{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 10: Hydroelecric power plant"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.10: Blade_angles.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"gh=35//Gross head in m\n",
"md=2//Mean diameter in m\n",
"N=145//Speed in rpm\n",
"a=30//Angle in degrees\n",
"oa=28//Outlet angle in degrees\n",
"x=7//Percentage of gross head lost\n",
"y=8//Reduction in relative velocity in percent\n",
"\n",
"//Calculations\n",
"H=((100-x)/100)*gh//Net haed in m\n",
"V1=sqrt(2*9.81*H)//Velocity in m/s\n",
"Vb=(3.14*md*N)/60//Velocity in m/s\n",
"b1=atand((V1*sind(a))/((V1*cosd(a))-Vb))//Angle in degrees\n",
"Vr1=((V1*sind(a))/sind(b1))//Velocity in m/s\n",
"Vr2=((100-y)/100)*Vr1//Velocity in m/s\n",
"Vw1=(V1*cosd(a))//Velocity in m/s\n",
"Vw2=(Vb-(Vr2*cosd(oa)))//Velocity in m/s\n",
"E=((Vb*(Vw1-Vw2))/9.81)//Workdone in kg.m/kg\n",
"nb=(E/gh)*100//Hydraulic efficiency in percent\n",
"\n",
"//Output\n",
"printf('Blade angle at inlet is %3.0f degrees \n Hydraulic efficiency is %3.0f percent',b1,nb)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.11: Speed_and_diameter.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"P=10000//Power in kW\n",
"h=12//Head in m\n",
"Nr=2//Speed ratio\n",
"Fr=0.65//Flow ratio\n",
"x=0.3//Diameter of hub is 0.3 times the eternal diameter of the vane \n",
"on=94//Overall efficiency in percent\n",
"\n",
"//Calculations\n",
"Q=(P/(9.81*h*(on/100)))//Discharge in m^3/s\n",
"Vr1=(Fr*sqrt(2*9.81*h))//Velocity in m/s\n",
"Ab=(Q/Vr1)//Area of flow in m^2\n",
"D=sqrt(((Ab*4)/3.14)/(1-x^2))//Diameter of runner in m\n",
"Vb=(Nr*sqrt(2*9.81*h))//Velocity in m/s\n",
"N=((Vb*60)/(3.14*D))//Speed in rpm\n",
"f=50//Taking frequency as 50 Hz\n",
"p=(120*50)/N//Number of poles\n",
"N1=(120*f)/int(p)//Speed in rpm\n",
"Ns=(N1*sqrt(P))/h^(5/4)//Specific speed\n",
"\n",
"//Output\n",
"printf('(a) the speed is %3.1f rpm \n (b) the diameter of the runner is %3.2f m \n (c) the specific speed is %3.0f',N1,D,Ns)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.12: Specific_speed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"P=10000//Power in kW\n",
"h=25//Head in m. In textbook it is given wrong as 2 m\n",
"N=135//Speed in rpm\n",
"h1=20//Head in m\n",
"\n",
"//Calculations\n",
"Ns=((N*sqrt(P))/h^(5/4))//Specific speed\n",
"N1=sqrt(h1/h)*N//Speed in rpm\n",
"P2=P/(h/h1)^(3/2)//Power in kW\n",
"\n",
"//Output\n",
"printf('Specific speed is %3.1f \n Normal speed is %3.1f rpm \n Output under a head of %i m is %3.0f kW',Ns,N1,h1,P2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.13: Number_of_turbines.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"Q=175//Discharge in m^3/s\n",
"h=18//Head in meter\n",
"N=150//Speed in rpm\n",
"oe=82//Overall efficiency in percent\n",
"Ns1=460//Maximum specific speed\n",
"Ns2=350//Maximum specific speed\n",
"d=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"P=(d*Q*9.81*h*(oe/100)*10^-3)//power in kW\n",
"P1=((Ns1*h^(5/4))/N)^2//Power in kW\n",
"n1=P/P1//No.of turbains\n",
"P2=((Ns2*h^(5/4))/N)^2//Power in kW\n",
"n2=ceil(P/P2)//No.of turbains\n",
"\n",
"//Output\n",
"printf('The number of turbines in \n (a) Francis turbine are%3.0f \n (b) Kaplan turbine are %i',n1,n2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.14: Speed_power_and_scale_ratio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"Ns=210//Specific speed \n",
"P=30//Power in MW\n",
"N=180//Speed in rpm\n",
"Q=0.6//Discharge in m^3/s\n",
"h=4.5//Head in m\n",
"e=88//Efficiency in percent\n",
"d=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"Pm=(d*Q*9.81*h*(e/100)*10^-3)//Power in kW\n",
"Nm=(Ns*h^(5/4))/sqrt(Pm)//Speed in rpm\n",
"Hp=((N*sqrt(P*1000))/Ns)^(4/5)//Head in m\n",
"Dpm=(Nm/N)*sqrt(Hp/h)//Scale ratio\n",
"Qp=(P*10^6)/(d*9.81*Hp*(e/100))//Discharge in m^3/s\n",
"\n",
"//Output\n",
"printf('Speed is %3.0f rpm \n Power is %3.3f kW \n Scale ratio is %3.3f \n Flow through the turbine is %3.1f m^3/s',Nm,Pm,Dpm,Qp)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.15: Speed_and_power.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"x=1/5//Scale model\n",
"h=1.5//Head in m\n",
"P=5//Power in kW\n",
"N=450//Speed in rpm\n",
"h1=30//Head in m\n",
"\n",
"//Calculations\n",
"N1=(x*N)/sqrt(h/h1)//Speed in rpm\n",
"Ns=(N*sqrt(P))/h^(5/4)//Specific speed\n",
"P1=((Ns*h1^(5/4))/N1)^2//Power in kW\n",
"\n",
"//Output\n",
"printf('Speed is %3.0f rpm \n Power is %3.0f kW',N1,P1)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.16: Efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"h=19//Head in m\n",
"Q=3//Flow rate in m^3/s\n",
"N=600//Speed in rpm\n",
"h1=5//Head in m\n",
"\n",
"//Calculations\n",
"N1=N/sqrt(h/h1)//Speed in rpm\n",
"Q1=Q/sqrt(h/h1)//Discharge in m^3/s\n",
"\n",
"//Output\n",
"printf('Speed of the turbine is %3.1f rpm \n Maximum flow rate is %3.1f m^3/s',N1,Q1)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.17: Least_number_of_turbines.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"Q=350//Discharge in m^3/s\n",
"h=30//Head in m\n",
"e=87//Turbine efficiency in percent\n",
"f=50//Frequency in Hz\n",
"p=24//Number of poles\n",
"Ns1=300//Specific speed\n",
"Ns2=820//Specific speed\n",
"d=1000//Dnsity of water in kg/m^3\n",
"\n",
"//Calculations\n",
"N=(120*f)/p//Speed in rpm\n",
"P=d*Q*9.81*h*(e/100)*10^-3//Power in kW\n",
"P1=((Ns1*h^(5/4))/N)^2//Power in kW\n",
"n1=P/P1//No.of turbines\n",
"P2=((Ns2*h^(5/4))/N)^2//Power in kW\n",
"n2=ceil(P/P2)//No.of turbines\n",
"\n",
"//Output\n",
"printf('Least number of machines required if using \n (a) Francis turbines are %3.0f \n (b) Kaplan turbines are %3.0f',n1,n2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.18: Power_developed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"h=27//Head in m\n",
"A=430//Area in sq.km\n",
"R=150//Rainfall in cm/year\n",
"pr=65//Percentage of rainfall utilised\n",
"pe=95//Penstock efficiency in percent\n",
"te=80//Turbine efficiency in percent\n",
"ge=86//Generator efficiency in percent\n",
"lf=0.45//Load factor\n",
"d=1000//Density of water in kg/m^3\n",
"\n",
"//Calculations\n",
"Q=A*10^6*(R/100)*(pr/100)//Discharge in m^3 per year\n",
"Qs=(Q/(365*24*3600))//Quantity of water per second in m^3\n",
"P=(pe/100)*(te/100)*(ge/100)*d*Qs*9.81*h*10^-3//Power in kW\n",
"plc=(P/lf)//Peak load capacity in kW\n",
"C=(plc/(2*(ge/100)))//Capacity of each unit in kW\n",
"\n",
"//Output\n",
"printf('(a) Power developed is %3.0f kW \n (b) As the available head is low, Kaplan turbines are suggested.\n Two turbines each of 3000kW capacity may be installed.',P)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.19: Power_developed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"q=[30,25,20,0,010,50,80,100,110,65,45,30]//Mean discharge in millions of cu.m per month respectively\n",
"h=90//Head in m\n",
"n=86//Overall efficiency in percent\n",
"\n",
"//Calculations\n",
"Qm=(q(1)+q(2)+q(3)+q(4)+q(5)+q(6)+q(7)+q(8)+q(9)+q(10)+q(11)+q(12))/12//Mean discharge in millions m^3/s\n",
"Q=[30,30,25,25,20,20,0,0,10,10,50,50,80,80,100,100,110,110,65,65,45,45,30,30,0]//Discharge(million m^3/month) on y-axis\n",
"y=[0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12]//Months on x-ais\n",
"D=[110,100,90,80,70,60,50,40,30,25,20,10,0]//Discharge per month in million m^3\n",
"pt=[8.3,16.7,25,25,25,33.3,41.7,50,66.7,75,83.3,91.7,100]//Percentage time \n",
"Po=((Qm*10^6*9.81*h*(n/100))/(30*24*3600*1000))//Power developed in MW\n",
"\n",
"//Output\n",
"subplot(121)\n",
"plot(y,Q)//Graph Discharge(million m^3/month) vs Month\n",
"xtitle('Discharge(million m^3/month) vs Month','Months','Discharge(million m^3/month)')\n",
"subplot(122)\n",
"plot(pt,D)//Graph percentage time vs Discharge(million m^3/month)\n",
"xtitle('percentage time vs Discharge(million m^3/month)','percentage time','Discharge(million m^3/month)')\n",
"printf('Power developed is %3.2f MW',Po)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.1: Efficiency_of_the_runner.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"P=4000//Power in kW\n",
"N=400//Speed in r.p.m\n",
"h=200//Head in m\n",
"e=90//Efficiency in percent\n",
"d=1.5//Diameter in m\n",
"vd=10//Percentage decrease in velocity\n",
"a=165//Angle with which jet is deflected in degrees\n",
"\n",
"//Calculations\n",
"V1=sqrt(2*9.81*h*(e/100))//Velocity in m/s\n",
"Vb=(3.14*d*N)/60//Velocity in m/s\n",
"nn=((2*(1-((e/100)*cosd(a)))*(V1-Vb)*Vb)/V1^2)*100//Efficiency in percent\n",
"p=(P/(nn/100))//Power developed in kW\n",
"pj=(p/2)//Power developed per jet in kW\n",
"dx=sqrt((pj*8)/(3.14*V1^3))//Diameter of each jet in m\n",
"\n",
"//Output\n",
"printf('(a) the efficiency of the runner is %3.2f percent \n (b) the diameter of each jet is %3.4f m',nn,dx)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.2: Number_of_jets.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"P=6000//Power in kW\n",
"h=300//Net head availabe in m\n",
"N=550//Speed in r.p.m\n",
"rd=(1/10)//Ratio of jet diameter to wheel diameter\n",
"nh=0.85//Hydraulic efficiency \n",
"Cv=0.98//Coefficient of velocity\n",
"f=0.46//Speed ratio\n",
"d=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"V1=Cv*sqrt(2*9.81*h)//Velocity in m/s\n",
"Vb=f*sqrt(2*9.81*h)//Velocity in m/s\n",
"Q=((P*10^3)/(nh*d*9.81*h))//Discharge in m^3/s\n",
"D=((Vb*60)/(3.14*N))//Diameter in m\n",
"d=(D/10)//Diameter of jet in m\n",
"n=(Q/((V1*(3.14/4)*d^2)))//Number of jets\n",
"\n",
"//Output\n",
"printf('(a) the number of jets are%3.0f \n (b) diameter of each jet is %3.3f m \n (c) diameter of the wheel is %3.2f m \n (d) the quantity of water required is %3.1f m^3/s',n,d,D,Q)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.3: Diameter_of_jet.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"P=10//Capacity in MW\n",
"h=500//Head in m\n",
"Ns=10//Specific speed of the turbine\n",
"on=80//Overall efficiency in percent\n",
"Cv=0.98//Coefficient of velocity\n",
"x=0.46//Speed of the bucket wheel to the velocity of jet\n",
"da=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"N=(Ns*h^(5/4))/sqrt(P*10^3)//Speed in r.p.m\n",
"V=(Cv*sqrt(2*9.81*h))//Velocity in m/s\n",
"Vb=(x*V)//Speed of bucket wheel in m/s\n",
"D=((60*Vb)/(3.14*N))//Diameter in m\n",
"d=sqrt((P*10^6)/((on/100)*(3.14/4)*da*V*9.81*h))//Diameter in m\n",
"\n",
"//Output\n",
"printf('Diameter of jet is %3.3f m \n Diameter of bucket wheel is %3.2f m',d,D)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.4: Specific_speed.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"Cv=0.97//Coefficient of velocity\n",
"f=0.45//Friction coefficient\n",
"h=0.85//Head in m\n",
"d=1000//Density in kg/m^3\n",
"n=1//For a single jet turbine\n",
"\n",
"//Calculations\n",
"Ns=((60/3.14)*(f*sqrt(2*9.8))*sqrt(n*(3.14/4)*Cv*sqrt(2*9.8)*9.8*h))//Specific speed in terms of d/D \n",
"\n",
"//Output\n",
"printf('The specific speed of a single jet Pelton wheel is about %3.0f (d/D) where d and D represent the jet and bucket wheel diameters respectively',Ns)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.5: Velocity_of_jet.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"n=4//Number of jets\n",
"d=60//Diameter of each jet in mm\n",
"a=165//Angle in degrees\n",
"v=45//Speed of the bucket wheel in m/s\n",
"de=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"v1=(2*v)//Jet velocity in m/s\n",
"Q=(3.14/4)*(d/1000)^2*v1//Discharge in m^3/s\n",
"P=(1-cosd(a))*(v1^2/4)*Q*de*10^-3//Power developed in kW\n",
"P4=(P*4)//For four jets in kW\n",
"nd=((1-cosd(a))/2)*100//Maximum efficiency in percent\n",
"\n",
"//Output\n",
"printf('Velocity of the jet for maximum efficiency is %3.0f m/s \n Power developed is %i kW \n Hydraulic efficiency is %3.1f percent',v1,P4,nd)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.6: Head_on_the_wheel.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"v=20//Peripheral velocity in m/s\n",
"vw=17//Velocity of whirl in m/s\n",
"vr=2//Radial velocity in m/s\n",
"Q=0.7//Flow in m^3/s\n",
"hn=80//Hydraulic efficiency in percent\n",
"d=1000//Density in kg/m^3\n",
"\n",
"//Calculations\n",
"H=((vw*v)/(9.81*(hn/100)))//Head on the wheel in m\n",
"P=(d*Q*9.81*H*(hn/100)*10^-3)//Power generated in kW\n",
"al=180-atand(vr/vw)//Angle of guide vanes in degrees\n",
"bl=atand(vr/(v-vw))//Inlet blade angle in degrees\n",
"\n",
"//Output\n",
"printf('Head on the wheel is %3.1f m \n The power generated by the turbine is %3.0f kW \n Eit angle of guide vanes is %3.2f degrees and Inlet blade angle is %3.1f degrees',H,P,al,bl)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.7: Outlet_and_inlet_blade_angles.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"od=1.5//Outer diameter in m\n",
"id=0.75//Inner diameter in m\n",
"h=150//Head in m\n",
"P=14000//Power in kW\n",
"Ns=120//Specific speed \n",
"vw2=0//Velocity in m/s\n",
"a=(11+(20/60))//Angle in degrees\n",
"hn=92//Hydraulic efficiency in percent\n",
"\n",
"//Calculations\n",
"N=(Ns*h^(5/4))/sqrt(P)//Speed in rpm\n",
"vb1=(3.14*od*N)/60//velocity in m/s\n",
"vw1=(((hn/100)*9.81*h)/vb1)//velocity in m/s\n",
"vf1=(tand(a)*vw1)//Velocity in m/s\n",
"vf2=vf1//Velocity in m/s\n",
"b1=atand(vf1/(vb1-vw1))//Angle in degrees\n",
"b1x=(b1-int(b1))*60//For output\n",
"vb2=(vb1/2)//Velocity in m/s\n",
"b2=atand(vf1/(vb2-vw2))//Angle in degrees\n",
"b2x=(b2-int(b2))*60//For output\n",
"\n",
"//Output\n",
"printf('Inlet blade angle is %3.0f degrees %3.0f minutes \n Outlet blade angle is %3.0f degrees %3.0f minute',b1,b1x,b2,b2x)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.8: The_guide_vane_angle.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"h=70//net head in m\n",
"N=700//speed in rpm\n",
"o=85//over all efficiency in %\n",
"P=350//shaft power in kW\n",
"he=92//hydraulic efficiency in %\n",
"fr=.22//flow ratio\n",
"b=.1//breadth ratio\n",
"s=2//outer diameter in terms of inner diametre\n",
"//Calculations\n",
"vf1=fr*sqrt(2*9.81*h)//velocity in m/s\n",
"q=(P/(9.81*h*(o/100)))//discharge in m^3/s\n",
"d1=sqrt(q/(.94*b*vf1*3.14))//diameter in metre\n",
"b1=d1*b//breadth in metre\n",
"d2=d1/2//diametre in metre\n",
"vb1=(3.14*d1*N)/60//velocity in m/s\n",
"vw1=((he/100)*9.81*h)/vb1//velcity in m/s\n",
"a=atand(vf1/vw1)//angle in degrees\n",
"bet=atand(vf1/(vw1-vb1))//angle in degrees\n",
"vb2=(d2/d1)*vb1//velocity in m/s\n",
"bet2=atand(vf1/vb2)//angle in degrees\n",
"\n",
"//Output\n",
"printf('(a)the guide vane angle is %3.1f degrees \n (b)the runner vane angle at inlet is %3.1f degrees and outlet is %3.2f degrees \n (c)the diametres of the runner at inlet is %3.1f metre and outlet is %3.2f metre\n (d)the width of the wheel at inlet is %3.2f metre',a,bet,bet2,d1,d2,b1)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.9: Discharge_of_turbine.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"n=4//Number of units\n",
"P=70000//Power in kVA\n",
"f=50//Frequency in Hz\n",
"p=10//No.of pair of poles\n",
"h=505//Gross head in m\n",
"tn=94//Transmission efficiency in percent\n",
"po=260//Power in MW\n",
"e=91//Efficiency in percent\n",
"nn=0.98//Nozzle efficiency\n",
"Cv=0.98//Coefficient of velocity\n",
"x=0.48//Vb=0.48 V\n",
"dd=25//Nozzle diameter is 25% bigger than jet diameter\n",
"a=165//Angle of buckets in degrees\n",
"de=99.75//Discharge efficiency in percent\n",
"\n",
"//Calculations\n",
"N=(120*f)/(p*2)//Synchronous speed in r.p.m\n",
"nh=((tn/100)*h)//Net head in m\n",
"pt=(po*10^3)/n//Power developed per turbine in kW\n",
"ip=(pt/(e/100))//Input water power in kW\n",
"Q=(ip/(9.81*nh))//Discharge in m^3/s\n",
"Qj=(Q/n)//Discharge per jet in m^3/s\n",
"V1=Cv*sqrt(2*9.81*nh)//Velocity in m/s\n",
"d=sqrt((4/3.14)*(Qj/V1))//Diameter of jet in m\n",
"nd=(1+(dd/100))*d//Nozzle tip diameter in m\n",
"Vb=(x*V1)//Velocity in m/s\n",
"D=((Vb*60)/(3.14*N))//Pitch circle diameter of the wheel in m\n",
"Ns=((N*sqrt(po*10^3))/nh^(5/4))//Specific speed\n",
"jr=(D/d)//Jet ratio\n",
"nob=(jr/2)+15//Number of buckets\n",
"nobb=ceil(nob)//Rounding off to next integer\n",
"W=((V1-Vb)*(1-(nn*cosd(a)))*Vb)/9.81//Workdone per kg in kg.m/kg\n",
"nth=((W/nh)*de)//Hydraulic efficiency in percent\n",
"\n",
"//Output\n",
"printf('(a) the discharge of the turbine is %3.2f m^3/s \n (b) the jet diameter is %3.3f m \n (c) the nozzle tip diameter is %3.3f m \n (d) the pitch circle diameter of the wheel is %3.2f m \n (e) the specific speed is %3.2f \n (f) the number of buckets on the wheel are %3.0f \n (g) the workdone per kg of water on the wheel is %3.2f kg.m/kg \n (h) the hydraulic efficiency is %3.0f percent',Q,d,nd,D,Ns,nobb,W,nth)"
   ]
   }
],
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		   "language": "scilab",
		   "name": "scilab"
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		   "help_links": [
			{
			 "text": "MetaKernel Magics",
			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
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