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-rw-r--r--[-rwxr-xr-x]Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb436
1 files changed, 221 insertions, 215 deletions
diff --git a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
index 75c1ae52..3e09ba67 100755..100644
--- a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
+++ b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11.ipynb
@@ -1,215 +1,221 @@
-{
- "metadata": {
- "name": "",
- "signature": "sha256:bac11063c240653dfd3c07e3907da1d648418ca108c3c127b610f8e4e00f83ef"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11:Thermodynamic relations Equilibrium and stability"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex11.3:pg-436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "Tb = 353.0 # boiling point of benzene in K\n",
- "T = 303.0 # Operational temperature in K\n",
- "R = 8.3143 #Gas constant\n",
- "P = 101.325*math.exp((88/R)*(1.0-(Tb/T)))\n",
- "\n",
- "print \"\\n Example 11.3\"\n",
- "print \"\\n Vapour pressure of benzene is \",P ,\" kPa\"\n",
- "#The answers vary due to round off error\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Example 11.3\n",
- "\n",
- " Vapour pressure of benzene is 17.6682592008 kPa\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex11.4:pg-436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "T = (3754-3063)/(23.03-19.49) # Temperature at triple point in K\n",
- "P = math.exp(23.03-(3754/195.2)) # Pressure at triple point\n",
- "R = 8.3143 # Gas constant\n",
- "Lsub = R*3754 # Latent heat of sublimation\n",
- "Lvap = 3063*R # Latent heat of vaporisation\n",
- "Lfu = Lsub-Lvap # Latent heat of fusion\n",
- "\n",
- "print \"\\n Example 11.4\"\n",
- "print \"\\n Temperature at triple point is \",T ,\" K\"\n",
- "print \"\\n Pressure at triple point is \",P ,\" mm Hg\"\n",
- "print \"\\n\\n Latent heat of sublimation is \",Lsub ,\" kJ/kg mol\"\n",
- "print \"\\n Latent heat of vapourization is is \",Lvap ,\" kJ/kg mol\"\n",
- "print \"\\n Latent heat of fusion is \",Lfu ,\" kJ/kg mol\"\n",
- "#The answers vary due to round off error\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Example 11.4\n",
- "\n",
- " Temperature at triple point is 195.197740113 K\n",
- "\n",
- " Pressure at triple point is 44.631622076 mm Hg\n",
- "\n",
- "\n",
- " Latent heat of sublimation is 31211.8822 kJ/kg mol\n",
- "\n",
- " Latent heat of vapourization is is 25466.7009 kJ/kg mol\n",
- "\n",
- " Latent heat of fusion is 5745.1813 kJ/kg mol\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex11.6:pg-438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "R = 8.3143 # Gas constant in kJ/kg-mol-K\n",
- "N1 = 0.5 # Mole no. of first system\n",
- "N2 = 0.75 # Mole no. of second system\n",
- "T1 = 200 # Initial temperature of first system in K\n",
- "T2 = 300 # Initial temperature of second system in K\n",
- "v = 0.02 # Total volume in m**3\n",
- "print \"\\n Example 11.6\\n\"\n",
- "Tf = (T2*N2+T1*N1)/(N1+N2)\n",
- "Uf_1 = (3.0/2.0)*(R*N1*Tf)*(10**-3)\n",
- "Uf_2 = (3.0/2.0)*(R*N2*Tf)*(10**-3)\n",
- "pf = (R*Tf*(N1+N2)*(10**-3))/v\n",
- "Vf_1 = R*N1*(10**-3)*Tf/pf\n",
- "Vf_2 = v-Vf_1\n",
- "print \"\\n Energy of first system is \",Uf_1 ,\" kJ,\\n Energy of second system is \",Uf_2 ,\" kJ,\\n Volume of first system is \",Vf_1 ,\" m**3,\\n Volume of second system is \",Vf_2 ,\" m**3,\\n Pressure is \",pf ,\" kN/m**2,\\n Temperature is \",Tf ,\" K.\"\n",
- "#The answers vary due to round off error\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Example 11.6\n",
- "\n",
- "\n",
- " Energy of first system is 1.6212885 kJ,\n",
- " Energy of second system is 2.43193275 kJ,\n",
- " Volume of first system is 0.008 m**3,\n",
- " Volume of second system is 0.012 m**3,\n",
- " Pressure is 135.107375 kN/m**2,\n",
- " Temperature is 260.0 K.\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex11.10:pg-446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "R = 0.082 # Gas constant in litre-atm/gmol-K\n",
- "m = 1.5 # Mass flow rate in kg/s\n",
- "p1 = 1.0 # Pressure in atm\n",
- "t2 = 300.0 # Temperature after compression in K\n",
- "p2 = 400.0 # Pressure after compression in atm\n",
- "Tc = 151.0 # For Argon in K\n",
- "pc = 48.0 # For Argon in atm\n",
- "print \"\\n Example 11.10 \"\n",
- "a = 0.42748*((R*1000)**2)*((Tc)**2)/pc\n",
- "b = 0.08664*(R*1000)*(Tc)/pc\n",
- "# By solving equation v2**2 - 49.24*v2**2 + 335.6*v2 - 43440 = 0\n",
- "v2 = 56.8 # In cm**3/g mol\n",
- "v1 = (R*1000)*(t2)/p1\n",
- "delta_h = -1790 # In J/g mol\n",
- "delta_s = -57 # In J/g mol\n",
- "Q = (t2*delta_s*(10**5)/39.8)/(3600*1000)\n",
- "W = Q - (delta_h*(10**5)/39.8)/(3600*1000)\n",
- "print \"\\n Power required to run the compressor = \",W ,\" kW, \\n The rate at which heat must be removed from the compressor = \",Q ,\" kW\"\n",
- "# Answers vary due to round off error.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Example 11.10 \n",
- "\n",
- " Power required to run the compressor = -10.6853713009 kW, \n",
- " The rate at which heat must be removed from the compressor = -11.9346733668 kW\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 11:Thermodynamic relations Equilibrium and stability"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex11.3:pg-436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Example 11.3\n",
+ "\n",
+ " Vapour pressure of benzene is 17.6682592008 kPa\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "Tb = 353.0 # boiling point of benzene in K\n",
+ "T = 303.0 # Operational temperature in K\n",
+ "R = 8.3143 #Gas constant\n",
+ "P = 101.325*math.exp((88/R)*(1.0-(Tb/T)))\n",
+ "\n",
+ "print \"\\n Example 11.3\"\n",
+ "print \"\\n Vapour pressure of benzene is \",P ,\" kPa\"\n",
+ "#The answers vary due to round off error\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex11.4:pg-436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Example 11.4\n",
+ "\n",
+ " Temperature at triple point is 195.197740113 K\n",
+ "\n",
+ " Pressure at triple point is 44.631622076 mm Hg\n",
+ "\n",
+ "\n",
+ " Latent heat of sublimation is 31211.8822 kJ/kg mol\n",
+ "\n",
+ " Latent heat of vapourization is is 25466.7009 kJ/kg mol\n",
+ "\n",
+ " Latent heat of fusion is 5745.1813 kJ/kg mol\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "T = (3754-3063)/(23.03-19.49) # Temperature at triple point in K\n",
+ "P = math.exp(23.03-(3754/195.2)) # Pressure at triple point\n",
+ "R = 8.3143 # Gas constant\n",
+ "Lsub = R*3754 # Latent heat of sublimation\n",
+ "Lvap = 3063*R # Latent heat of vaporisation\n",
+ "Lfu = Lsub-Lvap # Latent heat of fusion\n",
+ "\n",
+ "print \"\\n Example 11.4\"\n",
+ "print \"\\n Temperature at triple point is \",T ,\" K\"\n",
+ "print \"\\n Pressure at triple point is \",P ,\" mm Hg\"\n",
+ "print \"\\n\\n Latent heat of sublimation is \",Lsub ,\" kJ/kg mol\"\n",
+ "print \"\\n Latent heat of vapourization is is \",Lvap ,\" kJ/kg mol\"\n",
+ "print \"\\n Latent heat of fusion is \",Lfu ,\" kJ/kg mol\"\n",
+ "#The answers vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex11.6:pg-438"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Example 11.6\n",
+ "\n",
+ "\n",
+ " Energy of first system is 1.6212885 kJ,\n",
+ " Energy of second system is 2.43193275 kJ,\n",
+ " Volume of first system is 0.008 m**3,\n",
+ " Volume of second system is 0.012 m**3,\n",
+ " Pressure is 135.107375 kN/m**2,\n",
+ " Temperature is 260.0 K.\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "R = 8.3143 # Gas constant in kJ/kg-mol-K\n",
+ "N1 = 0.5 # Mole no. of first system\n",
+ "N2 = 0.75 # Mole no. of second system\n",
+ "T1 = 200 # Initial temperature of first system in K\n",
+ "T2 = 300 # Initial temperature of second system in K\n",
+ "v = 0.02 # Total volume in m**3\n",
+ "print \"\\n Example 11.6\\n\"\n",
+ "Tf = (T2*N2+T1*N1)/(N1+N2)\n",
+ "Uf_1 = (3.0/2.0)*(R*N1*Tf)*(10**-3)\n",
+ "Uf_2 = (3.0/2.0)*(R*N2*Tf)*(10**-3)\n",
+ "pf = (R*Tf*(N1+N2)*(10**-3))/v\n",
+ "Vf_1 = R*N1*(10**-3)*Tf/pf\n",
+ "Vf_2 = v-Vf_1\n",
+ "print \"\\n Energy of first system is \",Uf_1 ,\" kJ,\\n Energy of second system is \",Uf_2 ,\" kJ,\\n Volume of first system is \",Vf_1 ,\" m**3,\\n Volume of second system is \",Vf_2 ,\" m**3,\\n Pressure is \",pf ,\" kN/m**2,\\n Temperature is \",Tf ,\" K.\"\n",
+ "#The answers vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex11.10:pg-446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ " Example 11.10 \n",
+ "\n",
+ " Power required to run the compressor = -10.6853713009 kW, \n",
+ " The rate at which heat must be removed from the compressor = -11.9346733668 kW\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "R = 0.082 # Gas constant in litre-atm/gmol-K\n",
+ "m = 1.5 # Mass flow rate in kg/s\n",
+ "p1 = 1.0 # Pressure in atm\n",
+ "t2 = 300.0 # Temperature after compression in K\n",
+ "p2 = 400.0 # Pressure after compression in atm\n",
+ "Tc = 151.0 # For Argon in K\n",
+ "pc = 48.0 # For Argon in atm\n",
+ "print \"\\n Example 11.10 \"\n",
+ "a = 0.42748*((R*1000)**2)*((Tc)**2)/pc\n",
+ "b = 0.08664*(R*1000)*(Tc)/pc\n",
+ "# By solving equation v2**2 - 49.24*v2**2 + 335.6*v2 - 43440 = 0\n",
+ "v2 = 56.8 # In cm**3/g mol\n",
+ "v1 = (R*1000)*(t2)/p1\n",
+ "delta_h = -1790 # In J/g mol\n",
+ "delta_s = -57 # In J/g mol\n",
+ "Q = (t2*delta_s*(10**5)/39.8)/(3600*1000)\n",
+ "W = Q - (delta_h*(10**5)/39.8)/(3600*1000)\n",
+ "print \"\\n Power required to run the compressor = \",W ,\" kW, \\n The rate at which heat must be removed from the compressor = \",Q ,\" kW\"\n",
+ "# Answers vary due to round off error.\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2024-12-19 15:10:06 +0000