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

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Heating, Ventilating and Air Conditioning Analysis and Design , 6th Edition Solution Manual - Page 1 preview image

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Chapter 11-1(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/hrx1.488 = 16.52Nsmμ(d) 1050BtuIbmx419.48x10−JBtux2.20462 Ibmkg= 2.44MJkg(e) 12,000BtuIbmx13.412= 3.52 kW(f) 14.72Ibfinx 6894.76 = 101 kPa1-2(a) 120 kPa x2lbf / in6.89476kPa= 17.4 lbf/in2(b) 100WmK−x 0.5778 = 57.8 Btu/hr-ft-F(c) 0.82WmK−x 0.1761 = 0.14 Btu/hr-ft2-F(d) 10-6N-s/m2x11.488= 6.7 x 10-7lbmftsec−(e) 1200 kW x 3412 = 4.1 x 10-6Btu/hr

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2(f) 1000kJkgx1 Btu1.055 kJx1 kg2.2046 lbm= 430Btulbm1-3Hp = 50 (ft) x 0.3048 (mft) = 15.2 m∆P =15.2 m1000 Pa/kPax9.8071(Nkg) x 1000 (kg/m3) = 149 kPa1-4P =∆412(ft) x 0.3048 (mft) x9.8071(Nkg) x 1000 (3kgm)∆P = 996 Pa1.0 kPa≈1-5TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE+ METER CHARGE()()()()96,000kw - hrs0.045 $ / kwhr +624 kw 1150 $ / kw−−+ $68 = $4,320 + $7,176 + $68 = $11,5641-67 AM to 6 PM11 hrs/day, 5 days/wkhrsdays(11)(22)242 hrs / monthdaymonths=

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3()()()624 kwratio =1.5796,000 kwhr242 hr=⎛⎞−⎜⎟⎝⎠1-7This is a trial and error solution since eq. 1-1 cannot be solvedexplicitly for i.Answer converges at just over 4.2% using eq. 1-11-8Determine present worth of savings using eq. 1-1()()()12120.012$10001-1+12P =0.01212P$134,000−⎡⎤⎛⎞⎢⎥⎜⎟⎝⎠⎢⎥⎣⎦⎛⎞⎜⎟⎝⎠=1-9(a)QVA== 2 x 3.08 x 10-3= 6.16 x 10-3m3/sm6.16 x 10Qρ==-3x 998 = 6.15 kg/s(b) A=4π(0.3)2= 7.07 x 10-2m2Q7.07x10=-2x 4 = 0.283 m3/ s;ρ=1.255 kq/m3m= 1.225 x 0.283 = 0.347 kg/s1-10V = 3x10x20 = 600m3

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4= 600 xiQ14x13600= 4.17 x 10-2m3/s1-11pp3q = mcTc = 4.183 kJ/(kg-K)= 983.2 kg/mρ∆1-11 (cont’d)( )()()()3c3mkgkJq =1983.24.183520,564skgKmq = 20,564 kw=−kJspermitted1-12=watqairq−11,200(1)(10) ==25000x60x14.7x144x0.24(t50)(53.35x510)−11,200 = 5601.5 (t2-50); t2= (11,200/5601.5) + 50 = 70 F1-13Diagram as in 1-12 above.qwat= -qair1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000)6279(90-t2) = 29,400

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5t2= 90 -29,4006279= 85.3 C1-14qhA(t=s-)t∞A=π(1/12) x 10 = 2.618 ft2st= tsur212 F≈= 10x2.618x(212-50) = 4241 Btu/hrq1-15A=πx 0.25x4 = 3.14 16 m2hA(tq=s-)t∞h=sqA(t -t)∞=12503.1416(10010)−; h = 4.42 W/(m2– C)1-16(tpqmc=2-t1) ;mQxρ=ρ=P/RT = 14.7x144/53.35(76+460)ρ= 0.074 lbm/ft3m= 5000x0.074x60 = 22,208 lbm/hr= 0.24 Btu/lbm-Fpcq= 22,208x0.24(58-76) = -95,939 Btu/hrNegative sign indicates cooling1-17(t1 pm c3-t1) +ofit basis fo

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6(t2p2m c3-t2) = 0=p1cp2ct3=1 12 212(m tm t )(mm )++12mQ1ρ== 1000x14.7x14453.35(46050)+= 73.5 lbm/min1-17 (cont’d)22mQ2ρ== 600x14.7x14453.35(46050)+= 46.7 lbm/min3(73.5x80)(46.7 x 50)t68.3 F(73.546.7)+==+

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7permittedRequests for permission or further information should be addressed to the Permission Department, JohnWiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.Chapter 22-1 through 2-20Solutions are not furnished since many acceptable responses existfor each problem. It is not expected that the beginning student can handlethese questions easily. The objective is to make the student think aboutthe complete design problem and the various functions of the system.These problems are also intended for use in class discussions to enlargethe text material.

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permittedRequests for permission or further information should be addressed to the Permission Department, JohnWiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.Chapter 33-1(a) Pv==srPφ0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psiaPa= 101 – 1.43 = 99.57 kPa or 14.696-0.196 = 14.5 psia(b)=vvPρRvT or===3vvvvP1430ρ;ρ0.0104 kg/mR T462.5(297)or=0.196(144)0.0006285.78(535)lbv/ft3(c) W =0.6219 (1.43)(99.57)= 0.00893 kgv/kgaor0.6219(0.196)0.00854 lbv/lba14.5=3-2(a) English Units – t = 80F; P = 14.696 psia;Pv= 0.507 psia Table A-1aW = 0.6219avPP=0.6219 (0.507)(14.6960.507)−= 0.0222 lbv/lbai = 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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8v =aaR T53.35(46080)P(14.6960.507)144+=−= 13.61 ft3/lbm(b) English Units – 32F, 14.696 psiaPv= 0.089 psia (Table A-1)3-2 (cont’d)W =0.6219(0.089)lbmv0.00379(14.6960.089)lbma=−i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbmav =53.35(492)(14.6960.089)144−= 12.48 ft3/lbma3-2(a) SI Units – 27C; 101.325 kPaPv= 3.60 kPa, Table A-1bW = 0.6219vaP0.6219(3.6)kgv0.0229P(101.3253.6)kga==−i = 1.0t + W(2501.3 + 1.86t) kJ/kgai = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kgav =3aaR T0.287(300)==0.88 m /kgaP(101.325 - 3.6)(b) SI Units 0.0C; 101.325 kPaPv= 0.61 kPa, Table A-1bW =0.6219(0.61)=0.00377 kgv/kga(101.325 - 0.61)

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9i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kgav =30.287(273)0.778 m /kga(101.325 - 0.61)=3-3(a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hgt = 80 F; Pv= 0.507 psia (Table A-1a)W = 0.6219vaP0.6219(0.507)=P(12.24 - 0.507)= 0.0269 lbv/lbai = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbmav =aaR T53.35(540)=P(12.24 - 0.507) 144= 17.05 ft3/ lbma(b) English Units – t = 32 F, Pv= 0.089 psia ( Table A-1a)W =0.6219(0.089)(12.240.089)−= 0.00456 lbmv/lbmai = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbmav =53.35(492)(12.240.089)144−= 15.00 ft3/lbma3-3(a) SI Units -27 C, 1500 m elevationP = 99.436 + 1500(-0.01) = 84.436 kPaPv= 3.60 kPa, Table A-1b

Heating, Ventilating and Air Conditioning Analysis and Design , 6th Edition Solution Manual - Page 12 preview image

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10W =0.6219x3.600.0277 kgv/kga(84.4363.60)=−i = 27 + 0.0277 (2501.3 + 1.86 x 27) = 97.68 kJ/kga3-3(cont’d)v =30.287x3001.065 m/ kga(84.436 - 3.60)=(b)SI Units – 0.0C; 1500m or 84.436 kPaPv= 0.61 kPa; Table A-1bW =0.6219 x 0.610.00453 kgv / kga(84.436 - 0.61)=i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kgav =0.287 x 273(84.436 - 0.61)= 0.935 m3/ kga3-4 (a)English Units – 70F, Pv= 0.363 psiaPv=φPg= 0.75(0.363) = 0.272 psiaW =0.6219 (0.272)0.0117 lbmv / lbma(14.696 - 0.272)=i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)]29.58 Btu / lbma=(b)Pv= 0.75 (0.363) = 0.272 psia; P = 12.24 psia

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11W =0.6219 (0.272)(12.24 - 0.272)= 0.0141 lbmv / lbmai = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)]32.20 Btu/ lbma=3-4SI Units –(a)20C, 75% RH, Sea Level3-4 (cont’d)Ps= 2.34 kPa; Pv= 0.75 x 2.34 = 1.755 kPa0.6219 x 1.755W ==(101.325 - 1.755)0.0110 kgv / kgai = 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, 1525mP = 99.436 – 0.01 x 1525 = 84.186 kPaPs= 2.34 KPa; Pv= 0.75 x 2.34 = 1.755 kPaW =0.6219 x 1.755(84.186 - 1.755)= 0.0132 kgv / kgai = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga3-5 English Units –t = 72 Fdb;psia14.696P%;50==φsvsvPPorPPφφ==; Pv= 0.5(0.3918) =0.196 psia

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12Air dewpoint = saturated temp. at 0.196 psia = 52.6 FMoisture will condense because the glass temp.40 F is below the dew point temp.3-5SI Units –t = 22C ; 50% ; P = 100 kPaPv=φPs ; Pv= 0.5(2.34) = 1.17 kPa3-5 (cont’d)Air dewpoint = sat.temp. at 1.17 kPa = 9.17 CGlass temp. of 4 C is below the dewpoint of 9.17 C, therefore,moisture will ccondense on the glass3-6English Units -(a) At 55F, 80% RH, va = 13.12 ft3/ lba andρa= 0.0752 lbma / ft3= 22,860 lbma / hram5000 (0.0762)381 lbma / min==(b) Using PSYCHρa= 0.0610 lbma / ft3or va= 16.4 ft3/ lba= 5000 (0.061) = 305 lbma / minam18,300 lbma / hr=3-6SI Units –(a) t = 13 C and relative humidity 80%then va0.820 m≈3/ kga;am2.36 / 0.822.88 kga / s==(b) Assuming same conditions;3av0.985 m/ kga=am2.36 / 0.9852.40 kga / s==

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133-7English Units – t = 80F, 60% RH(a)vsPP0.6 (0.507)0.304 psiaφ==== 64.5 Fdpsatvt(t@ P=)(b) Same as (a) above3-7SI Units –(a) 27 C, 60% RH, Sea LevelPs= 3.57 kPa; Pv= 0.6 x 3.57 = 2.14 kPadpsatvt=(tat P )18.4 C≈(b) Same as (a) above3-8dpt9C (48F)≤42%φ≤;W0.0071 kgv / kga (lbv / lba)≤Chart 1a & 1b

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1410152025303540455055556060ENTHALPY - BT U PER POUND OF DRY AIR1520253035404550ENTHALPY - BT U PER POUND OF DRY AIRSATURATION TEMPERATURE - °F35404550556 065707 58 08 59 09510010511011512 0DR Y BUL B T EMPERAT UR E - °F.002.004.006.008.010.012.014.016.018.020.022.024.026.02810% RELATIVE H UMIDITY20%30%40%50%60%70%80%90%353540404545505055556060656570707575808085 WET BULB TEMPERAT URE - °F859012.513.013.514.0 VOLUME - CU.FT . PER LB. DRY AIR14.515.0HUMIDITY RATIO - POUN DS MOISTURE PER PO UND DRY AIRdpRoomProblem 3-8W=0.007172 (22)48 (9)42 %RRASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATUREBAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.SEA LEVEL01.01.02.04.08 .0-8.0-4.0-2.0-1.0-0 .5- 0.4-0.3-0 .2-0 .10 .10.20.30.40 .50.60 .8-2 00 0-1 00 005 001 0001 50 02000300 05000-SENSIB LE HEATQsTOTAL HEATQtENTHALPYHU MIDITY RATIO'h'W3-9(a,b,d) Using the Properties option of PSYCH:Relative Humidity = 0.59 or 59%Enthalpy = 30.4 Btu/lbmaHumidity Ratio = 0.0114 lbu/lba(c) Again using the Properties optionAt W=0.0114 lbv/lba; RH = 1.00 or 100%The dew point = tdbor twb= 59.9 F

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