path: root/Fundamentals_of_Turbomachinery_by_W_W_Peng/9-Steam_Turbines.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 9: Steam Turbines"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.1: ST.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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	   },
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"source": [
"clear all; clc;\n",
"\n",
"rm=8/12\n",
"N=7500\n",
"U=rm*N*%pi/30\n",
"printf('The peripheral velocity is calculated as U= %0.1f ft/s\n\n',U)\n",
"\n",
"disp('From equation 9.1 we have U/V1=sinα1/4')\n",
"//let x= U/V1\n",
"alpha1=70*%pi/180\n",
"x=(sin(alpha1))/4\n",
"printf('\n Thus U/V1= %0.4f',x)\n",
"\n",
"V1=U/x\n",
"printf('\n Thus V1= %0.1f ft/s',V1)\n",
"\n",
"disp('From velocity diagram at station1 we have V1sinα1-W1sinß1=U and V1cosα1=W1cosß1 or W1sinß1')\n",
"//let y= W1sinß1\n",
"V1=2228.8\n",
"U=523.6\n",
"y=V1*sin(alpha1)-U\n",
"printf('\n Hence W1sinß1= %0.1f ft/s',y)\n",
"\n",
"//Let z=W1cosß1\n",
"z=V1*cos(alpha1)\n",
"printf('\n Thus W1cosß1= %0.1f ft/s',z)\n",
"\n",
"disp('Hence tanß1=2.06')\n",
"tanbeta1=2.06\n",
"beta1=(atan(tanbeta1))*180/%pi\n",
"printf('\n Thus beta1= %0.1f degrees and W1=1746 ft/s',beta1)\n",
"\n",
"disp('At station 2 we have W2sinß2-V2sinα2=U and V2cosα2=W2cosß2,with W2=W1=1746ft/s and ß1=ß2=64.1 degrees')\n",
"//Let l=V2sinα2\n",
"l=1746*sin(64.1*%pi/180)-523.6\n",
"printf(' Thus V2sinα2=%0.0f ft/s',l)\n",
"\n",
"//m=V2cosα2\n",
"m=1746*cos(64.1*%pi/180)\n",
"printf('\n V2cosα2 %0.2f ft/s',m)\n",
"\n",
"disp('Hence tanα2=1.373')\n",
"tanalpha2=1.373\n",
"alpha2=((atan(tanalpha2)*180/%pi))\n",
"printf(' Hence α2= %0.2f degrees',alpha2)\n",
"\n",
"disp('Hence V2=1295.2 ft/s')\n",
"\n",
"disp('At station 3 we have V3sinα3-W3sinα3=U=523.6ft/s')\n",
"disp('Also W3cosß3=V3cosα3')\n",
"//let n=V3cosα3\n",
"V3=1295.2\n",
"alpha3=53.9*%pi/180\n",
"n=V3*cos(alpha3)\n",
"printf(' Thus W3cosß3= %0.1f ft/s',n)\n",
"\n",
"disp('Hence tanß3=0.685')\n",
"tanbeta3=0.685\n",
"beta3=((atan(tanbeta3))*180/%pi)\n",
"printf(' Hence ß3= %0.1f degrees',beta3)\n",
"\n",
"disp('Thus W3=925.1 ft/s')\n",
"\n",
"disp('Also W4=W3=925.1ft/s')\n",
"disp('ß4=ß3=34.4 degrees')\n",
"disp('And V4=VaV1cosß4')\n",
"beta4=34.4*%pi/180\n",
"//let y=Va*V1\n",
"y=925.0848\n",
"V4=y*cos(beta4)\n",
"printf(' Thus V4= %0.1f ft/s',V4)\n",
"disp('α4=0 degrees')\n",
"\n",
"disp('From these velocities,the energy transfers of the rotors can be calculated')\n",
"\n",
"U=523.6\n",
"V1=2228.8\n",
"alpha1=70*%pi/180\n",
"V2=1295.2\n",
"alpha2=53.9*%pi/180\n",
"delta_E1=U*(V1*sin(alpha1)+V2*sin(alpha2))\n",
"printf(' Thus delta_E1= %0.1f ((ft/s)^2)',delta_E1)\n",
"\n",
"delta_E1=1.643*(10^6)/(32.2*778)//converting units from (ft/s)^2 to Btu/lb\n",
"printf('\n On converting to Btu/lb we have delta_E1=%0.1f Btu/lb',delta_E1)\n",
"\n",
"V3=1295.2\n",
"alpha3=alpha2\n",
"delta_E2=U*(V3*sin(alpha3))\n",
"printf('\n delta_E2=%0.1f ((ft/s)^2)',delta_E2)\n",
"delta_E2=0.546*(10^6)/(32.2*778)\n",
"printf('\n On converting to Btu/lb we have delta_E2=%0.1f Btu/lb',delta_E2)\n",
"\n",
"delta_Ec=65.6+21.8\n",
"printf('\n Hence the total energy transfer is delta_Ec= %0.1f Btu/lb',delta_Ec)\n",
"disp('To compare with that calculated with equation9.3,we have delta_Ec=8*U^2')\n",
"delta_Ec=8*(U^2)\n",
"printf(' delta_Ec= %0.2f ((ft/s)^2)',delta_Ec)\n",
"delta_Ec=2.19*10^6/(32.2*778)//converting units\n",
"printf('\n On converting we have delta_Ec= %0.2f Btu/lb',delta_Ec)//answer given in the book is 87.5,however 87.42 is more accurate\n",
"\n",
"disp('The difference is due to round off error.')\n",
"disp('The static enthalpies and pressure at stations 1,2,3 and 4 are same for the ideal case and can be calculated from h1=h01-((V1)^2)/2 ')\n",
"disp('Where h01=h0i=1405Btu/lb from the Mollier diagram for p0i=3000 psia,T01=950 degrees Farenheit')\n",
"//let l=(V1^2)/2\n",
"V1=2228.8\n",
" l=(V1^2)/(2*32.2*778)\n",
"printf(' Thus (V1^2)/2 = %0.0f Btu/lb',l)\n",
"\n",
"disp('Hence we have h1=1306 Btu/lb and p1=1400psia')\n",
"\n",
"\n",
"\n",
"\n",
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""
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{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.2: ST.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('To use Figure 9.8,with Qf=ΣU^2/delta_Hs=2*g_c*lambda^2*R_H')\n",
"disp('The value of R_H can be estimated with equation 8.4.')\n",
"disp('Using k=1.3 for steam and suusming ETA_p=0.90 we have ETAad=[1-(p_e/p_i)^(ETAp*(k-1)/k)]/[1-(p_e/p_i)^(k-1)/k]=0.931')\n",
"\n",
"ETA_ad=0.931\n",
"ETA_p=0.90\n",
"R_H=ETA_ad/ETA_p\n",
"printf(' R_H=ETA_ad/ETA_p= %0.3f',R_H)\n",
"\n",
"disp('For impulse stages,the optimal efficiencies occur at lambda=U/V2=sinα2/2=0.47 with alpha2=70 degrees')\n",
"QF=2*25052*(0.47^2)*1.035\n",
"printf(' So Qf can be calculated as %0.0f',QF)\n",
"\n",
"disp('From figure 9.8, the efficiency can be estimated as ETA=83%')\n",
"\n",
"disp('From the Mollier diagram in figure A1 we have hi=1525 Btu/lbm,hse=1150 Btu/lbm,with s_i=s_es=1.8Btu/lb-R')\n",
"delta_Hs=1525-1150\n",
"printf(' Hence delta_Hs=%0.0f Btu/lbm',delta_Hs)\n",
"\n",
"summation_sqr(U)=11455*375\n",
"printf('\n So we have ΣU^2=%0.0f ((ft/s)^2)',summation_sqr(U))\n",
"\n",
"disp('With 10 identical stages,we have U^2=429562')\n",
"sqr(U)=429562\n",
"U=sqrt(sqr(U))\n",
"printf(' Thus U= %0.0f ft/s',U)\n",
"\n",
"omega=3600*%pi/30\n",
"D=2*U/omega\n",
"printf('\n The turbine diameter D= %0.3f ft',D)//The answer has been incorrectly rounded off to 3.47 in the book. A more accurate answer is provided here.\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n",
"\n",
""
   ]
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,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.3: ST.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('The tangential velocity at the rotor mean radius is Um')\n",
"rm=1.5\n",
"N=3600\n",
"Um=rm*N*%pi/30\n",
"printf('Um=rm*N*pi/30= %0.1f',Um)\n",
"\n",
"disp('From the velocity diagram in figure 8.11 for the impulse stages we have delta_h0')\n",
"disp('delta_h0=delta_Et=UmVu2-UmVu3/g_c=UmVu2/g_c=2((Um)^2)/gc')\n",
"Um=565.5\n",
"gc=32.2\n",
"delta_h0=2*((Um)^2)/gc\n",
"printf('\n delta_h0= %0.0f lbf-ft/lbm=25.5 Btu/lbm',delta_h0)\n",
"\n",
"disp('From the Mollier diagram in appendix A, we have hoi=1565 Btu/lbm')\n",
"disp('For the stages with constant mean radi,we have hoi-hoe=n_s*delta_hoe or hoe=h0i-n_s*delta_hoe')\n",
"h_oe=1565-(12*25.5)\n",
"printf(' hoe= %0.0f Btu/lbm.',h_oe)\n",
"\n",
"disp('Also from ETA_ad=(hoi-hoe)/(hoi-hsoe), we have hsoe=hoi-(hoi-hoe)/ETA_ad')\n",
"h_soe=1565-306/0.85\n",
"printf(' hsoe= %0.0f Btu/lbm',h_soe)\n",
"\n",
"disp('From s_soe=Soi=1.69 Btu/(lbm-R),we have poe=50 psia and Toe=450 degrees Farenheit')\n",
"disp('Also from given efficiencies we have Ps=ETAv*ETAm*m*n_s*delta_ho')\n",
"Ps=0.92*0.90*7*12*19863/550\n",
"printf(' Ps=%0.0f hp',Ps)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.4: ST.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('Based on the law of Willian we have m_t=0.5+CP_s')\n",
"C=(7-0.5)/2512\n",
"printf(' Where C= %0.4f lbm/(hp-s)\n',C)\n",
"\n",
"disp('So we have SR=mt/Ps=0.5/Ps+(2.6*10^-3)')\n",
"disp('And HR=Q_h*SR')\n",
"\n",
"\n",
"disp('OR at full load,')\n",
"SR_1=(0.5/2512)+(2.6*10^-3)\n",
"printf('\n SR_1= %0.4f lbm/(hp-s)=10.1 lbm/(hp-h)',SR_1)\n",
"\n",
"HR_1=1750*10.1\n",
"printf('\n HR_1=%0.0f Btu/(hp-h)',HR_1)\n",
"\n",
"\n",
"disp('at 50% load,')\n",
"SR_2=(0.5/1256)+(2.6*10^-3)\n",
"printf('\n SR_2= %0.4f lbm/(hp-s)=10.8 lbm/(hp-h)',SR_2)\n",
"\n",
"HR_2=1750*10.8\n",
"printf('\n HR_2=%0.0f Btu/(hp-h)',HR_2)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.5: ST.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('To calculate the thermal efficiency,the units have to be consisitent. With hp=0.707 Btu/s=2545Btu/h, we have ETAth=Ps+Qe/Qin')\n",
"ETA_th=(2512*0.707+1259*7)/(7*1750)\n",
"printf(' Thus ETA_h= %0.3f',ETA_th)\n",
"\n",
"ETA_th=2512*0.707/(7*1750)\n",
"printf('\n For the simple shaft power system,we have ETA_th= %0.3f',ETA_th)\n",
"\n",
"ETA_th=2545/17675\n",
"printf('\n From the heat rate,ETA_th=2545/HR %0.3f',ETA_th)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.6: Steam_Turbines.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('The enthalpies at various points have t be determined first. For the steam turbine cycle,fro the Mollier diagram or steam tables,we have h1=hf1=83.6')\n",
"\n",
"h2=83.6+(0.0185*(1000-1.5)*144)/(778*0.80)\n",
"printf(' h2=h1+nu*deltap/ETA_p =%0.1f Btu/lbm',h2)\n",
"\n",
"disp('h3=1447 Btu/lbm')\n",
"\n",
"disp('s_s4=s3=1.61 Btu/(lbm-R)')\n",
"\n",
"disp('hs4=925 Btu/lbm')\n",
"\n",
"disp('Hence from ETAst=(h3-h4)/(h3-hs4) we have h4=h3-(h3-hs4)*ETAst')\n",
"h3=1447\n",
"hs4=925\n",
"ETA_st=0.85\n",
"h4=h3-(h3-hs4)*ETA_st\n",
"printf(' h4= %0.0f Btu/lbm',h4)\n",
"\n",
"h4=1003\n",
"h2=87.9\n",
"h1=83.6\n",
"ETA_ths=[(h3-h4)-(h2-h1)]/(h3-h2)\n",
"printf('\n The thermal efficiency of the steam turbine cycle is then obtained as ETA_th,s=(Wst-Wp)/(Qin,s)=%0.4f ',ETA_ths)//it has been rounded off to 32.3 in the book\n",
"\n",
"disp('For the gas turbine cycle, an ideal gas with constant Cp is assumed for the working gas. With Cp=0.24 Btu/(lbm-R) and k=1.4 we have T6')\n",
"T5=540\n",
"//Let n=p6/p5 and m= (k-1)/k\n",
"n=15\n",
"m=0.2857\n",
"ETAc=0.82\n",
"T6=T5+[T5*({(n)^(m)}-1)]/ETAc\n",
"printf('\n T6= %0.0fR=849 degrees Farenheit',T6)\n",
"\n",
"T7=2560\n",
"//let b=(1/15)^0.2857\n",
"b=0.461\n",
"ETA_gt=0.85\n",
"T8=T7-T7*[1-b]*ETA_gt\n",
"printf('\n T8=%0.0f R= 928 degrees Farenheit',T8)\n",
"\n",
"disp('Which should be greater than T3')\n",
"T7=2560\n",
"T8=1388\n",
"T6=1309\n",
"T5=540\n",
"ETA_thg=[(T7-T8)-(T6-T5)]/(T7-T6)\n",
"printf('\n The thermal efficiency of the gas turbine is obtained as %0.3f=32.2 percent',ETA_thg)\n",
"\n",
"disp('From the energy balance equation across  the HRSG,we have m_g*Cp*(T8-T9)')\n",
"disp('ms/mg=[Cp*(T8-T9)]/(h3-h2)')\n",
"//let x=ms/mg\n",
"Cp=0.24\n",
"T8=928\n",
"T9=450\n",
"h3=1447\n",
"h2=87.9\n",
"x=[Cp*(T8-T9)]/(h3-h2)\n",
"printf('\n Thus ms/mg=%0.3f',x)\n",
"\n",
"disp('Hence the thermal efficiency of the combined cycle is obtained as ETA_th,c=[(Wgt-Wc)+(ms/mg)*(Wst-Wp)]/[Cp*(T7-T6)]')\n",
"ETA_thc=[0.24*(1172-769)+0.084*(439.7)]/(0.24*1251)\n",
"printf('\n ETA_th,c= %0.3f=44.5 percent',ETA_thc)\n",
"\n",
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""
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