Heating, Ventilating and Air Conditioning Analysis and Design , 6th Edition Solution Manual

Name: Heating, Ventilating and Air Conditioning Analysis and Design , 6th Edition Solution Manual
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Author: Theodore Long

Heating, Ventilating and Air Conditioning Analysis and Design, 6th Edition Solution Manual provides expert solutions to textbook questions, making complex problems easier to solve.

Theodore Long

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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

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 =

3
( )
( )
( )
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 10-3m3/s
m 6.16 x 10Q
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

4
= 600 xiQ 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 Km
q = 20,564 kw
=
−

kJ
s
permitted
1-12 =watq airq−
11,200(1)(10) =
= 25000x60x14.7x144x0.24(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

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
st = tsur 212 F≈
= 10x2.618x(212-50) = 4241 Btu/hrq
1-15 A=
π x 0.25x4 = 3.14 16 m2
hA(tq = s- )t∞
h=
s
q
A(t -t )∞
= 1250
3.1416(100 10)− ; h = 4.42 W/(m2 – C)
1-16 (tpq 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-Fpc
q = 22,208x0.24(58-76) = -95,939 Btu/hr
Negative sign indicates cooling
1-17 (t1 pm c 3-t1) +
ofit basis fo

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7
permitted
Requests for permission or further information should be addressed to the Permission Department, John
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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permitted
Requests for permission or further information should be addressed to the Permission Department, John
Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.
Chapter 3
3-1 (a) Pv = =sr P
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 = = = 3v
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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8
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 = 3a
a
R T 0.287(300)
= =0.88 m /kga
P (101.325 - 3.6)
(b) SI Units 0.0C; 101.325 kPa
Pv = 0.61 kPa, Table A-1b
W = 0.6219(0.61) =0.00377 kgv/kga
(101.325 - 0.61)

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9
i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga
v = 30.287(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
Pv = 3.60 kPa, Table A-1b

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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 = 30.287x300 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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11
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; psia14.696P%;50 ==
φ
sv
s
v PPor
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 / hram 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 / minam 18,300 lbma / hr=
3-6 SI Units –
(a) t = 13 C and relative humidity 80%
then va 0.820 m≈ 3 / kga; am 2.36 / 0.82 2.88 kga / s= =
(b) Assuming same conditions
;3
av 0.985 m / kga= am 2.36 / 0.985 2.40 kga / s= =

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14
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
ENTHALPY - BT U PER POUND OF DRY AIR
SATURATION TEMPERATURE - °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 WET BULB TEMPERAT URE - °F
85
90
12.5
13.0
13.5
14.0 VOLUME - CU.FT . PER LB. DRY AIR
14.5
15.0
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

2.04.0
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
-2 00 0
-1 00 0
0
5 00
1 000
1 50 0
2000
300 0
5000
-

SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPY
HU MIDITY 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

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Heating, Ventilating and Air Conditioning Analysis and Design , 6th Edition Solution Manual

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