AC 6Ed.pdf

SOLUTIONS MANUAL
Heating, Ventilating and Air Conditioning
Analysis and Design
Sixth Edition
Faye C. McQuiston
Jerald D. Parker
Jeffrey D. Spitler
John Wiley & Sons
2004
M16FMR
m16fmr@hotmail.co.uk
USE IT WITH YOUR MIND
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PREFACE
This manual consists of solutions for the text problems, Chapters 1
through 15. Many of the problems with a design orientation have more
than one acceptable solution. Therefore, when an acceptable solution
may be quite variable or closely follows an example, the solution may be
omitted or only general guidelines given.
The degree of difficulty and length of the problems vary
considerably. In many cases the problem statement should be discussed
in class prior to assignment. This manual should be helpful in this
respect.
The use of computer software is strongly encouraged. Many
problems can be solved by hand or computer calculations. Most of these
problems are in Chapters 3, 7, 8, 10, 12 and 14. A number of problems
are designed for solution using the software distributed with the website.
A great effort has been made to eliminate errors in the solutions;
however, some probably still exist. Please see the addendum to this
manual for assumptions used in problems 6-10, 7-9, 7-14, 8-25, and 8-26.
The authors would appreciate notification of any errors discovered.
Faye C. McQuiston
Jerald D. Parker
Jeffrey D. Spitler
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CONTENTS
Chapter 1 Introduction 1-6
Chapter 2 Air-Conditioning Systems 7
Chapter 3 Moist Air Properties and Conditioning 8-65
Processes
Chapter 4 Comfort and Health – Indoor Environmental 66-78
Quality
Chapter 5 Heat Transmission in Building Structures 79-92
Chapter 6 Space Heat Load 93-105
Chapter 7 Solar Radiation 106-129
Chapter 8 The Cooling Load 130-183
Chapter 9 Energy Calculations and Building Simulation 184-188
Chapter 10 Flow, Pumps, and Piping Design 189-210
Chapter 11 Space Air Diffusion 211-219
Chapter 12 Fans and Building Air Distribution 220-259
Chapter 13 Direct Contact Heat and Mass Transfer 260-265
Chapter 14 Extended Surface Heat Exchangers 266-298
Chapter 15 Refrigeration 299-310
Addendum 311-312
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Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for
testing or instructional purposes only to students enrolled in courses for which the textbook has been
adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or
108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Requests for permission or further information should be addressed to the Permission Department,
John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.
PREFACE
This manual consists of solutions for the text problems, Chapters 1
through 15. Many of the problems with a design orientation have more
than one acceptable solution. Therefore, when an acceptable solution
may be quite variable or closely follows an example, the solution may be
omitted or only general guidelines given.
The degree of difficulty and length of the problems vary
considerably. In many cases the problem statement should be discussed
in class prior to assignment. This manual should be helpful in this
respect.
The use of computer software is strongly encouraged. Many
problems can be solved by hand or computer calculations. Most of these
problems are in Chapters 3, 7, 8, 10, 12 and 14. A number of problems
are designed for solution using the software distributed with the website.
A great effort has been made to eliminate errors in the solutions;
however, some probably still exist. Please see the addendum to this
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Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes
only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work
beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner
is unlawful.
manual for assumptions used in problems 6-10, 7-9, 7-14, 8-25, and 8-26.
The authors would appreciate notification of any errors discovered.
Faye C. McQuiston
Jerald D. Parker
Jeffrey D. Spitler
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311
ADDENDUM
to
Solutions Manual for McQuiston, HVAC 6e
Problem 6-10
For the floor, it is unclear what 2 in. vertical edge insulation means (whether 2 in. is the
thickness of insulation or the depth of the edge insulated).
• The solution assumes that the insulation has R-value of 5.4 hr-ft²-°F/Btu and the
depth of the edge is 2 ft.
For the door, Table 5-8 in the 6th
edition does not have U-value for the wood storm door
and there are three types of the wood door with 1 ¾ in. thickness.
• The solution assumes that the doors are panel doors with metal storm door; hence,
its U-value is 0.28 But/hr-ft²-°F.
Problem 7-9
The standard time zone for Ottawa, Ontario is Eastern Standard Time instead of Central
Standard Time.
• The solution uses Eastern Standard Time.
Problem 7-14
For the specified location, the sunset occurs before 9:00 p.m. CDST on June 21.
• The solution uses 8:00 p.m. CDST instead of 9:00 p.m.
Problems 8-25 and 8-26
Both problems do not specify the window orientation.
• The solutions assume the west-facing window for both problems.
Table 8-20
Recommended radiative and convective fractions for solar heat gains should be revised
since the 6th
edition uses the SHGC values in the calculation of the (combined) solar heat
gain for the RTS method.
Example 8-16
The example actually uses 90%/10% of radiative/convective split of the combined solar
heat gain. However, the text (page 270) says 100%/0% for the transmitted solar heat gain
and 63%/37% for the absorbed solar heat gain.
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312
Problems 8-25 and 8-26
The solutions for both problems use 90%/10% for the combined solar heat gain.
Example 9-1
The calculation for this example should be
606
,
122
)
1000
)(
0
70
)(
55
.
0
(
)
66
.
0
)(
000
,
80
)(
3725
)(
24
(
=
−
=
F .
(Changing 13 to 24 and 122790 to 122606).
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adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108
of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Requests for permission or further information should be addressed to the Permission Department, John
Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.
Chapter 1
1-1 (a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K)
(b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K
(c)
0.04 Ibm/(ft-hr)
3600 sec/hr
x1.488 = 16.5 2
Ns
m
µ
(d) 1050
Btu
Ibm
x 4
1
9.48x10−
J
Btu
x
2.20462 Ibm
kg
= 2.44
MJ
kg
(e) 12,000
Btu
Ibm
x
1
3.412
= 3.52 kW
(f) 14.7 2
Ibf
in
x 6894.76 = 101 kPa
1-2 (a) 120 kPa x
2
lbf /in
6.89476kPa
= 17.4 lbf/in2
(b) 100
W
m K
−
x 0.5778 = 57.8 Btu/hr-ft-F
(c) 0.8 2
W
m K
−
x 0.1761 = 0.14 Btu/hr-ft2
-F
(d) 10-6
N-s/m2
x
1
1.488
= 6.7 x 10-7 lbm
ft sec
−
(e) 1200 kW x 3412 = 4.1 x 10-6
Btu/hr
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2
(f) 1000
kJ
kg
x
1 Btu
1.055 kJ
x
1 kg
2.2046 lbm
= 430
Btu
lbm
1-3 Hp = 50 (ft) x 0.3048 (
m
ft
) = 15.2 m
∆ P =
15.2 m
1000 Pa/kPa
x
9.807
1
(
N
kg
) x 1000 (kg/m3
) = 149 kPa
1-4 P =
∆
4
12
(ft) x 0.3048 (
m
ft
) x
9.807
1
(
N
kg
) x 1000 ( 3
kg
m
)
∆ P = 996 Pa 1.0 kPa
≈
1-5
TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE
+ METER CHARGE
( ) ( ) ( ) ( )
96,000 kw - hrs 0.045 $/kw hr + 624 kw 11 50 $/kw
− −
+ $68 = $4,320 + $7,176 + $68 = $11,564
1-6 7 AM to 6 PM 11 hrs/day, 5 days/wk
hrs days
(11) (22) 242 hrs/month
day months
=
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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( )
( )
( )
624 kw
ratio = 1.57
96,000 kw hr
242 hr
=
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
1-7 This is a trial and error solution since eq. 1-1 cannot be solved
explicitly for i.
Answer converges at just over 4.2% using eq. 1-1
1-8 Determine present worth of savings using eq. 1-1
( )
( )( )
12 12
0.012
$1000 1- 1+
12
P =
0.012
12
P $134,000
−
⎡ ⎤
⎛ ⎞
⎢ ⎥
⎜ ⎟
⎝ ⎠
⎢ ⎥
⎣ ⎦
⎛ ⎞
⎜ ⎟
⎝ ⎠
=
1-9 (a) Q VA
=
= 2 x 3.08 x 10-3
= 6.16 x 10-3
m3
/s
m 6.16 x 10
Q
ρ
=
= -3
x 998 = 6.15 kg/s
(b) A=
4
π
(0.3)2
= 7.07 x 10-2
m2
Q 7.07x10
=
-2
x 4 = 0.283 m3
/ s; ρ = 1.255 kq/m3
m = 1.225 x 0.283 = 0.347 kg/s

1-10 V = 3x10x20 = 600m3
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4
= 600 x
i
Q
1
4
x
1
3600
= 4.17 x 10-2
m3
/s
1-11
p p
3
q = mc T c = 4.183 kJ/(kg-K)
= 983.2 kg/m
ρ
∆

1-11 (cont’d)
( ) ( ) ( ) ( )
3
c
3
m kg kJ
q = 1 983.2 4.183 5 20,564
s kg K
m
q = 20,564 kw
=
−

kJ
s
1-12 =
wat
q
air
q
−
11,200(1)(10) =
= 2
5000x60x14.7x144x0.24(t 50)
(53.35x510)
−
11,200 = 5601.5 (t2-50); t2 = (11,200/5601.5) + 50 = 70 F
1-13 Diagram as in 1-12 above.
q
wat = -q
air
1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000)
6279(90-t2) = 29,400
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permissi
5
t2 = 90 -
29,400
6279
= 85.3 C
1-14 q hA(t
=
s- )
t∞
A= π (1/12) x 10 = 2.618 ft2
s
t = tsur 212 F
≈
= 10x2.618x(212-50) = 4241 Btu/hr
q

1-15 A= π x 0.25x4 = 3.14 16 m2
hA(t
q =
s- )
t∞
h=
s
q
A(t -t )
∞

=
1250
3.1416(100 10)
−
; h = 4.42 W/(m2
– C)
1-16 (t
p
q mc
=
2-t1) ; m Q x ρ
=

ρ = P/RT = 14.7x144/53.35(76+460)
ρ = 0.074 lbm/ft3
m = 5000x0.074x60 = 22,208 lbm/hr

= 0.24 Btu/lbm-F
p
c
q= 22,208x0.24(58-76) = -95,939 Btu/hr

Negative sign indicates cooling
1-17 (t
1 p
m c
3-t1) +
for-profit basis for testing or instructiona
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(t
2 p2
m c
3-t2) = 0
=
p1
c p2
c
t3 = 1 1 2 2
1 2
(m t m t )
(m m )
+
+

1 2
m Q 1
ρ
=
= 1000x
14.7x144
53.35(460 50)
+
= 73.5 lbm/min
1-17 (cont’d)
2 2
m Q 2
ρ
=
= 600x
14.7x144
53.35(460 50)
+
= 46.7 lbm/min
3
(73.5x80) (46.7 x 50)
t 68.3 F
(73.5 46.7)
+
= =
+
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Wiley Sons, Inc, 111 River Street, Hoboken, NJ 07030.
Chapter 2
2-1 through 2-20
Solutions are not furnished since many acceptable responses exist
for each problem. It is not expected that the beginning student can handle
these questions easily. The objective is to make the student think about
the complete design problem and the various functions of the system.
These problems are also intended for use in class discussions to enlarge
the text material.
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Chapter 3
3-1 (a) Pv = =
s
rP
φ 0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psia
Pa = 101 – 1.43 = 99.57 kPa or 14.696-0.196 = 14.5 psia
(b) =
v
v
P
ρ
RvT or = = = 3
v
v v
v
P 1430
ρ ; ρ 0.0104 kg/m
R T 462.5(297)
or =
0.196(144)
0.00062
85.78(535)
lbv/ft3
(c) W =
0.6219 (1.43)
(99.57)
= 0.00893 kgv/kga
or
0.6219(0.196)
0.00854 lbv/lba
14.5
=
3-2 (a) English Units – t = 80F; P = 14.696 psia;
Pv = 0.507 psia Table A-1a
W = 0.6219
a
v
P
P
=
0.6219 (0.507)
(14.696 0.507)
−
= 0.0222 lbv/lba
i = 0.24t + W(1062.2 + 0.444t)
i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm
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v =
a
a
R T 53.35(460 80)
P (14.696 0.507)144
+
=
−
= 13.61 ft3
/lbm
(b) English Units – 32F, 14.696 psia
Pv = 0.089 psia (Table A-1)
3-2 (cont’d)
W =
0.6219(0.089) lbmv
0.00379
(14.696 0.089) lbma
=
−
i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbma
v =
53.35(492)
(14.696 0.089)144
−
= 12.48 ft3
/lbma
3-2 (a) SI Units – 27C; 101.325 kPa
Pv = 3.60 kPa, Table A-1b
W = 0.6219 v
a
P 0.6219(3.6) kgv
0.0229
P (101.325 3.6) kga
= =
−
i = 1.0t + W(2501.3 + 1.86t) kJ/kga
i = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kga
v = 3
a
a
R T 0.287(300)
= =0.88 m /kga
P (101.325 - 3.6)
(b) SI Units 0.0C; 101.325 kPa
W =
0.6219(0.61)
=0.00377 kgv/kga
(101.325 - 0.61)
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i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga
v =
3
0.287(273)
0.778 m /kga
(101.325 - 0.61)
=
3-3 (a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hg
t = 80 F; Pv = 0.507 psia (Table A-1a)
W = 0.6219
v
a
P 0.6219(0.507)
=
P (12.24 - 0.507)
= 0.0269 lbv/lba
i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma
v = a
a
R T 53.35(540)
=
P (12.24 - 0.507) 144
= 17.05 ft3
/ lbma
(b) English Units – t = 32 F, Pv = 0.089 psia ( Table A-1a)
W =
0.6219(0.089)
(12.24 0.089)
−
= 0.00456 lbmv/lbma
i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma
v =
53.35(492)
(12.24 0.089)144
−
= 15.00 ft3
/lbma
3-3 (a) SI Units -27 C, 1500 m elevation
P = 99.436 + 1500(-0.01) = 84.436 kPa
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10
W =
0.6219x3.60
0.0277 kgv/kga
(84.436 3.60)
=
−
i = 27 + 0.0277 (2501.3 + 1.86 x 27) = 97.68 kJ/kga
3-3 (cont’d)
v =
3
0.287x300
1.065 m / kga
(84.436 - 3.60)
=
(b) SI Units – 0.0C; 1500m or 84.436 kPa
Pv = 0.61 kPa; Table A-1b
W =
0.6219 x 0.61
0.00453 kgv / kga
(84.436 - 0.61)
=
i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kga
v =
0.287 x 273
(84.436 - 0.61)
= 0.935 m3
/ kga
3-4 (a) English Units – 70F, Pv = 0.363 psia
Pv = φ Pg = 0.75(0.363) = 0.272 psia
W =
0.6219 (0.272)
0.0117 lbmv / lbma
(14.696 - 0.272)
=
i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)] 29.58 Btu / lbma
=
(b) Pv = 0.75 (0.363) = 0.272 psia; P = 12.24 psia
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W =
0.6219 (0.272)
(12.24 - 0.272)
= 0.0141 lbmv / lbma
i = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)] 32.20 Btu/ lbma
=
3-4 SI Units –
(a) 20C, 75% RH, Sea Level
3-4 (cont’d)
Ps = 2.34 kPa; Pv = 0.75 x 2.34 = 1.755 kPa
0.6219 x 1.755
W = =
(101.325 - 1.755)
0.0110 kgv / kga
i = 1.0 t + W(2501.3 + 1.86t)
i = 20 + 0.0110(2501.3 + 1.86 x 20) = 47.92 kJ / kga
(b) 20C, 75% RH, 1525m
P = 99.436 – 0.01 x 1525 = 84.186 kPa
Ps = 2.34 KPa; Pv = 0.75 x 2.34 = 1.755 kPa
W =
0.6219 x 1.755
(84.186 - 1.755)
= 0.0132 kgv / kga
i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga
3-5 English Units –
t = 72 Fdb; psia
14.696
P
%;
50 =
=
φ
s
v
s
v
P
P
or
P
P
φ
φ =
= ; Pv = 0.5(0.3918) = 0.196 psia
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12
Air dewpoint = saturated temp. at 0.196 psia = 52.6 F
Moisture will condense because the glass temp.
40 F is below the dew point temp.
3-5 SI Units – t = 22C ; 50% ; P = 100 kPa
Pv = φ Ps ; Pv = 0.5(2.34) = 1.17 kPa
3-5 (cont’d)
Air dewpoint = sat.temp. at 1.17 kPa = 9.17 C
Glass temp. of 4 C is below the dewpoint of 9.17 C, therefore,
moisture will ccondense on the glass
3-6 English Units -
(a) At 55F, 80% RH, va = 13.12 ft3
/ lba and ρ a = 0.0752 lbma / ft3
= 22,860 lbma / hr
a
m 5000 (0.0762) 381 lbma / min
= =

(b) Using PSYCH ρ a = 0.0610 lbma / ft3
or va = 16.4 ft3
/ lba
= 5000 (0.061) = 305 lbma / min
a
m
18,300 lbma / hr
=
3-6 SI Units –
(a) t = 13 C and relative humidity 80%
then va 0.820 m
≈ 3
/ kga; a
m 2.36 / 0.82 2.88 kga / s
= =

(b) Assuming same conditions
;
3
a
v 0.985 m / kga
= a
m 2.36 / 0.985 2.40 kga / s
= =

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3-7 English Units – t = 80F, 60% RH
(a) v s
P P 0.6 (0.507) 0.304 psia
φ
= = =
= 64.5 F
dp sat v
t (t @ P
= )
(b) Same as (a) above
3-7 SI Units –
(a) 27 C, 60% RH, Sea Level
Ps = 3.57 kPa; Pv = 0.6 x 3.57 = 2.14 kPa
dp sat v
t =(t at P ) 18.4 C
≈
(b) Same as (a) above
3-8 dp
t 9C (48F)
≤
42%
φ ≤ ; W 0.0071 kgv / kga (lbv / lba)
≤
Chart 1a 1b
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14
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY - BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE - °F
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
dp Room
Problem 3-8
W=0.0071
72 (22)
48 (9)
42 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1
NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY
Copyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
00
500
0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-9 (a,b,d) Using the Properties option of PSYCH:
Relative Humidity = 0.59 or 59%
Enthalpy = 30.4 Btu/lbma
Humidity Ratio = 0.0114 lbu/lba
(c) Again using the Properties option
At W=0.0114 lbv/lba; RH = 1.00 or 100%
The dew point = tdb or twb = 59.9 F
M16FMR
M
1
6
F
M
R

15
3-9 (cont’d)
(e) Using the Density of Dry Air option:
Mass Density = 0.070 lba/ft3
3-10 Using program PSYCH
(a) tdb = 102.6; twb = 81.1F
75 Fdb; 65 fwb; 14.2 psia
(b) m 58.7
ν =
lbm/hr
Q
2 = 1027 cfm
3-11 t1 = 80 / 67 F; t2 = 55 F and sat.; assume std. barometer
(a) W1 – W2 = 0.0112 – 0.0092 = 0.002 lbv / lba
(b) l
q 31.5 - 29.3 2.2 Btu / lba
= =
(c) qs = 29.3 – 23.2 = 6.1 Btu / lba
(d) q = l s
q q 8.3 Btu / lba
+ =
3-12 (a) *
2
0.6219 (0.3095)
W 0.0134 kgv / kga
(14.696 0.3095)
= =
−
1
0.24 (65 - 80) ( 0.0134 x 1056.5)
W 0.00993 lbv / lba
(1096 - 33)
+
= =
also W1 = 0.6219 Pv1 / (P – Pv1)
M16FMR
M
1
6
F
M
R

16
Pv1 = (0.00993 x 14.696) / ( 0.6219 + 0.00993) = 0.231 psia
3-12 (cont’d)
1
0.231
0.46 or 46%
0.507
φ = =
(b) P = 29.42 – (0.0009 x 5000) = 24.92 in.Hg. or P = 12.24 psia
*
2
0.6219 x (0.3095)
W 0.01613
(12.24 - 0.3095)
= = lbv/lba
W1 =
0.24(65 80) (0.01613 x 1056.5)
0.01265 lbv / lba
( 1096 - 33)
− +
=
or kgv / kga
Pv1 = 0.01265 x 12.24 / ( 0.6219 + 0.01265) = 0.244 psia
1
0.244
0.48 or 48%
0.507
φ = =
3-13 (a) Sea Level
Dry
Bulb, F
Wet
Bulb, F
Dew
point
F
Humid.
Ratio, lba/lbv
Enthalpy
Btu/lba
Rel.
Humid., %
Mass
Density
lba/ft3
85 60 40.6 0.0053 26.6 21 0.072
75 59.6 49.2 0.0074 26.1 40 0.073
74.6 65.1 60.1 0.0111 30 60 0.073
88.6 70 60.9 0.01143 33.8 40 0.071
100 85.8 81.7 0.0235 50 56 0.068
M16FMR
M
1
6
F
M
R

17
(a) 5000 ft.
Dry
Bulb, F
Wet
Bulb, F
Dew
point
F
Humid.
Ratio, lba/lbv
Enthalpy
Btu/lba
Rel.
Humid., %
Mass
Density
lba/ft3
85 60 45.1 0.0076 28.7 25 0.060
75 58.6 49.2 0.0089 27.7 40 0.061
71.2 61.6 56.7 0.0118 30 60 0.061
102.7 70 55.8 0.01143 37.3 22 0.058
100 81.3 76.1 0.0235 50 47 0.057
(c) Note effect of barometric pressure.
M16FMR
M
1
6
F
M
R

18
3-14
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
dp Room
Problem 3-14
72 (22)
52 (11)
Max RH=49.6 %
W=0.0083
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-15 Use Chart 1b, SI
(a) td = 10 C; SHF = 0.62
(b) 1 2
2.4
q m (i i ) (57.1 - 34)
0.867
= − =
= 63.95 kJ / s = 63.95 k W
s
q 63.95 (0.62) 39.65 kW
= =

3-15 Use Chart 1a, IP
(a) td = 52 F; SHF = 0.63
M16FMR
M
1
6
F
M
R

19
3-15 (cont’d)
(b)
5000(60)
q = (32 - 22.6)= 203,317. Btu/hr
13.87

10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
ADP
2
Problem 3-15
80 (27)
55 (13)
52 (10)
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
s
q 203,317 (0.63) 128,089. Btu/hr
= =

3-16 (a) i1 = 30 Btu / lba; v1 = 13.78 ft3
/ lba; W = 0.0103
lba
lbv
; 50%
1 =
φ
(b) i1 = 51.6 kJ / kga
v1 = 0.86 m3
/ kga
M16FMR
M
1
6
F
M
R

20
3-16 (cont’d)
W1 =
kga
kgv
0103
.
0
50%
1 =
φ
3-17 Use the Heat Transfer option of program PSYCH:
q = 148,239 Btu/hr

s
q 102,235 Btu/hr
=

SHF = 0.69
3-18 Use the Heat Transfer option of program PSYCH for sensible heat
transfer only:
s
q 178,911 Btu/hr
= −

Negative sign indicates heating.
3-19 Use the program PSYC to compute the various
properties at 85/68 F; sea level and
6000 ft elevation.
Elevation
ft
Enthalpy
Btu/lbm
Rel. Hum
percent
Hum. Ratio
lbv/lba
Density
lba/ft3
0 32.2 42 0.0107 0.072
6000 36.3 45 0.0144 0.058
At sea level: a
m 5000 x 0.072 x 60 21,600 lba/hr
= =

M16FMR
M
1
6
F
M
R

21
3-19 (cont’d)
At 6000 feet:: a
m 5000 x 0.057 x 60 17,100 lba/hr
= =

Percent Decrease at 6000 ft:
(21
,600 17,100)100
PD 20.8%
21,600
−
= =
3-20 Use the program PSYC to compute the heat transfer
rates at 1000 and 6000 feet elevation:
(a) at 1000 ft, q 200,534 Btu/hr
=

(b) at 6000 ft, q 190,224 Btu/hr
=

(c) PD =
(200,534 190,224)100
5.1 %
200,543
−
=
3-21 (a) English Units –
; = 0
in.Hg.
29.92
PB =
q
w
i
i 180.2 0.8 (970.2)
W
∆
= = +
∆
iw = 956.4 Btu / lbv
From chart 1a; t2 = 91.5 F
M16FMR
M
1
6
F
M
R

22
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2
Problem 3-21
98 (38)
91.5 (32)
60 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-21 (a) SI Units –
PB = 101.325 kPa
w
i
i 419.04 (0.8x2257)
W
∆
= = +
∆
iW = 2224.6 kJ / kg
From chart 1b; t2 = 32 C
(b) Use Humidification (adiabatic) option to obtain 91.5 F db.
3-22 PB = 29.92 in.Hg.; 0
q =

(a) Using chart 1a
M16FMR
M
1
6
F
M
R

23
3-22 (cont’d)
w
i
i 1090 Btu / lbm
W
∆
= =
∆
From table A-1
f
fg
i-i 1090 - 196.1
x =
i 960.
=
1
x = 0.931 or about 93 %
(b) x will be the same
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
1
a
b
Problem 3-22
80
60
1090
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
M16FMR
M
1
6
F
M
R

24
3-23 Assume PB = 101.325 kPa; 0
q =

w
∆i 272.1
i kJ / kg
∆W 1000
= =
iw = 0.272 (on scale)
t2 = 22.6 C
10 20 30 40 50
60
70
80
90
100
110
110
120
120
ENTHALPY - KJ PER KILOGRAM OF DRY AIR
10
20
30
40
50
60
70
80
90
100
E
N
T
H
A
L
P
Y
-
K
J
P
E
R
K
I
L
O
G
R
A
M
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
C
5
10
15
20
2
5
30
35
40
45
50
D
RY
BU
LB
TEMPERA
TU
RE
-
°C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
10% RELATIVE HUMID ITY
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 WET BULB TEMPERATURE - °C
30
0
.7
8
0
.8
0
0
.8
2
0
.
8
4
0
.8
6
V
O
L
U
M
E
-
C
U
B
IC
M
E
T
E
R
P
E
R
k
g
D
R
Y
A
IR
0
.8
8
0
.9
0
0
.9
2
0
.9
4
HUMI
DITY
RATIO
-
GRAM
S
M
OISTURE
PER
KI
LOGR
AM
D
RY
AIR
1
2
Problem 3-23
38
22.6
20
80 %
0.272
R R
NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 101.325 kPa
Copyright 1992
SEA LEVEL
0
1.0 1.0

1.5
2.0
4.0
-4.0
-2.0
-1.0
-
0
.5
-0
.2
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2
.0
2.5
3
.0
4
.
0
5
.0
10
.0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-24 For adia. humidification
(a) w
∆i
= i 1131 Btu / lbw
∆W
=
M16FMR
M
1
6
F
M
R

25
3-24 (cont’d)
c a 2
q = m (i - i )
1
a
m 2000x60/13.14
=

a
m 9132 lba / hr
=

1 2
i 18.1 Btu / lba ; i 29.7 Btu / hr
= =
c
q 9132 (29.7 - 18.1) 105,931 Btu / hr
= =

w a 3 2 3 2
m m (W - W ) ; W = 0.0167; W 0.0032 lbv/lba
= =

w
m 9132 (0.01 67 - 0.0032) 123.3 lbw / hr
= =

M16FMR
M
1
6
F
M
R

26
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2
3
Problem 3-24
1131
30 %
110 (43)
60 (16)
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
(b) Solution similar to (a)
See diagram for construction on chart 1a.
1
3
Q
32 2000 2
=
3000 3
Q
12
= =

Layout 2L/3 on the chart and read:
W3 = 0.007 lbv/lba
M16FMR
M
1
6
F
M
R

27
I3 = 22.2 Btu/lba
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2
3
Problem 3-25
40 (4) 100 (38)
58.4 (15)
35
77
52
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-25 SI Units – Same procedure as above, read:
3
i 34 kJ / kga
=
3
W 0.007 kgv / kga
=
Layout the given data on Chart 1a as shown for problem 3-25.
a1
m 2000(60) 12.66 9,479lba hr
= =

M16FMR
M
1
6
F
M
R

28
3-26 (cont’d)
a2
m 1000(60) 14.44 4,155lba hr
= =

a1
a1 a2
m
32 9479
= 0.695
m +m 9479 4155
12
= =
+

Layout distance 32 on line from 1 to 2 to locate point 3 for the
mixture.
Read: i3 = 21.5 Btu/lbm
W3 = 0.0067 lbu/lba
For W, % Error =
(0.007 0.0067)100
4.5
0.0067
−
=
For I, % Error =
(22.2 21.5)100
3.3
21.5
−
=
3-27
250,000
SHF 0.8
200,000
= =
or SHF =
59
.81
73
=
M16FMR
M
1
6
F
M
R

29
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2
Problem 3-27
75 (24)
50 %
53 (12)
0.8
21.5
28.2
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.
4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-28 Refer to diagram for 3-27
(a) a 1 2 1 2
q = m (i - i ); i 28.2; i 21.5
= =

a
m 250,000 / (28.2 - 21.5) 37,313 lba / hr
= =

3
a 2
Q = m v 37,313 x 13.09 / 60 8,140 ft / min
= =

(b) similar procedure;
3
Q 3.85 m / s
=

M16FMR
M
1
6
F
M
R

30
3-29 (a) Use the AirQuantity option of program PSYCH, iterating on the
relative humidity and setting the minimum outdoor Air Quantity to
0.01, NOT ZERO.
Use the properties option to find the entering wet bulb
temperature of 62.6F. Then
φ = 0.852 (iterated)
ts = 56F
= 9,360 cfm
s
Q

(b) Proceed as above
φ = 0.882
ts = 56F
= 10,014 cfm
s
Q

3-30 Proceed as in 3-29 above.
φ = 0.92
ts = 56.1 56 F
≈
= 11,303 cfm
s
Q ≤

3-31 (a) 91
.
0
000
,
550
000
,
500
SHF =
=
M16FMR
M
1
6
F
M
R

31
3-31 (cont’d)
(b) a 2 1
q = m (i -i )

or a 2
m = q/(i -i )
1
a
550,000
m
(34.3 22.8)
=
−

a
m =47,826lba hr

a 2
2
m v 47,826
Q = = x 14.62=11,654 cfm
60 60

or 5.5 m3
/s
M16FMR
M
1
6
F
M
R

32
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
1
2
Problem 3-31
0.91
115 (46)
72 (22)
30 %
22.8
34.3
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.
4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
500
0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-32 a 2 1
q = m (i -i )

2
a
q
i = +i
m

1
a
1400 x 60
m 5,915.5
14.2
=

2
-5 x 12,000
i = +38.5
5,915.5
Btu/lba
2
i 28.3
= 6
M16FMR
M
1
6
F
M
R

33
Then from Chart 1a, t2= 67F
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
1
2
ADP
Problem 3-32
90
75
67
28.4
55
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-33 Use Adiabatic Mixing option of PSYCH with the Properties option to
enter requested data. Assume volume flow rates of 3 to 1 to obtain.
Tmix,db = 84.2 F
Tmix,wb = 71.3 F
3-34 Use Program PSYCH at Sea Level elevation
Iteration on the supply volume flow rate is required. This is the same as the
leaving air quantity for the coil.
M16FMR
M
1
6
F
M
R

34
3-34 (cont’d)
(a) Supply air quantity is 9,384 cfm.
(b) The outdoor air quantity is 938 cfm.
(c) Air enters the coil at 74.6 F db, 60.5 F wb at a rate of 9,740 cfm
(d) The coil capacity is 248,256 Btu/hr.
The amount of air returned is: (9,740 – 939) = 8,802 cfm.
3-35 Use Program PSYCH at 5,000 ft elevation
Iteration on the supply volume flow rate is required. This is the same as the
leaving air quantity for the coil.
(a) Supply air quantity is 11,267 cfm.
(b) The outdoor air quantity is 1,127 cfm.
(c) Air enters the coil at 74.6 F db, 62.1 F wb at a rate of 11,697 cfm
(d) The coil capacity is 334,143 Btu/hr.
The amount of air returned is: (11,697 – 1,127) = 10,570 cfm.
3-36 cfm
1000
Q0 =

(a) From Chart 1a
s
t =120/74 F
s
s r
q 200,000
m =
(i -i ) (37.2 22.8)
=
−

M16FMR
M
1
6
F
M
R

35
= 13,889 lb/hr = 1
m

3
s s s s
Q = m v = m (14.78)/60 = 3,421 ft /min

(b) o o o
m = Q /v 1000 x 60 / 12.61 4758 lb/hr
= =

r
1
m 13,889 4758
0.66;
m 13,889
−
= =

1
From Chart 1a t 61/ 47 F
=
3 1
t - t (119 61)
= −
(c) w s s 2
m = m (W -W ) 13,889 (0.0075 - 0.0036)
=
= 54.2 lbm/hr
(d) f 1 3 1
q = m (i -i ) =13,889 (32.8 18.6) 197,224 Btu/hr
− =

M16FMR
M
1
6
F
M
R

36
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
r
s
0
1
3
1 3
Problem 3-36
120
72
30 %
40 61
47
0.8
1150
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-37 (a) s
t 120 / 71.4 F Use Chart 1Ha
=
s 1
m 200,000 /(38.7 24.0) 13,605 lba/hr m
= − =

=
s
Q 13,605 x 17.85 / 60 4048 cfm
= =

(b) lba/hr
3947
60
x
)
2
.
15
/
1000
(
m0 =
=

r
1
1
m 13,605 3947
0.71; t 62.8/ 47 F
m 13,605
−
= = =

3 1
t -t (119.5 62.8)
= −
(c) w s s 1
m =m (w -W ) 13,605 (0.0088 - 0.0046) 57.14 lbw/hr
= =

M16FMR
M
1
6
F
M
R

37
(d) f
q 13,605 (33.8 - 20.2) 185,028 Btu/hr
= =

3-38 Assume fan power and
heat gain are load on the space
s
9384
m x 60 = 42,915 lbm/hr; Prob 3-34
13.12
=

fan duct s s c
W q m (i i
+ = −
)
= (4 x 2545) + 1000 = 11,180 Btu / hr
c
11
,180
i 20.8 20.54 Btu/lbm
42,915
= − =
State c is required condition leaving coil
Part a, b, and c are same as prob. 3-34;
(d) coil 1 1 c
q =m (i -i ) 42,915 (26.8 - 20.54) 268,648 Btu/hr
= =

M16FMR
M
1
6
F
M
R

38
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
s
r
0
1
c
1
Problem 3-38
100
72
55
20.54
50 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-39 )
i
i
(
m
W
);
i
i
(
m
q c
s
s
fan
s
r
s
r −
=
−
=

(a) c r
i 28 Btu/lbm; i 33.7 Btu/lbm
= =
Using Chart 1Ha
r
q 1,320,000 Btu/hr
=

fan
W 30 x 2545 76350 Btu/hr
= =

fan a s c
W 30 x 2545 76,350 = m (i -i )
= =

M16FMR
M
1
6
F
M
R

39
s a
q = 1,320,000 = m r s
(i -i )

10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
50
55
55
60
60
65
65
70
70
75
75
80
85
1
5
.5
1
6
.0
1
6
.5
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
7
.0
1
7
.5
1
8
.0
HUMIDITY
RATIO
-
POUNDS
MOISTURE
PER
PO
UND
DRY
AIR
r
0
c
s
s
Problem 3-39
90 (32)
80 (27)
50 %
59 (15)
62.5 (17)
0.8
R R
NORMAL TEMPERATURE
Copyright 1992
5000 FEET
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-0
.4
-
0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
10
00
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
Two unknowns two equations
Solve simultaneous:
fan s a r c
a
a
W +q = m (i -i )
1,320,00+76,3 50
m =
(33.7-28)
m =244,974lba/hr

M16FMR
M
1
6
F
M
R

40
s r s a
i = i - (q m )

s
1,320,000
i = 33.7 - =28.3
244,974
Btu/lba
Locate points on the condition line on Chart 1 Ha and point c is on
cooler process line horz. to left of points.
Read ts = 62.5 F, tc = 61.6F.
(a) s
244,974
Q = x16.2 = 66,143cfm
60

(b) 3
s
Q 31.2 m
=
s
H
3-40 English Units –Tucson, Arizona, Elevation 2,556 ft.
;
min 0
i =i =31.1 Btu/lba and sat. air min
t =64.5 F; PSYCH
Shreveport, Louisiana, Elevation 259 ft.
;
min 0
i =i = 42.5 Btu/lba and sat. air min
t 76.8 F; PSYC
=
SI Units – Tucson, Arizona
;
min 0
i =i 51.5 kJ/kga
= min
t =18.1 C; Chart 1b
Shreveport, Louisiana
;
min 0
i =i =75.5 kJ/kga min
t =24.8 C; Chart 1b
M16FMR
M
1
6
F
M
R

41
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
3
.0
1
3
.
5
1
4
.
0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
I
R
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
SL
TLO
Problem 3-40
Shreveport, LA
95
76.8
R R
NORMAL TEMPERATURE
Copyright 1992
259 FEET
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-0
.4
-
0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
10
00
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
M16FMR
M
1
6
F
M
R

42
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.00 2
.00 4
.00 6
.00 8
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40 45
45 50
50 55
55
60
60
65
65
70
70
75
75
80
85
85
1
4
.0
1
4
.
5
1
5
.
0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
I
R
1
5
.5
1
6
.0
1
6
.
5
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
TA
TLO
Problem 3-40
Tucson, Arizona
102
64.6
R R
NORMAL TEMPERATURE
Copyright 1992
2556 FEET
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-0
.4
-
0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
10
00
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-41 )
i
i
(
m
q s
r
s −
=

(a) = 1,319 lba/hr ton
s
m 12,000 /(28.2 19.1)
=
−
s
1319 x 15.6
Q 343 cfm/ton
60
= =

o
s
m r1 13
= 0.55 or 55%
m 23.5
r0
= =

(b) 3
s
Q 0.046 m / s - kW
≈

0 s
m /m 55%
≈

M16FMR
M
1
6
F
M
R

43
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
50
55
55
60
60
65
65
70
70
75
75
80
85
1
5
.5
1
6
.0
1
6
.5
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
7
.0
1
7
.5
1
8
.0
HUMI
DITY
RATIO
-
POUNDS
MOISTURE
PER
POUND
DRY
AIR
r
s
0
1
Problem 3-41
100 (38)
75 (24)
50 (10)
10 %
40 %
0.7
R R
NORMAL TEMPERATURE
Copyright 1992
5000 FEET
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-0
.4
-
0
.3
-0
.2
-0.
1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
10
00
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-42 2 2 3 2
500,000
q m (i i ); m
(41.1 21.9)
= − =
−

lba/hr
042
,
26
m2 =

= 6315 cfm
14.55/60
x
26042
Q2 =

lba/hr
6511
26,042
x
25
.
0
m0 =
=

F
5
.
49
/
5
.
67
t
;
25
.
0
m
/
m mix
3
0 =
=

M16FMR
M
1
6
F
M
R

44
3-42 (cont’d)
Preheat Coil:
ph 0 p 4 0
q = m c (t -t ) 6511 x 0.24 (60-6) 84,383 Btu/hr
= =

Heat Coil:
h 2 5 1
q = m (i -i ) 26,042 (28.4 - 20) 218,753 Btu/hr
= =

Humidifier:
w 2 2 5
m = m (W -W ) 26,042 (0.0144 - 0.0035)
=

283.9 lbw/hr
=
(b) 3
2 ph h
Q 2.98 m / s; q 24.7 kW; q 64.1 kW;
= = =

kg/s
036
.
0
mw =

M16FMR
M
1
6
F
M
R

45
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
3
2
4
1
5
2
Problem 3-42
1153
70 (21)
30 %
105 (40)
60 (16)
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HU MIDITY RATIO
'h
'W
3-43 Use Chart 1a; d a r
q m (i is)
= −

or )
i
i
/(
q
m s
r
d
a −
=

(a) m
a = 150 x 12,000 / (28.4-22) = 28,125 lbm/hr
d
Q 28,125 x 13.25/60 61,211 cfm
= =

m d
Q 0.20 Q 1,242 cf
= =
m
m m
m = 1
,242 x 60/13.5 5,521 lbm/hr [v assumed]
=

m r m
i =i 1.8 x 12,000/5,521 24.5 Btu/lbm; t 62 / 57 F
− = =
M16FMR
M
1
6
F
M
R

46
(b) 3 3
d m m
Q = 2.93 m /s; Q = .59 m /s; t = 17/14 C

10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
r
m
s
Problem 3-43
0.8
0.6
75 (24)
60 (16)
62 (17)
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-44 (a) a
15.0 x 12,000
m 29,508 lba/hr
(31.2 - 25.1)
= =

d m
Q 29,508 x 16.0/60 7,869 cfm; Q = 0.2 x Q
= =

s

1,574 cfm
=
m m
m =1
,574 x 60/16.2 5,829 lba/hr (v assumed)
=

m
i 35.7 1.8 x 12,000/5,829 27.5 Btu/lba;
= − =
M16FMR
M
1
6
F
M
R

47
m
t 62.5/58 F
=
(b) 3 3
s m m
Q =3.7 m / s; Q 0.74 m /s; t 17 /14.4 C
= =

3-45 Use Chart 1a; r
1
m 10
0.8
m 0r
= =

[Both design and min. load condition]
is = ir - /
m
q
s
m

s
r
d
s
i
i
Q
m
−
=

=
22.3)
-
(29.35
12,000
x
50
conditions
all
for
constant
is
m
lba/hr;
106
,
85
m s
s
=
Btu/lba
25.83
106
12,000/85,
x
25
35
.
29
i '
s =
−
=
(a) From Chart 1a; F
64
t '
s =
(b) s'
b
c
'
1
b
s
s i
)
m
m
(
i
m
i
m

+
=
+
271
.
0
7
.
31
8
.
25
8
.
25
2
.
24
)
i
i
(
)
i
i
(
m
m
'
1
'
s
'
s
s
c
b
=
−
−
=
−
−
=

(b) From chart 1a; cases
both
for
F
49
td =
M16FMR
M
1
6
F
M
R

48
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
r
s
0
1
s
0'
1'
s
s'
Problem 3-45
0.9
95 (35)
85 (29)
77 (25)
64 (18)
55 (13)
50 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-46 Refer to problem 3-45. Results are similar.
3-47 (a) It is probably impossible to cool the air from 1 to 2 in one
process. The extension of line 12 does not intersect the
saturation curve.
(b) Cool the air to state 1' and then heat to state 2.
M16FMR
M
1
6
F
M
R

49
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
1
2
1'
2
Problem 3-47
80 (27)
60 (16)
52 (11)
67
54
90 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-48 (a)
c
s
m sh
= =.83
m ch

7
h
s
m cs
= 0.16
m ch
=

3
c
h
m 0.837
5.14
m 0.163
= =

s r s
q m (i i
= −
)
s
50 x 12,000
m 93,750 lba/hr
(28.2-21.8)
= =

M16FMR
M
1
6
F
M
R

50
s
Q 93,750 x 13.2/60 20,625 cfm
= =

(b) 3
s
Q 9.7 m /
=
s
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
r
c
h
s
Problem 3-48
90 (32)
75 (24)
52 (11)
90 %
0.65
20 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0.
1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-49 See diagram of problem 3-48
(a) c h
s s
m m
36 10.1
0.9; 0.10
m 46.3 m 46.3
= = = =

; c
h
m 0.9
9.0
m 0.10
= =

s
50 x 12,000
m 83,333 lba/hr
(30.1 - 22.9)
= =

s
Q =83,333 x 15.67/60 21,763 cfm
=

M16FMR
M
1
6
F
M
R

51
(b) 3
s
Q =10.3 m /s

3-50 (a) See diagram for problem 3-48
c
c c r c c s
s
m
= 0.837; q = m (i -i ); m 0.714 x m 0.837 x 93,750
m
= =

;
c
m 78,469 lba/hr
=
c
Q 78,469 x 13.04/60 17,054 cfm
= =

c
q 78,469 (28.2-20.6) 596,364 Btu/hr
= =

(b) 3
c c
Q =8.1 m /s; q 175 kW
=

3-51 SI Units
(a) On the basis of volume flow rate using Chart 1b:
2 3
13
Q = Q 0.69 x 1.18 0.815
12
= =
m3
/s
and m
1 3 2
Q = Q - Q = 1.18 0.815 0.365
− =
3
/s
(b)
3
34 a3 4 3 4 3
3
34
Q
q = m (i -i ) = (i -i )
v
1.18
q = (47.8-41.0) = 9.6 kW
0.835

M16FMR
M
1
6
F
M
R

52
10 20 30 40 50
60
70
80
90
100
110
110
120
120
10
20
30
40
50
60
70
80
90
100
E
N
T
H
A
L
P
Y
-
K
J
P
E
R
K
I
L
O
G
R
A
M
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
C
5
10
15
20
2
5
30
35
40
45
50
D
RY
BU
LB
TEMPERA
TU
RE
-
°C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30
0
.7
8
0
.8
0
0
.8
2
0
.
8
4
0
.8
6
V
O
L
U
M
E
-
C
U
B
IC
M
E
T
E
R
P
E
R
k
g
D
R
Y
A
IR
0
.8
8
0
.9
0
0
.9
2
0
.9
4
HUMIDITY
RATIO
-
G
RAM
S
M
OISTURE
PER
KILOGR
AM
D
RY
AIR
1
2
3 4
Problem 3-51
Problem 3-51
29
24
17.2
12
50 %
11
14.7
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

1.5
2.0
4.0
-4.0
-2
.0
-1.0
-0
.5
-0
.2
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2
.0
2.5
3
.0
4
.
0
5
.0
10
.0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
English Units
(a) 1 34
Q = 640 cfm; q = 33,684 Btu/hr

3-52 (a),(b)
From Chart 1b, states 1.4 and ADP are known. Based on approx.
11.8 C db, 11.2 C wb, and 90% RH locate state 2.
Then for full load design condition air is cooled from 1 to 2 and the
room process proceeds from 2 to 4.
M16FMR
M
1
6
F
M
R

53
For the high latent load condition, the air at 2 is reheated to state 3
where it enters the space and the process proceeds to state 4.
(c) 2
24 a 4 2 4 2
2
Q
q = m (i -i ) = (i -i )
v

2
Q =35 x 0.817 (47.7-32)
; m
2
Q 1.8
=
2 3
/s
12 a 1 2
12
1.82
q = m (i -i ) = (60.6-32)
0.817
q = 63.7 kW

34 a 4 3
34
23 24 34
1.82
q = m (i -i )= (47.7-39.4)
0.817
q = 18.5 kW
q = q - q = 35-18.5=16.5 kW

M16FMR
M
1
6
F
M
R

54
10 20 30 40 50
60
70
80
90
100
110
110
120
120
10
20
30
40
50
60
70
80
90
100
E
N
T
H
A
L
P
Y
-
K
J
P
E
R
K
I
L
O
G
R
A
M
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
C
5
10
15
20
2
5
30
35
40
45
50
D
RY
BU
LB
TEMPERA
TU
RE
-
°C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30
0
.7
8
0
.8
0
0
.8
2
0
.8
4
0
.8
6
V
O
L
U
M
E
-
C
U
B
IC
M
E
T
E
R
P
E
R
k
g
D
R
Y
A
IR
0
.8
8
0
.9
0
0
.9
2
0
.9
4
HUMIDITY
RATIO
-
GRAM
S
M
OISTURE
PER
KILOGR
AM
D
RY
AIR
1
ADP 2 3
4
3
Problem 3-52
Problem 3-52 21
27
23
17
19
14
11.8
11
9
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

1.5
2.0
4.0
-4.0
-2.0
-1.0
-
0
.5
-0
.2
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2
.0
2.5
3
.0
4
.
0
5
.0
10
.0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-52 English Units
(a),(b) See above
(c) Btu/hr
2
Q = 4103cfm ;

12
q =221,243

Btu/hr;
34
q 67,49
=
8 2
23
q 52,50
=
Btu/hr
3-53 English Units
(a) s r s s
q=m (i -i ); m 5000 x 60/13.2 22,727 lba/hr
= =

(specific volume value of 13.2 ft3
/lbm is assumed.)
M16FMR
M
1
6
F
M
R

55
s r s
i = i - q /m =28.2 10 x 12,000 / 22,727 22.9 Btu/lba
− =

s o s o
t = t = 57.5 F; W =W 0.0083 lbv/lba
=
(b) r r
m s
m m
0m
= = 0.462
m m
0r

r
m =0.462 x 22,727 10,500 lba/hr
=

o
m 22,727 10,500 12,227 lba/hr
= − =

r
Q 10,500 x 13.68/60 2,394 cfm
= =

o
Q 12,227 x 12.11/60 2,468 cfm
= =

(c)
r
m'
m 0'm'
= =0.578
m 0'r

r o
m =0.578 x 22,727 13,131 lba/hr; m 9,596 lba/hr
= =
'
r o'
Q =13,131 x 13.68/60 2,994 cfm; Q 9,596 x 13.48/60
= =

= 2,156 cfm
(d) c s m' s
q = m (i -i ) 22,727 (28.4 - 22.8) 127,271 Btu/hr
= =

M16FMR
M
1
6
F
M
R

56
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMIDITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
r
0
m
s
0'
r
m'
ADP
Problem 3-53
0.8
1150
75 (24)
70 (21)
65 (18)
57.5 (14)
40 (4)
43 (6)
90 %
50 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-53 SI Units
(a) ts = 14.2C; Ws = 0.0083 kgv/kga
(b)
3
r
Q =1.13m s
;
3
o
Q =1.17m s

(c)
3
r
Q =1.41m s
;
3
o'
Q =1.02m s

(d) c
q = 37.3 kW

3-54 (a) Any combination that will yield
an enthalpy less than 57.0 kJ/kga or 33 Btu/lba
M16FMR
M
1
6
F
M
R

57
(b) r
s m
kga/s
95
.
5
84
.
0
/
5
m
=
=
=
o
r
m mr
= =0.36
m 0r

o
m 0.36 x 5.95 2.14 kga/s
= =

3
o
Q = 2.14 x 0.852 = 1.82 m /s = 3,857cfm

(c) ad
t 15.4 C or 60F
=
(d) (Essentially, no difference)
o n m s r s
q /q = (i -i )/(i -i ) = 1.0

10 20 30 40 50
60
70
80
90
100
110
110
120
120
10
20
30
40
50
60
70
80
90
100
E
N
T
H
A
L
P
Y
-
K
J
P
E
R
K
I
L
O
G
R
A
M
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
C
5
10
15
20
2
5
30
35
40
45
50
D
RY
BU
LB
TEMPERA
TU
RE
-
°C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30
0
.7
8
0
.8
0
0
.8
2
0
.
8
4
0
.8
6
V
O
L
U
M
E
-
C
U
B
IC
M
E
T
E
R
P
E
R
k
g
D
R
Y
A
IR
0
.8
8
0
.9
0
0
.9
2
0
.9
4
HUMIDITY
RATIO
-
G
RAM
S
M
OISTURE
PER
KI
LO
GR
AM
D
RY
AIR
r
s
0
m
s
Problem 3-54
25 (77)
18 (64)
20 (68)
0.6
57
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

1.5
2.0
4.0
-4.0
-2.0
-1.0
-0
.5
-0
.2
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2
.0
2.5
3
.0
4
.
0
5
.0
10
.0
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
M16FMR
M
1
6
F
M
R

58
3-55
424,000
SHF
530,000 424,000
−
=
−
= -4
Construct condition line on
Chart 1a with preheat and
mixing processes.
(a) sen s p r s
q = -424,000 = m c (t -t )

s
424,000
m 88,333 lba/hr
0.24 (75 95)
−
= =
−

3
s
Q =88,333 x 14.07/60 20,714 cfm or 9.8 m /s
=

(b)
r
r
m
m hm
= =0.33; m 0.33 x 88,333 lba/hr
m hr
=

r r
m =29,150 lba/hr; Q 29,150 x 13.68/60
=

3
6,646 cfm or 3.14 m /s
=
h
h
m
m
=1 0.33 0.67; m 0.67 x 88,333
m
− = =

h h
m 59,183 lba/hr; Q 59,183 x 13.1/60
= =

3
h
Q 12,922 cfm or 6.1 m /s (at heated condition)
=

(c) ph h p h o
q =m c (t -t ) = 59,183 x 0.24 (60-35)

q=355,098 Btu/hr or 104 kW

(d) m
q =88,333 x 0.24 (95 - 65) 635,998 Btu/hr or 186 kW
=

M16FMR
M
1
6
F
M
R

59
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
POUND
DRY
AIR
r
0 h
m
s
s
Problem 3-55
-4
95 (35)
75 (24)
50 %
60 (16)
35 (2)
20 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-56 Refer to chart 1a.
(a)
34 a3 4 3 3 4 3
3
34 3
3
4 3
3
3
60
q = m (i -i ) = Q x (i -i )
v
q v (1750 x 13.23)
Q x =
60(i -i ) 60(28.1-23)
Q = 75.7 or 76 cfm = 0.040 m /s

M16FMR
M
1
6
F
M
R

60
(b) t3db = 58.5 F and 80% RH or 15 C
(c)
3
2 3
31
Q = ; Q = 0.754 x 75.7 = 57 cfm or 0.028 m /s
12

3
1
Q = 76 - 57 = 19 cfm or 0.012 m /s

10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY - BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
T
EMPERAT
UR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2
3
4
Problem 3-56
84
70
75
62
58.5
50
90 %
50 %
0.8
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-57 (a) Refer to Chart 1
A reheat system is required. Process 1-2 is for the coil. Process 3-4
is defined by the SHF = 0.5
Process 2-3 represents the required heat.
M16FMR
M
1
6
F
M
R

61
State 3 is defined by the intersection of the reheat and space
condition lines.
(b)
3
34 a3 4 3 4 3
3
34 3
3
4 3
3
3
Q x 60
q = m (i -i ) = (i -i )
v
q v 100,000 x 13.4
Q = =
60(i -i ) 60(28.2-23.9)
Q = 5,194 cfm or 2.5 m /s

(c)
12 a 1 2
12
23
23
5,194 x 60
q = m (i -i ) = (34.2-20.2)
13.4
q = 325,594 Btu/hr or 95.4 kW
5,194 x 60
q = (23.9-20.2)
13.4
q =86,050 Btu/hr or 25.2 kW

M16FMR
M
1
6
F
M
R

62
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BULB
TEMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35
40
40
45
45 50
50 55
55 60
60
65
65
70
70
75
75
80
80
85
90
1
2
.
5
1
3
.
0
1
3
.5
1
4
.0
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
4
.5
1
5
.0
HUMI
DITY
RATIO
-
POUN
DS
MOISTURE
PER
PO
UND
DRY
AIR
1
2 3
4
ADP
Problem 3-57
85
70
75
62
66
56
51
45
50 %
R R
NORMAL TEMPERATURE
Copyright 1992
SEA LEVEL
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-0
.5
-
0
.4
-0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
1
000
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
3-58 Assume room temperature humidity of 50%
and layout the state
processes on
required from point c to s.
Supply Air:
=
sen
q 120,000 x 0.5 60,000 Btu/hr
= =
s p s r
m c (t -t )

s
60,000
m 53,192 lba/hr
0.24 (75-70.3)
= =

M16FMR
M
1
6
F
M
R

63
3
s
Q =53,192 x 16.33/60 14,477 cfm or 6.8 m /s
=

Mixed Air:
o
m 53,192 x 0.333 17,703 lba/hr
= =

3
o
Q 17,713 x 17.2/60 5,078 cfm or 2.4 m /s
= =

r
m 53,192 17,713 35,479 lba/hr
= − =

3
r
Q =35,479 x 16.5/60 9,757 cfm or 4.6 m /s
=

Reheat:
rh c p s c
q = m c (t -t ) 53,192 x 0.24 (70.3-55.2)
=

192,768 Btu/hr or 56.5 kW
=
Coil:
c m m c
q =m (i -i ) 53,192 (34.4 - 24.2)
=
542,558 Btu/hr or 159 kW
=
= %
1
.
5
412
,
200
100
)
109
,
190
412
,
200
(
=
−
M16FMR
M
1
6
F
M
R

64
10 15 20 25
30
35
40
45
50
55
55
60
60
15
20
25
30
35
40
45
50
E
N
T
H
A
L
P
Y
-
B
T
U
P
E
R
P
O
U
N
D
O
F
D
R
Y
A
I
R
S
A
T
U
R
A
T
I
O
N
T
E
M
P
E
R
A
T
U
R
E
-
°
F
35
40
45
50
55
6
0
65
70
7
5
8
0
8
5
9
0
95
100
105
110
115
12
0
DR
Y
BUL
B
T
EMPERATUR
E
-
°F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
50
55
55
60
60
65
65
70
70
75
75
80
85
1
5
.5
1
6
.0
1
6
.5
V
O
L
U
M
E
-
C
U
.F
T
.
P
E
R
L
B
.
D
R
Y
A
IR
1
7
.0
1
7
.5
1
8
.0
HUMIDITY
RATIO
-
POUNDS
MOISTURE
PER
PO
UND
DRY
AIR
r
0
m
c s
s
Problem 3-58
90 (32)
75
75 (24)
50 %
70 (21)
55 (13)
90 %
0.6
0.5
R R
NORMAL TEMPERATURE
Copyright 1992
5000 FEET
0
1.0 1.0

2.0
4.0
8.0
-8.0
-4.0
-2.0
-1.0
-
0
.5
-0
.4
-
0
.3
-0
.2
-0
.1
0
.1
0
.2
0
.
3
0.4
0.5
0.6
0.8
-2000
-1000
0
500
10
00
1
5
0
0
2
0
0
0
30
0
0
50
00
-

SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HUMIDITY RATIO
'h
'W
M16FMR
M
1
6
F
M
R

Chapter 4
4-1 (a) comfortable
(b) too warm
(c) comfortable
(d) too dry
4-2 (a) comfortable
(b) too warm
(c) comfortable
(d) too dry
4-3 (a) Assume sedentary dry bulb of 78 F, clo = o.5, met. = 1.8;
using equation 4-4a, to,act. = 75 – 5.4(1 + 0.5)(1.8 – 1.2) = 71 F
Relative humidity should be less than 50%
(b) Should wear a sweater or light jacket and slacks.
(clo = 0.8)
4-4 Use fig 4-1
(a) Summer, to = 76 F or 24 C; Winter, to = 72 F or 22 C
(b) Use equation 4-4a as a guide, with clo = 0.2,
met = 3.0, db
t 76
= F
M16FMR
M
1
6
F
M
R

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AC 6Ed.pdf