{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 12: Non steady flow friction and availability"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.1: Work_done_calculation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"p1=100 //psia\n",
"p2=14.7 //psia\n",
"k=1.4\n",
"T1=700 //R\n",
"R=10.73/29\n",
"V=50\n",
"cv=0.171\n",
"cp=0.24\n",
"R2=1.986/29\n",
"//calculations\n",
"T2=T1/ (p1/p2)^((k-1)/k)\n",
"m1=p1*V/(R*T1)\n",
"m2=p2*V/(R*T2)\n",
"Wrev= cv*(m1*T1 - m2*T2) - (m1-m2)*(T2)*cp\n",
"//results\n",
"printf('Work done in case 1 = %d Btu',Wrev)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.2: Friction_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"p1=100 //psia\n",
"p2=10 //psia\n",
"n=1.3\n",
"T1=800 //R\n",
"cv=0.172\n",
"R=1.986/29\n",
"T0=537 //R\n",
"cp=0.24\n",
"//calculations\n",
"T2=T1*(p2/p1)^((n-1)/n)\n",
"dwir=cv*(T1-T2)\n",
"dwr=R*(T2-T1)/(1-n)\n",
"dq=dwr-dwir\n",
"dI=-T0*(cp*log(T2/T1) - R*log(p2/p1))\n",
"//results\n",
"printf('The friction of the process per pound of air = %.1f Btu/lbm',dq)\n",
"printf('\n Loss of available energy = %.2f Btu/lbm',dI)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.3: Energy_loss_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"ms=10 //lbm\n",
"den=62.3 //lbm/ft^3\n",
"A1=0.0218 //ft^2\n",
"A2=0.00545 //ft^2\n",
"p2=50 //psia\n",
"p1=100 //psia\n",
"gc=32.2 //ft/s^2\n",
"dz=30 //ft\n",
"T0=537 //R\n",
"T1=620 //R\n",
"T2=420 //R\n",
"//calculations\n",
"V1=ms/(A1*den)\n",
"V2=ms/(A2*den)\n",
"df=-144/den*(p2-p1) - (V2^2 -V1^2)/(2*gc) - dz\n",
"dI=-T0/T1 *df\n",
"dI2= -T0/T2 *df\n",
"//results\n",
"printf('Friction = %.1f ft-lbf/lbm',df)\n",
"printf('\n Available energy loss in case a = %.1f ft-lbf/lbm',dI)\n",
"printf('\n Available energy loss in case b = %.1f ft-lbf/lbm',dI2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.4: Pressure_drop_calculatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"r=2.5 //in\n",
"mf=160 //cfm\n",
"rho=1/14\n",
"mu=0.0000121\n",
"v=14 //ft^3/lbm\n",
"g=32.2 //ft/s^2\n",
"z=100 //ft\n",
"//calculations\n",
"A=3.14*(r/12)^2\n",
"V=mf/A /60\n",
"Re=(2*r/12)*V*rho/mu\n",
"disp('From fig 12.4,')\n",
"f=0.0225/4\n",
"dp=4*f*(rho)*(V/v)^2 /(2*g*(2*r/12)) *z\n",
"//dp=2.32\n",
"//results\n",
"printf('Pressure drop = %.2f lbf/ft^2 100 ft',dp)\n",
"disp('The answer in the textbook is wrong. Please use a calculator to verify it.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.5: Mass_rate_calculatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"D=0.0724 //ft\n",
"gc=32.2 //ft/s^2\n",
"rho=1/14\n",
"L=100 //ft\n",
"mu2=1.46*10^(-10)\n",
"dp=2.32\n",
"dia=5 //in\n",
"rho2=48500\n",
"vol=14 //ft^3/lbm\n",
"//calculations\n",
"ref=D^3 *2*dp*gc*rho/(mu2*L)\n",
"mf=rho2*%pi/4 *(dia/12) *sqrt(mu2)\n",
"mfr=mf*vol*60\n",
"//results\n",
"printf('Mass rate of air flow = %d cfm',mfr)\n",
"disp('The answer is a bit different due to rounding off error in textbook')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.6: Loss_and_effectiveness_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"cp=0.25\n",
"T=3460 //R\n",
"T0=520 //R\n",
"dG=1228 //Btu/lbm\n",
"//calculations\n",
"hf=cp*(T-T0)-T0*cp*log(T/T0)\n",
"dC=hf-dG\n",
"Ec=hf/dG\n",
"//results\n",
"printf('Loss of available energy = %d Btu/lbm mixture ',dC)\n",
"printf('\n Effectiveness of combustion = %.3f ',Ec)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.7: Loss_and_effectiveness_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"cp1=0.25\n",
"T=3460 //R\n",
"T0=946.2 //R\n",
"T00=520 //R\n",
"dG=1228 //Btu/lbm\n",
"cp=0.45\n",
"//calculations\n",
"dqa=cp1*(T-T0)\n",
"w=cp*dqa\n",
"hf=cp1*(T-T00)-T00*cp1*log(T/T00)\n",
"heat=w-hf\n",
"eff=w/hf\n",
"epower=w/dG\n",
"//results\n",
"printf('Loss of available energy = %.1f Btu/lbm mixture ',heat)\n",
"printf('\n Efficiency of cycle = %.3f ',eff)\n",
"printf('\n Effectiveness of overall cycle = %.2f',epower)\n",
"disp('The answer is a bit different due to rounding off error in textbook')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.8: Loss_and_effectiveness_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"p1=400 //psia\n",
"t1=600 //F\n",
"h1=1306.9 //Btu/lbm\n",
"b1=480.9 //Btu/lbm\n",
"p2=50 //psia\n",
"h2=1122 //Btu/lbm\n",
"h3=1169.5 //Btu/lbm\n",
"b3=310.9 //Btu/lbm\n",
"//calculations\n",
"disp('All the values are obtained from Mollier chart,')\n",
"dw13=h1-h3\n",
"dw12=h1-h2\n",
"dasf=b3-b1\n",
"etae=dw13/dw12\n",
"eta=abs(dw13/dasf)\n",
"dq=dw13+dasf\n",
"//results\n",
"printf('Engine efficiency = %.1f percent',etae*100)\n",
"printf('\n Effectiveness = %.1f percent',eta*100)\n",
"printf('\n Loss of available energy  = %.1f Btu/lbm',dq)"
   ]
   }
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			 "text": "MetaKernel Magics",
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