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SOLUTION MANUAL

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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) µNs x1.488 = 16.5 3600 sec/hr m2

(d) 1050

Btu J 1 2.20462 Ibm MJ x x = 2.44 Ibm 9.48x10−4 Btu kg kg

(e) 12,000

(f) 14.7

1-2

Ibf in2

Btu 1 x = 3.52 kW Ibm 3.412 x 6894.76 = 101 kPa

(a) 120 kPa x

(b) 100

(c) 0.8

lbf / in2 = 17.4 lbf/in2 6.89476kPa

W x 0.5778 = 57.8 Btu/hr-ft-F m −K

W 2

m −K

x 0.1761 = 0.14 Btu/hr-ft2-F

(d) 10-6 N-s/m2 x

1 lbm = 6.7 x 10-7 1.488 ft − sec

(e) 1200 kW x 3412 = 4.1 x 10-6 Btu/hr

2

(f) 1000

1-3

1-4

1 Btu 1 kg Btu kJ x x = 430 1.055 kJ 2.2046 lbm lbm kg

Hp = 50 (ft) x 0.3048 (

m ) = 15.2 m ft

∆P =

15.2 m 9.807 N x ( ) x 1000 (kg/m3) = 149 kPa 1000 Pa/kPa 1 kg

∆P =

m kg 4 9.807 N (ft) x 0.3048 ( ) x ( ) x 1000 ( 3 ) ft 12 1 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)

11 hrs/day, 5 days/wk

hrs days (22) = 242 hrs / month day months

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3

ratio =

1-7

( 624 ) kw = 1.57 ⎛ ( 96,000 ) kw − hr ⎞ ⎜ ⎟ 242 hr ( ) ⎝ ⎠

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

⎡ $1000 ( ) ⎢1⎢⎣ P=

0.012 ⎞ ⎛ ⎜ 1+ ⎟ 12 ⎠ ⎝

−(12 )(12 ) ⎤

⎛ 0.012 ⎞ ⎜ ⎟ ⎝ 12 ⎠

⎥ ⎥⎦

P = $134,000

1-9

 = VA = 2 x 3.08 x 10-3 = 6.16 x 10-3m3/s (a) Q  = 6.16 x 10-3 x 998 = 6.15 kg/s  = ρQ m (b) A=

π 4

(0.3)2 = 7.07 x 10-2 m2

 = 7.07x10-2 x 4 = 0.283 m3 / s; ρ = 1.255 kq/m3 Q  = 1.225 x 0.283 = 0.347 kg/s m

1-10

V = 3x10x20 = 600m3

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4

 = 600 x 1 x 1 = 4.17 x 10-2 m3/s Q i 4 3600

1-11

 p ∆T q = mc

c p = 4.183 kJ/(kg-K)

ρ = 983.2 kg/m3

1-11 (cont’d) 3 q = (1) m ( 983.2 ) kg ( 4.183 ) kJ ( 5 )c = 20,564 kJ s kg − K s m3

q = 20,564 kw

1-12

q wat = −q air 11,200(1)(10) =

=

5000x60x14.7x144x0.24(t 2 − 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

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.

5 t2 = 90 -

29,400 = 85.3 C 6279

1-14 q = hA(ts- t ∞ ) A= π (1/12) x 10 = 2.618 ft2

t s = t ≈ 212 F sur q = 10x2.618x(212-50) = 4241 Btu/hr 1-15

A= π x 0.25x4 = 3.14 16 m2

q = hA(ts- t ∞ ) q 1250 = ; h = 4.42 W/(m2 – C) h= A(t s -t ∞ ) 3.1416(100 − 10)

 xρ  =Q  p (t2-t1) ; m 1-16 q = mc

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

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

 1cp (t3-t1) + 1-17 m 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.

6

 2cp2 (t3-t2) = 0 m c p1 = c p2 t3 =

 1t1 + (m 1+ (m

 2t 2 ) m  2) m

 ρ = 1000x 1 = Q m 2 1

14.7x144 = 73.5 lbm/min 53.35(460 + 50)

1-17 (cont’d)

 ρ = 600x 2 =Q m 2 2 t3 =

14.7x144 = 46.7 lbm/min 53.35(460 + 50)

(73.5x80) + (46.7 x 50) = 68.3 F (73.5 + 46.7)

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.

7 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.

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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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.

Chapter 3 3-1

(a) Pv = φ r Ps = 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)

P 1430 Pv = 0.0104 kg/m3 = RvT or ρv = v ; ρv = Rv T 462.5(297) ρv 0.196(144) = 0.00062 lbv/ft3 85.78(535)

or

(c) W =

or

3-2

0.6219 (1.43) = 0.00893 kgv/kga (99.57)

0.6219(0.196) = 0.00854 lbv/lba 14.5

(a) English Units – t = 80F; P = 14.696 psia; Pv = 0.507 psia Table A-1a W = 0.6219

Pv 0.6219 (0.507) = = 0.0222 lbv/lba Pa (14.696 − 0.507)

i = 0.24t + W(1062.2 + 0.444t) i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm

8

Ra T 53.35(460 + 80) = = 13.61 ft3/lbm Pa (14.696 − 0.507)144

v=

(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

53.35(492) = 12.48 ft3/lbma (14.696 − 0.089)144

v= 3-2

(a) SI Units – 27C; 101.325 kPa Pv = 3.60 kPa, Table A-1b W = 0.6219

Pv 0.6219(3.6) kgv = = 0.0229 Pa (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=

Ra T 0.287(300) = =0.88 m3 /kga Pa (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

0.287(273) = 0.778 m3 /kga (101.325 - 0.61)

v=

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

Pv 0.6219(0.507) = = 0.0269 lbv/lba Pa (12.24 - 0.507)

i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma v=

RaT 53.35(540) = = 17.05 ft3 / lbma Pa (12.24 - 0.507) 144

(b) English Units – t = 32 F, Pv = 0.089 psia ( Table A-1a) W=

0.6219(0.089) = 0.00456 lbmv/lbma (12.24 − 0.089)

i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma v= 3-3

53.35(492) = 15.00 ft3/lbma (12.24 − 0.089)144

(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=

0.287x300 (84.436 - 3.60)

= 1.065 m3 / kga

(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=

3-4 (a)

0.287 x 273 = 0.935 m3 / kga (84.436 - 0.61)

English Units – 70F, Pv = 0.363 psia

Pv = φ Pg = 0.75(0.363) = 0.272 psia W=

0.6219 (0.272) (14.696 - 0.272)

= 0.0117 lbmv / lbma

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) = 0.0141 lbmv / lbma (12.24 - 0.272)

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

W=

0.6219 x 1.755 = 0.0110 kgv / kga (101.325 - 1.755)

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 = 0.0132 kgv / kga (84.186 - 1.755)

i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga 3-5 English Units – t = 72 Fdb; φ = 50 %; P = 14.696 psia

φ=

Pv or Pv = φPs ; Pv = 0.5(0.3918) = 0.196 psia Ps

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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

 a = 5000 (0.0762) = 381 lbma / min = 22,860 lbma / hr m (b) Using PSYCH ρ a = 0.0610 lbma / ft3 or va = 16.4 ft3 / lba

 a = 5000 (0.061) = 305 lbma / min m

= 18,300 lbma / hr

3-6 SI Units – (a) t = 13 C and relative humidity 80%

 a = 2.36 / 0.82 = 2.88 kga / s then va ≈ 0.820 m3 / kga; m (b) Assuming same conditions

 a = 2.36 / 0.985 = 2.40 kga / s v a = 0.985 m3 / kga ; m 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.

13

3-7 English Units – t = 80F, 60% RH (a) Pv = φ Ps = 0.6 (0.507) = 0.304 psia

t dp = (t sat @ Pv ) = 64.5 F (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

t dp =(t sat at Pv ) ≈ 18.4 C (b) Same as (a) above 3-8 t dp ≤ 9C (48F)

φ ≤ 42% ;

W ≤ 0.0071 kgv / kga (lbv / lba)

Chart 1a & 1b

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.

14

ASHRAE PSYCHROMETRIC CHART NO.1 R

90

AMERICAN S OCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.



SEA LEVEL

1.0

60

55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

50

.028

60 1 .0



85 15

-2 00 0 -1 00 0

2. 0

0. 6

SENSIB LE HEAT TOTAL HEAT

5000 3000

0 .5

4 .0 8 .0

Qs Qt -2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

45

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

55

RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDITY RATIO

'W

75

35

50

14 .5

.018

R

D

A

N

ER

U

M

P

PO

TE N TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU SA

25

E

% 90

ME

% 60

-C

60

.5

50

% 50

45

A IR

40 35

W=0.0071

13

45

15

Room

%

4 0%

40 1 3.

30 %

0

35

20 % 1 2. 10

3-9

48 (9)

15

72 (22)

20

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

5

Y I VE H UMIDIT 10% RE LAT

45

.012

.010

40

.008

.006

35 .004

.002

115

55

RY .D LB ER

50

T. P

dp

U.F

42 %

% 70

DR Y BUL B T EMPERAT UR E - °F

80

55

LU VO

20

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

TU

F

30

O

Problem 3-8

E

D

-°

R

F

Y

A

IR

75

70

25

ENTHALPY - BT U PER POUND OF DRY AIR

(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

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.

30

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

 ν = 58.7 lbm/hr (b) m

 2 = 1027 cfm Q 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) ql = 31.5 - 29.3 = 2.2 Btu / lba (c) qs = 29.3 – 23.2 = 6.1 Btu / lba (d) q = ql + qs = 8.3 Btu / lba

3-12

(a) W2* = W1 =

0.6219 (0.3095) = 0.0134 kgv / kga (14.696 − 0.3095) 0.24 (65 - 80) + ( 0.0134 x 1056.5) = 0.00993 lbv / lba (1096 - 33)

also W1 = 0.6219 Pv1 / (P – Pv1) 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.

16 Pv1 = (0.00993 x 14.696) / ( 0.6219 + 0.00993) = 0.231 psia 3-12 (cont’d)

0.231 = 0.46 or 46% 0.507

φ1 =

(b) P = 29.42 – (0.0009 x 5000) = 24.92 in.Hg. or P = 12.24 psia

W2* =

0.6219 x (0.3095) = 0.01613 lbv/lba (12.24 - 0.3095)

0.24(65 − 80) + (0.01613 x 1056.5) = 0.01265 lbv / lba ( 1096 - 33) or kgv / kga

W1 =

Pv1 = 0.01265 x 12.24 / ( 0.6219 + 0.01265) = 0.244 psia

φ1 = 3-13

0.244 = 0.48 or 48% 0.507

(a) Sea Level

Dry Bulb, F

Wet Bulb, F

85 75 74.6 88.6 100

60 59.6 65.1 70 85.8

Dew point F 40.6 49.2 60.1 60.9 81.7

Humid. Ratio, lba/lbv 0.0053 0.0074 0.0111 0.01143 0.0235

Mass Enthalpy Rel. Density Btu/lba Humid., % lba/ft3 26.6 21 0.072 26.1 40 0.073 30 60 0.073 33.8 40 0.071 50 56 0.068

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.

17 (a) 5000 ft. Dry Bulb, F

Wet Bulb, F

85 75 71.2 102.7 100

60 58.6 61.6 70 81.3

Dew point F 45.1 49.2 56.7 55.8 76.1

Humid. Ratio, lba/lbv 0.0076 0.0089 0.0118 0.01143 0.0235

Mass Enthalpy Rel. Density Btu/lba Humid., % lba/ft3 28.7 25 0.060 27.7 40 0.061 30 60 0.061 37.3 22 0.058 50 47 0.057

(c) Note effect of barometric pressure.

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.

18 3-14 ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt



-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 - 0.4 -0.3

0. 3

-1 .0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHAL PY HU MIDITY RAT IO

'W

75

35

50 14 .5

.018

F

R ER

U

M

P

PO

TE N TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

.0 V

% 90

Max RH=49.6 % ME

% 60

-C

U.F

60

13

50

40

% 50

45

35

IR YA

.5

45

15

W=0.0083

Room

%

4 0%

40 30%

1 3. 0

35

20 %

3-15

52 (11)

15

72 (22)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.010

40

.008

.006

35

.004

ENTHALPY - BT U PER POUND OF DRY AIR

Use Chart 1b, SI

 (i1 − i2 ) = (b) q = m

.012

25

(a) td = 10 C; SHF = 0.62 2.4 (57.1 - 34) = 63.95 kJ / s = 63.95 k W 0.867

q s = 63.95 (0.62) = 39.65 kW 3-15

45

.002

115

55

dp

R B. D RL PE

50

T.

% 70

DR Y BUL B T EMPERATUR E - °F

80

55

U OL

20

14

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

TU

Problem 3-14

E

D

-°

R

F

Y

A

IR

75

70

Use Chart 1a, IP (a) td = 52 F; SHF = 0.63

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.

30

19 3-15 (cont’d)

5000(60) (32 - 22.6)= 203,317. Btu/hr 13.87

q =

(b)

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt



-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

80

.024

BU

LB

TE MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

A

IR

75

R TU TI O A R

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

U -B T

TU

Y

25

1

SA

AL P

.014

E

%

L B. ER

55 %

DR

1 3 .5

4 0%

40 1 3.

30%

0

35

20 %

52 (10) 55 (13)

15

20

80 (27)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

.006

35

.004

.002

115

% 50

DR Y BUL B T EMPERAT UR E - °F

IR YA

50 45

35

T. P

2

45 40

U.F

% 70

60

15

60

-C

ADP

ME

80

LU VO

% 90

55

50

1 4 .0

65

60

20

45

65

N

TH

70

N

PE

TE

R

M

P

PO

ER

U

A

N

D

O

30

.016

70

E

F

D

-°

R

F

Y

Problem 3-15

12 0

R

30

25

ENTHALPY - BT U PER POUND OF DRY AIR

q s = 203,317 (0.63) = 128,089. Btu/hr 3-16

(a) i1 = 30 Btu / lba; v1 = 13.78 ft3 / lba; W = 0.0103

lbv ; φ1 = 50% lba

(b) i1 = 51.6 kJ / kga v1 = 0.86 m3 / kga 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.

20

3-16 (cont’d) W1 = 0.0103

kgv kga

φ1 = 50% 3-17

Use the Heat Transfer option of program PSYCH:

q = 148,239 Btu/hr q s = 102,235 Btu/hr SHF = 0.69

3-18 Use the Heat Transfer option of program PSYCH for sensible heat transfer only:

q s = − 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 0 6000

Enthalpy Btu/lbm 32.2 36.3

Rel. Hum percent 42 45

Hum. Ratio lbv/lba 0.0107 0.0144

Density lba/ft3 0.072 0.058

 a = 5000 x 0.072 x 60 = 21,600 lba/hr At sea level: m 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.

21 3-19 (cont’d)

 a = 5000 x 0.057 x 60 = 17,100 lba/hr At 6000 feet:: m Percent Decrease at 6000 ft:

PD = 3-20

(21,600 − 17,100)100 = 20.8% 21,600

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

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

(200,534 − 190,224)100 = 5.1 % 200,543

3-21 (a) English Units – PB = 29.92 in.Hg. ; q = 0

∆i = iw = 180.2 + 0.8 (970.2) ∆W iw = 956.4 Btu / lbv From chart 1a; t2 = 91.5 F

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.

22

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT TOTAL HEAT

50 00 300 0

0 .5

4 .0 8 .0

Qs Qt



-8 -4 .0.0 -2.0

0.4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDITY RATIO

'W

2

75

35

50 14 .5

F

R

O D

A

N

ER

U

M

P

PO

TE N

70

TI O

.014

A R TU SA

25

14

E

65

.0 V ME

% 60

-C U.F

55 60

.5

50

% 50

45

A IR

40 35

% 13

45

15

RY .D LB ER

50

T. P

% 70

1

4 0%

40 30%

1 3. 0

35

20 %

15

20

110

105

100

95

90

85

80

65

60

55

50

45

40

35

75

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

.006

35

.004

.002

25

91.5 (32) 98 (38)

ENTHALPY - BT U PER POUND OF DRY AIR

3-21 (a)

SI Units – PB = 101.325 kPa

∆i = iw = 419.04 + (0.8 x 2257) ∆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

40

.008

115

80

55

U OL

% 90

DR Y BULB T EMPERATUR E - °F

60

20

45

65 HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

R PE U -B T Y AL P

.016

N

TH

60 %

TU

Problem 3-21

70

E

D

-°

R

F

Y

A

IR

75

30

.018

12 0

R

PB = 29.92 in.Hg.; q = 0 (a) Using chart 1a

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.

30

23

3-22 (cont’d)

∆i = iw = 1090 Btu / lbm ∆W

From table A-1

i-if 1090 - 196.1 = ifg 960.1

x=

x = 0.931 or about 93 % (b)

x will be the same

ASHRAE PSYCHROMETRIC CHART NO.1 R

90

AMERICAN S OCIETY OF HEATING , REFRIGERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

50

.028

60



85 15

-2 00 0 -1 00 0

2.0

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0  -8 -4 .0.0 -2. 0

Qs Qt

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

a

55

RA TU

40

1090

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RATIO

'W

75

35

50 14 .5

.018

R A

N

ER

U

M

P

PO

TE N

70

TI O

.014

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

% 90

ME

%

60

-C U.F

55

IR

% 50

YA

1

.5

50

45

DR

13

45 40

35

%

4 0%

40 1 3.

30%

0

35

20 %

15

80

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5 10

Y I VE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

.006

35 .004

.002

115

60 15

L B. ER

50

T. P

% 70

DR Y BUL B T EMPERAT UR E - °F

80

55

LU VO

20

65

60

1 4 .0

60

45

65

12 0

R PE U -B T Y AL P

b

N

TH

.016

TU

O

30

D

Problem 3-22

E

F

D

-°

R

F

Y

A

IR

75

70

25

ENTHALPY - BT U PER POUND OF DRY AIR

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.

30

24

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

∆i 272.1 = iw = kJ / kg ∆W 1000 iw = 0.272 (on scale) t2 = 22.6 C ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

30

SEA LEVEL

10 0

0.7

0. 6

SENSIB LE HEAT T OTAL HEAT

10 .0

0.5

1 .0



1 .5 2. 0

-5. 0

4 .0

Qs Qt

0.0

80

1 .0

EM

PE R

24

AT

UR

E-

°C

11 0 22

0 .9

0

5.0

4. 0

0 .1

WE TB UL BT

-0.2

0.2

30

0.272

-4.0 -2 .0

-0 .5

0. 3

-1 .0

12 0 26

90

-2.0



0. 4

4

0 .8

-

28

30 0 .9

1.0

0

R

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



12

2

2 .0

3 .0

2.5

70

25 20

'h

ENTHALPY HU MIDITY RAT IO

'W

10 0 25

18

C

R

-°

D

16

R

E

O F

TU

20

A EM

P

ER

20

A R TU SA

HUMI DITY RATIO - GRAM S M OISTURE PER KI LOGR AM D RY AIR

TI O

P J -K LP Y A E

N

20

8

80 %

0 .8

40

TH

14

N

E

T

R

90

2

90

15

%

% 80 0 .8 6V

15

%

U OL -C UB IC M

5

0%

4 0. 8

10

ET

5

ER

0 .8

5

2

3 0%

YA IR

20

22.6

30

38

8

70

6

4

60

40

ENTHALPY - KJ PER KILOGRAM OF DRY AIR

For adia. humidification (a)

80 10

2

40

20

15

10

5

8

10

HU MID ITY 25

0

0 .7

IVE 10% RE LAT

35

0 .8

20 %

30

10

R gD Rk PE

40 %

3-24

1

ME

% 60

20

45

70

10

D RY BU LB TEMPERA TU RE - °C

30

12

50

50

K

IL O G

R

AM

Problem 3-23

0

Y

0 .9

AI R

60

∆i = iw = 1131 Btu / lbw ∆W

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.

50

25

3-24 (cont’d)

 a (i2 - i1) q c = m

 a = 2000 x 60 / 13.14 m  a = 9132 lba / hr m

i1 = 18.1 Btu / lba ; i2 = 29.7 Btu / hr q c = 9132 (29.7 - 18.1) = 105,931 Btu / hr w = m  a (W3 - W2 ) ; W3 = 0.0167; W2 = 0.0032 lbv/lba m

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

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.

26

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -200 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

4 .0 8 .0

Qs Qt

-4 .0.0 -2.0

0.4

0 200

0 .1

0.2

- 0.4 -0.3

-0 .1

-0 .2

15 00

0

1131

85

0

WE T

0

-0 .5

0. 3

-1.

.026

45

 -8

50 0

.024

BU

80

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

3

E

F

D

-°

R

F

Y

A

IR

75

R TU ER M

P

PO

TE N TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU SA

25

E

% 90

ME

% 60

U.F

4 0%

40

1

1 3.

30%

0

35

20 %

15

60 (16)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

Y VE H UM IDIT 70

1 2.

10% REL ATI

10

45

25

.012

.010

40

.008

.006

35

.004

2 .002

115

% 50

DR Y BULB TEMPERATUR E - °F

IR YA

50

45

35

% 1 3 .5

45 40

DR L B. ER

55 60

15

T. P

% 70

50

30 %

-C

80

55

LU VO

20

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

U

Problem 3-24

70

110 (43)

ENTHALPY - BT U PER POUND OF DRY AIR

(b) Solution similar to (a)

3-25 English Units – See diagram for construction on chart 1a.

 32 2000 Q 2 = = 1= 12 3000 Q3 3 Layout 2L/3 on the chart and read: W3 = 0.007 lbv/lba 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.

30

27 I3 = 22.2 Btu/lba ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -200 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

4 .0 8 .0

Qs Qt

-4 .0.0 -2.0

0.4

0 200

0 .1

0.2

- 0.4 -0.3

-0 .1

-0 .2

15 00

0

85

0

WE T

0

-0 .5

0. 3

-1.

.026

45

 -8

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

1 000

77

RE

- °F

.022

80

.020

'h

ENTHALPY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

R A ER

U

M

P

PO

TE N

70

TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU SA

25

E

% 90

U.F T. P DR Y BULB TEMPERATUR E - °F

IR YA

4 0%

40 30%

1 3. 0

35

20 %

58.4 (15)

20

110

105

100

95

90

85

80

75

65

15

40 (4)

SI Units –

60

55

50

45

40

35

5

10

Y VE H UM IDIT 70

1 2.

10% REL ATI

.012

.010

.008

.006

35

.004

25

100 (38)

ENTHALPY - BT U PER POUND OF DRY AIR

Same procedure as above, read:

i3 = 34 kJ / kga W3 = 0.007 kgv / kga 3-26

40

.002

115

%

% 50

1

3-25

DR L B. ER

50

45

35

-C

60

1 3 .5

45

35

60

55

3

40

%

% 70

50

15

ME

80

55

LU VO

52

1 4 .0

65

60

20

45

65

12 0

R PE U -B T Y AL P

2

N

TH

.016

TU

O N

D

Problem 3-25

70

E

F

D

-°

R

F

Y

A

IR

75

30

English Units – Layout the given data on Chart 1a as shown for problem 3-25.

 a1 = 2000(60) 12.66 = 9,479lba hr m 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.

30

28 3-26 (cont’d)

 a2 = 1000(60) 14.44 = 4,155lba hr m  a1 m 32 9479 = = = 0.695  a1+m  a2 9479 + 4155 12 m

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 =

For I, % Error =

3-27 SHF =

(0.007 − 0.0067)100 = 4.5 0.0067

(22.2 − 21.5)100 = 3.3 21.5

250,000 = 0.8 200,000

or SHF =

59 = .81 73

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.

29

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

50



.028

60

85 1 5 .0

0.8

1 .0

60

0 .8 -2 00 0 -1 00 0

2. 0

-

0. 6

SENSIB LE HEAT TOTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt



-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 4 - 0. -0.3

0. 3

-1. 0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDITY RATIO

'W

75

35

50 14 .5

.018

E

F

D

-°

R

F

Y

A

IR

75

R TU TE

M

P

28.2

N

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

R TU SA

25

E

ME

% 60

-C U .F

1

60

50

% 50

45

35

IR YA

1 3 .5

45 40

%

4 0%

40 30%

1 3. 0

35

20 %

3-28

53 (12)

15

75 (24)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND O F DRY AIR

Refer to diagram for 3-27

 a (i1 - i2 ); i1 = 28.2; i2 = 21.5 (a) q = m

 a = 250,000 / (28.2 - 21.5) = 37,313 lba / hr m  =m  a v 2 = 37,313 x 13.09 / 60 = 8,140 ft 3 / min Q

 = 3.85 m3 / s (b) similar procedure; Q

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.

40

.002

115

55

2 15

DR L B. ER

50

T. P

% 70

DR Y BULB T EMPERATUR E - °F

80

55

LU VO

20

% 90

1 4 .0

21.5

50 %

65

60

12 0

A

-B T Y AL P

.014

65

N

TH

70

TI O

U

PE

R

PO

ER

U

Problem 3-27

A

N

D

O

30

.016

70

30

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 Q s (b) Proceed as above

φ = 0.882 ts = 56F

 = 10,014 cfm Q s 3-30

Proceed as in 3-29 above.

φ = 0.92 ts = 56.1 ≈ 56 F

 ≤ = 11,303 cfm Q s 3-31

(a) SHF =

500,000 = 0.91 550,000

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.

31

3-31 (cont’d) (b)

 a (i2 -i1) q = m

 a = q/(i  2 -i1) or m a = m

550,000 (34.3 − 22.8)

 a =47,826lba hr m  a v 2 47,826  =m Q = x 14.62=11,654 cfm or 5.5 m3/s 2 60 60

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.

32

ASHRAE PSYCHROMETRIC CHART NO.1 R

R

90

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.



SEA LEVEL

1.0

50



.028

60

85 1 5 .0

0.91

1.0

60

55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

0.8 -2000

2. 0

-

0. 6

SENSIBLE HEAT TOTAL HEAT

5000

300 0

0 .5 0 200

0 .1

15 00

-0 .1

-0 .2

0.2

-8 -4 .0.0 0

85

0

WE T

0

-0 .5 4 - 0. -0.3

0. 3

-1.

.026

45



-2.

0.4

0

-1 00 0

4 .0 8 .0

Qs Qt

50 0

.024

BU

LB

TE MP E

80

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHALPY HU MIDITY RATIO

'W

75

34.3 35

50 14 .5

.018

D

-°

R

F

Y

A

IR

75

E R TU

D

O

F

30

TI O

E

% 90

ME

% 60

U.F

2

1 4 0%

40 1 3.

30 %

0

35

20 % 1 2.

10

15

72 (22)

20

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

5

Y IVE H UMIDIT 10% RE LAT

25

1400 x 60 = 5,915.5 14.2

i2 =

-5 x 12,000 +38.5 5,915.5

.010

40

.008

.006

35

.004

30

115 (46)

q +i a 1 m

a m

.012

ENTHALPY - BT U PER POUND OF DRY AIR

 =m  a (i2 -i1) 3-32 q

i2 =

45

.002

115

50 %

DR Y BULB TEMPERATUR E - °F

IR YA

50

45

35

% 1 3 .5

45 40

DR L B. ER

55 60

15

T. P

% 70

50

30 %

-C

80

55

LU VO

20

1 4 .0

65

60

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

22.8

12 0

U -B T Y AL P

.014

65

N

TH

70

N

PE

TE

R

M

P

PO

ER

U

A

N

Problem 3-31

.016

70

i 2 = 2 8 .3 6 Btu/lba 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.

33 Then from Chart 1a, t2= 67F ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

50



.028

60

85 1 5 .0

0 .8

-2 00 0 -1 00 0

2. 0

-

0. 6

SENSIB LE HEAT TOTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

-8 -4 .0.0

85

0

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

45



-2.

0. 4

.026

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80 'h

ENTHALPY HU MIDIT Y RAT IO

'W

75

35

.020

75 50 14 .5

.018

F

R ER

U

P M TE N TI O

.014

A R

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

U -B T

TU

Y

25

SA

AL P

70

14

E

65

.0 V U OL

% 90

2

ME

% 60

-C U .F

55 60

IR YA

.5

50

40

% 50

45

35

% 13

45

15

R .D LB ER

50

T. P

% 70

4 0%

40 1 3.

30%

0

35

20 %

55

15

67

20

90

110

105

100

95

90

85

80

70

65

60

55

50

45

40

35

5

10

Y I VE H UM IDIT 75

1 2.

10% RE LAT

.012

.010

.008

.006

35

.004

25

ENTHALPY - BT U PER PO UND O F DRY AIR

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. 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.

40

.002

115

80

ADP55

DR Y BULB T EMPERAT UR E - °F

60

20

45

65

12 0

PO R PE

28.4

N

TH

.016

1

A

N

D

O

30

TU

Problem 3-32

E

D

-°

R

F

Y

A

IR

75

70

30

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

 = 1000 cfm Q 0 (a) From Chart 1a

t s =120 / 74 F  s= m

q 200,000 = (is -ir ) (37.2 − 22.8)

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.

35

1 = 13,889 lb/hr = m

 =m  sv s = m  s (14.78)/60 = 3,421 ft 3 / min Q s  /v = 1000 x 60 / 12.61 = 4758 lb/hr  o= Q (b) m o o  r 13,889 − 4758 m = = 0.66; From Chart 1a t1 = 61/ 47 F 1 m 13,889

t 3 - t1 = (119 − 61)

 w= m  s (Ws -W2 ) = 13,889 (0.0075 - 0.0036) = 54.2 lbm/hr (c) m

 1(i3 -i1) =13,889 (32.8 − 18.6) = 197,224 Btu/hr (d) q f = m

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.

36

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

50



.028

60

85 1 5 .0

50 00

0.8

1 .0

60

R

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

300 0

0 .5

4 .0 8 .0

Qs Qt



-8 -4 .0.0 -2.0

0.4

85

0

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

80

1 000

1150

-0 .5 - 0.4 -0.3

0. 3

-1 .0

0

.026

45

.020

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

E

F

D

-°

R

F

Y

A

IR

75

R TU ER

U

P M TE N TI O

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

14

65

.0 V ME

% 60

U.F

55

IR YA

.5

DR

13

50

40

% 50

r

45

35

%

4 0%

1

40

30%

1 3.

1

0

35

20 %

(a) t s = 120 / 71.4 F

61

72

20

110

105

100

95

90

85

80

65

60

55

50

45

40

35

3-37

15

75

5

40

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

0

25

.012

.010

.008

40

s

.006

35

.004

3 .002

115

60

45

15

B. RL PE

50

T.

% 70

47

30 %

-C

80

55

U OL

% 90

DR Y BUL B T EMPERAT UR E - °F

60

20

45

.014

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

PO

Problem 3-36

70

30

120

ENTHALPY - BT U PER POUND OF DRY AIR

Use Chart 1Ha

 s = 200,000 /(38.7 − 24.0) = 13,605 lba/hr = m 1 m  = 13,605 x 17.85 / 60 = 4048 cfm Q s

 0 = (1000 / 15 .2) x 60 = 3947 lba/hr (b) m  r 13,605 − 3947 m = = 0.71; t1 = 62.8 / 47 F 1 m 13,605

t 3 -t1 = (119.5 − 62.8)  w =m  s (w s -W1) = 13,605 (0.0088 - 0.0046) = 57.14 lbw/hr (c) m 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.

37 (d)

3-38

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

Assume fan power and heat gain are load on the space s = m

9384 x 60 = 42,915 lbm/hr; Prob 3-34 13.12

   W fan + qduct = ms (is − ic ) = (4 x 2545) + 1000 = 11,180 Btu / hr ic = 20.8 −

11,180 = 20.54 Btu/lbm 42,915

State c is required condition leaving coil Part a, b, and c are same as prob. 3-34;

 1(i1-ic ) = 42,915 (26.8 - 20.54) = 268,648 Btu/hr (d) q coil =m

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.

38

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00

300 0

0 .5

4 .0 8 .0

Qs Qt



-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 0

-0 .5 - 0.4 -0.3

0. 3

-1 .0

.026

45 85

0

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

0 1 000

80

.020

'h

ENTHAL PY HU MIDITY RAT IO

'W

75

35

50 14 .5

.018

R TU

O D

A

N

ER

U

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

M TE N TI O R

A

.014

TU SA

25

E

% 90

1

% 60

-C U.F

13

IR

% 50

45

35

YA

.5

s 50

40

%

4 0%

40 30%

1 3. 0

35

20 %

3-39

55

15

72

20

100

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

40

.008

.006

35

.004

.002

115

60

45

15

r

55

c

DR B. RL PE

50

T.

% 70

DR Y BULB T EMPERAT UR E - °F

80

55

ME

20

U OL .0 V

20.54

14

65

60

12 0

P

PO R PE U -B T Y AL P

50 %

70

65

N

TH

.016

70

E

F

D

Problem 3-38

-°

R

F

Y

A

IR

75

30

25

ENTHALPY - BT U PER PO UND O F DRY AIR

  s (i r − i s ); W  s (i s − i c ) q r = m fan = m

(a) ic = 28 Btu/lbm; ir = 33.7 Btu/lbm Using Chart 1Ha

q r = 1,320,000 Btu/hr

 W fan = 30 x 2545 = 76350 Btu/hr

  W fan = 30 x 2545 = 76,350 = ma (is -ic )

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.

30

39

 a(ir -is ) q s = 1,320,000 = m ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE

60 85

BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURY Copyright 1992

R

R

50

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

0.8

1.0

.028

80 1 .0

60

1 8 .0



5000 FEET



0 .8

-

SENSIBL E HEAT TOTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0.2

0 .1

ET

0

BU

LB

TE MP

.024

ER

75 50 0

AT U

40

RE

55 -° F .022

17 .5

15 00

-0 .1

0

80 W

.0

-0 .5 -0 .4 - 0.3 -0 .2

0. 3

-1

45

-1 000

4 .0 8 .0  -8 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2. 0

0 .6

75 10 00

.020 'h

ENTHALPY HU MIDIT Y RATIO

35

'W

70 50

UR AT

D F O

ER

.016

M

U

TE

PO

50 %

TI O

N RA TU

65

.014

r

s

HUMIDITY RATIO - POUNDS MOISTURE PER PO UND DRY AIR

SA

59 (15) c

60

% 90

55

16

% 80

.5 V

20

-C

% 60

U .F

50

45

%

ME

70

U OL

55

50

T. P

30 %

20%

10

15

62.5 (17)

20

80 (27)

90 (32)

110

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

I TY TIV E HUMID 10% RELA

.010

.006

35

.004

25

Solve simultaneous:

   W fa n + q s = m a (ir -i c ) 1 ,3 2 0 ,0 0 + 7 6 ,3 5 0 (3 3 .7 -2 8 )

 a = 2 4 4 ,9 7 4 lb a /h r m 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.

40

.008

ENTHALPY - BT U PER POUND OF DRY AIR

Two unknowns & two equations

 a= m

.012

.002

115

0

IR YA

4 0%

1 5 .5

35

%

R .D LB

50

DR Y BUL B T EMPERATUR E - °F

45

40

ER

40 35

16 .0

15

45

60

12 0

PE R TU -B

LP Y A TH

25

EN

65

P

ND

30

17 .0

Problem 3-39

70

E

RY

-°

F

AI R

.018

30

40

 a) is = ir - ( q s m is = 33.7 -

1,320,000 =28.3 Btu/lba 244,974

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.

 = 244,974 x16.2 = 66,143cfm (a) Q s 60  = 31.2 m 3 s (b) Q s

3-40

English Units –Tucson, Arizona, Elevation 2,556 ft.

imin =i0 =31.1 Btu/lba and sat. air ; t min =64.5 F; PSYCH Shreveport, Louisiana, Elevation 259 ft. imin =i0 = 42.5 Btu/lba and sat. air ; tmin = 76.8 F; PSYCH

SI Units – Tucson, Arizona

imin =i0 = 51.5 kJ/kga ; tmin =18.1 C; Chart 1b Shreveport, Louisiana imin =i0 =75.5 kJ/kga ; tmin =24.8 C; Chart 1b

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.

41

ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.642 INCHES OF MERCURY Copyright 1992

R

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



259 FEET

1.0

60

50

.028

60 1 .0

85



15 .0

0 .8

-

SENSIBL E HEAT T OTAL HEAT

50 00 300 0

0 .5

-8 -4.0.0 -2. 0

-0 .4 - 0.3

-0 .1

-0 .2

0 200

15 00

0 .1

85 0

.0

-0 .5

0. 3

-1

0.2

45



0. 4

0

-1 000

4 .0 8 .0

Qs Qt

WE TB UL

80

50 0

.024

BT EM

PE R

40

55 AT

UR

E°F

.022

10 00

80 .020 'h

ENT HAL PY HU MIDIT Y RAT IO

.026

-2 000

2.0 0 .6

75

'W

TLO

35

50 .018

-° F E AT U R

U

TE M

PE

PO

O N

70

.014

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

AT I R U AT S

25

E

65

60

% 90 UM OL 0V

%

60

%

E-

70

CU

55

.FT

50

%

1 3.

50

5

40 35

IR YA

45

4 0%

40 30% 1 3 .0

35

20 %

10

15

76.8

20

95

110

105

100

95

90

85

80

75

HUMIDITY

70

65

60

55

50

45

40

35

IVE 10% RE LAT

.012

.010

40

.008

.006

35

.004

.002

115

45

15

DR LB . ER .P

% 60

DR Y BUL B T EMPERATUR E - °F

50

1 4.

80

55

20

45

65

12 0

ER P TU

-B Y LP A

.016

N

TH

SL

R

30

N

D

O

F

D

R

Problem 3-40 Shreveport, LA

14 .5

Y

AI R

75

70

25

ENTHALPY - BT U PER PO UND O F DRY AIR

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.

30

42

ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 27.259 INCHES OF MERCURY Copyright 1992

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R

85 50

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

2556 FEET

1.0

1 .0

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HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

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Problem 3-40 Tucson, Arizona

25

ENTHALPY - BT U PER PO UND O F DRY AIR

 s (ir − i s ) q = m

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

 o r1 m 13 = = = 0.55 or 55%  s r0 23.5 m  ≈ 0.046 m3 / s - kW (b) Q s

 0 /m  s ≈ 55% m 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.

30

43

ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURY Copyright 1992

R

60 85

R

50

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

5000 FEET

1.0

.028

80 1 .0

60

1 8 .0





0 .8

50 00

-

300 0

0.7

SENSIBL E HEAT TOTAL HEAT

0 .5

-0.1

BU

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0. 3

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0.2

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 -8

0. 4

0

45

-1 000

4 .0 8 .0 -4.0.0 -2. 0

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2.0

0 .6

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35

'W

70 50

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O

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TU

DR

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% 90

40 %

60

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% 60

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10 %

1

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4 0%

0

1 5 .5

30 %

20%

10

50 (10)

 2 (i2 − i3 ); m 2 = 3-42 q = m

15

75 (24)

20

110

105

100

95

90

85

80

75

70

HUMIDI TY

65

60

55

50

45

40

35

IVE 10% REL AT

45

.012

.010

40

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35

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DR

DR Y BUL B T EMPERAT UR E - °F

%

L B.

50

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45

40

16 .0

40

-C

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50

45

r

%

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55 70

35

1 6 .5

% 80

50

35

HUMI DITY RATIO - POUNDS MOISTURE PER POUND DRY AIR

55

.014

12 0

AT SA T

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65

60

25

20

15

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65

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30

17 .0

Problem 3-41

70

25

100 (38)

ENTHALPY - BT U PER POUND OF DRY AIR

500,000 (41.1 − 21.9)

 2 = 26,042 lba/hr m

 = 26042 x 14.55/60 = 6315 cfm Q 2  0 = 0.25 x 26,042 = 6511 lba/hr m  0 /m  3 = 0.25; t mix = 67.5 / 49.5 F m

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30

44 3-42 (cont’d) Preheat Coil:

 0c p (t 4 -t 0 ) = 6511 x 0.24 (60-6) = 84,383 Btu/hr q ph = m

Heat Coil:

 2 (i5 -i1) = 26,042 (28.4 - 20) = 218,753 Btu/hr q h = m Humidifier:

 w= m  2 (W2 -W5 ) = 26,042 (0.0144 - 0.0035) m = 283.9 lbw/hr  = 2.98 m3 / s; q = 24.7 kW; q = 64.1 kW; (b) Q 2 ph h

 w = 0.036 kg/s m

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.

45 ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

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AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50

.028

60



85 1 5 .0

0 .8 -2 000

2.0

-

0. 6

SENSIB LE HEAT TOTAL HEAT

50 00 300 0

0 .5 0 200

0 .1

0.2

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-0 .2

15 00

50 0

WE T

.024

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80

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55 RA TU

40

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1 000

1153

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0

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-0 .5 - 0.4 -0.3

0. 3

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45



-8 -4 .0.0 -2.0

0.4

0

-1 00 0

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Qs Qt

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75

35

50 14 .5

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U

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70

TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

1 4 .0

E

65

ME

% 60

-C U.F

55 60

50 %

45

35

3

4 0%

40 1 3.

1

30%

0

35

IR YA

50

40

% 1 3 .5

45

15

30 %

DR L B. ER

50

T. P

% 70

5 20 %

3-43

15

60 (16)

70 (21)

20

110

105

100

95

90

85

80

75

65

4

60

55

50

45

40

35

5

10

70

1 2.

10%

H UM IDITY REL ATI VE

.012

.010

40

.008

.006

35

.004

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115

80

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DR Y BUL B TEMPERATUR E - °F

60

20

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65

12 0

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2

N

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.016

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Problem 3-42

70

E

F

D

-°

R

F

Y

A

IR

75

30

30

25

105 (40)

ENTHALPY - BT U PER PO UND O F DRY AIR

 a (ir − is ) Use Chart 1a; q d = m  a = q d /(ir − i s ) or m  a = 150 x 12,000 / (28.4-22) = 28,125 lbm/hr (a) m

 = 28,125 x 13.25/60 = 61,211 cfm Q d  = 0.20 Q m

 = 1,242 cfm Q d

 m = 1,242 x 60/13.5 = 5,521 lbm/hr [vm assumed] m im =ir − 1.8 x 12,000/5,521 = 24.5 Btu/lbm; tm = 62 / 57 F

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.

46

 = 2.93 m3 /s; Q  = .59 m3 /s; t = 17/14 C (b) Q d m m ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

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AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



1 .0



.028

60

85 1 5 .0

0.8

0 .8 -2 00 0 -1 00 0

2. 0

-

0. 6

50 00 300 0

SENSIB LE HEAT T OTAL HEAT

0 .5

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

0. 4

0 200

0.6

SEA LEVEL

1.0

50

50 0

.024

BU

LB

80

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55 RA TU

40

RE

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.022

1 000

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75

35

50 14 .5

R TU

O

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U

M

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.014

A R

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

R PE U -B T

TU

Y

25

SA

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70

E

U OL ME

60

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% 70

T. P L B. ER

55

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50

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1 3 .5

45

%

% 50

45

4 0%

40 1 3.

30%

0

35

62 (17) 20 %

3-44

15

60 (16)

75 (24)

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.012

.010

40

.008

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115

60

DR Y BUL B T EMPERAT UR E - °F

50

35

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15

14

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60

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F

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D

Problem 3-43

.018

12 0

R

60

25

ENTHALPY - BT U PER PO UND O F DRY AIR

15.0 x 12,000 a = (a) m = 29,508 lba/hr (31.2 - 25.1)

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

= 1,574 cfm

 m =1,574 x 60/16.2 = 5,829 lba/hr (v massumed) m im = 35.7 − 1.8 x 12,000/5,829 = 27.5 Btu/lba; 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.

30

47

tm = 62.5 / 58 F  =3.7 m3 / s; Q  = 0.74 m3 /s; t = 17 /14.4 C (b) Q s m m

3-45

Use Chart 1a;

r m 10 = 0.8 = 1 m 0r

[Both design and min. load condition]

s is = ir - q m / m s = m

 Q 50 x 12,000 d = ir − i s (29.35 - 22.3)

 s = 85,106 lba/hr; m  s is constant for all conditions m

i s' = 29.35 − 25 x 12,000/85, 106 = 25.83 Btu/lba (a) From Chart 1a; t s' = 64 F

 si s + m  b i1' = (m  c +m  b ) i s' (b) m  b (i s − i s' ) 24.2 − 25.8 m = = 0.271 =  c (i s' − i1' ) 25.8 − 31.7 m

(b) From chart 1a; t d = 49 F for both cases

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.

48

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

SEA LEVEL

1.0



.028

60 85 1 5 .0

0.9

1 .0

50

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

50 0

80

.024

BU

LB

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MP E

55 RA TU

0

40

RE

- °F

.022

1 000

80

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ENT HAL PY HU MIDIT Y RATIO

'W

75

35

50 14 .5

0'

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75

R TU ER

U

M

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PO

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R

N

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A

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25

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65

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50 %

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30

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70

E

F

D

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F

Y

A

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Problem 3-45

.026

45

 -8

0. 4

0

4 .0 8 .0 -4 .0.0 -2. 0

Qs Qt

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR



60

R

LU VO

1

% 90

1 4 .0

65

% 60

-C U.F

r

55

s

60

IR YA

% 50

45

35

DR

50

40

% 1 3 .5

45

15

L B. ER

50

T. P

% 70

4 0%

40 30%

1 3. 0

35

20 %

3-46

15

55 (13) 64 (18)

20

77 (25) 85 (29)

95 (35)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y I VE H UM IDIT 70

1 2.

10% RE LAT

45

.012

.010

40

.008

.006

35

.004

.002

115

80

55

ME

s' 20

DR Y BUL B T EMPERAT UR E - °F

60

12 0

R

25 ENTHALPY - BT U PER POUND OF DRY AIR

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.

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.

30

49

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

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AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENTHAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

R A

N

ER

U

M

P

PO

TE N

70

TI O

.014

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

E

U OL .0 V

% 90

% 60

-C U .F

55

50

% 50

45

35

A IR

40

% .5

45

15

60

13

2

4 0%

40 1 3.

30%

0

35

20 %

3-48

15

52 (11) 60 (16)

20

80 (27)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y IVE H UM IDIT 70

1 2.

10% RE LAT

.010

40

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND OF DRY AIR

 c sh m = =.837 (a)  s ch m

 h cs m = = 0.163  s ch m  c 0.837 m = = 5.14  h 0.163 m  s (ir − is ) q = m s = m

.012

.002

115

1'

RY B. D RL PE

50

T.

% 70

DR Y BUL B TEMPERATUR E - °F

80

55

ME

54

14

1

65

60

20

45

65

12 0

R PE U -B T Y AL P

67

N

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.016

90 %

TU

O

30

D

Problem 3-47

E

F

D

-°

R

F

Y

A

IR

75

70

50 x 12,000 = 93,750 lba/hr (28.2-21.8)

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.

30

50

 = 93,750 x 13.2/60 = 20,625 cfm Q s  = 9.7 m3 /s (b) Q s ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

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AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

R

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

0 .5

50 00 300 0

0.65

4 .0 8 .0

Qs Qt



0. 4

0 200

0. 1

0.2

-0 .1

-0 .2

15 00

0

-8 -4 .0.0 -2. 0

85

0

WE T

0

.5 -0 - 0.4 -0.3

0. 3

-1.

.026

45

50 0

.024

BU

80

LB

TE

MP E

55 RA TU

RE

40

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

R TU

O D

A

N

ER

U

M

P

PO

N

70

HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

TI O A R TU SA

25

.014

90 %

65

E

.0 V

% 90

%

ME

r

60

-C U.F

s % 50

45

35

20 % h

IR YA

50

.5

45 40

% 13

c

4 0%

40 30%

1 3. 0

35

20 %

3-49

52 (11)

15

75 (24)

20

90 (32)

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

Y VE H UM IDIT 70

1 2.

10% REL ATI

.012

.010

c  h 10.1 m m 36 = = 0.9; = = 0.10 ;  s 46.3  s 46.3 m m

s = m

40

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND O F DRY AIR

See diagram of problem 3-48 (a)

45

.002

115

55 60

15

R .D LB ER

50

T. P

% 70

DR Y BUL B T EMPERAT UR E - °F

80

55

U OL

20

14

65

60

12 0

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R PE U -B T Y AL P

.016

N

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70

E

F

D

Problem 3-48

-°

R

F

Y

A

IR

75

30

c m 0.9 = = 9.0  h 0.10 m

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

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

30

51

 =10.3 m3 /s (b) Q s

3-50

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

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

q c = 78,469 (28.2-20.6) = 596,364 Btu/hr  =8.1 m3 /s; q = 175 kW (b) Q c c

3-51 SI Units (a) On the basis of volume flow rate using Chart 1b:

 = 13 Q  = 0.69 x 1.18 = 0.815 m3/s Q 2 3 12  =Q  -Q  = 1.18 − 0.815 = 0.365 m3/s and Q 1 3 2

(b)

 a3 (i4 -i3 ) = q 34 = m q 34 =

 Q 3 (i -i ) 4 3 v3

1.18 (47.8-41.0) = 9.6 kW 0.835

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.

52 ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

SEA LEVEL

10 0

0 .8 0. 7 0. 6

Problem 3-51 SENSIB LE HEAT T OTAL HEAT

10 .0

0.5

1 .0



1 .5 2. 0

-5.0

4 .0

Qs Qt

-4.0

80

1 .0

MP ER

24

AT

UR

E-

°C

11 0 22

0 .9

0

5.0

4. 0

0 .1

WE TB UL BT E

-0.2

0.2

30

0.0

.0

-0 .5

0. 3

-1 .0

12 0 26

90

-2.0



-2

0. 4

4

-

28

30 0 .9

1.0

0 30

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



12

R

2

2 .0

3 .0

2.5

70

25 20

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

10 0 25

C

R

-°

D

R TU A

50

90 14

E

N

8

SA

1

HUMIDITY RATIO - G RAM S M OISTURE PER KILOGR AM D RY AIR

R

20

TU

LP Y

40

%

%

U OL ME IC M

%

UB

50

4 0.8

10

-C

% 60

20

ET

5

ER

0 .8

g Rk PE

40 % 2

DR

3 0%

IR YA

20

15

10

5

8

10

12

HUM ID ITY 25

0

0 .7

IVE 10% RE LAT

20

17.2

24

29

35

0 .8

20 %

30

5

10

30

12

80 10

8

70

6

4

60

2

45

10

4

15

6V

11

70

2

0 .8

% 80 3

30

D RY BU LB TEMPERA TU RE - °C

90

40

14.7 15

50

A

-K

J

TI O

P

N

E

T

R

K

EM

P

ER

20

0 .8

A

16

50 %

E

O F AM R IL O G

0

Y

0 .9

AI R

Problem 3-51

TH

18

60

40

ENTHALPY - KJ PER KILO GRAM O F DRY AIR

English Units

 = 640 cfm; q = 33,684 Btu/hr (a) Q 1 34 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. 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.

50

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.  Q  a (i4 -i2 ) = 2 (i4 -i2 ) (c) q 24 = m v2

 =35 x 0.817 (47.7-32) ; Q 2

 a (i1-i2 ) = q 12 = m

 = 1.82 m3/s Q 2

1.82 (60.6-32) 0.817

q 12 = 63.7 kW

 a (i4 -i3 )= q 34 = m

1.82 (47.7-39.4) 0.817

q 34 = 18.5 kW q 23 = q 24 - q 34 = 35-18.5=16.5 kW

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.

54 ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE

SEA LEVEL

10 0

0. 6

SENSIB LE HEAT T OTAL HEAT

10 .0

0.5

-4.0

30

0.0

.0

0

WE TB UL BT E

MP ER

80

1 .0

-0 .5

5.0

0. 3

-1.

24

AT

UR

E-

°C

11 0

-0.2

22

0 .9

0

4. 0

0 .1

26

90

-2.0



-2

0. 4

0.2

12 0

-5. 0

4 .0

Qs Qt

4



1 .5 2.0

0.7

0 .9

1 .0

0 .8

-

28

30

Problem 3-52

1.0

0

R

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



12

30

BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

2

2 .0

3 .0

2.5

70

25 20

'h

ENTHALPY HU MIDIT Y RAT IO

'W

10 0 25

18

C

R

-°

D

R

E

O F

TU

AM R

A

16

ER

20

90

EM T N R

%

ME -C

3

UB IC M ET

5

ER

0 .8

R gD Rk PE

40 % 2

3 0%

19

20

23

30

20

15

5

10

8

11.8

10

HU MID ITY 25

0

0 .7

IVE 10% RE LAT

9

3-52

IR YA

0 .8

20 %

27

35

5

10

30

12

80 10

8

70

6

4

60

2

45

4 0 .8

% 50

U OL

% 60

10

6V

%

10

2

0 .8

4

15

70

D RY BU LB TEMPERA TU RE - °C

SA

% 80

40

TU

LP Y A TH N E

90

11 ADP

8

17

14

30

HUMIDITY RATIO - GRAM S M OISTURE PER KILOGR AM D RY AIR

A

-K

J

TI O

P

1 20

50

K R E

14

0 .8

40

15

20

21

P

IL O G

50

0

Y

Problem 3-52

0 .9

AI R

60

40

ENTHALPY - KJ PER KILO GRAM O F DRY AIR

English Units (a),(b) See above

 = 4103cfm ; q  12 =221,243 Btu/hr (c) Q 2

q 34 = 67,498 Btu/hr; q 23 = 52,502 Btu/hr 3-53 English Units (a)

  s (ir -is ); m  s = 5000 x 60/13.2 = 22,727 lba/hr q=m (specific volume value of 13.2 ft3/lbm is assumed.)

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.

50

55

 s =28.2 − 10 x 12,000 / 22,727 = 22.9 Btu/lba is = ir - q /m

t s = to = 57.5 F; Ws =Wo = 0.0083 lbv/lba (b)

 r 0m  m m = = 0.462 r  m 0r s m m

 r =0.462 x 22,727 = 10,500 lba/hr m  o = 22,727 − 10,500 = 12,227 lba/hr m  = 10,500 x 13.68/60 = 2,394 cfm Q r  = 12,227 x 12.11/60 = 2,468 cfm Q o

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

 r =0.578 x 22,727 = 13,131 lba/hr; m  o' = 9,596 lba/hr m  =13,131 x 13.68/60 = 2,994 cfm; Q  = 9,596 x 13.48/60 Q r o' = 2,156 cfm

 s (im' -is ) = 22,727 (28.4 - 22.8) = 127,271 Btu/hr (d) q c = m

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.

56

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



1.0

1 .0

50



.028

60

85 1 5 .0

50 00

0.8

SEA LEVEL

60

R

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

300 0

0 .5

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

85

0

50 0

WE T

.024

BU

LB

80

TE

MP E

55 RA TU

40

RE

- °F

.022

80

1 000

1150

.026

45

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0 -8 -4 .0.0 -2. 0

Qs Qt

.020

'h

ENTHAL PY HU MIDITY RAT IO

'W

75

35

50 14 .5

A

IR

75

R TU M TE

R

N

PE

TI O

U

.014

R TU SA

0'

E

ME

% 60

-C U.F

r

60

% 13

50

YA

.5

45

IR

% 50

45

4 0%

m

40

43 (6)

1 3.

30%

0

35

55

s

ADP

35

DR L B. ER

50

T. P

% 70

20 %

10

40 (4)

15

70 (21)

57.5 (14) 65 (18)

75 (24)

20

110

105

100

95

90

85

80

75

Y IVE H UM IDIT 70

65

60

55

50

40

35

5

10% RE LAT 45

1 2.

0

.012

.010

40

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND OF DRY AIR

3-53 SI Units (a) ts = 14.2C; Ws = 0.0083 kgv/kga

 =1.17m3 s  =1.13m3 s ; Q (b) Q r o

 =1.41m3 s ; Q  =1.02m3 s (c) Q r o' (d) q c = 37.3 kW 3-54

45

.002

115

80

55

U OL .0 V

20

50 %

m'

14

65 % 90

DR Y BUL B TEMPERATUR E - °F

25

60

40

HUMIDITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A

-B T Y AL P

70

65

N

TH

90 %

P

PO

ER

U

A

N

D

O

30

.016

70

E

F

D

-°

R

F

Y

Problem 3-53

15

.018

12 0

R

(a) Any combination that will yield an enthalpy less than 57.0 kJ/kga or 33 Btu/lba

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.

30

57

 s = 5 / 0.84 = 5.95 kga/s = m r (b) m

 o mr m = =0.36  r 0r m

 o = 0.36 x 5.95 = 2.14 kga/s m  = 2.14 x 0.852 = 1.82 m3 /s = 3,857cfm Q o (c) tad = 15.4 C or 60F

 o /q n = (im -is )/(ir -is ) = 1.0 (Essentially, no difference) (d) q ASHRAE PSYCHROMETRIC CHART NO.1 11 0

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 101.325 kPa Copyright 1992

R

SEA LEVEL

10 0

0.7

0. 6

SENSIB LE HEAT TOTAL HEAT

10 .0

0.5



1 .5 2. 0

-5.0

4 .0

Qs Qt

5.0

0. 3

-1. 0

-4.0 -2 .0

WE TB UL BT

80

1 .0

EM

PE R

24

AT

UR

E-

°C

11 0 22

0 .9

0

4. 0

0 .1

30

0.0

-0.2

0.2

12 0 26

90

-2.0



0. 4 -0 .5

0.6

1 .0

4

0 .8

-

28

30

0 .9

1.0

0 30

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



12

R

2

2 .0

3 .0

2.5

70

25 20

'h

ENT HAL PY HU MIDIT Y RATIO

'W

10 0

25

18

60

-°

C

R D

16

R

E

O F

TU

AM

A

50

20

90

r

SA

E

N

8

TH

0

HUMIDITY RATIO - G RAM S M OISTURE PER KI LO GR AM D RY AIR

TU

A

R

LP Y

A

m2

0 .8

40

s%

90

15

% 80 0 .8 6V

15

%

U OL ME

10

UB IC M ET

5

ER

0 .8 2

3 0%

18 (64)

20

25 (77)

35

20

15

10

8

HUM ID ITY 25

0 .7 5

A IR

0

20 (68)

IVE 10% RE LAT

10

RY

0 .8

20 %

30

5

10

gD Rk PE

40 %

30

80 10

8

70

6

4

60

2

45

5

0%

4 0. 8

10

-C

% 60

20

D RY BU LB TEMPERA TU RE - °C

70

40

30

12

50

J -K

14

0

TI O

P

N

E

T

R

K

EM

P

ER

R IL O G

0

Y

AI R

57

0 .9

Problem 3-54

40

ENTHALPY - KJ PER KI LO GRAM O F DRY AIR

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.

50

58

SHF =

3-55

−424,000 = -4 530,000 − 424,000

Construct condition line on Chart 1a with preheat and mixing processes.

 sen = -424,000 = m  sc p (t r -t s ) (a) q s = m

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

 =88,333 x 14.07/60 = 20,714 cfm or 9.8 m3 /s Q s (b)

 r hm m  r = 0.33 x 88,333 lba/hr = =0.33; m  mm hr

 = 29,150 x 13.68/60 = 6,646 cfm or 3.14 m3 /s  r =29,150 lba/hr; Q m r h m  h = 0.67 x 88,333 =1 − 0.33 = 0.67; m m m

 = 59,183 x 13.1/60  h = 59,183 lba/hr; Q m h  = 12,922 cfm or 6.1 m3 /s (at heated condition) Q h

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

 355,098 Btu/hr or 104 kW q= (d) q m =88,333 x 0.24 (95 - 65) = 635,998 Btu/hr or 186 kW 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.

59

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

R

R

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.



SEA LEVEL

1.0

1 .0

60

50



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

-

0. 6

SENSIB LE HEAT T OTAL HEAT

50 00 300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

-4

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0  -8 -4 .0.0 -2. 0

Qs Qt

50 0

WE T

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HAL PY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

.018

F

R ER

U

M

P

PO

TE N TI O

.014

HUMI DITY RATIO - POUN DS MOISTURE PER POUND DRY AIR

A R TU SA

25

-C

r

T. P U.F

% 70

55 60

IR

50 %

45

35

YA

50

40

%

1 3 .5

45

15

20 %

DR L B. ER

50

4 0%

40

m 1 3.

30%

0

35

s 20 %

15

60 (16)

75 (24)

20

95 (35)

110

105

100

95

90

85

80

75

Y IVE H UM IDIT 70

60

55

50

45

40

35

5

35 (2)

10

10% RE LAT 65

1 2.

h

0

.012

.010

40

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND OF DRY AIR

3-56 Refer to chart 1a. (a)

 x 60 (i -i )  a3 (i4 -i3 ) = Q q 34 = m 3 4 3 v3  x Q 3

45

.002

115

60

DR Y BUL B T EMPERATUR E - °F

E

%

ME

80

55

LU VO

20

50 %

% 90

1 4 .0

65

60

12 0

R PE U -B T Y AL P

70

65

N

TH

.016

A

N

D

O

30

TU

Problem 3-55

E

D

-°

R

F

Y

A

IR

75

70

q 34v 3 (1750 x 13.23) = 60(i4 -i3 ) 60(28.1-23)

 = 75.7 or 76 cfm = 0.040 m3 /s Q 3

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.

30

60 (b) t3db = 58.5 F and 80% RH or 15 C

31  ; Q3 = 0.754 x 75.7 = 57 cfm or 0.028 m3 /s 12

 = (c) Q 2

 = 76 - 57 = 19 cfm or 0.012 m3 /s Q 1 ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

1 .0



.028

60

85 1 5 .0

0 .8 -2 00 0 -1 00 0

2.0

0. 6

SENSIB LE HEAT T OTAL HEAT

300 0

0 .5

4 .0 8 .0

Qs Qt



-8 -4 .0.0 0

-2.

0. 4

0 200

0 .1

0.2

-0 .1

-0 .2

15 00

0

85

0

50 0

80

1 000

RE

- °F

.022

.020

50

90 %

.018

70

R TU

O

.016

50 % 70

.014

R

1

TU

62

E

N

SA

25

65

ME

% 60

-C

80

LU VO

% 90

55

T. P L B. ER

55

3

60

%

DR

1 3 .5

IR YA

50

2

U.F

4

% 70

50

1 4 .0

20

% 50

45

4 0%

40 30%

1 3. 0

35

20 %

58.5

75

84

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

3-57 (a)

50

15

Y I VE H UM IDIT 70

1 2.

10% RE LAT

10

45

65

A

-B T

TI O

U

N

PE

TE

R

M

P

PO

ER

U

A

N

D

70

E

F

D

-°

R

F

Y

A

IR

75

30

Y AL P

55 RA TU

14 .5

TH

MP E

75

60

35

TE

80 'W

Problem 3-56

45

LB

40

35

40

.024

BU

'h

ENT HAL PY HU MIDIT Y RATIO

15

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

.026

45

HUMI DITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

-

.012

.010

40

.008

.006

35

.004

.002

12 0

50 00

0.8

SEA LEVEL

1.0

50

115



60

R

DR Y BULB T EMPERAT UR E - °F

R

25

ENTHALPY - BT U PER PO UND O F DRY AIR

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. 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.

30

61 State 3 is defined by the intersection of the reheat and space condition lines.

(b)

 a3 (i4 -i3 ) = q 34 = m

 x 60 Q 3 (i4 -i3 ) v3

 = q 34 v 3 = 100,000 x 13.4 Q 3 60(i4 -i3 ) 60(28.2-23.9)  = 5,194 cfm or 2.5 m3 /s Q 3 (c)

 a (i1-i2 ) = q 12 = m

5,194 x 60 (34.2-20.2) 13.4

q 12 = 325,594 Btu/hr or 95.4 kW q 23 =

5,194 x 60 (23.9-20.2) 13.4

q 23 =86,050 Btu/hr or 25.2 kW

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.

62

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

90

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

SEA LEVEL

1.0



1 .0

60

R

50



.028

60 85 15

-2 00 0 -1 00 0

2.0 0. 6

SENSIB LE HEAT TOTAL HEAT

50 00

300 0

0 .5

0. 4

0 200

0 .1

15 00

-0 .1

-0 .2

0.2

.026

45 85

0

WE T

0

-0 .5 - 0.4 -0.3

0. 3

-1.

0

4 .0 8 .0  -8 -4 .0.0 -2. 0

Qs Qt

.0

0 .8

-

50 0

80

.024

BU

LB

TE

MP E

55 RA TU

40

RE

- °F

.022

1 000

80

.020

'h

ENT HALPY HU MIDIT Y RAT IO

'W

75

35

50 14 .5

75

R TU A

.016

ER

U

50 %

M

P

PO

TE N

70

TI O

1 4 .0 ME

% 60

-C U.F

4

E

55 60

IR

% 50

45

35

3

YA

50

40

% .5

2

13

15

R B. D RL PE

50

T.

% 70

ADP45

4 0%

40 30%

1 3. 0

35

20 %

66

15

75

20

85

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

45 51

Y IVE H UM IDIT 70

1 2.

10% RE LAT

10

.010

40

.008

.006

35

.004

.002

25

ENTHALPY - BT U PER PO UND OF DRY AIR

3-58 Assume room temperature humidity of 50% and layout the state & processes on required from point c to s.

Supply Air:

 sc p (t s -tr ) q sen = 120,000 x 0.5 = 60,000 Btu/hr = m s = m

45

.012

115

80

55

LU VO

% 90

56

20

1 65

60

HUMI DITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR

A R TU

62

SA

25

.014

65

DR Y BULB TEMPERATUR E - °F

R PE U -B T Y AL P

70

E

D F N

D

O

30

N

TH

70

-°

R

F

Y

A

IR

Problem 3-57

.018

12 0

R

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

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.

30

63

 =53,192 x 16.33/60 = 14,477 cfm or 6.8 m3 /s Q s Mixed Air:

 o = 53,192 x 0.333 = 17,703 lba/hr m  = 17,713 x 17.2/60 = 5,078 cfm or 2.4 m3 /s Q o

 r = 53,192 − 17,713 = 35,479 lba/hr m  =35,479 x 16.5/60 = 9,757 cfm or 4.6 m3 /s Q r Reheat:

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

= 192,768 Btu/hr or 56.5 kW Coil:

 m (im -ic ) = 53,192 (34.4 - 24.2) = 542,558 Btu/hr or 159 kW q c =m

=

( 200,412 − 190,109 )100 = 5 .1 % 200,412

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.

64

ASHRAE PSYCHROMETRIC CHART NO.4 55

NORMAL TEMPERATURE R

60 85

BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURY Copyright 1992

R

50

AMERICAN S OCIETY OF HEATING , REFRIG ERATING AND AIR-CONDITIO NING ENGINEERS, INC.

5000 FEET

1.0

.028

80 1 .0

60

1 8 .0





0 .8 0 .6

SENSIBL E HEAT TOTAL HEAT

50 00 300 0

0 .5

0. 4 0. 3

0 200

0.2

0 .1

75

50 0

40

BU

TE MP

.024

ER

AT U

RE

55 -° F .022

50 %

75 10 00

.020 'h

ENTHALPY HU MIDIT Y RATIO

LB

75

17 .5

15 00

-0 .1

0

80 WE T

0

.0

-0 .5 -0 .4 - 0.3 -0 .2

0.6 0.5

-1

45

-1 000

4 .0 8 .0  -8 -4.0.0 -2. 0

Qs Qt

.026

-2 000

2. 0

-

35

'W

70

0 50

UR ER P M

U

TE

PO

.016

TI O

N RA TU

65

SA

.014

m

60

HUMIDITY RATIO - POUNDS MOISTURE PER PO UND DRY AIR

PE R TU -B LP Y A

90 %

EN

TH

25 60

% 90

55

%

ME -C

% 60

U .F

50

45

s

U OL

70

.5 V

55

c

50

16

r

% 80

20

T. P IR YA

30 %

10

55 (13)

15

75 (24)

20

90 (32)

110

105

100

95

90

85

70 (21) 80

65

60

55

50

45

40

35

I TY TIV E HUMID 10% RELA

75

20%

70

1 5 .5

35

4 0%

.010

.008

.006

35

.004

25

ENTHALPY - BT U PER POUND OF DRY AIR

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.

40

.002

115

40

%

R .D LB

50

DR Y BUL B T EMPERATUR E - °F

45

ER

40 35

16 .0

15

45

.012

12 0

AT

D F O ND

30

65

17 .0

Problem 3-58

70

E

RY

-°

F

AI R

.018

30

Exοerpts from this wοrk may be reproduced by instructors for distribution on a not-for-proΓrt basis for testing or instructional purposes only tο students enrolled in courses for which the textboοk has been adopted. Αny other reproduction or trαnsΙαtiοn ofthis work beyond thαt permitted by Sections ]07 or ]08 of the 1976 United Stαtes Copyright Αct withοut the permissiοn of the copyright owner is unΙαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment, John Wiley & Sons, Ιnc, Ι ] ] Riνer Street' Hoboken, NJ 07030'

Chapter 4

4-1

(a) comfortable (b) too warm

(c) comfortabΙe (d) too dry

4-2

(a) comfortable (b) too warm

(c) comfortable (d) too dry

4-3

(a) Assume sedentary dry using equation 4-4a,

bulb of 78

to,act =

75

-

F,

clo

=

o.5, met. = 1 .8,

5.4(1 + 0.5)(1.8

-

1

Relative humidity should be less than 50% (b) Should wear a S\Ι/eater or light jacket and slacks.

(clo = 0.8)

4-4

Use fig 4-1

(a) Summer,

to =

76 F or 24 C; Winter,

(b) Use equation 4-4a aS a guide, met =

3.0,

tdb

:76

F

\Λ/ith

to =

72 F or 22 C

clo = 0.2,

.2) = 71 F

o/

t

to =76

4-5

-5.4 (1+0.2)(3-1.2) = 64 F [winterorSummer]

From fig 4-3 temperature can rise about3.2 F.(j.g C)

t=68 +3.2=71.2 Fort=20+ 1.8= 21.8C

4-6 4-7

From fig 4-3 @200 fpm, temp rise ρ 5.3 F (2.9 C) with t,,"-t _ 9 F (5 c), temp rise ε 6.5 F (3.6 c) to = (t,

T,fn

+t^r)|2, then using Eq.

='6* Cν\l2

4-1

σg _Tr) = (53s)4 +(O.103 x 109) (4o)1Ι2(78_74)

tmft:82For27'8C to=(74+82)Ι2 = 78F or25.6C

4-8

Compute the operative temperature, Τ,xn

= φ4q4

to = (84

+

to

(O.103 x 1o911eo11/'(εo _76)= 83.5 F or 28.6 C

+76)12= 79.8 F or 26.5 C

From Fig 4-1, to

=

79'8 F and 50 % R.Η. is out of the comfort

zone. Recommend lowering to to about 77 F or 25 C. tu

4-9

x72 F

Use Eq. 4-4 to estimate a value of the operative temperature

to, active, assuming to for sedentary activities is 78 F (25.6 C) with met = 2.0. to, active = 78

-

5.4 (1 + 0.5) (2

-

1.2) = 71.5 F, (22C)

Exοerpts fiom this work may be reprοduοed by instruοtors for distribution on a not-1br-proΓit basis for testing or instruοtionaΙ puφoses only to students enrolled in οοurses for whiοh the textbook has been adopted' Αny οther reproduction οr trαnsιαtion of this wοrk beyοnd ιhaι permiιted by Secιions 107 οr ]08 ofthe Ι976 Uniιed Stqιes CopνriPhι Αcιwithouι ιhe oermission οfthe cοpyrighι οwner is unlωυful.

Αs an approximation Tmrt = 2To

_Τ,

and

Tflx =

Tno

* ci1Ι21Tg

_ Τ,

)

Eq '

(4-1)

eliminating Tmrt between the 2 equations

2(Τo_T3)4

= Tno

*CV1/21τn _Tr)

where all temperatures are absolute

Solve by trial and error with T, =72+ 460

=

532 R

and Te =(71.5+460)=531.5 R, C=0.103 x 1Oe, V=30

ta=85F(30C) Cold surroundings require high ambient air temperature for comfort, even with high activity level.

4-10

(a) Most occupants will be uncomfortable because the relative humidity is more than 60%, even with

trx

=

t,

(b) The lightest weight possible. Short sleeves, shorts, open neck, etc.

(c) Lower relative humidity

if possible by adjusting the cooling

system to remove more moisture. CouΙd also increase the relative air motion to highest values, perhaps use fans.

4-11

(a) Even

if the suit

was heavy weight, many executives would be

ΕXceφts from this work may be reproduοed by instτuctors for distribution on a not-for-pro1'it basis for testing or instructional purpοses onΙy to students enrollοd in οourses for which thΘ teΧtbook has been adοpted. Αny other reproducιion or trαnsιαtion οf ιhis νοrk beyond ιhαt permitted by Secιions ] 07 or 108 o{ the ] 97 6 United Stαtes Copyrighι Αcι ινithout the permissiοn of the coρyright oινner is unΙωνfuΙ.

cool if sedentary.

(b) Would definitely be cold, especiaΙly hands and feet.

(c) Probably would be comfortable

in typical work cΙothes

(d) Probably would be comfortable since they would keep their coats on and would be walking around.

(e) Cold to very cold

4-12

Determine relative temperatures difference between inside and outside.

68 - 45 23 = 7 4 _ 45 29 Costs 74 68

4-13

- 45 29 - 45 23

are79o/o of that for increased setting, or

Costs are increased by

26o/o if

thermostat is raised.

Too much air motion in the cold winter months tends to cause drafts and make people uncomfortabΙe. Air velocity just sufficient to prevent large temperature gradients from floor to ceiling is best for winter. Τhe opposite is true for hot summer months. Higher air velocity tends to compensate for high temperature and humidity.

4-14

(a) Raising the chiΙled water temperature will cause the cooling coil to operate with a higher surface temperature and the relative humidity in the space will tend to rise if the latent heat gain is signifΙcant such as would be the case with many occupants, this could lead to u

ncomfortable cond itions.

(b) Yes, during the unoccupied hours the space load may be almost totalΙy ΕΧοerpts from thrs work may be reproduced by instructors for distribution on a not-fοr-profit basis for testing or instruοtional puφoses only to students enrolled in οourses for whiοh the textbοok has been adopled' Αny other reproducιion or ιrαnsιαιion of this νοrk beyοnd ιhαι permiιιed by Sections ] 07 or 108 οf the Ι 97 6 United Stαtes Copyright Αcι τν ithοuι ιhe permissiοn οf the cοpyright oτνner is unlατνful.

SensibΙe heat gain and the load is much less than the design value. ln this case the chiΙled water temperature may be increased.

4-15

Τhese fans may bring air down in the Summer, increasing the velocity of air in the occupied zone and providing improved comfort. ln the winter, air may be drawn upward, pushing the warm air at the

ceiling downward where it can increase the temperature in the

occupied zone without increasing significantly the air motion below the fan.

4'16

(a) Τable 4-2 gives a minimum required amount of ventilation air of 15 ft3 /min per occupant. this is the minimum amount of outdoor air that should be used under any circumstances.

Therefore, (Qo)rin = 15(30)

=

450

ft3/min

(b) on the basis of floor area, the occupancy wouΙd be 25 and the minimum ventilation requirement would be

Q,

=

15 (25) = 375 ft3 /min. lt would be better to design for

floor area if lowest air flow is desired. With 30 actuaΙ student air flow is such a case wouΙd be insufficient.

4-17

Use Eq. 4-5, Solving for C, Cs

=

(QtC"

+

N)/Qt

=

C"

+

(N/at)

= (2001196 + (O.25l9oο)

Ξ

478 x 1o-6 = 478 ppm

ΕXcΘφtS from this work may be reprοduced by instruοtors for distribution on a not_tbr-profit basis for tosting or instruοtional purposes only to students enΙoιled in courses for which the textbook has been adopted. Αny οιher reprοduction or ιrαnsιαιiοn of ιhis νork beyond ιhaι permiιιed by Secιions ] 07 or 1 08 of ιhe Ι 97 6 Uniιed Stαιes Copyrighι Αcι w ithοul the permission of the cοpyrighι oιυner is unlαινful.

or using Sl Units

c, =

4-18

=

(2oo / 106)+ (0. 118 t 0.472x 9OO)

(2OO /

106) + (278 t 106):478 ppm

n = number of people to occupy a room

N=n(5.Oml/s) Solving Eq. 4-5 for Ν N = Qt (C, n

- C") = n (5.0) ml/s-Person

: Qt (C, - C")

:

/ (5.0)

2.8 (1000-280) / 5.ο

n = 403 persons or 0.0069 m3 /s

- person

For English Units: n=

6000 (1oOO - 28Ox 10-6) / O.O107

= 404 persons

4-19

or 14.8 cfm/person

Use the M-100 media of fig. 4-8. From table 4-3, select a

12x24 x 8

unit; 650 cfm, ΔP = 0.4 in. wg

At ΔP = 0.25 in. wg. each unit will handle Q

=

Ql |o'25 Ι o.40]1l2

=

650 [O.25 t o'4oJ1l2

:514 cfm/unit. Then the number

of units

EΧcerpts fiom this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructiοnal puφoses only to

StudΘnts enroιιed in οourses for which the teΧtbook has been adopted. Αny oιher reproduction or ιrαnsιαtιon οf ιhis νork beyond ιhαt by Secιions 1 07 οr 1 08 of ιhe Ι 976 Uniιed Stαtes Copyrighι Αcι lυithοuι ιhe permission οf the copyright oνner is unΙανful.

permiιιed

;g μ=(2000

4-20

l

514) = 3.89 or

4. This

is a satisfactory number.

Use the M-100 media from TabΙe 4-3 select a O.3 x O.6 x-O.2 unit. This is rated at 0.3 m'/s with 1oo pa pressure drop. 60 pa the alΙowabte flow rate for each unit would be Q = (0.3) (60/1 OOf tz = 0.23 m3/s

Αt ΔP

=

require 1'Oolo'23 = 4'34 units. This requires at Ιeast 5 filter units, but since this is an odd number, recommend 1.OO m3/s wouΙd

using six units. Trying the 0.6 x 0.6 x 0.2 filter the allowable flow per unit would be

Q

= (0.62) (60/1}q1t2 =

0.48 This would require more than two units of this size. Εconomies would determine the best choice.

4-21

Solving Εq' 4-1O for Q Q

=

Qr [ΔP / ΔP,]1'2

N = ss00/235 = VeΙ = Q/Α =

4-22

= 9OO

[o'1 l0'15]1|2 =735 cfm/module

T.4g [must be integer] Use g modules

ΨΨ (2)(8)= 344fpm '|-

=

5.7 fps

Solving Εq' 4-1O for Q Q = Qr [ΔP

/

ΔP,J1l2 = (o'42) |24

l 37

'4]1t2 =

0.336

m=(2.8)/0.336=8.3 Use 9 modules, a 3 x 3 arrangement.

Exceφts from this work may be τeproduοed by instructors for'distribution on a not_1br-proΓit basis for testing or instructionaΙ purposes only tο students enrolled in courses for which the textbook has been adopted. Αny other ,rpr:orl'u"Ιιon or trαnsιαιion of this work beyond thαι permilιecΙ by Sectiοns 1 07 or 108 οf the ] 976 t]niιed StαιeS Copyrι?ht Αcι w-ithοut thi permis'ιδ" oj ιn, copyrighι οwner ιi unlανful.

/3

VelocitY

4-23

=

a

(2.8)m3

FAcΕ AREA

M-200; 0.6x0.6 xO.2;

/s

(0.3)(0.6)(e)m2

O.4Om3

=1.73m1s

/s/module

Use Eq. 4-10 ΔP

=

Δη ta / Qι.]' :1OO

Velocity

4-24

=

a

A

lO,4OtO.42l2 =90.7 Pa

0.4 =2.22m1s (0,3)(o 6)

No solution exists due to the fixed air quantity for the unit. This part of the problem is intended to show the student that typical direct expansion equipment cannot be used in this \May. lt also shows that the load due to outdoor air is very large.

4-25

exhaust

sHF= 0.7

Γho

:

o'25 rh"; Locate point 1 on psychrometric Chart at82'4 F db and

66.8 F wb it = 31.4 Btu / lbm and v1 = 13.9 ft3 /lbm fbr tΘSting oΙ instructional puφosΘS only to Exοerpts from this wοrk ιnay be reproduced by instructors foΙ distribution on a nοt-for-proftt basis or ιrαnsιαtion οf ιhis wοrk beyοnd ιhαι permiιted students errrolled in courses fbr whiοh the textbook has been adopted. Αny other reprodicrion is unlανful' by Sectiοns 107 οr ]08 ofthe t976 Llnited Stαtes Copyrilht Αctv,ithοut ιhe peιmissiοn ofιhe copyright oνner

74

Q1, =

Φl

rhi (ii - is) = at

=

/V

35ο Ι 12,ooo Qls is = 31

.-

,

(60)

(i1

-

is)

= (350 l 12'000) (6ο / v1)

''''J8?rε;

=

''

(i1

_i.)

23'46 Btu / lbm

Locate on psychrometric chart' ts = 65'6 F db' 55'5 F wb Q.r =

lil,

tr,

= Γil1

O,

=

(ir.

=

- is) = 36'000;

^ -ΨΨ- 23.46)

(27 .6

rh, (vr,

=

=

:8695'7

Ψ(13.4)

Qt" = 8695.7 (31.4 -

Qr

ir =27

=

'6 Btu / lbm lb / hr

1940 cfm

23.46) = 69,000 Btu / hr = 5.75 tons

5.75 (350) = 2014 cfm

(ο) Design filters for 2014 cfm, use M-200 media

of fig 4-8.

Try the 24x24x8 units of table 4-3. 920 cfm @0.4 in. wg.

For max. ΔP of 0'125 in.wat. Q = 920 [0. 125 tO.4O]

1t2

=514 cfm / module;

n = 2014 I 514 = 3'92, use 4 modules

4-26

Use the M-15 media, η

=

93 % from fig' 4-3'

From table 4-2,60 cfm / person is required, outdoor air. purpοses only to on a not-for-profit basis for testing or instruοtional Excerpts tiom this work ιnay be reproduced by instructors fοr distribution thαι permiιιed beyond work thιs or trαnslαιionλf Αny oιher reprοduction students enrolled in οourses for which tl-ιe textbook ι-,u, υ..n uαopt"J ' copyright oινner is unlαwfuΙ' ofιhe permission thi \,'ithout Αcι Copyrighι Stαtes tJnited ιhe-|97|6 108 o7 by Sectiοns ] 07 οr

75

-A fresh air balance on the filter gives QrEt

+

Qo

=

Qs

where Q,. is recirculated air, Qs is outdoor air and

Q. is supply

air.

8, = (60 - 20) 10.93 = 43.0; Q,

=

43.0 +20

=

63.0 cfm / person

or the total amount of air supplied is

Qτ

=

63.0 x 55 = 3465 cfm; Try the 12x24x8 unit of table 4-3

Q/unit=9oO[O'1 /O.35]1Ι2=481 cfm; n =4755 l481 = 7 .2

modules

Use 8 modules [Note: The M-24 media could also be used]

4-27

Q, = (25 - 15) / 0.S = 12.5 cfm / person

Φ.

4-28

= 15 +

12.5 =

27 .5 cfm / person

Filter location is B, figure 4-9

Use Eq. 4-12, solve for RQ.. since RQΓ = { -QoEv[C,

_(1_Et)Co] + N}/ (EvEfcS)

RQr={-2OOxO.85[180-(1-0.8)0.0]+(10x150x35'32)]l (0.85 x 0.8 x '180) where Co

RQr

=

=

0.0

185 ft3 / min or cfm

for-testing or instruοtional puφoses only tο Exοerpts from this work may be reproduced by instructors for distribution on a not-ibr-profit basis of this νork beyond ιhαι permitted ιrαnslαιiοn or reprodiction Αny other adopted. been has the textbook students enroΙΙed in οourses fοr which is unlιτνν.ful' oνner copyrighι of ιhe permissiοn the Αcι ιiithout by Sectiοns Ι 07 or 1 08 οf the Ι 976 tJnited Sιαtes Copyrιght

76

Qo

4-2g

= 2OO

cfm, Qs

:

(1S5 + 2OO) = 385 cfm

Solve Εq' 4-11 for RQ, RQr

=

(_Qo)(Eu)(cr)+ N / ΕrEiC,

RQr = [ (-20) (0.65) (220) + (125) (35.32 ft3/m3)] (0.65X0.7)(220)

-Ψ9-*_!1!5^, ' =(0.65)(0.7)(220)

RQ.

4-3o

= 15.53 cfm/person

For filter location A, use Εq. 4-1 1, solving for RQ,

RQr

=

(-QoEvCs

+

N) / (EvEfCs)

RQr = t (-2OO (0.85) 180) + (10 x 120 x 35.32 ft3/m3 (0.85 x 0.8 x '180

RQr

4-31

=

)l

t

)

183 cfm, Qo

= 2OO

cfm;

d,

=

383 cfm

(a) This type of space will require a high ventilation (supply air)

rate to handle the load, air cleanliness is not the main criterion.

Therefore, a low efficiency filter with low pressure drop is acceptable. From table 4-2, assume occupancy will be about 30 persons / l

OOO

ft2. So the total design occupancy is 90

persons. Τhe design will be based on this occupancy although the cooling requirements may dictate a larger supply air rate. 1br distribution οn a not-fοr-prοfit basls for testing οr instructiοnal purposes only to ιhαt permitted students enrolΙed in courses Γor which the textboοk has been adopted. Αny οther reprοducιion or trαnsΙαtion of ιhis νork beyond by Secιiοns 1 07 or ] 08 of the Ι 97 6 [Jniιed Stαtes Cοpyright Αcι νιthout the permissiοn οf ιhe copyright oνner is unlανful.

Exceφts from this work may be reproduced by instruοtors

77

A ''fresh air'' balance on the filter d," = (20

gives

Φ,

=

(Q" _ Qo) / Ef

-15) / 0.5 = 10 cfm / person recirculation rate

4-31 (continued)

Φ, :

1O +

1

5 =25 cfm / person supply rate

Qτ = 25 x 90 = 2250 cfm total supply rate Net face area, (b)

Αf

= 2250

/ 35o

= 6'43 ft2

A higher efficiency would reduce the total amount of air and

reduce the required face area. However this is not desirable in this case. First the filter system would have to be enlarged to handle the greater amount of air. A lower filter efficiency could be used and still maintain the required air quality. For example, suppose the load dictates 4000 cfm instead of 2250 cfm, then for 90 PeoPle

Φ,

=

4ooo / 90 = 44'4 cfm / person

Using a minimum of 15 cfm / person of outdoor air. Qr. = 44.4

8. Et

4-32

(a)

-

15 = 29 '4 cfm / Person

29.4: (20 - 1s) / Er = 5 Ι29'4 : 0'17 or 17oλ required =

Q=

(Q, / v) 60

(i|.

l5x

-ir)

Εxοerpts frοm this work may be reproduοed by instructors for distribution on a not-for-profit bι studeπs enrolled in courses for which the textbook has been adopted. Αny oιher reprοduction ο by Sections ] 07 or ] 08 of ιhe 1976 Uniιed SιαteS Copyrighι Αcι ν)ithouι ιhe permission οfιhe cop;

I

225 people

75F

RH=5ο%

125,ο0ο

78

Υ x13 ft3

/ Ιba

Φ. = (125,oOo x13)

Qs (b)

=

Ι [

60 x (28 - 1e.4

3,149 cfm

Φ,

= Φo

Φ,

=

= 15 x225

3,375 cfm

-D (c)

)]

Q. must be 3,375

cfm, find

50 52

ne\Λ/

Supply air condition

125,000 = (3,375 I 13) 60 (28 - i.)

i'

= 28

-

(125,000 x 13 ) / ( 3,375x 60)

=

20 Btu i lba

Locate new condition on chart aS Sho\Λ/n' Coil must cool oDA down to this new condition.

Exceφts frοm this work may be reproduοed by instruοtors Γor distribution on a not-tbr-profit basis for testing οr instructional purposes only to students enroΙΙed in courses for which the textbook has been adopted. Αny οιher reproduction or ιrαnsιation of ιhis work beyond thαι permitted by Sections 107 οr ]08 ofιhe )976 Uniιed SιαιeS Cοpyright Αct withοut the permissiοn οfιhe cοpyright oνι]ner Β unlα:wful.

Exοeφts from this work may be reproduοed by instructors for distribution on a not-for-profit basis for testing or instruοtional purposes only to students enrοlled in courses for which the textbook has been adopted. Αny other reproduction or trαnslαtiοn of this work beyond thαt permitted by Sections Ι07 or ]0B of the Ι976 United Stαtes Copyright Αct without the permission of the copyright owner is unlαwful. Requests for permission or further informαtion should be αddressed to the Permission Depαrtment' John Wiley & Sοns, Ιnc, ] ] ] Riνer Street, Hoboken, NJ 07030.

Chapter 5

5-'1

Χ4 =O.8 (Btu

(a)

k = CΔx =0.2

(b)

k = 1 .14 x 0.1 = 0.114 W / (m-C)

_

in) / ( hr - ft2

_

l

5-2

(a)

C=k

5-3

(a)

R = 1lC= 1/0.055 = 18.3 (ft'-

hr- F)/

R' = R l A= 1 l CA= 18.3/ 1ο0

=

Δx=o.3o / 5.5 = o.o55 Btu / 1ft2-nr_ (b) C = O.O43 / ο. 14 = .307 W/ (m2 _ C)

F)

11

Btu

0.183 (

hr-F)/Btu

(b) R ='1 I .307 = 3.26(m'-C)/W R'=3.26l9.3=0.35C/W 5-4

ΣRi , Rgyp =1ΙC=1l3'1=o'32 Rbtd = !0.33 = 3.03; Rair = 0.68

R=

R = 0.68

+

0.32+ 3.03

+

R-0.68

R=0.68

0.32+ 0.68

R=0.32

R = 5.03 (hr- ft2- F) / Btu

5-5

tnb Γ2

ιn2

+ η R'= 2πk| 2πkoL AssumeL=1ft

kι:0'2

Btu - in' t(ft2

_hr_F);

kp =314 Btu-in

tσe _hr_F)

81

lnside Surface (7 m/s

0120

)

=

overalI Τhermal Resis.

0.652 m2clW

Between Frame

Αt Framinq

Outside Surface

0.17

0.17

4 in. Face Brick

0.65

0.65

Sheathing

1.32

1.32

lnsulation

1

5-10

1.0

4.27

2x4 stud

Gypsum board

0.32

0.32

lnside surface

ο.68

0.68

14.14

7.41

Τotal

UA:U1Α; +U1Α1, U = UiAi Αr

A

=

14'5

A'

16 16^nd A =lΞanα

/Α + U1Α1/A

υ=L R

0 o77Btu , =ΓΨ "14.14) ΓΨ " +1= : 7.41J " -)116 110 5-11

/ (r',..-tt'

-r) A

An ordinary walt with ε

=

O.9 has a unit resistance of 0.68.

highIy reflective wall, ε

=

O.05, has a unit resistance of 1'70.

Assume radiation heat transfer is zero for reflective wall. Τhen the resistance due to convection alone is approximately

Rc=1.7; hc=1/Rc=0.59; h.*r:1/0'68 = 1'47 Frac. Conv. = ha lh. *,

=

0.59 I 1'47 = 0'4

by instructors fοr distribution on a not-for-profit basis for testing or instructional purposes only tο has been adopted' Αny οther reprοclucliοn the texibook *nιοh

from this work may be reproduοed Εxοerpts "'.Φ;.;;;1l;l,,.o,rr.r'r-

o^:.:::Ei::::":!,::iχ:,i''Ψ'y''o"!:*!

-,

-

-

,

82

RzQ x 4\ Rr (2 x 6) wind. 0'17 0'17 1. Outside surface, 15 mPh 0'79 0'79 2. Siding 1'32 1'32 3. Sheathing 11'0 19'0 4. lnsulation, 4'27 2x4 6.7 2x6 0'32 0'32 5. GYPsum wall board 0'68 0.68 6. lnside surface 18'55 28.98 Total

5-12

Αssume 15 mph

Ut

=

0.035 Btu / (hr - ft2 - F)

υ2 __O.o54 Btu / 1ιlr - ft2 - F1 % DiffereΠCΘ = [o'O5-4r_0-035) ι'' )'

5-13

0.0sη

[

ool = 35'2

Air space will be near the indoor temperature with small Δt across the air Space.

Use t."rn R

5-14

1.oz(rrr

50 F and Δt

=

=

10 F and read

-f( -r) I Atu

Assume tr"rn R

5-15

=

=

=

50

F;

3.55 (hr - ft2 _F) /

[Tabte

S-3a]

or 0.18

m2clW

Δt = 10 F

Btu

or 0.62

(.2

_c/W)

[Τabte 5-3a]

qc/Α = U"Δt Find U for highly reflective surfaces because radiation will be minimal. This will give a good approximation for the convection component. From Table 5-2a,l1orz', heat flow down

83

Uc=1/R = 1l(2x4.55) =

0'11

q./Α or

" --l' (2x0

U^ _

= o.625; q.

/Α

0.625(63 - 43) = 12'5 W/m2

=

/ a '.4l 'l

Γ, _ '4 |l

_[ l' l _ι1οo]

l( =σ'n'Lι1oO]

Q/A..

for ε1= t2=O'9, (q/A),.

= O.1

-]'

E : 0.82,

713 x9.s2 t635

l'

σ' o

_s.τol

m

Radiation heat transfer is about 10 times greater'

5-16

U*

= O.O7 Btu / (hr - tt2

Ud = O.4O Btu / (hr - ft2

-

- F)

Uwin = O'81 Btu / (hr - ft2

Ad

Α*

=

11

g'

- F)

17 '78 ft2; Awin =25'0 ft2;

= 117 '2

20'

f(

Parallel heat flow Paths

UΑ

ι

Iι I_

=

5-17

=

U*A*

+

U6Α6

+

U*;nΑ*in

(o.O7 x117'2) + (O.4 x 17.78) + (0'81 x25'0) 117.2

O.3O Btu /(hr -

q/Α

f( _F)

or about

1

'72\Ν t(m' _

c)

= U(ti _to)

a not-for-profit basis foι testing or instructiona] purposes onιy to Exοerpts from this work may be reproduced by instruοtors for distribution on reprολucιion or ιrαnsιαtion of ιhis work beyond ιhαι peπnitted Αny οιher uJo|t.α. students enrolled in courses tbr wbich the texibook has been : : -'-ι.41-^

^^^'',.:/Ιa!

^''''aΔv;"

"'-Ι6!$ιΙ

84

From Table 5-4b, construction 2, R = 8.90 (hr

Assume insulation does not

fitl the

-

ft2

-

F) / Btu

airspace'

tt' - f)/Btu and Remove R for metal bath and plaster of 0.47 (f''ι. ) \"' add R for acoustical tile and insulation' Ceiling, R" = 1 / 0'8; insulation'

q/Α = o'o48 (72 - 5) =

1

9!+=oe71wr(m2-c) \ 0.1761

Ud = 2.27 Uwin =

-a)'

Wl(m2

4'62w1(m2 -

/

Table 5-8

")'

Table 5-5b

Αw

=

35 m2;Αwin =8m2;Αd = 2m2

UΑ

=

U*A*

+

U6

1x35) u _ Q.e7

+

+

Αα

+

-Fi

3'22 Bτυ / (hr _ ft2)

From Table 5-4a, Construction

Uw=

1'00;

0'048 Btu / (hr - ft2

R1e61=20.68; U = 1/R =

5-18

R='1

UwinΑwin

(2.27x2)

+

35

(4'62x8)

-_

2.16 w

I

(m2

- c)

5-19 U = O.14 Btu / (hr

_

ft2 _ F1τable5 _ 4a, Construction No. 2

R=1ΙO.14 = 7'14, Rn =7'14_(1 to'44) + (1/0.55) =6.69 Un = 0.15 Btu/(h r

5-20

-

ft2

- F) or about ο.85 W

l

1m2-c1

Αssume Ηardwood, k = 1.25 (Btu-in) / (hr - ft2 - F) Winter Summer Rι =

0.68

Ri = 0.68

not-for_profit basis for testing or ιnstructionaι purposes only to ExcΘφts from thts work may be reproduοed by instruοtors for distribution on a reproλuction or ιrαfisιαtion of ιhis ινork beyond thαι permiιιed students enτolled in courses for which the teΧtboοk has been adopted. Αny οther

85 Rα =

'1

'375

l

0.25 R, = 2.03

1.25

Rα = 1 '375

l

1'25

Ro = 0.17

Ro =

R* = '1.95

-F)

U, =0.49 Btu / (hr-ft2

U* =0.51 Btu /(hr-ft2-F)

Both values are greater than the value given in

Τable 5-8 of O.39 Btu / (hr _

5-21

Computed: Ri Ri

-

+RgaRo=0.96=R

:

Utub

-r);

\Ν

l(m2

\

From Table

_c); /'

5-5

Rn

l l,.Jfl

=

Table 5-5a

U=1.08 Btu / 1nr - ft2 - F)

-f( -r)

=U-R', ' ., = ++ 1.og

Uw

computed

Table 5_5b; Same result

(b) Αssume tr"rn : 50 F; Δt = Ras = 1.ol (nr

-c)

or 5.92 wl(m2

_r);

1.O4 Btu / (hr - tι2

or 5.91

5-23 (a)

F), but acceptable.

= 0.68, Rs = 0.03 (estimate); Ro = 0.25

U = 1.04 Btu / (hr - tt2

5-22 (a)

ft2

1

10 F

r atu

.O1

o.o89 Btu /(hr - tt2 _

_ ο.o29 Btu lJιur\ιll/ (hr --ιLft2 -ν'νLζ)

= 1'94, Un = O.52 Btu/1ιlr-ft2-F)

11

_F) _| l

or O.51 W/(m2 _ c)τaοle 5-9

(-'_c) -)

ιJlv'l\J vYl1ιιl or O'16 W/

τante 5-1o

(b) Q=UΑ(ti _tg); tr= t"ur-A

Εxοerpts from this work may be reproduοed by instruοtors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in οourses for which the textbook has been adopted. Αny οther reproducιion or trαnslαtion οf ιhis νork beyond thαt permitιed ..-_''r-] ^_':'_^'-

!

^'.'.ι ^'.4'ι

86 tavg

A

=35'8

F

:22

F

(2'44

(12

C)

5-23 (continued)

=

R"

=

R1,

Rrin =

5-25

*

)

0.ο89 (4 x 20 x7) (72 - 1 3'8) = 2,900 Btu / hr or 0.85 kW

qn = 0.029 (20

5-24

Figure 5-7

tι:72re2c)

tg__35'8-22 = 13.8; q*

C) Table 5-11}Chicaαo.lllinois

Un =

x20) (72 - 13.8) = 675 Btu /hr or 0.2 kW

0.029 Table

5-10

1l 0.48 Table 5-1a (Fibrous Pad) 2.08

36.6

Re

=

#"+

U"

=

0.027 Btu / (hr - ft2 - F) or 0.16 Wl(m'z - C)

(a) R*=: "0089+11+

=

(1 t \

g.1) =

22.6

U* = 0.044 Btu/(hr-ftu-F) or 0.25 w1m2-c)

Rfι

=

+ (1 tO.4s): -: 0029

36.6

\

Un = 0.027 Btu/(hr - tt2

-f)

or 0.155

Wl(m'z-c)

(b) Refer to problem solution 5-23 Q* = 0.044 (4 x20 x7) (72 - 13.8) = 1434 Qn =

5-26

0 027 (20 x20) (72 - 13.8) = 629 Btu / hr or 0.18 kW

Rins =

C

=

Btu / hr or 0'42k\^Ι

#

=

4'1z

O.24Btu/(hr

(rrr

-'f(

_f( _r)l

εtu

-F) or 1.36 Wl(m'z - c)

Excerpts fτom this work may be reproduced by instruοtors for distribution on a not-for-prοfit basis for testing οr instructional purposes only to

studentsenrolledincoursesforwhichthetextbοokhasbeenadopted. 'l-''':'''

bνSectiοη.s1[]7ny ιnρn{t]ιo 1076'ΙΙb;l.)ζl-l-"/-^4'-')-ι"}

ΑnyοιherreproductionοrιrαnsΙαtionofιhisνοrkbeyondιhαιpermiιιed :"'''.Ι^''Ω'Ι

l

λ / .*'

87

Then from Fig. 5-8, U' Q=

5-27

U'P

=

0.85 Btu/(hr

- to) = 0.85 x 300172 -101

(ti

tι =

72 F (22 C) Assumed

R"

= R5 +

Rfi,.,,

-f(

=R1

R1η, Ub

=

=

-F) or 1.47 W(m-C)

15,8'10 Btu/hr or 4.63 kW

o.ο52 Btu / 1nr -tt2 _

11 Table

5-9

+Ru6*Rqyp=(5.0) + 0.0 +(1 12.22) = 7.22

1 +7.22=26.5 R^- 0.052 5-28

o.o38 Btu / (hr -ft2 _F) or o'22ννl(m'z_c)

U"

=

Ub

: 1.14

Rn

=

from Table 5-9

++ 1.14

O.7 + (1

t

12'6)= 1.66 (m2-c)Λ//, Un = 0.60 W(m2-

c)

or

Un = O'1OO Btu / 1nr -ft2

-F)

This does not account for the walls above grade.

5-29

U = 0'16 Τable 5-'10 (no finish)

Rn:++(t69)+ ' 0.16 \ Un

5-30

(1 t4.6) =6.611m2-c1

= 0.15 W/(m2-C) or 0.027 Btu / (hr

q/A=

Un(ti

-ts)=

(ti

= ti _UnR1(ti

- r)

-tt)/Rt=(tt-t)lR2

Rl=Rgyp+R1nr+R1, R1 = tl

-ft2

lw

(112.6)+ 0.7 +0.12=

0.90

_tg) =20 - [1.05 x 0.9 (2ο - 10)]

EΧcerpts frοm this work may be reproduced by instructors for distribution on a not-for_pΙofit basis fοr testing or instruοtional purposes only to students enτolled in courses for which the textbook has been adopted . Αny οther reproduction or trαnsltιιion of ιhis νοrk beyοnd ιhαt permiιιed '.''-:-ι'ι ^'"-^-:' "'-ι5υ{ιi ' hιi 9λ"};.-" ιηa ^'' ]Λ9 ^.}L- ιoaA ι L-;.^) c'-''^'

. . .

88

t1= 14.6 C or 58F R2

=

Rrr, +Ri

=

(

I 12.6) +0'12='20

tz = 20_ [O.60 x O'2 (2o-1ο)] = 18.8

5-31

q/A=Un

(ti

c

or 65.8 F

_tg)=(tι _t'')/R1; Rl=Ri +Rc

= 0.12 + (1 I 4.6) -- 0.34 tι=20_ (O.15) (O.34) (20-10) = 19.5 c or 67 F

5-32

C

= 0.2 Btui(hr-ft2-F);

U'P

8=

U'ni

=

(t1

_to);

Figure 5-8

Ui, = 0.81 Btu/ (hr-ft-F) or 1.a

1.37 Btu / (hr - ft -F)

(a) q/P = 0.81 (70 - 5) = 52'7 (b) q/P = 1.37 (70 - 5) = 89'1

5-33

Q=

W(m-C)

Δt / R'

;

R'=

Eq. 5-25; L>>Ζ'

Btu / (hr - ft) or 50.7 W/m Btu / (hr - ft) or 85.6 W/m

L=

100ft

2πkL

. Γzoo x121Γ^ tn(12x1OO/2x3o)l

_'nL-o--@J _3"100Ι12) R'

= 8.

12

x

1O-3 thr - F) / Btu

Which neglects the resistance

of pipe wall.

purposes only to for distribution on a not_for-profit basis for testing or instruοtional Excerpts from this work may be reproduοed by instructors οf this νork beyond ιhαt permitιed οr ιrαnslαιioi reprολucιiοn oιher Αny uJo|i"a. υ"., hu, '' : '1-"4'ι Students enΙollod in courses fbr ινh'iοh the textbook "' ^' L- Q.nl;n-"

ιnη ^.' 'Λο

70-42 - =3,4488tu 6' 8.12 x 1O-'

/ hr;

q/L

= 34.488tu/(hr-ft)

1W/m orq=1.01 kW; 9=33 L 5-34

Q=Δt /R'

R'g

=

=4.98x10-3 C/W

2π (1.4) 100

film and the tube wall' Neglect resistance of the inside

.

'

O=

5-35

60-5

----------c =11.04kw 4.98 x 10--

Moisturewillmovetowardtheinside.Locatethevapor

insulation' retardent on the outer side of the retardent is placed on the The insulation will beοome wet if the plywood would probably \Λ/arp inside or left out entirely and the

and rot.

5-36 (a) Q/A = Uo(ti Ro = O'68

+

_to)=(tι _t1)/R1 =(ti

O'45 +1

1+

1'O +O'8 + O'17 = 14'1

Uo

=

_F1 O'o71 Btu /( ιrr - ft2

Rr

=

0'68

R2

=

O'68

+ +

O'45 = 1'13( hr O'45 +11

_t)lR2

f(

-F) / Btu

-12'13(hr

- ft2

-F)

/ Btu

tt=tι_R1Uo(tι_to)=7O_(1.13xO'O71)(7O-1O)=65.2F purposes only to basis for ιesting or instructiona' for distribution οn a noι-for-proΓlι ιyt.pern:ted insιructors by reproduced be lnay Excerpιs Γrom ιhis \νοrk j:*tbook has *9,i'"Ιl'Ji."o"'i,-:';;;;;;;;;;;"Ζii"111i117; γ,:y:γ:'::'|"*o students enrolled in courses foΙ ^γηι:ι lh.

90

tz=70-(12'13 x 0.071) (70-10) = 18'3 F of air to surfaces

(b) At 70 Fοο, 3O%o R.Η. and possible leakage

1

or 2-

tdp=37F 2 in. (e) 2OO gpm; g %in., .(,'1< 4 ftllOO pUsec; dia > 2in.

(η

2ooo gpm; 8 in., /1 slightly

>

4 ftl1oo

ft

Exοerpts 1iom this work may be reprοduced by instructors fbr distribution on a not-fοr-profit basis for testing or instructional puφosos only to students enrolΙed in courses tbr which the textbook has been adopted' Αny οιher reproduction or trαnsιαιion of this νork beyond ιhαι permiιιed by Secιions ] 07 or Ι 08 of ιhe ] 97 6 Unιted Sιαιes Copyrighι Αcι ν ithοuι the permission of ιhe cοpγ,ι'ighι ονner is unlαwful.

197

1ο-18. (a) K = 30 ft, ft = 0.019; K = 0.57 (Table

V

(b) K

10-2; Figure 10-22a) = 3.82 fVsec; ! r = 0'57(3'822lβ2.2x2)= O.13 ft

= 340 ft, ft= 0.017;

V

= 5.0

f/sec; !

r

K

= 5.78

= 5.78 x 5.02t132.2

x2)

(c) K= 60ft, ft= 0.018; K = 1.08 V = 6.5 ft/sec; ! r= 1.08 x 6.52t(2x32.2) '10-1e.

!r=

2.31

(#)'

=

2.24 ft

= 0.71 ft

= 10.8 ft of water or 4.7 psi.

10-20. Assume com. stl. pipe Q

= O.O3

mt/s = 108 m3/hr, size pipe for about 4 mllOO m

From Fig. 10-20, use 5 inch pipe, lD = 130 mm a, f -_ ..25 ml100 m; [1= (3'251100)200

nt λ'

ΔPg = 35 kPa

Γ o-99-1' ' | "

For strainer. ΔP"

1 0.00722 J

=

ι7

=

'27

6.5 m of water or 63.7 kPa

kPa

Then for the pump: ΔPp = 63.7 + 35 +

'17

.3 + 3(1000)(9 .807)11000 = 145'4

kPa

Ηp = 145.419'807 = 14.8 m

Q

= O.O3 mu/s

-

30 L/s

10-21. Size the pipe using Fig. 10-20 or program PIPE. Fitting equivalent lengths found using Fig. 10-22a; 10-22b and Τable 10-2' Program PιPE could be used to solve the complete problem including fitting losses. Data for hard calculations are summarized below: Εxcerpts from this work may be reproduced by instruοtors for distribution on a not-for-pro1it basis for testing or instruοtional puφoses οnly to students enrolΙed in οourses Γor whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsΙαtion of ιhis νork beyond ιhαt permitted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed SιζlιeS Cοpyright Αcιlνiιhοuι ιhe permission ofιhe cοpyrighι oνner is unΙανful.

L

198

lO(3)

1ot3] Θ

Θ

ro(3)

I

/o\

\:-/

o(3)

]

Sec. No. 1

5

6 7

4 10 2 3

I 9

ch

qpm

stze

tn.

120 3

70 2.5 40 2 40 2

90 2.5 120 3

5ο 2 50 2 30 1.5 30 1.5

120

5ι5)

it

Le

ιt

ft./'100 ft

ft.

ft.

3.38

45

1.5

1.5

3.64

'15

0.6

0.6

3.'1

24

0.7

3.'1

13

0.4

5.84

27

1.6

1.6

3.38

42

1.4

1.4

4.7

22

1

4.7

26

1.2

6.3

28

1.8

6.3

13

0.8

Coil

Con.

Valve

ft.

ft.

11.4

Τotal ft.

12.1

12.0

10.0

12.4

11.0

10.0 14.4

11.2 16.2

15.0

15.8

20

Exοerpts lrom this work may be reproduced by instruοtors for distribution on a not-1br-profit basis for testing or instructional purposes only to students enrol]ed in courses Γor whiοh the tΘXtbook has been adopted' Αny οιher reproducιiοn or trαislαtiοn of ιhis νori beyoncl ιhαι permitted by Secιions ] 07 or Ι 08 of ιhe ] 976 (Ιnιted SιαιeS Copyrighι Αct wiιhοut the peimission οf ιhe cοpyrighι owλer is unΙcnνful.

199

The head losses for the three parallel runs are approximately the

same.

For run (1-5-6-Z-4-10), Hp = 49.6 ft For run (1-5-S-g-10), Hp = 55.5 ft For run (1-2-3-4-10), Hp = 46J ft Therefore, a pump should be selected to provide about 56 ft of head at 120 gpm.

10-22. 500 gpm, Use 5 inch pipe; !'f

V

= 8.0

=

4.17 fil1OO

ft

ftlsec

Length of pipe = 160 + 3O + 12 = 202 6-5 in elbows = ,lS

ft

ft

(Figure 10_22)

3-5 in gate valve = 12 ft 1-5 in gΙobe valve = 130 ft; Total equivalent length = 419 ft

. 4.17(41e\ /, = -Ι_1}-J "/ = For strainer: !

|

!

Τhen Hp =

17

For cond

"o

17

'5 ft of water

2.31 = e.24ft of water "= = 20 ft of water

[#)'

'5 + 9.24 + 20 + (3o

_

12) = 64.7 ft at 5OO gpm

10-23'Use Εq. 10-33

Exοerpts frοm this work may be reproduοed by instructors for distribution on a not-Γor-profit basis fbr testing or instructional puφosοs only to students enrolled in οourses fοr which the textbook has been adopted Αny οιher Ιeprοdactιon or ιrq'nsιαιion of ιhis νork beyοnd thαι permiιιed by Sections ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyright Αct νiιhλut ιhe peimission of the cοpyτ.ighι owner is unlανfuΙ.

L

200 ,,_

".

_ -

6oott*:;*g I

-,l) - 3x6 sxlo-u (1,0

gz.οgο

10-24' Use Eq. 10-34 v, Vr-_8.Ζ-gal

=

-oull

= 19.4 9a1. = 74 L

og.οgο ]

Ψξg L[ 0'ο1ο0zz

_,'l _ 3x6.5x1

οoo[[9

=

)

o_u (1,, o _

ou1l

'-___-_-J

ι '_ 69:ο%,

33 L

10-25. Use Eq.

10-33 tl = 60oF, P2= 50 psig, P1 = 20 psig, v1 = O.O16ο53 ft3/lbm

vz = O.O1 6772 ft3/lbm, tz = 220"F

Vτ=

= _11_5_gal.

-

435 L

10-26

(a) Use Eq.

'10-16

P,+ PτPzι= Pzgzz +ρνv+ρg

'g"9cι

''

=

ff

lL

Τ 24o

9c

(zz-zι)+ ρw

+

P9nLE

9cl

-

Exοerpts from this work may be reproduced by instruοtors for distribution on a nοt_for-profit basis for testing οr instructional purposes only to students enrolled in οourses for which the textbook has been adopted,' Αny other reproductιon or ffαλιαιion of ιhis wori beyond-ιhαι permiιιed by Secιiοns ]07 οr ]08 οfιhe Ι976 L]niιed Sιαιes CopyrightΑctνΙιhouι ιhe peimission οfthe copyrighι owλer is untιιυful'

Ι

tΙ,

=

#

e4o)

#(60)

+

-

#

es1=Be psis or

61

2 kpa201

(b)

P,'+ρg!= P2+ρg2 9c - --gc Pz=Pι-

t

(zι-zz)=89.o

Pz = -15 psig

=

-effi

-i03 kpa or about o absorute

(c) No, makeup water is not available to overcome a pressure of gg psig. However, the domestic water system probabry has a

booster pump.

10-27

(a)

Pι

=

=

Pz *

(zz-zι) + ρνν * ρ9ι, o* gc 9c

62.4(240) 5+ 144

62.4(60)

144

.

62.4(25) 144

'n0.8

Pr = 93.8 psig or 647 kpa

(b) η =Pz+ Pl

!9 9c

(zz-zι)

= 109 psig or 752

=

5 *62|(,2ra0) = 5 + 104 =.t09 psig

kPa

Excerpts from this work may be repτoduced by instructors for distributiοn οn a not-for-profit basis for testing or instructiοnal puφoses only to students enrolled in courses for which the iextbook t-,u. υ".n permiιιed by Sections ] 07 or Ι 08 of the ] 976 1dγνωj Ατ-ιl λrir, Ιrpr"a""rιon or ιrαnslαιion οf ιhis work beyond ιhαι United Stαιes copyrighi Αcι withλuι ιhe p,e1miιsιon of ιhe copyright ονner is unlαινful.

202 (c) This location is at Ιeast workable. Ηowever the pressure at the pump is still very high. The domestic service water

pressure would have to be boosted to a higher pressure at the 2oth

floor.

10-28

(oo , sz) Qo

*Q.

=

+

(a,

x +o)=

(ao x

ιτ)

Qα= 100

Solve Simultaneous 57Qb + (O x 1OO) - 4OQb = 100 x47

ab = (b) (c)

Ψ 17

Q.

= 41.2 say 41gpm

= Q.. = 1OO -

41 = 59 gpm

Size all pipe for 100 gpm D = 3 in. from Fig.

'1

0-20 or PIPE

10-29. (a) Each chiller requires 600 gpm. Since chiller 2 is partially loaded must have the full flow of 600 gpm. Therefore, Q"p = 1200 - 750 = 450 gpm

(b) (150 - 60)

(c) LR

+ (450

x 42)

=

600 ts, ts

=

46.5 F

= 150/600 = 0.25

(d) Main pipe to and from sec. Circuits: D = 8 in. com.

stl.

Excerpts from this work ιnay be reproduοed by instructors for distribution on a not-for_profit basis fbr testing or instructional puφoses only to students enτοlled in courses for whiοh the textbook has been adopted' Αny οther reproductiοn or trαnslαιion of this work beyond ιhαι permitιed by Sectiοns ]07 or Ι08 οfιhe ]976 United Snιes CοpyrighιΑcιιυiιhοuι the permissiοn ofιhe copyright oνner is unlαννfuΙ'

it

203

Dns = Dco = O in. com. sfl. (S in. a litile small) D"p =

6 in. com. sfl.

Dsc = DRo = 6 Ιn. com. stl. ( could be 5 in. but easier to make all 6

in.)

(e) Rpm,

(η ιW or

10-30. (a)

= ΓPIΠl

=

050l12oo)

*#=1-ff

= 35OO(75o/12o0) =

2188

=,l_[ffiJ' =, (##)

= O 756

75o/o

Q"n = 12OO

-750

=

450 gpm

(450x42)+ (750x60) = 53.3 F 1200 Both chillers receive the same temp. water

(b) Qrtr

+ Qztz = Qsts; ,. =

(c) Load

ratios are the same:

LR= 10-31.

##=0628or63%

Εxcerpts from this work may be reproduοed by instructors fοr distribution on a not-for_profit basis fοr testing or instruοtionaΙ puφoses only to students enrolled in courses for whiοh the textbook has been adopted. Αny οιher )eproclucιion or ιrαnslαιiοn of this ιυοrk beyoncl thαι permitted by Sectiοn's Ι 07 or 108 of ιhe ] 976 ΙJnitecl Stαtes Cοpyrιghι Αcι wιthλuι ιhe peλιssion of the copyright ονner is untωνful.

204 10

(13)

2

2Ο(6) 3

*{s) T'r^;n ι }iJJυα' ^

!

**ntΓ*if Φr 1*{S}

u?

ι$

Ω*{s} s 2Ο(s) 8

BaΙ*n*s ve,v* {typi*aii

**rnnr** pip*

Note: Piping is type L copper Αll |engths are total equivalent lengths

10-31.

FΙow rate Coil opm (L/s) A 40(2.5) B 40(2.5) c 50(3.2) (continued)

Section No.

gpm 130 90 50

1-2

2-3 3-4 ^,^il

Dia.

C

in.

3

2% 2

' he reproduced

Lost head ft (m) Coil Con. valve 12(3.7) 10(3) 15(4.6) 12(3.7) 18(5.5) 15(4.6)

ir ftl1ο0' 3.7 4.8 5.0

L"

!.r ft

ft

60 20 30

2.2

r;

)

355

by instruοtors for distribution on a not_Γor-profit basis fortiting or instructionai puφoses onΙy to -ιhαι hiοh the textbook has been adopted' Αny oιher λeprοduction οr ιrαislαιiοn of ιhis wori beyond ''/ United Sιαιes Cοpyrighι Αcι ιιithouι ιhe permission of ιhe copyv.ight oνλer is unlωυful.

205

Con. C 4-5 Com. oi 2-6 Coil A Con. A

130

3-7

Coil B Con. B 7-8

3.7

0

3 3

40

2

3.4

15 1.5 0.0 39.2 1.0 12

40 30

10

40

2

3.4

10

80

2%

3.9

20

23

0.5 15 12

ft

ft

0.8 28.3

ft

(tote

(total)

(totat)

Circuit 1-2-3-4-5-1 is the path of greatest lost head. From Fig. 40 ftof head and 130 gpm the 7 in., 1750 rpm model which

choose at froduces about 43

ft of head.

1O-1 1

Ξp^q 2ΟΟ 2ΟΟ lbu] 2(rΟ)3(rΟ} t p,ih-*:y +ο* PumP {12)l 4Q t1i

r

f

ι*

{}

Ιs*ianc* vxlγ* {typi*ai}

*hill*r

p ιl Π]$} $

*οntr*i . valv*

ιJ

{

typi*xi}

10-32. Notes: PΙpe is schedule 40, commercial steel. Αll lengths are total equivalent lengths exc|uding control valves.

Circuit

Α B

c

Flow rate gpm (L/s) 60(3.8) 70(4.4\ 70(4.4\

Control valve head loss ft (m) 40(12) 5ο(15) 50(15)

'' be reproduοed by instruοtors for distribution on a not-for_profit basis for testing or instructional puφoses on1y to "hiοh the textboοk has been adopted. Αny oιh-er )eproducιiοn or ιrαλhιiοn o7 tnπ beyond thαι '976 United Stαtes Cοpyrighι Αcι ννithouι the peλission of ιhe copyright ονner 'orκ is unlαwful.

206

Section

if

Dia.

gpm

No.

in.

200 140 70

4 3 2%

4-1

200

4

2-5 Con. A 5-4

60

2%

1-2 2-3

3-4 Con.

L"

(.t

ft

ft

ft/'l00' 2.4 4.2 3.5

200 200 240

2.4 2.5

400 240

4.8 8.4 8.4 50

9.6

81 .2

ft

(total)

54.4

tt

(total)

51.4

ft

(total)

o

40 140 70

3-6

3 2%

200 40

4.2 3.5

Con. B

8.4 1.4

50

Circuit 1-2-3-4-1 has the largest head loss of alΙ paths. Select pump for 2OO gpm at 81 ft of head. From Fig. 10-1 1, use: 5Υ' in., 3500 rpm model. Will operate at 96 ft at 200 gpm. 10-33

(a)

qst = 20

/1

_

x 12,000 x2= 480,000 Btu

gst

ρc, (t,.-t.)

480000 62.4(1) (60-45)

=

512.8 ft3

orQ=3,8369a1

(b) Vol

= 513 ft3 10.2 ft

ora Space

8

ftx 8 ftx 8 ft ora cyΙindrical tank

8 ftdia. x

10-33. (continued) Solution - Sl: (a)

q

=

Qg1

=

(352_280) (2)= 144k\^l-hr= mc, (trt.) _ Qρcp(t-t.)

Qst ρο, (t1ts )

' 1le reproduοed

144 x3600 = 14 m3 980(4. 184) (1 6-7)

by instruοtors for distribution on a not-for-profit basis for testing or instruοtional purposes only to '.'hiοh the textboοk has been adopted. Αny oιher reproduction οr trαnslαtion of ιhis wοrk beyond thαt ^76 (Ιnιιed Sιαιes Copyrighι Αcινιιhout ιhe permission οfthe cοpyrighι oιυner is unlανful'

207

(b) Vol.

=

2Αmx2.4mx2.4m

10-34. Solutions may/can Vary. Α typical solution

(a)

(b)

Use 2 chillers of '15 tons total capacity in a reverse return system similar to figure 10-32. The piping would be routed overhead around the complex with supply and return running parallel, starting and returning to the equipment room. Total flow rate is

Qτ =16 x2'25

(c)

is:

=

Estimated length 3600 ft

36 gpm Using PIPE or Fig 10-21; Dia. = 2in'

=225x4x2= 1s00ft.

Τotal Eq. Length=2x 18OO

=

Assuming an average Ιoss of about 2.5 ft1100 ft; The pump head required would be: Hp = 2.5 x 3600/100 = g0 ft with flow rate of 36 gpm

'10-35 Solutions may vary (a)Figure 10-34 is a schematic of what the system wouΙd be. However, there would be 3 chillers and the secondary piping would be routed in a square fashion around the outside of the parking garage in reverse return. (b)Τhe primary system would appear as in Figure 10-34 with the

common pipe as shown because of the expected variable and light load at night.

(c)The tertiary circuits would be as shown in Figure 10-34 and piped in a reverse return manner. (d)For each building: Excerpts Γrom this work may be reproduοed by instructors Γor distribution on a not-fοr-profit basis for testing or instruοtiοnal puφoses only to students enrolΙed in οourses for whiοh the textboοk has been adopted' Αny other reprοductiοn οr ιrαislαιion of ιhis νork' beyοnd thαt peιmitted by Sections Ι 07 or Ι 08 of the ] 976 tJnited Sιαιes Cοpyrighι Αcι 1ψiιhouι the permission of ιhe cοpyrighι owner is unΙα:wful.

208

Qi

1500 x 't2000

600 gpm

4x500(60-45) = QΤ=4x600 =2400 gpm =

(e) Dia. = 10

in., Figure 10-20 or plpE

10-36. ;ControΙ valve (Typical)

I

Y φ a_Air Vent (Typical) ,+9*Π4 Heating Device (Typical) r-€

T