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"source": [
"# Chapter 10 :Stream Line Flow and Heat Convection"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 10.1 pgno:203"
]
},
{
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"collapsed": false
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{
"name": "stdout",
"output_type": "stream",
"text": [
"\t example 10.1 \t\n",
"\t approximate values are mentioned in the book \t\n",
"\t 1.for heat balance \t\n",
"\t for crude \t\n",
"\t total heat required for crude is : Btu/hr \t388000.0\n",
"\t for steam \t\n",
"\t total heat required for steam is : Btu/hr \t387655.0\n",
"\t delt1 is : F \t155.0\n",
"\t delt2 is : F \t105.0\n",
"\t LMTD is : F \t128.381317817\n",
"\t On the assumption that the fluids are mixed between passes each pass must be solved independently Since only two passes are present in this exchanger it is simply a matter of assuming the temp at the end of the first pass More than half the heat load must be transferred in the first pass therefore assume ti at the end of the first pass is 125 degres f \n",
"\n",
"\t caloric temperature of cold fluid is : F \t110.0\n",
"\t hot fluid:shell side,steam \t\n",
"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \t1500\n",
"\t cold fluid:inner tube side,crude \t\n",
"\t flow area is : ft**2 \t0.177375\n",
"\t mass velocity is : lb/(hr)*(ft**2) \t90395.480226\n",
"\t reynolds number is : \t918.009849613\n",
"\t reynolds number is : \t564.193553408\n",
"\t prandelt number is : \t72.6936774194\n",
"\t Hi is : Btu/(hr)*(ft**2)*(F) \t1.98827586207\n",
"\t phyt is : \t1.2141948844\n",
"\t Correct hi to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \t2.41415438049\n",
"\t tp is : F \t249.775040982\n",
"\t delt is : F \t139.775040982\n",
"\t total surface area is : ft**2 \t117.6\n",
"\t delt3 is : F \t5.1\n",
"\t ti is : F \t100.1\n",
"The oil now enters the second pass at given 126.9 f\n"
]
}
],
"source": [
"print\"\\t example 10.1 \\t\"\n",
"print\"\\t approximate values are mentioned in the book \\t\"\n",
"#given\n",
"T1=250.; # inlet hot fluid,F\n",
"T2=250.; # outlet hot fluid,F\n",
"t1=95.; # inlet cold fluid,F\n",
"t2=145.; # outlet cold fluid,F\n",
"W=16000.; # lb/hr\n",
"w=410.; # lb/hr\n",
"#solution\n",
"from math import log\n",
"print\"\\t 1.for heat balance \\t\"\n",
"print\"\\t for crude \\t\"\n",
"c=0.485; # Btu/(lb)*(F)\n",
"Q=((W)*(c)*(t2-t1)); # Btu/hr\n",
"print\"\\t total heat required for crude is : Btu/hr \\t\",Q\n",
"print\"\\t for steam \\t\"\n",
"l=945.5; # Btu/(lb)\n",
"Q=((w)*(l)); # Btu/hr\n",
"print\"\\t total heat required for steam is : Btu/hr \\t\",Q\n",
"delt1=T2-t1; #F\n",
"delt2=T1-t2; # F\n",
"print\"\\t delt1 is : F \\t\",delt1\n",
"print\"\\t delt2 is : F \\t\",delt2\n",
"LMTD=((delt2-delt1)/((1)*(log(delt2/delt1))));\n",
"print\"\\t LMTD is : F \\t\",LMTD\n",
"\n",
"print\"\\t On the assumption that the fluids are mixed between passes each pass must be solved independently Since only two passes are present in this exchanger it is simply a matter of assuming the temp at the end of the first pass More than half the heat load must be transferred in the first pass therefore assume ti at the end of the first pass is 125 degres f \\n\"\n",
"ti=125; # F\n",
"tc=((t1)+(ti))/2; # caloric temperature of cold fluid,F\n",
"print\"\\t caloric temperature of cold fluid is : F \\t\",tc\n",
"print\"\\t hot fluid:shell side,steam \\t\"\n",
"ho=(1500); # condensation of steam Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",ho\n",
"print\"\\t cold fluid:inner tube side,crude \\t\"\n",
"Nt=86;\n",
"n=2; # number of passes\n",
"L=12; #ft\n",
"at1=0.594; # flow area, in**2,from table 10\n",
"at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48\n",
"print\"\\t flow area is : ft**2 \\t\",at\n",
"Gt=(W/(.177)); # mass velocity,lb/(hr)*(ft**2)\n",
"print\"\\t mass velocity is : lb/(hr)*(ft**2) \\t\",Gt\n",
"mu2=2.95*2.42; # at 145F,lb/(ft)*(hr)\n",
"D=(0.87/12); # ft\n",
"Ret1=((D)*(Gt)/mu2); # reynolds number\n",
"print\"\\t reynolds number is : \\t\",Ret1\n",
"mu3=4.8*2.42; # at 110F,lb/(ft)*(hr)\n",
"D=(0.87/12); # ft\n",
"Ret2=((D)*(Gt)/mu3); # reynolds number\n",
"print\"\\t reynolds number is : \\t\",Ret2\n",
"c=0.485; # Btu/(lb)*(F),at 120F,from fig.2\n",
"k=0.0775; # Btu/(hr)*(ft**2)*(F/ft), from table 4\n",
"Pr=((c)*(mu3)/k); # prandelt number\n",
"print\"\\t prandelt number is : \\t\",Pr\n",
"Hi=((1.86)*(k/D)*((Ret2*(D/L)*Pr)**(1/3))); # using eq.6.1,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t Hi is : Btu/(hr)*(ft**2)*(F) \\t\",Hi\n",
"muw=1.2*2.42; # lb/(ft)*(hr),at 249F from fig.14\n",
"phyt=(mu3/muw)**0.14;\n",
"print\"\\t phyt is : \\t\",phyt # from fig.24\n",
"hi=(Hi)*(phyt); # from eq.6.37\n",
"print\"\\t Correct hi to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \\t\",hi\n",
"tp=(tc)+(((ho)/(hi+ho))*(T1-tc)); # from eq.5.31\n",
"print\"\\t tp is : F \\t\",tp\n",
"delt=tp-tc; #F\n",
"print\"\\t delt is : F \\t\",delt\n",
"Ai1=0.228 # internal surface per foot of length,ft\n",
"Ai=(Nt*L*Ai1/2); # ft**2\n",
"print\"\\t total surface area is : ft**2 \\t\",round(Ai,1)\n",
"delt3=((hi*Ai*delt)/(W*c)); # delt3=ti-t1, F\n",
"print\"\\t delt3 is : F \\t\",round(delt3,1)\n",
"ti=t1+delt3; # F\n",
"print\"\\t ti is : F \\t\",round(ti,1)\n",
"print\"The oil now enters the second pass at given 126.9 f\"\n",
"# end\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 10.2 pgno:207"
]
},
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{
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"text": [
"\t example 10.2 \t\n",
"\t approximate values are mentioned in the book \t\n",
"\t 1.for heat balance \t\n",
"\t for kerosene \t\n",
"\t total heat required for kerosene is : Btu/hr \t400000.0\n",
"\t for steam \t\n",
"\t total heat required for steam is : Btu/hr \t399946.5\n",
"\t delt1 is : F \t155.0\n",
"\t delt2 is : F \t105.0\n",
"\t LMTD is : F \t128.525612445\n",
"\t caloric temperature of cold fluid is : F \t120.0\n",
"\t hot fluid:shell side,steam \t\n",
"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \t1500\n",
"\t cold fluid:inner tube side,kerosene \t\n",
"\t flow area is : ft**2 \t0.177375\n",
"\t mass velocity is : lb/(hr)*(ft**2) \t90395.480226\n",
"\t reynolds number is : \t1991.27136497\n",
"\t reynolds number is : \t1547.50231792\n",
"\t Hi is : Btu/(hr)*(ft**2)*(F) \t10.2620689655\n",
"\t Hio is : Btu/(hr)*(ft**2)*(F) \t8.928\n",
"\t tw is : F \t249.23081817\n",
"\t phyt is : \t1.16189787585\n",
"\t Correct hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \t10.3734242356\n",
"\t delt is : F \t129.23081817\n",
"\t Since the kerosene has a viscosity of only 1.75 cp at the caloric temperature and delt=129F, free convection should be investigated. \t\n",
"\t beta is : /F \t0.000449494949495\n",
"\t G is : \t1289776.29615\n",
"\t psy is : \t0.712464789946\n",
"\t corrected hio1 is : Btu/(hr)*(ft**2)*(F) \t7.39069951902\n",
"\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \t14.7\n",
"\t total surface area is : ft**2 \t270.1776\n",
"\t actual design overall coefficient is : Btu/(hr)*(ft**2)*(F) \t11.5\n",
"\t actual Rd is : (hr)*(ft**2)*(F)/Btu \t-0.0079\n"
]
}
],
"source": [
"print\"\\t example 10.2 \\t\"\n",
"print\"\\t approximate values are mentioned in the book \\t\"\n",
"#gien\n",
"T1=250.; # inlet hot fluid,F\n",
"T2=250.; # outlet hot fluid,F\n",
"t1=95.; # inlet cold fluid,F\n",
"t2=145.; # outlet cold fluid,F\n",
"W=16000.; # lb/hr\n",
"w=423.; # lb/hr\n",
"#solution\n",
"from math import log10\n",
"print\"\\t 1.for heat balance \\t\"\n",
"print\"\\t for kerosene \\t\"\n",
"c=0.5; # Btu/(lb)*(F)\n",
"Q=((W)*(c)*(t2-t1)); # Btu/hr\n",
"print\"\\t total heat required for kerosene is : Btu/hr \\t\",Q\n",
"print\"\\t for steam \\t\"\n",
"l=945.5; # Btu/(lb)\n",
"Q=((w)*(l)); # Btu/hr\n",
"print\"\\t total heat required for steam is : Btu/hr \\t\",Q\n",
"delt1=T2-t1; #F\n",
"delt2=T1-t2; # F\n",
"print\"\\t delt1 is : F \\t\",delt1\n",
"print\"\\t delt2 is : F \\t\",delt2\n",
"LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));\n",
"print\"\\t LMTD is : F \\t\",LMTD\n",
"tc=((t1)+(t2))/2; # caloric temperature of cold fluid,F\n",
"print\"\\t caloric temperature of cold fluid is : F \\t\",tc\n",
"print\"\\t hot fluid:shell side,steam \\t\"\n",
"ho=(1500); # condensation of steam Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",ho\n",
"print\"\\t cold fluid:inner tube side,kerosene \\t\"\n",
"Nt=86;\n",
"n=2; # number of passes\n",
"L=12; #ft\n",
"at1=0.594; # flow area, in**2,from table 10\n",
"at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48\n",
"print\"\\t flow area is : ft**2 \\t\",at\n",
"Gt=(W/(.177)); # mass velocity,lb/(hr)*(ft**2)\n",
"print\"\\t mass velocity is : lb/(hr)*(ft**2) \\t\",Gt\n",
"mu2=1.36*2.42; # at 145F,lb/(ft)*(hr)\n",
"D=(0.87/12); # ft\n",
"Ret1=((D)*(Gt)/mu2); # reynolds number\n",
"print\"\\t reynolds number is : \\t\",Ret1\n",
"mu3=1.75*2.42; # at 120F,lb/(ft)*(hr)\n",
"D=(0.87/12); # ft\n",
"Ret2=((D)*(Gt)/mu3); # reynolds number\n",
"print\"\\t reynolds number is : \\t\",Ret2\n",
"Z1=331; # Z1=(L*n/D)\n",
"jH=3.1; # from fig 24\n",
"mu4=1.75; # cp and 40 API\n",
"Z2=0.24; # Z2=((k)*(c*mu4/k)**(1/3)), from fig 16\n",
"Hi=((jH)*(1/D)*(Z2)); # using eq.6.15a,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t Hi is : Btu/(hr)*(ft**2)*(F) \\t\",Hi\n",
"ID=0.87; # ft\n",
"OD=1; #ft\n",
"Hio=(Hi*(ID/OD)); #Btu/(hr)*(ft**2)*(F), from eq.6.5\n",
"print\"\\t Hio is : Btu/(hr)*(ft**2)*(F) \\t\",Hio\n",
"tw=(tc)+(((ho)/(Hio+ho))*(T1-tc)); # from eq.5.31\n",
"print\"\\t tw is : F \\t\",tw\n",
"muw=1.45; # lb/(ft)*(hr),at 249F from fig.14\n",
"phyt=(mu3/muw)**0.14;\n",
"print\"\\t phyt is : \\t\",phyt # from fig.24\n",
"hio=(Hio)*(phyt); # from eq.6.37\n",
"print\"\\t Correct hio to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \\t\",hio\n",
"delt=tw-tc; #F\n",
"print\"\\t delt is : F \\t\",delt\n",
"print\"\\t Since the kerosene has a viscosity of only 1.75 cp at the caloric temperature and delt=129F, free convection should be investigated. \\t\"\n",
"s=0.8;\n",
"row=50; # lb/ft**3, from fig 6\n",
"s1=0.810; # at 95F\n",
"s2=0.792; # at 145F\n",
"bita=((s1**2-s2**2)/(2*(t2-t1)*s1*s2)); # /F\n",
"print\"\\t beta is : /F \\t\",bita\n",
"G=((D**3)*(row**2)*(bita)*(delt)*(4.18*10**8)/(mu3**2));\n",
"print\"\\t G is : \\t\",G\n",
"psy=((2.25)*(1+(0.01*G**(1/3)))/(log10(Ret2)));\n",
"print\"\\t psy is : \\t\",psy\n",
"hio1=(hio*psy);\n",
"print\"\\t corrected hio1 is : Btu/(hr)*(ft**2)*(F) \\t\",hio1\n",
"Uc=2*((hio1)*(ho)/(hio1+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",round(Uc,1)\n",
"A2=0.2618; # actual surface supplied for each tube,ft**2,from table 10\n",
"A=(Nt*L*A2); # ft**2\n",
"print\"\\t total surface area is : ft**2 \\t\",A\n",
"UD=((Q)/((A)*(delt)));\n",
"print\"\\t actual design overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",round(UD,1)\n",
"Rd=-0.407*((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu\n",
"print\"\\t actual Rd is : (hr)*(ft**2)*(F)/Btu \\t\",round(Rd,4)\n",
"# end\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 10.3 pgno:211"
]
},
{
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{
"name": "stdout",
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"text": [
"\t example 10.3 \t\n",
"\t approximate values are mentioned in the book \t\n",
"\t 1.for heat balance \t\n",
"\t for gas oil \t\n",
"\t total heat required for gas oil is : Btu/hr \t587500.0\n",
"\t for steam \t\n",
"\t total heat required for steam is : Btu/hr \t588101.0\n",
"\t delt1 is : F \t145.0\n",
"\t delt2 is : F \t120.0\n",
"\t LMTD is : F \t132.254461931\n",
"\t caloric temperature of cold fluid is : %.1f F \t117.5\n",
"\t hot fluid:shell side,steam \t\n",
"\t flow area is : ft**2 \t0.317708333333\n",
"\t mass velocity is : lb/(hr)*(ft**2) \t19577.704918\n",
"\t reynolds number is : \t37409.6272319\n",
"\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \t1500\n",
"\t cold fluid:inner tube side,crude oil \t\n",
"\t at1 is : in**2 \t0.3979165\n",
"\t flow area is : ft**2 \t0.118822288194\n",
"\t mass velocity is : lb/(hr)*(ft**2) \t420796.474801\n",
"\t De is : ft \t0.0485536398467\n",
"\t reynolds number is : \t1223.42517882\n",
"\t Hi is : Btu/(hr)*(ft**2)*(F) \t22.3464194121\n",
"\t Correct Hi0 to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \t19.4413848885\n",
"\t phyt is : \t1.18932885404\n",
"\t Correct hi0 to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \t23.1222000104\n",
"\t tw is : F \t247.988545173\n",
"\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \t22.7711867211\n",
"\t total surface area is : ft**2 \t270\n",
"\t actual design overall coefficient is : Btu/(hr)*(ft**2)*(F) \t16.4694016372\n",
"\t actual Rd is : (hr)*(ft**2)*(F)/Btu \t0.0168035136331\n",
"\t pressure drop for annulus \t\n",
"\t number of crosses are : \t9\n",
"\t delPs is : psi \t1.19559186672\n",
"\t pressure drop for inner pipe \t\n",
"\t dt is : ft \t0.0308333333333\n",
"\t Ret2 is : \t776.919639103\n",
"\t phyt is : \t1.35\n",
"\t delPt is : psi \t2.0\n",
"\t delPr is negligible \t\n"
]
}
],
"source": [
"print\"\\t example 10.3 \\t\"\n",
"print\"\\t approximate values are mentioned in the book \\t\"\n",
"#given\n",
"T1=250.; # inlet hot fluid,F\n",
"T2=250.; # outlet hot fluid,F\n",
"t1=105.; # inlet cold fluid,F\n",
"t2=130.; # outlet cold fluid,F\n",
"w=50000.; # lb/hr\n",
"W=622.; # lb/hr\n",
"#solution\n",
"from math import log10\n",
"print\"\\t 1.for heat balance \\t\"\n",
"print\"\\t for gas oil \\t\"\n",
"c=0.47; # Btu/(lb)*(F)\n",
"Q=((w)*(c)*(t2-t1)); # Btu/hr\n",
"print\"\\t total heat required for gas oil is : Btu/hr \\t\",Q\n",
"print\"\\t for steam \\t\"\n",
"l=945.5; # Btu/(lb)\n",
"Q=((W)*(l)); # Btu/hr\n",
"print\"\\t total heat required for steam is : Btu/hr \\t\",Q\n",
"delt1=T2-t1; #F\n",
"delt2=T1-t2; # F\n",
"print\"\\t delt1 is : F \\t\",delt1\n",
"print\"\\t delt2 is : F \\t\",delt2\n",
"LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));\n",
"print\"\\t LMTD is : F \\t\",LMTD\n",
"tc=((t1)+(t2))/2; # caloric temperature of cold fluid,F\n",
"print\"\\t caloric temperature of cold fluid is : %.1f F \\t\",tc\n",
"print\"\\t hot fluid:shell side,steam \\t\"\n",
"ID=15.25; # in\n",
"C=0.25; # clearance\n",
"B=15; # baffle spacing,in\n",
"PT=1.25;\n",
"As=((ID*C*B)/(144*PT)); # flow area,ft**2, eq 7.1\n",
"print\"\\t flow area is : ft**2 \\t\",As\n",
"Gs=(6220/As); # mass velocity,lb/(hr)*(ft**2), calculation mistake\n",
"print\"\\t mass velocity is : lb/(hr)*(ft**2) \\t\",Gs\n",
"mu1=0.0314; # at 250F,lb/(ft)*(hr), from fig.15\n",
"De=0.060; # from fig.29,ft\n",
"Res=((De)*(Gs)/mu1); # reynolds number, calculation mistake\n",
"print\"\\t reynolds number is : \\t\",Res\n",
"ho=1500; #Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t individual heat transfer coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",ho\n",
"print\"\\t cold fluid:inner tube side,crude oil \\t\"\n",
"d1=0.5; # in\n",
"d2=0.87; # in\n",
"at1=((3.14*(d2**2-d1**2))/4);\n",
"print\"\\t at1 is : in**2 \\t\",at1\n",
"Nt=86;\n",
"n=2; # number of passes\n",
"L=12; #ft\n",
"at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48\n",
"print\"\\t flow area is : ft**2 \\t\",at\n",
"Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)\n",
"print\"\\t mass velocity is : lb/(hr)*(ft**2) \\t\",Gt\n",
"De=(d2**2-d1**2)/(12*d2);\n",
"print\"\\t De is : ft \\t\",De\n",
"mu2=16.7; # at 117F,lb/(ft)*(hr)\n",
"Ret=((De)*(Gt)/mu2); # reynolds number\n",
"print\"\\t reynolds number is : \\t\",Ret\n",
"jH=3.1; # from fig.24\n",
"Z=0.35; # Z=(K*(c*mu3/k)**(1/3)),Btu/(hr)(ft**2)(F/ft), at mu3=6.9cp and 28 API\n",
"Hi=((jH)*(1/De)*(Z)); #Hi=(hi/phyp),using eq.6.15a,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t Hi is : Btu/(hr)*(ft**2)*(F) \\t\",Hi\n",
"ID=0.87; # ft\n",
"OD=1; #ft\n",
"Hio=((Hi)*(ID/OD)); #Hio=(hio/phyp), using eq.6.5\n",
"print\"\\t Correct Hi0 to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \\t\",Hio\n",
"muw=4.84; # lb/(ft)*(hr), from fig.14\n",
"phyt=(mu2/muw)**0.14;\n",
"print\"\\t phyt is : \\t\",phyt # from fig.24\n",
"hio=(Hio)*(phyt); # from eq.6.37\n",
"print\"\\t Correct hi0 to the surface at the OD is : Btu/(hr)*(ft**2)*(F) \\t\",hio\n",
"tw=(tc)+(((ho)/(hio+ho))*(T1-tc)); # from eq.5.31\n",
"print\"\\t tw is : F \\t\",tw\n",
"Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",Uc\n",
"A=270; # ft**2\n",
"print\"\\t total surface area is : ft**2 \\t\",A\n",
"UD=((Q)/((A)*(LMTD)));\n",
"print\"\\t actual design overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",UD\n",
"Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu\n",
"print\"\\t actual Rd is : (hr)*(ft**2)*(F)/Btu \\t\",Rd\n",
"print\"\\t pressure drop for annulus \\t\"\n",
"f=0.0016; # friction factor for reynolds number 25300, using fig.29\n",
"s=0.00116; # for reynolds number 25300,using fig.6\n",
"Ds=15.25/12; # ft\n",
"phys=1;\n",
"N=(12*L/B); # number of crosses,using eq.7.43\n",
"print\"\\t number of crosses are : \\t\",N\n",
"delPs=((f*(19600**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys)))/(2); # using eq.7.44,psi\n",
"print\"\\t delPs is : psi \\t\",delPs\n",
"print\"\\t pressure drop for inner pipe \\t\"\n",
"dt=(d2-d1)/(12); # ft\n",
"print\"\\t dt is : ft \\t\",dt\n",
"Ret2=(dt*Gt/mu2);\n",
"print\"\\t Ret2 is : \\t\",Ret2\n",
"f=0.00066; # friction factor for reynolds number 8220, using fig.26\n",
"phyt=1.35; # fig 6\n",
"print\"\\t phyt is : \\t\",phyt\n",
"s=0.85;\n",
"delPt=((f*(420000**2)*(L)*(n))/(5.22*(10**10)*(0.0309)*(s)*(phyt))); # using eq.7.45,psi\n",
"print\"\\t delPt is : psi \\t\",round(delPt)\n",
"print\"\\t delPr is negligible \\t\"\n",
"#end\n"
]
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"source": [
"## Example 10.4 pgno:217"
]
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"\t example 10.4 \t\n",
"\t approximate values are mentioned in the book \t\n",
"\t delt is : F \t100\n",
"\t convection loss is : Btu/(hr)(ft**2)(F) \t0.948683298051\n",
"\t radiation loss is : Btu/(hr)(ft**2)(F) \t0.510696\n",
"\t combined loss is : Btu/(hr)(ft**2)(F) \t1.45937929805\n",
"\t total tank area is : ft**2 \t227.65\n",
"\t total heat loss : Btu/hr \t33222.7697201\n",
"\t This heat must be supplied by the pipe bundle,Assuming exhaust steam to be at 212 degree F \t\n",
"\t X is : \t75.7575757576\n",
"\t tf is : F \t156\n",
"\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \t46\n",
"\t design overall coefficient is : Btu/(hr)*(ft**2)*(F) \t23.9583333333\n",
"\t total surface is : ft**2 \t12.3811564174\n",
"\t number of pipes are : \t7.0\n"
]
}
],
"source": [
"print\"\\t example 10.4 \\t\"\n",
"print\"\\t approximate values are mentioned in the book \\t\"\n",
"#given\n",
"t1=100; # F\n",
"t2=0; # F\n",
"T1abs=100+460; # R\n",
"T2abs=460; #R\n",
"#solution\n",
"delt=t1-t2;\n",
"from math import ceil\n",
"print\"\\t delt is : F \\t\",delt\n",
"hc=0.3*(delt**0.25); # convection loss, Btu/(hr)*(ft**2)*( degree F)\n",
"print\"\\t convection loss is : Btu/(hr)(ft**2)(F) \\t\",hc\n",
"e=0.8; # emissivity\n",
"hr=((0.173*e*((T1abs/100)**4-(T2abs/100)**4))/(T1abs-T2abs)); # radiation rate, from 4.32, Btu/(hr)(ft**2)(F)\n",
"print\"\\t radiation loss is : Btu/(hr)(ft**2)(F) \\t\",hr\n",
"hl=hc+hr; # combined loss, Btu/(hr)(ft**2)(F)\n",
"print\"\\t combined loss is : Btu/(hr)(ft**2)(F) \\t\",hl\n",
"D=5; # ft\n",
"L=12; # ft\n",
"A1=((2*3.14*D**2)/(4))+(3.14*D*L); # total tank area\n",
"print\"\\t total tank area is : ft**2 \\t\",A1\n",
"Q=(hl*A1*delt); # total heat loss\n",
"print\"\\t total heat loss : Btu/hr \\t\",Q\n",
"print\"\\t This heat must be supplied by the pipe bundle,Assuming exhaust steam to be at 212 degree F \\t\"\n",
"d0=1.32;\n",
"X=(delt/d0);\n",
"tf=((t1+212)/2); # F\n",
"print\"\\t X is : \\t\",X\n",
"print\"\\t tf is : F \\t\",tf\n",
"hio=48; # from fig 10.4, Btu/(hr)(ft**2)(F)\n",
"ho=1500; # condensation of steam,Btu/(hr)(ft**2)(F)\n",
"Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t clean overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",Uc\n",
"Rd=0.02; # dirt factor, (hr)(ft**2)(F)/Btu\n",
"UD=((Uc)/((1)+(Uc*Rd))); # design overall coefficient,Btu/(hr)*(ft**2)*(F)\n",
"print\"\\t design overall coefficient is : Btu/(hr)*(ft**2)*(F) \\t\",UD\n",
"A2=((Q)/((UD)*(212-100))); # total surface,ft**2\n",
"print\"\\t total surface is : ft**2 \\t\",A2\n",
"A3=2.06; # area/pipe\n",
"N=(A2/A3);\n",
"print\"\\t number of pipes are : \\t\",ceil(N)\n",
"#end\n"
]
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