Solucionario De Refrigeracion Y Aire Acondicionado De Stoecker

solucionario de refrigeracion y aire acondicionado de stoecker

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-,Humedad,relativa,exterior,60,%,-,Temperatura,interior,de,la,cmara,frigorfica,1,Cb),,,El,,,flujo,,,msico,,,del,,,agua,,,de,,,reemplazo,,,requerido,,,,en,,,kg/minJones,WWby,SBWTurbomaquinas,Hidraulicas,-,Claudio,Mataix,3ra,Edicionby,Enrique,MelgarMore,From,Michael,OCEtiquetado,Frontalby,Michael,OCCacaoby,Michael,OCBpmby,Michael,OC,Footer,MenuAboutAbout,ScribdPressOur,blogJoin,our,team!Contact,UsJoin,todayInvite,FriendsGiftsLegalTermsPrivacyCopyrightSupportHelp,/,FAQAccessibilityPurchase,helpAdChoicesPublishersSocial,MediaCopyright,,2017,Scribd,IncCiclo,,de,,la,,figura,,del,,problemaSolo,,analizar,,el,,calor,,removido,,en,,aire,,de,,enfriamiento,,,en,,las,,condiciones,,de,,la,,cmara,,de,,almacenamientoProblemas,,,Resueltos,,,-,,,Tecnologa,,,Frigorfica,,,t,,,w,ent,,,,,,t,,,w,sal,,,15C,,,,,,10.43Ct,,,evap,,,=,,,t,,,w,ent,,,,,,UA,,,lim,,,pio,,,=,,,15C,,,,,,800,,,W,,,/,,,K,,,,,,0C,,,0.5267,,,kg,,,/,,,s,,,4180,,,J,,,/,,,kg,,,1e,,,&,,,mw,,,cp,,,1eCon,,,esta,,,temperatura,,,de,,,evaporacin,,,ya,,,podemos,,,dibujar,,,el,,,ciclo,,,estndar,,,decompresin,,,sobre,,,un,,,diagrama,,,P-h,,,del,,,amoniaco,,,(R-717):,,,P,,,(kPa),,,3,,,40C,,,2s,,,0C,,,4,,,1,,,h,,,(kJ/kg)Las,,,entalpas,,,de,,,los,,,puntos,,,son:h1,,,=,,,1460.66,,,kJ,,,/,,,kg,,,h2s,,,=,,,1646,,,kJ,,,/,,,kg,,,h3,,,=,,,h4,,,=,,,386.43,,,kJ,,,/,,,kgRealizando,,,un,,,balance,,,de,,,energa,,,en,,,el,,,lado,,,del,,,refrigerante,,,del,,,evaporador:,,,qf,,,10.061,,,kWqf,,,=,,,mR,,,(h1,,,,,,h4,,,);,,,mR,,,=,,,&,,,&,,,=,,,=,,,0.0094,,,kg,,,/,,,s,,,h1,,,,,,h4,,,1460,,,.66,,,,,,386,,,.43,,,kJ,,,/,,,kgDe,,,la,,,definicin,,,de,,,rendimiento,,,de,,,la,,,compresin,,,podemos,,,obtener,,,el,,,trabajo,,,realde,,,compresin:,,,P,,,P,,,m,,,(h,,,,,,h1,,,),,,0.0094,,,kg,,,/,,,s(1646,,,,,,1460.66,,,kJ,,,/,,,kg),,,&c,,,=,,,c,s,,,;,,,Pc,,,=,,,c,s,,,=,,,R,,,2s,,,=,,,=,,,2.480,,,kW,,,Pc,,,c,,,c,,,0.7El,,,coeficiente,,,de,,,eficiencia,,,energtica,,,valdr,,,para,,,el,,,caso,,,limpio:,,,qf,,,10.061,,,kWCOPlim,,,pio,,,=,,,=,,,=,,,4.057,,,Pc,lim,,,pio,,,2.480,,,kWPara,,,el,,,caso,,,del,,,intercambiador,,,sucio,,,,el,,,primer,,,paso,,,es,,,calcular,,,el,,,valor,,,del,,,nuevo,,,1,,,1UA:,,,UA,,,sucio,,,=,,,=,,,=,,,740.741,,,W,,,/,,,K,,,1,,,R,,,1,,,0.001,,,mK,,,/,,,W,,,+,,,ensu,,,+,,,UA,,,lim,,,pio,,,A,,,ext,,,800,,,W,,,/,,,K,,,10,,,mEn,,,este,,,caso,,,la,,,temperatura,,,de,,,evaporacin,,,cambiar,,,puesto,,,que,,,la,,,vlvula,,,deexpansin,,,mantiene,,,la,,,potencia,,,frigorfica:,,,t,,,w,ent,,,,,,t,,,w,sal,,,15C,,,,,,10.43Ct,,,evap,,,=,,,t,,,w,ent,,,,,,UA,,,sucio,,,=,,,15C,,,,,,740.741,,,W,,,/,,,K,,,,,,1C,,,&,,,mw,,,cp,,,0.5267,,,kg,,,/,,,s,,,4180,,,J,,,/,,,kg,,,1e,,,1eLas,,,entalpas,,,de,,,los,,,puntos,,,son:,,,40,,,4123Problema,,7

&,,air,,m,,10.391,,kg,,/,,sAhora,,podemos,,colocar,,el,,punto,,de,,salida,,del,,aire,,sobre,,el,,diagrama,,psicromtricoen,,el,,punto,,de,,interseccin,,entre,,la,,lnea,,de,,entalpa,,igual,,a,,la,,anterior,,y,,detemperatura,,seca,,igual,,a,,5C.Si,,miramos,,en,,el,,eje,,de,,humedades,,absolutas,,obtenemos:wa,ent,,=,,0.006,,kg,,H2O,,/,,kg,,a.sProblemas,,,Resueltos,,,-,,,Tecnologa,,,FrigorficaEl,,,factor,,,de,,,espacio,,,muerto,,,o,,,factor,,,de,,,huelgo,,,C=0.048,,,,y,,,los,,,volmenesespecficos,,,en,,,la,,,succin,,,y,,,la,,,descarga:v,,,suc,,,=,,,v1,,,=,,,43.2,,,l,,,/,,,kg,,,v,,,des,,,=,,,v,,,2,,,=,,,14.13,,,l,,,/,,,kg,,,,,,43.2,,,l,,,/,,,kg,,,vol,t,,,=,,,1,,,,,,0.048,,,,,,14.13,,,l,,,/,,,kg,,,,,,1,,,=,,,0.9012,,,,,,,,,Desplazamiento,,,volumtrico,,,del,,,compresor:,,,2&,,,DcVt,,,=,,,Nc,,,,,,Lc,,,4Donde:Nmero,,,de,,,cilindros:,,,Nc,,,=,,,6Velocidad,,,de,,,giro:,,,,,,=,,,1740,,,r.p.mUnformatted,text,preview:,Refrigeracin,y,Aire,Acondicionado,2014,UNIVERSIDAD,TCNICA,DE,ORURO,FACULTAD,NACIONAL,DE,INGENIERIA,INGENIERIA,MECNICA-ELECTROMECNICA,0,Refrigeracin,y,Aire,Acondicionado,2014,Contenido,11st,,ed2

36Problema,12,m.],,N,,=,,1000,,[personas],,,,,,[,,[,,],,[,,],,[,,],,[,,],,,,,,SOLUCIN:,,La,,temperatura,,de,,mezcla,,es:,,,,,,,,,,,,,,,,,,[,,],,Clculo,,de,,la,,humedad,,especifica,,de,,la,,mezcla:,,47,,[,,],,Refrigeracin,,y,,Aire,,Acondicionado,,2014,,Siendo:,,[,,Si,,[,,],,[,,[,,[,,[,,Siendo,,la,,Pz,,para,,ciudad,,CbbaAre,,you,,sure,,you,,want,,to,,continue?CANCELOKProblemas,Resueltos,-,Tecnologa,FrigorficaProblema,4Ciclos,mltiples,de,compresin,mecnicaEn,un,sistema,de,amoniaco,con,dos,evaporadores,y,un,compresor,el,evaporador,de,bajatemperatura,suministra,180,kW,de,refrigeracin,con,una,temperatura,de,evaporacin,de-30C,y,el,otro,evaporador,suministra,200,kW,a,4CEl,dimetro,de,la,tubera,exterior,es,5,cm,,Despreciando,el,espesor,del,tubo39Problema,13,Problemas,,Resueltos,,-,,Tecnologa,,Frigorfica,,p,,cond,,1533.5,,kPaRelacin,,de,,compresin:,,rc,,=,,=,,=,,9.38,,p,,evap,,163.5,,kPaEntalpas:,,h1,,=,,393.147,,kJ,,/,,kg,,h2,,=,,451.021,,kJ,,/,,kg,,h3,,=,,h4,,=,,249.674,,kJ,,/,,kgCalculemos,,el,,caudal,,de,,refrigerante,,a,,partir,,del,,balance,,en,,el,,evaporador:,,qf,,116.28,,kWqf,,=,,mR,,(h1,,,,h4,,);,,mR,,=,,&,,&,,=,,=,,0.8105,,kg,,/,,s,,(h1,,,,h4,,),,393.147,,,,249.674,,kJ,,/,,kgTrabajo,,de,,compresin:Pc,,=,,mR,,(h2,,,,h1,,),,=,,0.8105,,kg,,/,,s(451.021,,,,393.147,,kJ,,/,,kg),,=,,46.907,,kW,,&Calor,,de,,condensacin:qc,,=,,mR,,(h2,,,,h3,,),,=,,0.8105,,kg,,/,,s(451.021,,,,249.674,,kJ,,/,,kg),,=,,163.192,,kW,,&,,qf,,116.28,,kWCoeficiente,,de,,eficiencia,,energtica:,,COP,,=,,=,,=,,2.479,,Pc,,46.907,,kWb.Refrigerantes,,,Codigo,,de,,Colores,,Gildardo,,Yaez,,Manual,,de,,mantenimiento,,del,,aire,,acondicionado,,guest4709eb,,English,,Espaol,,Portugus,,Franais,,Deutsch,,About,,Dev,,&,,API,,Blog,,Terms,,Privacy,,Copyright,,Support,,LinkedIn,,Corporation,,,,2017,,Problemas,Resueltos,-,Tecnologa,FrigorficaPresiones:,p,cond,=,1500,kPa,p,evap,=,163.5,kPa,p,cond,1500,kPaRelacin,de,compresin:,rc,=,=,=,9.1743,p,evap,163.5,kPaTemperaturas:,t,cond,=,39.1C,t,evap,=,30CEntalpas:,h1,=,393.15,kJ,/,kg,h2,=,450.38,kJ,/,kg,h3,=,h4,=,248.48,kJ,/,kgCalculemos,el,caudal,de,refrigerante,a,partir,del,balance,en,el,evaporador:,qf,180,kWqf,=,mR,(h1,,h4,);,mR,=,&,&,=,=,1.2442,kg,/,s,(h1,,h4,),393.15,,248.48,kJ,/,kgTrabajo,de,compresin:Pc,=,mR,(h2,,h1,),=,1.2442,kg,/,s(450.38,,393.15,kJ,/,kg),=,71.206,kW,&,qf,180,kWCoeficiente,de,eficiencia,energtica:,COP,=,=,=,2.528,Pc,71.206,kWb

Lacapacidad,,frigorfica,,del,,evaporador,,es,,180,,kW,,a,,una,,temperatura,,de,,-30CProblemas,Resueltos,-,Tecnologa,FrigorficaEl,trabajo,de,compresin,puede,obtenerse,como:,Pc,=,mR,(h2,,h1,),&Despejando,de,aqu,la,entalpa,del,punto,2,tendremos:,Pc,10.0,kWh2,=,h1,+,=,400.049,kJ,/,kg,+,=,457.586,kJ,/,kg,&,mR,0.1738,kg,/,sEl,ciclo,sobre,el,cual,queremos,instalar,nuestro,compresor,es,el,siguiente:,3,2s,4,1,h,(kJ/kg)Los,valores,de,las,entalpas,de,los,puntos,son:h1,=,391.321,kJ,/,kg,h2s,=,430.328,kJ,/,kg,h3,=,h4,=,271.418,kJ,/,kgAl,mantenerse,la,presin,de,succin,y,presin,de,descarga,entre,las,cuales,trabajael,compresor,,tenemos,que,la,relacin,de,presiones,es,la,misma,que,en,el,casoanterior,y,puede,considerarse,una,buena,hiptesis,suponer,que,el,rendimientovolumtrico,y,de,la,compresin,del,compresor,se,mantienenCalcular,la,potencia,consumida,por,cada,uno,de,los,compresores,y,el,COP,de,la,instalacinDeterminar:,,,a),,,El,,,flujo,,,volumtrico,,,del,,,aire,,,hacia,,,la,,,torre,,,de,,,enfriamiento,,,,en,,,m,,,3/sEn,,,la,,,torre,,,entra,,,aire,,,atmosfrico,,,a,,,nivel,,,del,,,mar,,,,en,,,las,,,siguientes,,,condiciones,,,temperatura,,,de,,,bulbo,,,seco,,,de,,,22,,,C,,,y,,,temperatura,,,de,,,bulbo,,,hmedo,,,de,,,16,,,C,,,,y,,,sale,,,por,,,la,,,parte,,,superior,,,a,,,34,,,C,,,con,,,humedad,,,relativa,,,de,,,90,,,%&Si,suponemos,que,el,caudal,volumtrico,de,aire,ha,sido,medido,a,la,entrada,alevaporador,,podemos,decir,que,su,densidad,a,10C,es,aproximadamente,1.247,&,&kg/m,,y,por,lo,tanto:,mair,=,Vair,,air,=,8.333,m,/,s,1.247,kg,/,m,=,10.391,kg,/,s,.Podemos,discutir,en,este,punto,si,este,caudal,es,de,aire,seco,o,aire,hmedo,,perola,diferencia,entre,ambos,ser,tan,pequea,que,puede,considerarse,que,ambosvalen,lo,mismo,y,son,iguales,al,valor,anterior.Si,colocamos,sobre,un,diagrama,psicromtrico,del,aire,a,presin,atmosfrica,elpunto,de,entrada,podremos,leer,en,el,eje,de,entalpas,cual,es,la,entalpa,del,aire,ala,entrada:,ha,ent,=,26,kJ,/,kg,a.s.Por,tanto,podemos,despejar,del,balance,de,energa,anterior,la,entalpa,a,la,salidadel,evaporador:,q,100,kWha,sal,=,ha,ent,,f,=,26,kJ,/,kg,a.s[,,[,,Si,,T,,=,,18,3,,[C],,[,,[,,[,,[,,Sustituyendo,,en,,(1):,,Si,,Tm=24,825,,[C],,[,,La,,presin,,del,,vapor,,de,,agua,,en,,la,,mezcla,,se,,calcula,,por:,,[,,Pr,,tanto,,la,,humedad,,relativa,,es:,,48,,[,,Refrigeracin,,y,,Aire,,Acondicionado,,2014,,La,,densidad,,del,,aire,,se,,calcula,,por:,,[,,Caudal,,volumtrico,,para,,1000,,personas:,,,,,,[,,,,[,,[,,a),,El,,calor,,sensible,,se,,calcula,,por:,,,,,,El,,calor,,latente,,se,,calcula,,por:,,,,,,La,,humedad,,especfica,,para,,el,,estado,,de,,confort:,,Donde:,,[,,[,,[,,Las,,entalpias,,se,,calculan,,por,,la,,siguiente,,relacin:,,[,,[,,Para,,el,,punto,,0,,con,,Tm,,=,,24,825,,[C],,49,,Refrigeracin,,y,,Aire,,Acondicionado,,2014,,[,,Para,,el,,punto,,Confort,,con,,TConfort,,=,,23,,[C],,[,,La,,entalpia,,de,,evaporizacin,,se,,calcula,,por:,,[,,Sustituyendo,,en,,(5),,y,,(6),,se,,tiene:,,,,,,[,,[,,,,,,30,,%,,60,,%,,90,,%,,100,,%,,b),,Se,,trata,,de,,un,,proceso,,de,,enfriamiento,,con,,des,,humidificacinb),,,,,,[,,,,,,[,,,,[,,[,,[,,,,,,],,SolLa,,,temperatura,,,de,,,condensacin,,,es,,,40,,,C,,,,y,,,a,,,la,,,salida,,,del,,,mismo,,,condensador,,,tiene,,,5C,,,de,,,subenfriamiento,,,,con,,,lo,,,que,,,es,,,lquido,,,subenfriado

Problemas,,Resueltos,,-,,Tecnologa,,FrigorficaProblema,,14Ciclo,,simple,,de,,compresin,,mecnica,,(evaporadores,,y,,condensadores)Se,,dispone,,de,,una,,mquina,,para,,enfriamiento,,de,,agua,,condensada,,por,,aire,,que,,realiza,,unciclo,,simple,,de,,compresin,,mecnica,,,sin,,recalentamiento,,del,,vapor,,ni,,subenfriamiento,,dellquido,,,utilizando,,R-22La,,eficiencia,,del,,enfriador,,intermedio,,se,,planteacomo:,,t,,,,t7,,=,,0.8,,=,,5,,;,,t,,7,,=,,t,,5,,,,(t,,5,,,,t,,6,,),,=,,40C,,,,0.8(40,,,,0.19C,,),,=,,8.152C,,t5,,,,t,,6Con,,esta,,temperatura,,del,,punto,,7,,y,,la,,presin,,de,,condensacin,,obtenemos,,laentalpa,,de,,este,,punto:h7,,=,,209.66,,kJ,,/,,kg,,=,,h8Del,,balance,,del,,evaporador,,podemos,,ahora,,despejar,,el,,caudal,,de,,refrigerante,,sobreel,,evaporador:,,&,,qf,,116.28,,kWmRB,,=,,=,,=,,0.6337,,kg,,/,,s,,(h1,,,,h8,,),,393.147,,,,209.66,,kJ,,/,,kgY,,volviendo,,al,,balance,,sobre,,el,,enfriador,,intermedio,,obtenemos,,el,,caudal,,derefrigerante,,sobre,,el,,condensador:,,&,,&mRA,,=,,mRB,,8,,(h,,,,h2,,),,=,,0.8561,,kg,,/,,s,,(h6,,,,h3,,)Trabajo,,de,,compresin:,,Pc,,=,,mRA,,(h4,,,,h3,,),,+,,mRB,,(h2,,,,h1,,),,=,,40.963,,kW,,&,,&Calor,,de,,condensacin:,,Q,,c,,=,,mRA,,(h4,,,,h5,,),,=,,157.231,,kW,,&,,&,,qf,,116.28,,kWCoeficiente,,de,,eficiencia,,energtica:,,COP,,=,,=,,=,,2.839,,Pc,,40.963,,kW,,8,,9Determnese,,,resultados,,,para,,,el,,,ciclo,,,real,,,de,,,refrigeracin-,,WAsumir,,,las,,,propiedades,,,del,,,agua,,,a,,,20,,,C,,,*,,,+,,,;,,,*,,,+;,,,*,,,+;,,,Las,,,propiedades,,,del,,,refrigerante,,,R-134a,,,para,,,40,,,C,,,es:,,,*,,,+;,,,*,,,*,,,+;,,,*,,,+;,,,*,,,+,,,R134a,,,40C,,,40C,,,AGUA,,,20C,,,SOLUCIN:,,,Balance,,,de,,,energa,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,De,,,tablas,,,de,,,R134a:,,,25,,,,,,(,,,),,,+;,,,Refrigeracin,,,y,,,Aire,,,Acondicionado,,,2014,,,,,,Sustituyendo,,,en,,,ecuacin,,,2,,,utilizando,,,la,,,igualdad,,,de,,,la,,,ecuacin,,,1,,,tenemos:,,,A,,,contracorriente:,,,(,,,),,,(,,,),,,Coeficiente,,,global,,,de,,,transferencia,,,de,,,calor:,,,Para,,,la,,,convectiva,,,del,,,refrigerante:,,,,,,[,,,,,,[,,,Calculo,,,de,,,la,,,convectiva,,,del,,,agua,,,(H2O),,,rea:,,,(,,,),,,Para,,,la,,,velocidad,,,del,,,flujo:,,,26,,,Refrigeracin,,,y,,,Aire,,,Acondicionado,,,2014,,,,,,Dimetro,,,equivalente:,,,Numero,,,de,,,Reynolds:,,,Para,,,flujo,,,turbulento,,,(Re,,,>,,,4000):,,,Numero,,,de,,,Nusselt:,,,De,,,esta,,,ecuacin,,,despejamos:,,,h,,,(convectiva,,,del,,,agua):,,,Todo,,,en,,,ecuacin,,,5,,,despreciando,,,el,,,espesor,,,del,,,tubo:,,,Todo,,,en,,,Ecuacin,,,4,,,despejando,,,el,,,rea,,,de,,,transferencia,,,de,,,calor,,,y,,,usando,,,la,,,ecuacin,,,1:,,,27,,,Refrigeracin,,,y,,,Aire,,,Acondicionado,,,2014,,,[,,,EJERCICIO,,,N,,,8,,,Realizar,,,el,,,clculo,,,de,,,carga,,,trmica,,,debido,,,a,,,la,,,apertura,,,de,,,puertas,,,e,,,infiltraciones,,,de,,,aire,,,en,,,24,,,horas,,,,de,,,una,,,cmara,,,de,,,refrigeracin,,,,para,,,enfriar,,,carne,,,de,,,ternero,,,,cuyos,,,datos,,,de,,,diseo,,,son,,,los,,,siguientes:,,,-,,,Carga,,,de,,,enfriamiento,,,o,,,potencia,,,frigorfica,,,12500,,,kcal/hProblemas,,Resueltos,,-,,Tecnologa,,Frigorfica,,qf,,116.28,,kWqf,,=,,mRB,,(h1,,,,h4,,),,&,,&,,mRB,,=,,=,,=,,0.6121,,kg,,/,,s,,(h1,,,,h4,,),,393.147,,,,203.18,,kJ,,/,,kgBalance,,de,,energa,,sobre,,el,,intercambiador,,intermedio:,,&,,&,,&,,&mRAh8,,+,,mRBh2,,=,,mRBh3,,+,,mRAh5despejando:,,&,,&,,(h,,,,h2,,),,=,,0.8658,,kg,,/,,smRA,,=,,mRB,,3,,(h8,,,,h5,,)Trabajo,,de,,compresin:,,Pc,,=,,mRA,,(h6,,,,h5,,),,+,,mRB,,(h2,,,,h1,,),,=,,43.804,,kW,,&,,&Calor,,de,,condensacin:,,qc,,=,,mRA,,(h6,,,,h7,,),,=,,160.078,,kW,,&,,qf,,116.28,,kWCoeficiente,,de,,eficiencia,,energtica:,,COP,,=,,=,,=,,2.655,,Pc,,43.804,,kWResumen,,de,,resultados:Caso,,qf,,(kW),,qc,,(kW),,Pc,,(kW),,COPCompresin,,mecnica,,simple,,116.28,,163.192,,46.907,,2.479Con,,enfriador,,intermedio,,,inyeccin,,parcial,,116.28,,157.231,,40.963,,2.839Con,,enfriador,,intermedio,,,inyeccin,,total,,116.28,,156.258,,39.978,,2.909En,,cascada,,116.28,,160.078,,43.804,,2.655,,11,,12Are,,,you,,,sure,,,you,,,want,,,to,,,continue?CANCELOK 487462e4f8
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