path: root/Process_Heat_Transfer_by_D_Q_Kern/9-Gases.ipynb

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	   "source": [
       "# Chapter 9: Gases"
	   ]
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		   "source": [
			"## Example 9.1: Calculation_of_an_Ammonia_Compressor_Aftercooler.sce"
		   ]
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"printf('\t example 9.1 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"T1=245; // inlet hot fluid,F\n",
"T2=95; // outlet hot fluid,F\n",
"t1=85; // inlet cold fluid,F\n",
"t2=95; // outlet cold fluid,F\n",
"W=9872; // lb/hr\n",
"w=78500; // lb/hr\n",
"printf('\t 1.for heat balance \n');\n",
"printf('\t for ammonia gas \n');\n",
"c=0.53; // Btu/(lb)*(F)\n",
"Q=((W)*(c)*(T1-T2)); // Btu/hr\n",
"printf('\t total heat required for ammonia gas is : %.2e Btu/hr \n',Q);\n",
"printf('\t for water \n');\n",
"c=1; // Btu/(lb)*(F)\n",
"Q=((w)*(c)*(t2-t1)); // Btu/hr\n",
"printf('\t total heat required for water is : %.2f Btu/hr \n',Q);\n",
"delt1=T2-t1; //F\n",
"delt2=T1-t2; // F\n",
"printf('\t delt1 is : %.0f F \n',delt1);\n",
"printf('\t delt2 is : %.0f F \n',delt2);\n",
"LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1))));\n",
"printf('\t LMTD is :%.1f F \n',LMTD);\n",
"R=((T1-T2)/(t2-t1));\n",
"printf('\t R is : %.0f \n',R);\n",
"S=((t2-t1)/(T1-t1));\n",
"printf('\t S is : %.4f \n',S);\n",
"printf('\t FT is 0.837 \n'); // from fig 18\n",
"delt=(0.837*LMTD); // F\n",
"printf('\t delt is : %.1f F \n',delt);\n",
"Tc=((T2)+(T1))/2; // caloric temperature of hot fluid,F\n",
"printf('\t caloric temperature of hot fluid is : %.0f F \n',Tc);\n",
"tc=((t1)+(t2))/2; // caloric temperature of cold fluid,F\n",
"printf('\t caloric temperature of cold fluid is : %.0f F \n',tc);\n",
"printf('\t hot fluid:shell side,ammonia at 83psia \n');\n",
"ID=23.25; // in\n",
"C=0.1875; // clearance\n",
"B=12; // baffle spacing,in\n",
"PT=0.937;\n",
"as=((ID*C*B)/(144*PT)); // flow area,ft^2,from eq 7.1\n",
"printf('\t flow area is : %.3f ft^2 \n',as);\n",
"Gs=(W/as); // mass velocity,lb/(hr)*(ft^2),from eq 7.2\n",
"printf('\t mass velocity is : %.2e lb/(hr)*(ft^2) \n',Gs);\n",
"mu1=0.012*2.42; // at 170F,lb/(ft)*(hr), from fig.15\n",
"De=0.55/12; // from fig.28,ft\n",
"Res=((De)*(Gs)/mu1); // reynolds number\n",
"printf('\t reynolds number is : %.2e \n',Res);\n",
"jH=118; // from fig.28\n",
"k=0.017; // Btu/(hr)*(ft^2)*(F/ft),from table 5\n",
"Z=0.97; // Z=(Pr*(1/3)) prandelt number\n",
"ho=((jH)*(k/De)*(Z)*1); // using eq.6.15,Btu/(hr)*(ft^2)*(F)\n",
"printf('\t individual heat transfer coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n',ho);\n",
"printf('\t cold fluid:inner tube side,water \n');\n",
"Nt=364;\n",
"n=8; // number of passes\n",
"L=8; //ft\n",
"at1=0.302; // flow area, in^2,from table 10\n",
"at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48\n",
"printf('\t flow area is : %.4f ft^2 \n',at);\n",
"Gt=(w/(at)); // mass velocity,lb/(hr)*(ft^2)\n",
"printf('\t mass velocity is : %.2e lb/(hr)*(ft^2) \n',Gt);\n",
"V=(Gt/(3600*62.5)); // fps\n",
"printf('\t V is : %.2f fps \n',V);\n",
"mu2=0.82*2.42; // at 90F,lb/(ft)*(hr),from fig 14\n",
"D=(0.62/12); // ft,from table 10\n",
"Ret=((D)*(Gt)/mu2); // reynolds number\n",
"printf('\t reynolds number is : %.2e \n',Ret);\n",
"hi=900; // using fig 25,Btu/(hr)*(ft^2)*(F)\n",
"printf('\t hi is : %.0f Btu/(hr)*(ft^2)*(F) \n',hi);\n",
"ID=0.62; // ft\n",
"OD=0.75; //ft\n",
"hio=((hi)*(ID/OD)); // using eq.6.5\n",
"printf('\t Correct hi0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n',hio);\n",
"Uc=((hio)*(ho)/(hio+ho)); // clean overall coefficient,Btu/(hr)*(ft^2)*(F)\n",
"printf('\t clean overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n',Uc);\n",
"A2=0.1963; // actual surface supplied for each tube,ft^2,from table 10\n",
"A=(Nt*L*A2); // ft^2\n",
"printf('\t total surface area is : %.0f ft^2 \n',A);\n",
"UD=((Q)/((A)*(delt)));\n",
"printf('\t actual design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n',UD);\n",
"Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu\n",
"printf('\t actual Rd is : %.3f (hr)*(ft^2)*(F)/Btu \n',Rd);\n",
"printf('\t pressure drop  for annulus \n');\n",
"f=0.00162; // friction factor for reynolds number 40200, using fig.29\n",
"Ds=23.25/12; // ft\n",
"phys=1;\n",
"N=(12*L/B); // number of crosses,using eq.7.43\n",
"printf('\t number of crosses are : %.0f \n',N);\n",
"rowgas=0.209;\n",
"printf('\t rowgas is %.3f lb/ft^3 \n',rowgas);\n",
"s=rowgas/62.5;\n",
"printf('\t s is %.5f \n',s);\n",
"delPs=((f*(Gs^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys))); // using eq.7.44,psi\n",
"printf('\t delPs is : %.0f psi \n',delPs);\n",
"printf('\t allowable delPs is 2 psi \n');\n",
"printf('\t pressure drop  for inner pipe \n');\n",
"f=0.000225; // friction factor for reynolds number 21400, using fig.26\n",
"s=1;\n",
"D=0.0517; //ft\n",
"phyt=1;\n",
"delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(D)*(s)*(phyt))); // using eq.7.45,psi\n",
"printf('\t delPt is : %.1f psi \n',delPt);\n",
"X1=0.090; // X1=((V^2)/(2*g)), for Gt 1060000,using fig.27\n",
"delPr=((4*n*X1)/(s)); // using eq.7.46,psi\n",
"printf('\t delPr is : %.1f psi \n',delPr);\n",
"delPT=delPt+delPr; // using eq.7.47,psi\n",
"printf('\t delPT is : %.1f psi \n',delPT);\n",
"printf('\t allowable delPT is 10 psi \n');\n",
"//end"
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		   "source": [
			"## Example 9.2: Calculation_of_the_Heat_Load_for_an_Air_Intercooler.sce"
		   ]
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"printf('\t example 9.2 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"V1=4670; // inlet air volume,cfm\n",
"Pp=0.8153; // Saturation partial pressure of water at 95F,psi,from table 7\n",
"Ps=404.3;//  Saturation specific volume of water at 95F,ft^3/lb, from table 7\n",
"printf('\t The air and water both occupy the same volume at their respective partial pressures \n');\n",
"Vw1=(V1*60/Ps); // water entering per hr,lb\n",
"printf('\t volume of water entering is : %.0f lb \n',Vw1);\n",
"printf('\t for first stage \n');\n",
"c=2.33; // compression ratio\n",
"P1=14.7; // psi\n",
"P2=(P1*c); // (c=(P2/P1)),psi\n",
"printf('\t P2 is : %.1f psi \n',P2);\n",
"gama=1.4; // for air\n",
"T1abs=95; // F\n",
"T2absr=((T1abs+460)*(P2/P1)^((gama-1)/gama));\n",
"printf('\t T2absr is : %.0f R \n',T2absr);\n",
"T2abs=(T2absr-459.67); // F\n",
"printf('\t T2abs is : %.0f F \n',T2abs);\n",
"printf('\t for intercooler \n');\n",
"V2=(V1*60*P1/P2); // ft^3/hr\n",
"printf('\t final gas volume is : %.1e ft^3/hr \n',V2);\n",
"Vw2=(V2/Ps); // water remaining in air, lb/hr\n",
"printf('\t water remaining in air is : %.0f lb/hr \n',Vw2);\n",
"C=(Vw1-Vw2); // condensation in inter cooler, lb/hr\n",
"printf('\t condensation in inter cooler is : %.0f lb/hr \n',C);\n",
"Vs=14.8; // Specific volume of atmospheric air,ft^3/lb\n",
"printf('\t Specific volume of atmospheric air is : %.1f ft^3/lb \n',Vs);\n",
"Va=(V1*60/Vs); // air in inlet gas, lb/hr\n",
"printf('\t air in inlet gas is : %.2e lb/hr\n',Va);\n",
"printf('\t heat load(245 to 95F) \n)');\n",
"printf('\t sensible heat \n');\n",
"Qair=((Va)*(0.25)*(245-T1abs)); // Btu/hr\n",
"printf('\t Qair is : %.2e Btu/hr \n',Qair);\n",
"Qwaters=(Vw1*0.45*(245-T1abs)); // Btu/hr\n",
"printf('\t Qwaters is : %.2e Btu/hr \n',Qwaters);\n",
"printf('\t latent heat \n');\n",
"l=1040.1; // latent heat\n",
"Qwaterl=(C*l); // Btu/hr\n",
"printf('\t Qwater1 is : %.2e Btu/hr \n',Qwaterl);\n",
"Qt1=Qair+Qwaters+Qwaterl;\n",
"printf('\t total heat is : %.3e Btu/hr \n',Qt1);\n",
"printf('\t for second stage \n');\n",
"c=2.33; // compression ratio\n",
"P3=(P2*c); // (c=(P3/P1)),psi\n",
"printf('\t P3 is : %.1f psi \n',P3);\n",
"V3=(V1*60*P1/P3); // ft^3/hr\n",
"printf('\t final gas volume is : %.2e ft^3/hr \n',V3);\n",
"Vw3=(V3/Ps); // water remaining in air, lb/hr\n",
"printf('\t water remaining in air is : %.1f lb/hr \n',Vw3);\n",
"C1=(297-Vw3); // condensation in inter cooler, lb/hr\n",
"printf('\t condensation in inter cooler is : %.1f lb/hr \n',C1);\n",
"printf('\t heat load(245 to 95F) \n)');\n",
"printf('\t sensible heat \n');\n",
"Qair=(Va*0.25*(245-T1abs)); // Btu/hr\n",
"printf('\t Qair is : %.2e Btu/hr \n',Qair);\n",
"Qwaters=(Vw2*0.44*(245-T1abs)); // Btu/hr\n",
"printf('\t Qwater is : %.2e Btu/hr \n',Qwaters);\n",
"printf('\t latent heat \n');\n",
"l=1040.1; // latent heat\n",
"Qwaterl=(C1*l); // Btu/hr, calculation mistake in book\n",
"printf('\t Qwater is : %.2e Btu/hr \n',Qwaterl);\n",
"Qt1=Qair+Qwaters+Qwaterl;\n",
"printf('\t total heat is : %.3e Btu/hr \n',Qt1);\n",
"// end"
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			"## Example 9.3: Calculation_of_the_Dew_Point_after_Compression.sce"
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"printf('\t example 9.3 \n');\n",
"printf('\t approximate values are mentioned in the book \n');\n",
"Va=18900; // air in inlet gas\n",
"Vw1=692; // water entering\n",
"Ma=(Va/29); // moles\n",
"Mw=(Vw1/18); // moles\n",
"M=(Ma+Mw); // moles\n",
"printf('\t total number of moles re : %.1f \n',M);\n",
"printf('\t Moles of air is : %.0f \n',Ma);\n",
"printf('\t Moles of water is : %.1f \n',Mw);\n",
"printf('\t after compression \n');\n",
"P=34.2; // pressure,psi\n",
"pw=(Mw/M)*(P); // partial pressure\n",
"printf('\t partial pressure is :%.1f psi \n',pw);\n",
"Td=124; // F, table table 7\n",
"printf('\t dew point is : %.0f F \n',Td);\n",
"// end"
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