
959.52
11
1036.28
12
1119.19
13
1208.72
14
1305.42
15
1409.85
Maint. Cost*
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Net Cash Flow
-532.88
-494.48
-453.00
-408.21
-359.84
-307.60
-251.18
-190.24
-124.43
-53.35
23.41
106.31
195.85
292.54
396.98
Main. Inflation
Rate (%)
5
Added Annual
Payment ($)
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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HVAC Simplified Solutions Manual
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)
64
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
Pres. Worth
-173000.00
-147858.27
-124447.18
-102647.82
-82349.47
-63448.96
-45850.22
-29463.75
-14206.22
0.02
Chapter 12—Energy, Costs, and Economics
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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)
65
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
HVAC Simplified Solutions Manual
Problem 12.16
Solution
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
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
66
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT
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
--`,,``,`,,,`,,`,````,`,`,,,,``,-`-`,,`,,`,`,,`---
Copyright ASHRAE
Provided by IHS under license with ASHRAE
No reproduction or networking permitted without license from IHS
Licensee=Kellogg Brown & Root Jakarta /3262700002, User=Rohana, Mumuh
Not for Resale, 09/09/2009 18:54:17 MDT