path: root/Fundamentals_of_Turbomachinery_by_W_W_Peng/3-Energy_Transfer_in_Turbomachines.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 3: Energy Transfer in Turbomachines"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.10: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_5S,page 53.\n",
"\n",
"//this numerical is based on numerical 3_4S\n",
"//values found in the book for numerical 3_4S will be used to solve this numerical(3.5S)\n",
"\n",
"delta_Et=40.85\n",
"U_1=6.9\n",
"U_2=2.92\n",
"V_m2=0.782\n",
"V_2=0.782//since V2=Vm2\n",
"\n",
"V_r1=0.508\n",
"W_u1=0.997\n",
"W_1=(V_r1^2+W_u1^2)^0.5\n",
"printf('\n W1 is equal to %0.2f ft/s',W_1)\n",
"\n",
"W_2=(U_2^2+V_m2^2)^0.5\n",
"printf('\n W_2 is equal to %0.2f ft/s',W_2)\n",
"\n",
"Rt=0.5*[(U_1^2-U_2^2)+(W_2^2-W_1^2)]/(delta_Et)\n",
"printf('\n Thus the value Rt is equal to %0.3f',Rt)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.11: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_6E,page 53.\n",
"disp('The velocity diagram is similar to that in figure 3.19')\n",
"\n",
"r_t=1.2\n",
"r_h=0.7\n",
"N=4800\n",
"V_1=600\n",
"\n",
"r_m=[0.5*(r_t^2+r_h^2)]^0.5\n",
"printf('\n r_m is equal to %0.3f ft',r_m)\n",
"\n",
"A=%pi*(r_t^2-r_h^2)\n",
"printf('\n A is equal to %0.2f ft^2',A)\n",
"\n",
"r_m=0.982//rounding off rm\n",
"U=(N*%pi*r_m)/30\n",
"printf('\n U is equal to %0.1f ft/s',U)\n",
"\n",
"V_a1=V_1*cos(60*%pi/180)//angle is given as 60 degrees\n",
"printf('\n V_a1 is equal to %0.2f ft/s',V_a1)\n",
"\n",
"V_u1=V_1*sin(60*%pi/180)//angle is given as 60 degrees\n",
"printf('\n V_u1 is equal to %0.1f ft/s',V_u1)\n",
"\n",
"//let tan(betaf1)=(Vu1-U)/Va1=x\n",
"x=(V_u1-U)/V_a1\n",
"printf('\n tan(ßf1) is equal to %0.4f',x)\n",
"beta_f1=(atan(x))*180/%pi\n",
"printf('\n Hence ßf1= %0.2f degrees',beta_f1)\n",
"\n",
"disp('ßf1=ßb1. Also ßb2=ßf2=alpha_1=60 degrees')\n",
"disp('From the velocity diagram we have V_u2=W_u1=V_u1-U')\n",
"W_u1=V_u1-U\n",
"printf('\n Thus Wu1 = %g ft/s',W_u1)\n",
"\n",
"//let l=Ps/m=U*(Vu1+Vu2)\n",
"V_u2=W_u1//already stated above\n",
"l=U*(V_u1+V_u2)\n",
"printf('\n P_s/m=U*(V_u1+V_u2) is equal to %g (ft/s)^2',l)\n",
"disp('Thus we can round it off to 29*10^5 (ft/s)^2')\n",
"\n",
"disp('Where m=rho*V_a1*A')\n",
"rho=0.085/32.2\n",
"m=rho*V_a1*A\n",
"printf('\n m= %0.2f slug/s',m)\n",
"\n",
"disp('Thus we can determine P_s')\n",
"l=2.69*10^5//rounded off value\n",
"m=2.36//rounded off value\n",
"P_s=(l*m)/550\n",
"printf(' P_s =%0.0f hp',P_s)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.12: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_6S,page 54.\n",
"\n",
"disp('The velocity diagram is similar to that in figure 3.19')\n",
"\n",
"r_t=0.36\n",
"r_h=0.21\n",
"N=4800\n",
"V_1=183\n",
"\n",
"r_m=[0.5*(r_t^2+r_h^2)]^0.5\n",
"printf('\n rm is equal to %0.3f m',r_m)\n",
"\n",
"A=%pi*(r_t^2-r_h^2)\n",
"printf('\n A is equal to %0.2f m^2',A)\n",
"\n",
"r_m=0.295//rounding off rm\n",
"U=(N*%pi*r_m)/30\n",
"printf('\n U is equal to %0.1f m/s',U)\n",
"\n",
"V_a1=V_1*cos(60*%pi/180)//angle is given as 60 degrees\n",
"printf('\n V_a1 is equal to %0.2f m/s',V_a1)\n",
"\n",
"V_u1=V_1*sin(60*%pi/180)//angle is given as 60 degrees\n",
"printf('\n V_u1 is equal to %0.1f m/s',V_u1)\n",
"\n",
"//let tan(betaf1)=(Vu1-U)/Va1=x\n",
"x=(V_u1-U)/V_a1\n",
"printf('\n tan(ß1) is equal to %g m/s^2',x)\n",
"\n",
"beta_f1=(atan(x))*180/%pi\n",
"printf('\n We have ß_b1= ß_f1= %0.2f degrees',beta_f1)//value in book is 6.35 degrees Difference is obtained because actual value of x is substituted. Value substituted in the book is not 0.111 or 0.11147.\n",
"\n",
"disp('ß_f1=ß_b1. Also ß_b2=ß_f2=alpha_1=60 degrees')\n",
"disp('From the velocity diagram we have V_u2=W_u1=V_u1-U')\n",
"W_u1=V_u1-U\n",
"printf('\n Thus W_u1 = %0.2f m/s',W_u1)\n",
"\n",
"//let l=Ps/m=U*(Vu1+Vu2)\n",
"V_u2=W_u1//already stated above\n",
"l=U*(V_u1+V_u2)\n",
"printf('\n P_s/m=U*(V_u1+V_u2) is equal to %g (m/s)^2',l)\n",
"disp('Thus we can round it off to 2.5*10^4 (m/s)^2')\n",
"\n",
"disp('Where m=rho*V_a1*A')\n",
"rho=1.36\n",
"V_a1=91.5//rounded off\n",
"A=0.269//rounded off\n",
"m=rho*V_a1*A\n",
"printf('\n m= %0.1f kg/s',m)\n",
"\n",
"disp('Thus we can determine P_s')\n",
"l=2.5*10^4//rounded off value\n",
"m=33.5//rounded off value\n",
"P_s=(l*m)\n",
"printf(' P_s =%0.0f W',P_s)\n",
"disp('Thus on rounding off P_s is equal to 837kW')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.13: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_7E,page 54.\n",
"\n",
"//Velocity diagrams are not drawn. This is with Scilab team's permission.\n",
"//The numerical part of the question has been solved.\n",
"\n",
"r_t=5.2\n",
"r_h=3.5\n",
"N=4500\n",
"\n",
"A=%pi*(r_t^2-r_h^2)\n",
"printf(' A is equal to %0.2f in^2',A)\n",
"disp('On converting to feet we get A= 0.322 ft^2')\n",
"\n",
"r_m=[0.5*(r_t^2+r_h^2)]^0.5\n",
"printf('\n rm is equal to %0.2f in',r_m)\n",
"disp(' On converting to feet we get rm =0.369ft')\n",
"\n",
"r_m=0.369//in feet\n",
"U_m=(N*%pi*r_m)/30\n",
"printf('\n U_m is equal to %0.0f ft/s',U_m)\n",
"\n",
"disp('From inlet velocity triangle we have V_1/sin(ß_1+pi/2)+U_m/sin(aplha_1-ß_1)')\n",
"disp('Hence V_1=174*(sin(120)/sin(25))')\n",
"\n",
"V_1=174*(sin(120*%pi/180)/sin(25*%pi/180))\n",
"printf(' V_1 = %0.2f ft/s',V_1)\n",
"\n",
"alpha_1=(55*%pi)/180//radians\n",
"V_1=356.5//rounded off\n",
"V_a=V_1*cos(alpha_1)\n",
"printf('\n Thus V_a= %0.1f ft/s',V_a)\n",
"\n",
"rho=0.095\n",
"V_a=204.5//rounded off\n",
"A=0.322//rounded off\n",
"m=rho*V_a*A\n",
"printf('\n m= %0.3f lb/s',m)\n",
"\n",
"disp('From delta_E=U_m*V_u1 we have P_s=m*delta_E')\n",
"m=6.25/32.2//in lbf\n",
"U_m=174//rounded off\n",
"V_u1=356.5*sin((55*%pi)/180)\n",
"delta_E=U_m*V_u1\n",
"P_s=m*delta_E\n",
"printf(' P_s is equal to %0.1f ft-lbf/s',P_s)\n",
"disp('On converting we get P_s = 17.9hp')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.1: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_1E,page 43.\n",
"N=800\n",
"Q=1750//in gallon per minute\n",
"r_1=3\n",
"b_1=4\n",
"r_2=9\n",
"b_2=3\n",
"\n",
"omega=N*%pi/30\n",
"printf('\n The angular velocity is %g rad/s',omega)\n",
"omega_r=83.7\n",
"disp('After rounding off the value of angular velocity is 83.7 rad/s')\n",
"\n",
"\n",
"U_1=omega_r*r_1/12\n",
"printf(' U_1=%g',U_1)\n",
"disp('After rounding off the value of U_1 is 20.9 ft/s')\n",
"U_1r=20.9//rounded value of U1\n",
"\n",
"\n",
"U_2=omega_r*r_2/12\n",
"printf(' U_2=%g',U_2)\n",
"disp('After rounding off the value of U_2 is 62.7 ft/s')\n",
"\n",
"\n",
"A_1=2*%pi*r_1*b_1/144\n",
"printf(' A_1=%g ft^2',A_1)\n",
"disp('After rounding off the value of A_1 is 0.523 ft^2')\n",
"\n",
"A_2=2*%pi*r_2*b_2/144\n",
"printf(' A_2=%g ft^2',A_2)\n",
"disp('After rounding off the value of A_2 is 1.18 ft^2')\n",
"A_1r=0.523//rounded off\n",
"A_2r=1.18//rounded off\n",
"\n",
"\n",
"V_r1=(Q*0.00223)/(A_1r)\n",
"printf(' The value of V_r1 is %g',V_r1)\n",
"disp('The value of V_r1 after rounding off is 7.47 ft/s')\n",
"\n",
"V_r2=(Q*0.00223)/(A_2r)\n",
"printf(' The value of V_r2 is %g',V_r2)\n",
"disp('The value of V_r2 after rounding off is 3.27 ft/s')//actual value is 3.30,however the value given in the book is 3.27 ft/s\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.2: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_1S,page 43.\n",
"N=800\n",
"Q=397//in meter cube per hour\n",
"r_1=7.6\n",
"b_1=10.2\n",
"r_2=22.9\n",
"b_2=7.6\n",
"omega=N*%pi/30\n",
"printf('The angular velocity is %g rad/s',omega)\n",
"omegar=83.7\n",
"disp('After rounding off the value of angular velocity is 83.7 rad/s')\n",
"U_1=omega*r_1/100\n",
"printf(' U_1=%g',U_1)\n",
"disp('After rounding off the value of U_1 is 6.36 m/s')\n",
"U_1r=6.36//rounded value of U1\n",
"U_2=omega*r_2/100\n",
"printf(' U_2=%g',U_2)\n",
"disp('After rounding off the value of U_2 is 19.2 m/s')\n",
"A_1=2*%pi*r_1*b_1\n",
"printf(' A_1=%g cm^2',A_1)\n",
"disp('After rounding off the value of A_1 is 487 cm^2')\n",
"A_2=2*%pi*r_2*b_2\n",
"printf(' A_2=%g cm^2',A_2)\n",
"disp('After rounding off the value of A_2 is 1093.5 cm^2')\n",
"A_1r=487//rounded off\n",
"A_2r=1093.5//rounded off\n",
"V_r1=(Q/3600)/(A_1r/10000)\n",
"printf(' The value of V_r1 is %g',V_r1)\n",
"disp('The value of V_r1 after rounding off is 2.26 m/s')\n",
"V_r2=(Q/3600)/(A_2r/10000)\n",
"printf(' The value of V_r2 is %g',V_r2)\n",
"disp('The value of V_r2 after rounding off is 1.01 m/s')\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.3: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_2E,page 44.\n",
"\n",
"V_r1=7.47\n",
"U_1=20.9\n",
"V_r2=3.27\n",
"U_2=62.7\n",
"\n",
"//let x=tanbeta1\n",
"x=V_r1/U_1\n",
"printf('\n The value of ß_f1 is equal to %0.3f degrees',x)\n",
"beta_f1=(atan(x))*180/%pi\n",
"printf('\n Thus the value of ß_f1 is %0.2f degrees',beta_f1)\n",
"\n",
"V_1=V_r1\n",
"W_1=(U_1^2+V_r1^2)^0.5\n",
"printf('\n Thus the value of W_1 is %0.2f ft/s',W_1)\n",
"\n",
"beta_f2=beta_f1-10\n",
"printf('\n Hence the value of beta_f2 is equal to %0.2f degrees',beta_f2)\n",
"\n",
"//rounding of value of betaf2 to be equal to 9.6\n",
"beta_f2=9.6\n",
"W_u2=V_r2/tan(beta_f2*%pi/180)\n",
"printf('\n Hence the value of W_u2 is %0.1f ft/s',W_u2)\n",
"\n",
"V_u2=U_2-W_u2\n",
"printf('\n Hence the value of V_u2 is equal to %0.1f ft/s',V_u2)\n",
"\n",
"\n",
"//rounding off Wu2\n",
"W_u2=19.3\n",
"W_2=(W_u2^2+V_r2^2)^0.5\n",
"printf('\n The value of W_u2 is equal to %0.3f ft/s',W_2)\n",
"\n",
"//rounding off Vu2\n",
"V_u2=43.4\n",
"V_2=(V_u2^2+V_r2^2)^0.5\n",
"printf('\n Thus he value of V_2 is equal to %0.2f ft/s',V_2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.4: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_2S,page 44.\n",
"\n",
"V_r1=2.26\n",
"U_1=6.36\n",
"V_r2=1.01\n",
"U_2=19.2\n",
"\n",
"//let x=tan(beta_1)\n",
"x=V_r1/U_1\n",
"printf('\n The value of ß_f1 is equal to %0.3f degrees',x)\n",
"beta_f1=(atan(x))*180/%pi\n",
"printf('\n Thus the value of ß_f1 is %0.1f degrees',beta_f1)\n",
"\n",
"V_1=V_r1\n",
"W_1=(U_1^2+V_r1^2)^0.5\n",
"printf('\n Thus the value of W_1 is %0.2f m/s',W_1)\n",
"\n",
"beta_f2=beta_f1-10\n",
"printf('\n Hence the value of ß_f2 is equal to %0.1f degrees',beta_f2)\n",
"\n",
"//rounding of value of betaf2 to be equal to 9.6\n",
"beta_f2=9.6\n",
"W_u2=V_r2/tan(beta_f2*%pi/180)\n",
"printf('\n Hence the value of W_u2 is %0.2f m/s',W_u2)\n",
"\n",
"V_u2=U_2-W_u2\n",
"printf('\n Hence the value of V_u2 is equal to %0.2f m/s',V_u2)\n",
"\n",
"\n",
"//rounding off W_u2\n",
"W_u2=5.97\n",
"W_2=(W_u2^2+V_r2^2)^0.5\n",
"printf('\n The value of W_2 is equal to %0.3f m/s',W_2)\n",
"\n",
"//rounding off V_u2\n",
"V_u2=13.23\n",
"V_2=(V_u2^2+V_r2^2)^0.5\n",
"printf('\n Thus he value of V_2 is equal to %0.2f m/s',V_2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.5: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_3E,page 46.\n",
"//the value given in the book for Um is 200.5,but on calculating the value comes out to be 200.3\n",
"\n",
"r_t=24\n",
"r_h=10\n",
"N=1250\n",
"Q=53000\n",
"\n",
"r_m=(0.5*(r_t^2+r_h^2))^0.5\n",
"printf('\n The mean radius rm is equal to %0.1f in',r_m)\n",
"disp('Converting to feet we have r_m equal to 1.53 ft')\n",
"\n",
"\n",
"//rm in feet equals 1.53\n",
"r_m=1.53\n",
"U_m=((N*%pi)/30)*r_m\n",
"printf(' Um = %0.1f ft/s',U_m)//the value given in the book for Um is 200.5,but on calculating the value comes out to be 200.3\n",
"\n",
"A=(%pi*(r_t^2-r_h^2))/144\n",
"printf('\n The value of A is %0.1f ft^2',A)\n",
"\n",
"//rounding off A to be 10.4\n",
"A=10.4\n",
"V_a2=Q/(A*60)\n",
"V_a1=V_a2\n",
"printf('\n V_a1=V_a2 %0.1f ft/s',V_a2)\n",
"\n",
"beta_f1=(atan(U_m/V_a1))*180/%pi\n",
"beta_b1=beta_f1\n",
"printf('\n beta_b1=beta_f1 %0.1f degrees',beta_b1)\n",
"\n",
"\n",
"rho_w=62.4\n",
"rho_a=0.0762\n",
"g=32.2\n",
"H_w=2/12\n",
"disp('We know that g*Ha=(rho_w/rho_a)*g*H_w=U_mV_u2=Um*(U_m-V_a2*tan(beta_f2))')\n",
"disp('Hence we can find out the value of tan(ß_f2).Thus we can determine value of  ß_f2.')\n",
"//Let n=U_m*(U_m-V_a2*tan(beta_f2)) and m=(rho_w/rho_a)*g*H_w\n",
"m=(rho_w/rho_a)*g*H_w\n",
"//therefore tan_betaf2=(U_m-m/U_m)/V_a2\n",
"tanbeta_f2=(U_m-m/U_m)/V_a2\n",
"beta_f2=(atan(tanbeta_f2))*180/%pi\n",
"printf('\n Thus the value of ß_f2 is equal to %0.1f degrees',beta_f2)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.6: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_3S,page 46.\n",
"\n",
"r_t=0.6\n",
"r_h=0.25\n",
"N=1250\n",
"Q=1500\n",
"\n",
"r_m=(0.5*(r_t^2+r_h^2))^0.5\n",
"printf('\n The mean radius rm is equal to %0.2f in. ',r_m)\n",
"\n",
"U_m=((N*%pi)/30)*r_m\n",
"printf(' On converting,U_m = %0.1f m/s',U_m)\n",
"\n",
"A=(%pi*(r_t^2-r_h^2))\n",
"printf('\n The value of A is %0.4f m^2',A)\n",
"\n",
"\n",
"V_a2=Q/(A*60)\n",
"V_a1=V_a2\n",
"printf('\n V_a1=V_a2 %0.1f m/s',V_a2)\n",
"\n",
"beta_f1=(atan(U_m/V_a1))*180/%pi\n",
"beta_b1=beta_f1\n",
"printf('\n beta_b1=beta_f1 %0.1f degrees',beta_b1)\n",
"\n",
"\n",
"rho_w=998\n",
"rho_a=1.22\n",
"g=9.8\n",
"H_w=5/100\n",
"U_m=60.2//rounding off Um\n",
"disp('We know that g*H_a=(rho_w/rho_a)*g*H_w=U_mV_u2=U_m*(U_m-V_a2*tan(beta_f2))')\n",
"disp('Hence we can find out the value of tanbet_f2')\n",
"//Let n=U_m*(U_m-V_a2*tan(beta_f2)) and m=(rho_w/rho_a)*g*H_w\n",
"m=(rho_w/rho_a)*g*H_w\n",
"//therefore tanbeta_f2=(U_m-m/U_m)/V_a2\n",
"tanbeta_f2=(U_m-m/U_m)/V_a2\n",
"beta_f2=(atan(tanbeta_f2))*180/%pi\n",
"printf('\n Thus the value of beta_f2 is equal to %0.1f degrees',beta_f2)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.7: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_4E,page 49.\n",
"\n",
"//the value of deltaE2 slightly differs from the value given in the book. However the calculated value is correct for te given substitutions\n",
"\n",
"N=120\n",
"r_1=1.8\n",
"b_1=0.3\n",
"\n",
"omega=(N*%pi)/30\n",
"printf('\n Omega is equal to %0.3f rad/s',omega)\n",
"\n",
"U_1=r_1*omega\n",
"printf('\n U_1 is equal to %0.1f ft/s',U_1)\n",
"\n",
"r_m2=(0.5*(1^2+0.4^2))^0.5\n",
"//rounding off the value of rm2\n",
"r_m2=0.761\n",
"\n",
"U_2=r_m2*omega\n",
"printf('\n U_2 is equal to %0.2f ft/s',U_2)\n",
"\n",
"A_1=2*%pi*r_1*b_1\n",
"printf('\n A_1 is equal to %0.2f ft^2',A_1)\n",
"\n",
"r_t2=1.0\n",
"r_h2=0.4\n",
"b_2=0.5\n",
"A_2=%pi*(r_t2+r_h2)*b_2\n",
"printf('\n A_2 is equal to %0.2f ft^2',A_2)\n",
"\n",
"disp('Assume swirl free flow at discharge that is V_u2=0 and ß_f1=ß_b1, ß_f2=ß_b2.')\n",
"\n",
"V_m2=U_2*tan(15*(%pi/180))\n",
"printf('\n So V_m2=U_2tan15 is equal to %0.2f ft/s',V_m2)\n",
"disp('Thus now we can determine Q')\n",
"\n",
"V_m2=2.56//rounding off\n",
"Q=V_m2*A_2\n",
"disp('Q=V_m2*A_2=V_r1*A_1')\n",
"printf('\n Thus Q is equal to %0.2f ft^3/s',Q)\n",
"\n",
"disp('Since the value of Q,A_1 is known,we can determine the value of V_r1')\n",
"V_r1=Q/A_1\n",
"printf(' The value of V_r1 is equal to %0.2f ft/s',V_r1)\n",
"\n",
"beta_f1=27\n",
"W_u1=V_r1/tan((beta_f1*%pi)/180)\n",
"printf('\n W_u1 is equal to %0.2f ft/s',W_u1)\n",
"\n",
"V_u1=U_1-W_u1\n",
"printf('\n The value of V_u1 is %0.2f ft/s',V_u1)\n",
"\n",
"U_1=22.6//rounding off\n",
"V_u1=19.3//rounding off\n",
"delta_Et=U_1*V_u1\n",
"printf('\n delta_Et is equal to %0.1f ft^2/s^2',delta_Et)//the value given in the book is 437.1,but the actual value is as calculated for the given values of U1 and Vu1\n",
"\n",
"m=(62.4/32.2)*5.63\n",
"P_s=m*delta_Et\n",
"printf('\n Thus P_s is equal to %0.1f ft*lbf/s',P_s)//since value of deltaEt differs from the one given in the book,so does the value of Ps\n",
"disp('Converting P_s to hp we get 8.65hp')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.8: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_4S,page 50.\n",
"//the value of deltaE2 slightly differs from the value given in the book. However the calculated value is correct for te given substitutions\n",
"\n",
"N=120\n",
"r_1=54.8\n",
"b_1=9.1\n",
"\n",
"omega=(N*%pi)/30\n",
"printf('\n Omega is equal to %0.3f rad/s',omega)\n",
"\n",
"U_1=r_1*omega/100\n",
"printf('\n U_1 is equal to %0.1f m/s',U_1)\n",
"\n",
"r_m2=(0.5*(30.5^2+ 12.2^2))^0.5\n",
"\n",
"U_2=r_m2*omega/100\n",
"printf('\n U_2 is equal to %0.2f m/s',U_2)\n",
"\n",
"A_1=2*%pi*r_1*b_1/10000\n",
"printf('\n A1 is equal to %0.3f m^2',A_1)\n",
"\n",
"r_t2=30.5\n",
"r_h2=12.2\n",
"b_2=15.2\n",
"A_2=%pi*(r_t2+r_h2)*b_2/10000\n",
"printf('\n A_2 is equal to %0.3f m^2',A_2)\n",
"\n",
"disp('Assume swirl free flow at discharge that is V_u2=0 and ß_f1=ß_b1, ß_f2=ß_b2.')\n",
"\n",
"V_m2=U_2*tan(15*(%pi/180))\n",
"printf('\n So V_m2=U_2tan15 is equal to %0.3f m/s',V_m2)\n",
"disp('Thus now we can determine Q')\n",
"\n",
"Q=V_m2*A_2\n",
"disp('Q=V_m2*A_2=V_r1*A_1')\n",
"printf('\n Thus Q is equal to %0.3f m^3/s',Q)\n",
"\n",
"disp('Since the value of Q,A_1 is known,we can determine the value of V_r1')\n",
"Q=0.159\n",
"V_r1=Q/A_1\n",
"printf(' The value of V_r1 is equal to %0.4f m/s',V_r1)//actual values are taken,hence a 0.0005 difference in answer is observed\n",
"\n",
"beta_f1=27//ßf1\n",
"V_r1=0.508//rounding off Vr1\n",
"W_u1=V_r1/tan((beta_f1*%pi)/180)\n",
"printf('\n W_u1 is equal to %0.4f m/s',W_u1)\n",
"\n",
"W_u1=0.997//rounding off\n",
"U_1=6.9//rounding off\n",
"V_u1=U_1-W_u1\n",
"printf('\n The value of V_u1 is %0.2f m/s',V_u1)//0.02 difference obtained because of substituting the values as they ahve been found out\n",
"\n",
"\n",
"V_u1=5.92//as substituted in the book\n",
"delta_Et=U_1*V_u1\n",
"printf('\n deltaEt is equal to %0.2f m^2/s^2',delta_Et)//\n",
"\n",
"m=998*0.782*0.204\n",
"P_s=m*delta_Et\n",
"printf('\n Thus P_s is equal to %0.0f W',P_s)\n",
"disp('Converting P_s to kW we get 6.503kW')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.9: ETT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//This numerical is Ex 3_5E,page 52.\n",
"\n",
"//this numerical is based on numerical 3.4E\n",
"//values found in the book for numerical 3.4E will be used to solve this numerical(3.5E)\n",
"\n",
"delta_Et=437.1\n",
"U_1=22.6\n",
"U_2=9.56\n",
"V_m2=2.56\n",
"V_2=2.56\n",
"\n",
"V_r1=1.66\n",
"W_u1=3.26\n",
"W_1=(V_r1^2+W_u1^2)^0.5\n",
"printf('\n W_1 is equal to %0.2f ft/s',W_1)\n",
"\n",
"W_2=(U_2^2+V_m2^2)^0.5\n",
"printf('\n W_2 is equal to %0.2f ft/s',W_2)\n",
"\n",
"V_u1=19.3\n",
"V_1=(V_r1^2+V_u1^2)^0.5\n",
"printf('\n V_1 is equal to %0.2f ft/s',V_1)\n",
"\n",
"Rt=0.5*[(U_1^2-U_2^2)+(W_2^2-W_1^2)]/(delta_Et)\n",
"printf('\n Thus the value Rt is equal to %0.3f',Rt)"
   ]
   }
],
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		   "display_name": "Scilab",
		   "language": "scilab",
		   "name": "scilab"
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		   "file_extension": ".sce",
		   "help_links": [
			{
			 "text": "MetaKernel Magics",
			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
			}
		   ],
		   "mimetype": "text/x-octave",
		   "name": "scilab",
		   "version": "0.7.1"
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