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

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

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    Chapter 1 1-1 (a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K) (b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K (c) 0.04 Ibm/(ft-hr) 3600 sec/hr x1.488 = 16.5 2 Ns m μ (d) 1050 Btu Ibm x 4 1 9.48x10 − J Btu x 2.20462 Ibm kg = 2.44 MJ kg (e) 12,000 Btu Ibm x 1 3.412 = 3.52 kW (f) 14.7 2 Ibf in x 6894.76 = 101 kPa 1-2 (a) 120 kPa x 2 lbf / in 6.89476kPa = 17.4 lbf/in 2 (b) 100 W m K − x 0.5778 = 57.8 Btu/hr-ft-F (c) 0.8 2 W m K − x 0.1761 = 0.14 Btu/hr-ft 2 -F (d) 10 -6 N-s/m 2 x 1 1.488 = 6.7 x 10 -7 lbm ft sec − (e) 1200 kW x 3412 = 4.1 x 10 -6 Btu/hr

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    2 (f) 1000 kJ kg x 1 Btu 1.055 kJ x 1 kg 2.2046 lbm = 430 Btu lbm 1-3 Hp = 50 (ft) x 0.3048 ( m ft ) = 15.2 m ∆ P = 15.2 m 1000 Pa/kPa x 9.807 1 ( N kg ) x 1000 (kg/m 3 ) = 149 kPa 1-4 P = ∆ 4 12 (ft) x 0.3048 ( m ft ) x 9.807 1 ( N kg ) x 1000 ( 3 kg m ) ∆ P = 996 Pa 1.0 kPa ≈ 1-5 TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE + METER CHARGE ( ) ( ) ( ) ( ) 96,000 kw - hrs 0.045 $ / kw hr + 624 kw 11 50 $ / kw − − + $68 = $4,320 + $7,176 + $68 = $11,564 1-6 7 AM to 6 PM 11 hrs/day, 5 days/wk hrs days (11) (22) 242 hrs / month day months =

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    3 ( ) ( ) ( ) 624 kw ratio = 1.57 96,000 kw hr 242 hr = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 1-7 This is a trial and error solution since eq. 1-1 cannot be solved explicitly for i. Answer converges at just over 4.2% using eq. 1-1 1-8 Determine present worth of savings using eq. 1-1 ( ) ( ) ( ) 12 12 0.012 $1000 1- 1+ 12 P = 0.012 12 P $134,000 − ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 1-9 (a) Q VA =  = 2 x 3.08 x 10 -3 = 6.16 x 10 -3 m 3 /s m 6.16 x 10 Q ρ =   = -3 x 998 = 6.15 kg/s (b) A= 4 π (0.3) 2 = 7.07 x 10 -2 m 2 Q 7.07x10 =  -2 x 4 = 0.283 m 3 / s; ρ = 1.255 kq/m 3 m = 1.225 x 0.283 = 0.347 kg/s  1-10 V = 3x10x20 = 600m 3

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    4 = 600 x i Q  1 4 x 1 3600 = 4.17 x 10 -2 m 3 /s 1-11 p p 3 q = mc T c = 4.183 kJ/(kg-K) = 983.2 kg/m ρ ∆   1-11 (cont’d) ( ) ( ) ( ) ( ) 3 c 3 m kg kJ q = 1 983.2 4.183 5 20,564 s kg K m q = 20,564 kw = −   kJ s permitted 1-12 = wat q  air q −  11,200(1)(10) = = 2 5000x60x14.7x144x0.24(t 50 ) (53.35x510) − 11,200 = 5601.5 (t 2 -50); t 2 = (11,200/5601.5) + 50 = 70 F 1-13 Diagram as in 1-12 above. q  wat = - q  air 1.5 (4186)(90-t 2 ) = 2.4 (1.225)(1.0)(30-20)(1000) 6279(90-t 2 ) = 29,400

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    5 t 2 = 90 - 29,400 6279 = 85.3 C 1-14 q hA(t =  s - ) t ∞ A= π (1/12) x 10 = 2.618 ft 2 s t = t sur 212 F ≈ = 10x2.618x(212-50) = 4241 Btu/hr q  1-15 A= π x 0.25x4 = 3.14 16 m 2 hA(t q =  s - ) t ∞ h= s q A (t -t ) ∞  = 1250 3.1416(100 10) − ; h = 4.42 W/(m 2 – C) 1-16 (t p q mc =   2 -t 1 ) ; m Q x ρ =   ρ = P/RT = 14.7x144/53.35(76+460) ρ = 0.074 lbm/ft 3 m = 5000x0.074x60 = 22,208 lbm/hr  = 0.24 Btu/lbm-F p c q = 22,208x0.24(58-76) = -95,939 Btu/hr  Negative sign indicates cooling 1-17 (t 1 p m c  3 -t 1 ) + ofit basis fo

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    6 (t 2 p2 m c  3 -t 2 ) = 0 = p1 c p2 c t 3 = 1 1 2 2 1 2 (m t m t ) (m m ) + +     1 2 m Q 1 ρ =   = 1000x 14.7x144 53.35(460 50) + = 73.5 lbm/min 1-17 (cont’d) 2 2 m Q 2 ρ =   = 600x 14.7x144 53.35(460 50) + = 46.7 lbm/min 3 (73.5x80) (46.7 x 50) t 68.3 F (73.5 46.7) + = = +

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    7 permitted Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030 . Chapter 2 2-1 through 2-20 Solutions are not furnished since many acceptable responses exist for each problem. It is not expected that the beginning student can handle these questions easily. The objective is to make the student think about the complete design problem and the various functions of the system. These problems are also intended for use in class discussions to enlarge the text material.

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    permitted Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030. Chapter 3 3-1 (a) P v = = s r P φ 0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psia P a = 101 – 1.43 = 99.57 kPa or 14.696-0.196 = 14.5 psia (b) = v v P ρ R v T or = = = 3 v v v v P 1430 ρ ; ρ 0.0104 kg/m R T 462.5(297) or = 0.196(144) 0.00062 85.78(535) lbv/ft 3 (c) W = 0.6219 (1.43) (99.57) = 0.00893 kgv/kga or 0.6219(0.196) 0.00854 lbv/lba 14.5 = 3-2 (a) English Units – t = 80F; P = 14.696 psia; P v = 0.507 psia Table A-1a W = 0.6219 a v P P = 0.6219 (0.507) (14.696 0.507) − = 0.0222 lbv/lba i = 0.24t + W(1062.2 + 0.444t) i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm

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    8 v = a a R T 53.35(460 80) P (14.696 0.507)144 + = − = 13.61 ft 3 /lbm (b) English Units – 32F, 14.696 psia P v = 0.089 psia (Table A-1) 3-2 (cont’d) W = 0.6219(0.089) lbmv 0.00379 (14.696 0.089) lbma = − i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbma v = 53.35(492) (14.696 0.089)144 − = 12.48 ft 3 /lbma 3-2 (a) SI Units – 27C; 101.325 kPa P v = 3.60 kPa, Table A-1b W = 0.6219 v a P 0.6219(3.6) kgv 0.0229 P (101.325 3.6) kga = = − i = 1.0t + W(2501.3 + 1.86t) kJ/kga i = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kga v = 3 a a R T 0.287(300) = =0.88 m /kga P (101.325 - 3.6) (b) SI Units 0.0C; 101.325 kPa P v = 0.61 kPa, Table A-1b W = 0.6219(0.61) =0.00377 kgv/kga (101.325 - 0.61)

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    9 i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga v = 3 0.287(273) 0.778 m /kga (101.325 - 0.61) = 3-3 (a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hg t = 80 F; P v = 0.507 psia (Table A-1a) W = 0.6219 v a P 0.6219(0.507) = P (12.24 - 0.507) = 0.0269 lbv/lba i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma v = a a R T 53.35(540) = P (12.24 - 0.507) 144 = 17.05 ft 3 / lbma (b) English Units – t = 32 F, P v = 0.089 psia ( Table A-1a) W = 0.6219(0.089) (12.24 0.089) − = 0.00456 lbmv/lbma i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma v = 53.35(492) (12.24 0.089)144 − = 15.00 ft 3 /lbma 3-3 (a) SI Units -27 C, 1500 m elevation P = 99.436 + 1500(-0.01) = 84.436 kPa P v = 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 = 3 0.287x300 1.065 m / kga (84.436 - 3.60) = (b) SI Units – 0.0C; 1500m or 84.436 kPa P v = 0.61 kPa; Table A-1b W = 0.6219 x 0.61 0.00453 kgv / kga (84.436 - 0.61) = i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kga v = 0.287 x 273 (84.436 - 0.61) = 0.935 m 3 / kga 3-4 (a) English Units – 70F, P v = 0.363 psia P v = φ P g = 0.75(0.363) = 0.272 psia W = 0.6219 (0.272) 0.0117 lbmv / lbma (14.696 - 0.272) = i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)] 29.58 Btu / lbma = (b) P v = 0.75 (0.363) = 0.272 psia; P = 12.24 psia

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    11 W = 0.6219 (0.272) (12.24 - 0.272) = 0.0141 lbmv / lbma i = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)] 32.20 Btu/ lbma = 3-4 SI Units – (a) 20C, 75% RH, Sea Level 3-4 (cont’d) P s = 2.34 kPa; P v = 0.75 x 2.34 = 1.755 kPa 0.6219 x 1.755 W = = (101.325 - 1.755) 0.0110 kgv / kga i = 1.0 t + W(2501.3 + 1.86t) i = 20 + 0.0110(2501.3 + 1.86 x 20) = 47.92 kJ / kga (b) 20C, 75% RH, 1525m P = 99.436 – 0.01 x 1525 = 84.186 kPa P s = 2.34 KPa; P v = 0.75 x 2.34 = 1.755 kPa W = 0.6219 x 1.755 (84.186 - 1.755) = 0.0132 kgv / kga i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga 3-5 English Units – t = 72 Fdb; psia 14.696 P %; 50 = = φ s v s v P P or P P φ φ = = ; P v = 0.5(0.3918) = 0. 196 psia

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    12 Air dewpoint = saturated temp. at 0.196 psia = 52.6 F Moisture will condense because the glass temp. 40 F is below the dew point temp. 3-5 SI Units – t = 22C ; 50% ; P = 100 kPa P v = φ Ps ; P v = 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 ft 3 / lba and ρ a = 0.0752 lbma / ft 3 = 22,860 lbma / hr a m 5000 (0.0762) 381 lbma / min = =  (b) Using PSYCH ρ a = 0.0610 lbma / ft 3 or v a = 16.4 ft 3 / lba = 5000 (0.061) = 305 lbma / min a m  18,300 lbma / hr = 3-6 SI Units – (a) t = 13 C and relative humidity 80% then v a 0.820 m ≈ 3 / kga; a m 2.36 / 0.82 2.88 kga / s = =  (b) Assuming same conditions ; 3 a v 0.985 m / kga = a m 2.36 / 0.985 2.40 kga / s = = 

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    13 3-7 English Units – t = 80F, 60% RH (a) v s P P 0.6 (0.507) 0.304 psia φ = = = = 64.5 F dp sat v t (t @ P = ) (b) Same as (a) above 3-7 SI Units – (a) 27 C, 60% RH, Sea Level P s = 3.57 kPa; P v = 0.6 x 3.57 = 2.14 kPa dp sat v t =(t at P ) 18.4 C ≈ (b) Same as (a) above 3-8 dp t 9C (48F) ≤ 42% φ ≤ ; W 0.0071 kgv / kga (lbv / lba) ≤ Chart 1a & 1b

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    14 10 15 20 25 30 35 40 45 50 55 55 60 60 ENTHALPY - BT U PER POUND OF DRY AIR 15 20 25 30 35 40 45 50 ENTHALPY - BT U PER POUND OF DRY AIR SATURATION TEMPERATURE - °F 35 40 45 50 55 6 0 65 70 7 5 8 0 8 5 9 0 95 100 105 110 115 12 0 DR Y BUL B T EMPERAT UR E - °F .002 .004 .006 .008 .010 .012 .014 .016 .018 .020 .022 .024 .026 .028 10% RELATIVE H UMIDITY 20% 30% 40% 50% 60% 70% 80% 90% 35 35 40 40 45 45 50 50 55 55 60 60 65 65 70 70 75 75 80 80 85 WET BULB TEMPERAT URE - °F 85 90 12.5 13.0 13.5 14.0 VOLUME - CU.FT . PER LB. DRY AIR 14.5 15.0 HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIR dp Room Problem 3-8 W=0.0071 72 (22) 48 (9) 42 % R R ASHRAE PSYCHROMETRIC CHART NO.1 NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992 AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC. SEA LEVEL 0 1.0 1.0  2.0 4.0 8 .0 -8.0 -4.0 -2.0 -1.0 -0 .5 - 0.4 -0.3 -0 .2 -0 .1 0 .1 0.2 0.3 0.4 0 .5 0.6 0 .8 -2 00 0 -1 00 0 0 5 00 1 000 1 50 0 2000 300 0 5000 -   SENSIB LE HEAT Qs TOTAL HEAT Qt ENTHALPY HU MIDITY RATIO ' h ' W 3-9 (a,b,d) Using the Properties option of PSYCH: Relative Humidity = 0.59 or 59% Enthalpy = 30.4 Btu/lbma Humidity Ratio = 0.0114 lbu/lba (c) Again using the Properties option At W=0.0114 lbv/lba; RH = 1.00 or 100% The dew point = t db or t wb = 59.9 F

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