ejercicios op2

datosquerosene 42° API Destilado 35° APIW= 60000 lb/hT1= 400 °F T3= 100T2= 225 °F T4= 200Cp= 0.61 Btu/lb°F Cp= 0.45viscosidad= 0.35 cp Viscosidad= 2.2kh= 0.18 Btu/ft h °F kc= 0.24

Primero se calcula la TcTc= T2+Fc(T1-T2) Tc= 302 °F Fluido Calientetc= T3+Fc(T4-T3) tc= 144 °F 400

225ΔTc/ΔTh= 0.625 175Fc= 0.44Ft= 0.94 R= (T1-T2)/(T4-T3) 1.75

S=(T4-T3)/(T1-T3) 0.33333333333333

Calcular Qkeroseno Qk= 6405000calculando Wdestilado W= 142333.333333333

Suponiendo Ud Ud= 15Calculando Area tubo At= 3.1408 ft2Calcular Area Transferencia Total AT= 2846.68864805573calcular numero tubos nt=AT/At nt= 906.357822228646total tubos son 506, nt= 938Dic= 33 inCalculando la nueva Area de Transferencia AT=nt*At AT=calculando la Ud Ud= 14.4939950249784

Lado corazaArea de Flujo B=Dic/5 B= 6.6 in

Afc=Dic*C*B/(144*Pt) 0.3025 ft2

Velocidad masica Gc= Wc/AfcGc= 198347.1074 lb/h ft2

Q=W*ΔTh CpW=Q/Cp ΔTc

AT= Q/Ud Δt

Ud=Q/AT Δt

s. ingCalcular NRE μ= 0.847 lb/ft h

De= 0.045833333 ftRec= 10733.06858

jh jhc 60k(Cvisc/k).33 0.13h0c=jh*kcvk*φ/De

φ h0/φ= 170.1818182

tw=Tc+(h0/φ)/(hi0/φ+h0/φ)tw 288.7702284 °Fμw= 0.9922φ= 0.978092179h0= 166.4535054hio= 17.51252207

Calcular UcUc=(hio*ho)/(hio+ho) Uc= 15.8454293

calcular RdRd=(Uc-Ud)/(Uc*Ud) Rd= 0.00588441

como Rd calculada es mayor Rd dada, se considera correcto

f= 0.0019s 0.76N+1=12L/B 29.09090909Ds=DI/12 2.75ΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φcΔP= 3.362404868

Tubo 16 BGW Arreglo triangularDE 0.75 in Pt= 0.9375 in

°F L 16 ft Claro 0.1875 in°F DI 0.62 inBtu/lb°F A/L 0.1963 ft2/ftcpBtu/ft H °F

Fluido Frio DiferenciaAlta Temp 200 200Baja Temp 100 125Diferencias 100 75

LMT=((T1-T4)-(T2-T3))/ln((T1-T4)/(T2-T3))LMT= 159.573236Δt= 149.998842

Btu/hlb/h

ft2

2946.0704 ft2

Lado TuboAft=nt*at/144n Aft= 0.98359722 ft2at= 0.302 in2

Gt=Wt/AftGt= 144706.929 lb/h ft2

sist ingμ= 5.324 lb/ft hD 0.05166667 ftRet= 1404.30591

jht 3.6 309.677419k(cvis/k).333 0.27

hi/φ= 18.8129032hio= 15.552

μw= 2.28φ 1.12606238hi= 21.1845025

f= 0.00036s 0.82

ΔP=f*G^2*L*n/(5.22*10^10 *D*s* φtΔP= 0.09686656

datosanilina bencenoW= 120000 lb/h W= 100000 lb/hT1= 275 °F T3= 100 °FT2= 200 °F T4= 200 °FCp= 0.54 Btu/lb°F Cp= 0.486 Btu/lb°Fviscosidad= 0.55 cp Viscosidad= 0.35 cpkh= 0.09822 Btu/ft h °F kc= 0.086673 Btu/ft H °F

Calculando las Tprom Fluido Caliente275 Alta Temp

Ta= 237.5 200 Baja Tempta= 150 75 Diferencias

Ft= 0.825 R= (T1-T2)/(T4-T3) 0.75S=(T4-T3)/(T1-T3) 0.57142857

Calcular Qanilina Qk= 4860000 Btu/hcalculando Q benceno Q= 4860000 Btu/h

Suponiendo Ud Ud= 50Calculando Area tubo At=A/L * L 3.1408 ft2Calcular Area Transferencia Total AT= 1355.76715 ft2calcular numero tubos nt=AT/At nt= 431.662999total tubos son460 nt= 460Dic= 27 inCalculando la nueva Area de Transferencia AT=nt*At AT= 1444.768calculando la Ud Ud= 46.9198912


Afc=Dic*C*B/(144*Pt) 0.253125 ft2



AT= Q/Ud Δt

Ud=Q/AT Δt


De= 0.07916667 ftRec= 36925.2895

jh jhc 130

h0c=jh*kcvk*φ/Deφ h0/φ= 257.116313

tw=tc+(h0/φ)/(hi0/φ+h0/φ) *(Tc-tc)tw 211.309481 °Fμw= 0.6292 lb/ft hφc= 1.04249326h0= 268.042024hio= 104.223167




f= 0.0016s 0.88N+1=12L/B 35.5555556Ds=DI/12 2.25 ftΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φc)ΔP= 5.26952048 psi

Tubo 14 BGW Arreglo cuadoDE 0.75 in Pt= 1 inL 16 ft Claro 0.25 inDI 0.584 inA/L 0.1963 ft2/ftAi/L 0.1529 ft2/ft

Fluido Frio Diferencia200 75100 100100 -25


ft2



sist ingμ= 1.331 lb/ft hDI 0.04866667 ftRet= 10250.2546Npr 7.31765425jht 36

hi/φ= 141.056955hio/φ= 109.836349

μw= 1.936 lb/ft hφt 0.94889504hi= 133.848245

f= 0.00024s 1.024

ΔP=f*G^2*L*n/(5.22*10^10 *D*s* φtΔP= 0.2445137 psi

datosAgua querosene 42° APIW= 242526.316 W= 84000 lb/hT3= 85 °F T1= 300 °FT4= 120 °F T2= 100 °FCp= 0.95 Btu/lb°F Cp= 0.48 Btu/lb°FViscosidad= 0.71 cp viscosidad= 0.83 cpkc= 0.4 Btu/ft H °F kh= 0.079 Btu/ft h °F

Primero se calcula la TcTc= T2+Fc(T1-T2) Tc= 160 °F Fluido Caliente

300 Alta Temp100 Baja Temp

ΔTc/ΔTh= 0.08333333 200 DiferenciasFc= 0.3Ft= 0.825 R= (T1-T2)/(T4-T3) 5.71428571

S=(T4-T3)/(T1-T3) 0.1627907Ta 102.5 °F

Calcular Qkeroseno Qk= 8064000 Btu/hcalculando Wagua W= 242526.316 lb/hSuponiendo Ud Ud= 80Calculando Area tubo At=A/L * L 3.1408 ft2Calcular Area Transferencia Total AT= 1840.06311 ft2calcular numero tubos nt=AT/At nt= 585.858095total tubos son 830 nt= 604Dic= 29 inCalculando la nueva Area de Transferencia AT=nt*At AT= 1897.0432calculando la Ud Ud= 77.5970987


Afc=Dic*C*B/(144*Pt) 0.29201389 ft2


s. ing


AT= Q/Ud Δt

Ud=Q/AT Δt

Calcular NRE μ= 2.0086 lb/ft hDe= 0.06083333 ftRec= 8712.12171Npr 0

jh jhc 50k(cvis/k).333 0.19h0c=jh*kcvk*φ/De

φ ho/φ= 156.164384






Tubo 16 BGW Arreglo triangularDE 0.75 in Pt= 1 inL 16 ft Claro 0.25 inDI 0.62 in at= 0.302 in2A/L 0.1963 ft2/ftAi/L 0.1623 ft/ft

Fluido Frio Diferencia120 180

85 1535 165


ft2



sist ing

μ= 1.7182 lb/ft hDI 0.05166667 ftRet= 11514.4767Npr 4.080725jht 41 309.677419k(cvis/k).333 0

hi/φ= 507.215079hio/φ= 419.297799

μw= 1.331 lb/ft hφt 1.0363952hi= 161.848017

f= 0.00028s 0.99


datosgas oil 28 api destilado 35 api Tubo 16 BGWW= 9428.57143 lb/h W= 22000 DET1= 500 °F T3= 200 °F LT2= 300 °F T4= 300 °F DICp= 0.63 Btu/lb°F Cp= 0.54 Btu/lb°F A/Lviscosidad= 0.73 cp Viscosidad= 0.86 cp Ai/Lkh= 0 Btu/ft h °F kc= 0 Btu/ft H °F

Primero se calcula la TcTc= T2+Fc(T1-T2) Tc= 385 °F Fluido Caliente Fluido Friotc= T3+Fc(T4-T3) tc= 242.5 °F 500 Alta Temp 300

300 Baja Temp 200ΔTc/ΔTh= 0.5 200 Diferencias 100Fc= 0.425Ft= 0.82 R= (T1-T2)/(T4-T3) 2

S=(T4-T3)/(T1-T3) 0.33333333

Calcular Q destilado Qk= 1188000 Btu/hcalculando Wgas oil W= 9428.57143 lb/hSuponiendo Ud Ud= 40Calculando Area tubo At=A/L * L 3.1408 ft2Calcular Area Transferencia Total AT= 251.054528 ft2calcular numero tubos nt=AT/At nt= 79.933306total tubos son 82 nt= 82Dic= 12 inCalculando la nueva Area de Transferencia AT=nt*At AT= 257.5456 ft2calculando la Ud Ud= 38.9918566

Lado corazaArea de FlujoB=Dic/5 B= 2.4 in

Afc=Dic*C*B/(144*Pt) 0.05 ft2

Velocidad maGc= Wc/AfcGc= 188571.429 lb/h ft2

s. ing


AT= Q/Ud Δt

Ud=Q/AT Δt

Calcular NRE μ= 1.7666 lb/ft hDe= 0.06083333 ftRec= 6493.50649


φ h0/φ= 120.986301





f= 0.0024s 0.78N+1=12L/B 80Ds=DI/12 1 ftΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φc)ΔP= 2.93750618 psi

Tubo 16 BGW Arreglo triangular0.75 in Pt= 1 in

16 ft Claro 0.25 in0.62 in at= 0.302 in2

0.1963 ft2/ft0.1623 ft/ft

Diferencia200100100




sist ing

μ= 2.0812 lb/ft hDI 0.05166667 ftRet= 6351.71495 309.677419

jht 25k(cvis/k).333 0.19

hi/φ= 91.9354839hio/φ= 76

μw= 1.21 lb/ft hφt 1.07888209hi= 99.1875467

f= 0.00033s 0.78





S=(T4-T3)/(T1-T3) 0.33333333

Calcular Q destilado Qk= 1210000 Btu/hcalculando Wgas oil W= 9453.125 lb/hSuponiendo Ud Ud= 50Calculando Area tubo At=A/L * L 3.1416 ft2Calcular Area Transferencia Total AT= 204.562948 ft2calcular numero tubos nt=AT/At nt= 65.1142566total tubos son 82 nt= 52Dic= 10 inCalculando la nueva Area de Transferencia AT=nt*At AT= 163.3632 ft2calculando la Ud Ud= 62.6098621

Lado corazaArea de FlujoB=Dic/5 B= 2 in

Afc=Dic*C*B/(144*Pt) 0.02777778 ft2

Velocidad maGc= Wc/AfcGc= 792000 lb/h ft2

s. ing


AT= Q/Ud Δt

Ud=Q/AT Δt



φ h0/φ= 1081.66667



calcular RdRd=(Uc-Ud)/(Uc*Ud) Rd= -0.03643747


f= 0.0018s 0.77N+1=12L/B 72Ds=DI/12 0.83333333 ftΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φc)ΔP= 24.6056557 psi

Tubo 14 BGW Arreglo triangular1 in Pt= 1.25 in


0.2618 ft2/ft0.2183 ft/ft

Diferencia200100100




sist ing

μ= 1.7666 lb/ft hDI 0.05166667 ftRet= 2804.42687 232.258065

jht 9.5k(cvis/k).333 0.17

hi/φ= 31.2580645hio/φ= 19.38

μw= 1.7666 lb/ft hφt 1hi= 31.2580645

f= 0.00035s 0.77


datosagua gasolina 56 api Tubo 14BGWW= 480533.333 W= 68000 lb/h DET3= 85 °F T1= 200 °F LT4= 100 °F T2= 100 °F DICp= 0.5 Btu/lb°F Cp= 0.53 Btu/lb°F A/LViscosidad= 0.87 cp viscosidad= 0.42 cp Ai/Lkc= 0.37 Btu/ft H °F kh= 0 Btu/ft h °F

Primero se calcula la TcTc= T2+Fc(T1-T2) Tc= 135 °F Fluido Caliente Fluido Friota= 92.5 °F 200 Alta Temp 100

100 Baja Temp 85ΔTc/ΔTh= 0.15 100 Diferencias 15Fc= 0.35Ft= 0.925 R= (T1-T2)/(T4-T3) 6.66666667

S=(T4-T3)/(T1-T3) 0.13043478

Calcular Q gasolina Qk= 3604000 Btu/hcalculando Wagua W= 480533.333 lb/hSuponiendo Ud Ud= 60Calculando Area tubo At=A/L * L 3.1416 ft2Calcular Area Transferencia Total AT= 1449.3313 ft2calcular numero tubos nt=AT/At nt= 461.335404total tubos son 82 nt= 460Dic= 33 inCalculando la nueva Area de Transferencia AT=nt*At AT= 1445.136 ft2calculando la Ud Ud= 60.1741831 #VALUE!


Afc=Dic*C*B/(144*Pt) 0.3025 ft2


s. ing


AT= Q/Ud Δt

Ud=Q/AT Δt


jh jht 82k(cvis/k).333 0.16

φ h0/φ= 159.030303






Tubo 14BGW Arreglo cuadrado1 in Pt= 1.25 in


0.2618 ft2/ft0.2183 ft/ft

Diferencia100

1585




sist ing

μ= 2.1054 lb/ft hDI 0.0695 ftRet= 18189.2888Npr 2.84513514jhc 55 172.661871

h0c=jh*kcvk*φ/Dehi/φ= 414.902191hio/φ= 346.028427

μw= 1.21 lb/ft hφt 1.08062969hi= 448.355625

f= 0.00021s 1


datosoxigeno 5 psi agua Tubo 16BGWW= 32000 lb/h W= 16984.6154 lb/h DET1= 300 °F T3= 85 °F LT2= 150 °F T4= 150 °F DICp= 0.23 Btu/lb°F Cp= 1 Btu/lb°F A/Lviscosidad= 0.024 cp Viscosidad= 0.6 cp Ai/Lkh= 0.019 Btu/ft h °F kc= 0.37 Btu/ft H °F

Calculando las Tprom Fluido Caliente Fluido Frio300 Alta Temp 150

Ta= 225 °F 150 Baja Temp 85ta= 117.5 °F 150 Diferencias 65

Ft= 0.85 R= (T1-T2)/(T4-T3) 2.30769231S=(T4-T3)/(T1-T3) 0.30232558

Calcular Qoxigeno Qk= 1104000 Btu/hcalculando W agua W= 16984.6154 Btu/h

Suponiendo Ud Ud= 30Calculando Area tubo At=A/L * L 2.3556 ft2Calcular Area Transferencia Total AT= 425.93671 ft2calcular numero tubos nt=AT/At nt= 180.818776total tubos son460 nt= 166Dic= 14.25 inCalculando la nueva Area de Transferencia AT=nt*At AT= 391.0296 ft2calculando la Ud Ud= 32.6780921


Afc=Dic*C*B/(144*Pt) 0.07050781 ft2



AT= Q/Ud Δt

Ud=Q/AT Δt


De= 0.07916667 ftRec= 618626.456Npr 0.70307368

jh jhc 500

h0c=jh*kcvk*φ/Deφ h0/φ= 106.705456






Tubo 16BGW Arreglo cuadrado0.75 in Pt= 1 in

12 ft Claro 0.25 in0.62 in

0.1963 ft2/ft 232.2580650.1623 ft2/ft

Diferencia150

6585




sist ingμ= 1.452 lb/ft hDI 0.05166667 ftRet= 3471.97829Npr 3.92432432jht 9

hi/φ= 101.661198hio/φ= 84.0399236

μw= 1.936 lb/ft hφt 0.96052479hi= 97.6481005

f= 0.00039s 1





S=(T4-T3)/(T1-T3) 0.33333333

Calcular Q destilado Qk= 1188000 Btu/hcalculando Wgas oil W= 9428.57143 lb/hSuponiendo Ud Ud= 50Calculando Area tubo At=A/L * L 3.1416 ft2Calcular Area Transferencia Total AT= 200.843622 ft2calcular numero tubos nt=AT/At nt= 63.930361total tubos son 82 nt= 66Dic= 13.25 inCalculando la nueva Area de Transferencia AT=nt*At AT= 207.3456 ft2calculando la Ud Ud= 48.4320917


Afc=Dic*C*B/(144*Pt) 0.04876736 ft2


s. ing


AT= Q/Ud Δt

Ud=Q/AT Δt



φ h0/φ= 120






Tubo 14 BGW Arreglo triangular1 in Pt= 1.25 in


0.2618 ft2/ft0.2183 ft/ft

Diferencia200100100




sist ing

μ= 2.0812 lb/ft hDI 0.07025 ftRet= 5934.86851 170.818505

jht 25k(cvis/k).333 0.19

hi/φ= 67.6156584hio/φ= 57

μw= 1.089 lb/ft hφt 1.09491406hi= 74.0333348

f= 0.00033s 0.78


datosvapor 15 lb/in yoduro de potacio W= 0 lb/h W= 100000T1= 249.7 °F T3= 80 °FT2= 249.7 °F T4= 100 °FCp= 0.5074 Btu/lb°F Cp= 0.34 Btu/lb°Fviscosidad= 0.03099 cp Viscosidad= 0.43 cpkh= 0.01607 Btu/ft h °F kc= 0.06 Btu/ft H °Fcalor latente= 945.65 btu*lbPrimero se calcula la TcTc= T2+Fc(T1-T2) Ta= 249.7 °F Fluido Calientetc= T3+Fc(T4-T3) ta= 90 °F 249.7 Alta Temp

249.7 Baja TempΔTc/ΔTh= 0 0 DiferenciasFc= 0Ft= 1 R= (T1-T2)/(T4-T3) 0

S=(T4-T3)/(T1-T3) 0.11785504

Calcular Q yoduro de potacio Qy= 680000 Btu/hcalculando Wgas oil W= 719.082113 lb/hSuponiendo Ud Ud= 0Calculando Area tubo At=A/L * L 3.1408 ft2Calcular Area Transferencia Total AT= 0 ft2calcular numero tubos nt=AT/At nt= 0total tubos son 50 nt= 50Dic= 10 inCalculando la nueva Area de Transferencia AT=nt*At AT= 157.04calculando la Ud Ud= 27.1495284

Lado coraza yoduro de potasioArea de Flujo B=Dic/5 B= 0 in

0.39201823 ft2

Velocidad masicaGc= Wc/AsGc= 255090.178 lb/h ft2

s. ing

Q=W*ΔTh CpW=Q/λ

AT= Q/Ud Δt

Ud=Q/AT Δt

As=1/144((πDI^2)/4 - (NUT*π*DE^2)/4


jh jhc 55Npr 5.89673333h0c=jh*kcvk*φ/De

φ h0/φ= 130.077742

tw=tc+(h0/φ)/(hi0/φ+h0/φ) *(Tc-tc)tw 0 °Fμw= 0 lb/ft hφc= 1h0= 130.077742hio= 1.02812608



f= 0.0018s 0.77N+1=12L/B #DIV/0!Ds=DI/12 0.83333333 ftΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φc)ΔP= #DIV/0! psi

Tubo 16 BGW Arreglo triangularDE 0.75 in Pt= 0.9375 inL 16 ft Claro 0.1875 inDI 0.62 in at= 0 in2A/L 0.1963 ft2/ftAi/L 0.302 ft/ft 2 pasos

Fluido Frio Diferencia100 149.7

80 169.720 -20


ft2

Lado Tubo vaporAft=nt*at/144n Aft= 0.09479167 ft2at= 0.546 in2


sist ing

μ= 0.0749958 lb/ft hDI 0.05166667 ftRet= 5226.14948 309.677419

jht 3Npr 2.36794455

hi/φ= 1.2437009hio/φ= 1.02812608

μw= 0 lb/ft hφt 1hi= 1.2437009

f= 0.00035s 0.77


datosvapor 15 lb/in yoduro de potacio Tubo 16 BGWW= 0 lb/h W= 100000 DET1= 249.7 °F T3= 80 °F LT2= 249.7 °F T4= 100 °F DICp= 0.53 Btu/lb°F Cp= 0.35 Btu/lb°F A/Lviscosidad= 0.99 cp Viscosidad= 0.43 cp Ai/Lkh= 0.01607 Btu/ft h °F kc= 0.07 Btu/ft H °Fcalor latente 945.65 btu*lbPrimero se calcula la TcTc= T2+Fc(T1-T2) Ta= 249.7 °F Fluido Caliente Fluido Friotc= T3+Fc(T4-T3) ta= 90 °F 249.7 Alta Temp 100

249.7 Baja Temp 80ΔTc/ΔTh= 0 0 Diferencias 20Fc= 0Ft= 1 R= (T1-T2)/(T4-T3) 0

S=(T4-T3)/(T1-T3) 0.11785504

Calcular Q yoduro de potac Qy= 700000 Btu/hcalculando Wgas oil W= 740.231587 lb/hSuponiendo Ud Ud= 0Calculando Area tubo At=A/L * L 3.1408 ft2Calcular Area Transferencia Total AT= 0 ft2calcular numero tubos nt=AT/At nt= 0total tubos son 50 nt= 50Dic= 10 inCalculando la nueva Area de Transferencia AT=nt*At AT= 157.04 ft2calculando la Ud Ud= 27.948044

Lado coraza yoduro de potasioArea de FlujoB=Dic/5 B= 0 in

0.39201823 ft2

Velocidad maGc= Wc/AsGc= 255090.178 lb/h ft2

s. ing

Q=W*ΔTh CpW=Q/λ

AT= Q/Ud Δt

Ud=Q/AT Δt

As=1/144((πDI^2)/4 - (NUT*π*DE^2)/4


jh jhc 55Npr 5.203h0c=jh*kcvk*φ/De

φ h0/φ= 145.55615

tw=tc+(h0/φ)/(hi0/φ+h0/φ) *(Tc-tc)tw 0 °Fμw= 0 lb/ft hφc= 1h0= 145.55615hio= 3.30994649



f= 0.0018s 0.77N+1=12L/B #DIV/0!Ds=DI/12 0.83333333 ftΔP=f*G^2*Ds*(N+1)/(5.22*10^10 *De*s* φc)ΔP= #DIV/0! psi

Tubo 16 BGW Arreglo triangular0.75 in Pt= 0.9375 in

16 ft Claro 0.1875 in0.62 in at= 0 in2

0.1963 ft2/ft0.302 ft/ft 2 pasos

Diferencia149.7169.7

-20


Lado Tubo vaporAft=nt*at/144n Aft= 0.09479167 ft2at= 0.546 in2


sist ing

μ= 2.3958 lb/ft hDI 0.05166667 ftRet= 168.405913 309.677419

jht 3Npr 79.0151836

hi/φ= 4.00396753hio/φ= 3.30994649

μw= 0 lb/ft hφt 1hi= 4.00396753

f= 0.00035s 0.77


metiletilcetona alcohol amilicow= 50000 lb/h T1= 275t3= 100 t2200 115t4= 200 cp 0.71cp 0.56 cp 0.71vic 0.24 visdcv 1.4

Q=w cp AT

Q= 2800000

w= 24647.88730.91428571 0.57142857

ejercicios op2

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