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diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Boiling_Heat_Transfer.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Boiling_Heat_Transfer.ipynb new file mode 100644 index 00000000..301b16bb --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Boiling_Heat_Transfer.ipynb @@ -0,0 +1,157 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Boiling Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 13.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient for nucleate boiling (W/m^2 C) = 1.76e+05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Water at 1 atm is boiled on a brass surface. Using the rohsenow correlation\n", + "#determine the heat transfer coefficient for nucleate boiling if the wall \n", + "#superheat is 11C?\n", + "import math\n", + "#initialisation of variables\n", + "P= 1 \t\t\t\t#atm\n", + "dt= 11 \t\t\t\t#C\n", + "Csf= 0.006\n", + "r= 1./3.\n", + "s= 1.\n", + "cl= 4.218 \t\t\t#J/gm K\n", + "hfg= 2257 \t\t\t#J/gm\n", + "Pr= 1.75\n", + "ul= 283.1/1000. \t#gm/m s\n", + "s= 57.78/1000. \t\t#N/m\n", + "pl= 958*1000. \t\t#gm/m^3\n", + "pv= 598. \t\t\t#gm/m^3\n", + "gc= 1000. \t\t\t#gm m/N s^2\n", + "g= 9.8 \t\t\t\t#m/s^2\n", + "#CALCULATIONS\n", + "p= pl-pv \t\t\t#change in density\n", + "q= ((math.pow(((cl*dt)/(hfg*Csf*Pr)),(1/r))*(ul*hfg))/math.pow((gc*s/(g*p)),(0.5))) #Heat transfer rate\n", + "h= q/dt \t\t\t#Heat transfer coefficient\n", + "#RESULTS\n", + "print '%s %.2e' % ('Heat transfer coefficient for nucleate boiling (W/m^2 C) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 13.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Condensation heat tranfer coefficient if the tube is vertical (Btu/h ft^2 F) = 758.79\n", + " \n", + " Condensation heat tranfer coefficient if the tube is horizontally (Btu/h ft^2 F) = 1926.98\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A tube, 1/2 in OD and 5 ft long, is used to condense steam at 6 psia; the\n", + "#average temperature of the tube is maintained at 130 F. Determine the \n", + "#condensation heat transfer coefficient if the tube is mounted\n", + "#(a) vertically and (b) horizontally\n", + "import math\n", + "#initialisation of variables\n", + "k= 0.384 \t\t\t\t\t\t\t\t\t\t\t#Btu/hr ft F\n", + "Tsat= 170.03 \t\t\t\t\t\t\t\t\t\t#F\n", + "hfg= 996.2 \t\t\t\t\t\t\t\t\t\t\t#Btu/lbm\n", + "T= 130 \t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "l= 5. \t\t\t\t\t\t\t\t\t\t\t\t#ft\n", + "P= 6. \t\t\t\t\t\t\t\t\t\t\t\t#psia\n", + "g= 4.17*math.pow(10,8) \t\t\t\t\t\t\t\t#ft/h^2\n", + "d= 0.042 \t\t\t\t\t\t\t\t\t\t\t#ft\n", + "p= 61.2 \t\t\t\t\t\t\t\t\t\t\t#lbm/ft^3\n", + "u= 1.05 \t\t\t\t\t\t\t\t\t\t\t#lbm/ft h\n", + "#CALCULATIONS\n", + "dt= Tsat-T \t\t\t\t\t\t\t\t\t\t\t#Change in temp.\n", + "Tf= (Tsat+T)/2. \t\t\t\t\t\t\t\t\t#Average temp.\n", + "hc= 0.943*math.pow(((k*k*k*p*p*g*hfg)/(l*u*dt)),(1./4.))\n", + "hc1= 0.725*math.pow(((k*k*k*p*p*g*hfg)/(d*u*dt)),(1./4.))\n", + "#RESULTS\n", + "print '%s %.2f' % ('Condensation heat tranfer coefficient if the tube is vertical (Btu/h ft^2 F) = ',hc)\n", + "print '%s %.2f' % (' \\n Condensation heat tranfer coefficient if the tube is horizontally (Btu/h ft^2 F) = ',hc1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Combustion_Processes.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Combustion_Processes.ipynb new file mode 100644 index 00000000..7fb09a1d --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Combustion_Processes.ipynb @@ -0,0 +1,227 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: Combustion Processes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of O2 required (lbm) = 96.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If 3 mol carbon are completely burned in a process, determine \n", + "#(a) the number of moles of O2 required and\n", + "#(b) the mass of O2 required\n", + "#initialisation of variables\n", + "n= 3 \t\t\t\t#lbm mol\n", + "Mo2= 32 \t\t\t#lbm/lbm mol\n", + "#CALCULATIONS\n", + "m= n*Mo2 \t\t\t#Mass of O2 required\n", + "#RESULTS\n", + "print '%s %.2f' % ('Mass of O2 required (lbm) = ',m)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air-fuel ratio (gm air/gm fuel) = 15.12\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the air-fuel ratio on a gram-molal and a mass basis if octane is \n", + "#completely burned in the presence of theoretical air\n", + "#initialisation of variables\n", + "n= 12.5 \t\t\t\t\t#mol\n", + "n1= 3.76 \t\t\t\t\t#mol\n", + "M= 114 \t\t\t\t\t\t#gm/gm mol\n", + "M1= 28.96 \t\t\t\t\t#gm/gm mol\n", + "#CALCULATIONS\n", + "n2= n*(1+n1) \t\t\t\t#Moles\n", + "m= n2*M1/M \t\t\t\t\t#Air-fuel ratio \n", + "#RESULTS\n", + "print '%s %.2f' % ('Air-fuel ratio (gm air/gm fuel) = ',m)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air-fuel ratio (kg mol air/kg mol fuel) = 89.25\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Develop the stoichiometric equation that expresses the process of octane \n", + "#being burned with 150% theoretical air. What is the air-fuel ratio on a \n", + "#kilogram molal basis?\n", + "#initialisation of variables\n", + "p= 150\n", + "nO2= 12.5 \t\t\t\t\t\t\t\t#mol\n", + "n1= 3.76\n", + "#CALCULATIONS\n", + "n2= (nO2*(p/100.))+(nO2*n1*(p/100.))\t#ir-fuel ratio \n", + "#RESULTS\n", + "print '%s %.2f' % ('Air-fuel ratio (kg mol air/kg mol fuel) = ',n2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Benzene required (gm) = 568.40\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Benzene is used in a heating process that is 65% efficient. How much benzene\n", + "#is required to heat 185 liter tank of water from 10 to 30C?\n", + "#initialisation of variables\n", + "P= 65 \n", + "T= 30 \t\t\t\t#C\n", + "T1= 10 \t\t\t\t#C\n", + "c= 4.19 \t\t\t#J/gm C\n", + "h= 41961\n", + "m= 185 \t\t\t\t#lt\n", + "#CALCULATIONS\n", + "Q= m*1000*c*(T-T1)\t#Heat required\n", + "M= (Q*100.)/(h*P) \t#Mass of benzene required\n", + "#RESULTS\n", + "print '%s %.2f' % ('Benzene required (gm) = ',M)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Conduction_Heat_Transfer.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Conduction_Heat_Transfer.ipynb new file mode 100644 index 00000000..a0581011 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Conduction_Heat_Transfer.ipynb @@ -0,0 +1,421 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: Conduction Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Outside wall temperature (F) = 23.08\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#THe inside surface of a plane wall is exposed to air at 76 F and the outside \n", + "#surface to air at 21 F. The inside surface conductance is 1.5., and the outside\n", + "#is 6.5. If a thermocouple indicates that the inside wall temperature is 67 F\n", + "#what is the outside wall temperature.?\n", + "#initialisation of variables\n", + "T= 76 \t\t\t\t\t#F\n", + "T1= 21 \t\t\t\t\t#F\n", + "Tw= 67 \t\t\t\t\t#W\n", + "h= 1.5 \t\t\t\t\t#Btu/hr ft^2 F\n", + "A= 1. \t\t\t\t\t#ft^2 \n", + "h0= 6.5 \t\t\t\t#Btu/hr\n", + "#CALCULATIONS\n", + "q= h*A*(T-Tw)\t\t\t#Heat flow\n", + "t= (q/(h0*A))+T1 \t\t#Outside wall temperature\n", + "#results\n", + "print '%s %.2f' % ('Outside wall temperature (F) = ',t)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal transmittance (Btu/hr ft^2 F) = 0.62\n", + " \n", + " Heat transfer rate (Btu/hr) = 31.25\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#The inside and outside surface conductances are 2.0 and 10.0 Btu/hr ft^2 F\n", + "#respectively and the thermal conductivity of the wall is 0.5 units. Determine\n", + "#(a)the thermal transmittance and (b) the hear transfer rate for 1 ft^2 of wall\n", + "#surfaces\n", + "#initialisation of variables\n", + "hi= 2. \t\t\t\t\t\t\t\t\t#Btu/hr ft^2 F\n", + "l= 6. \t\t\t\t\t\t\t\t\t#in\n", + "k= 0.5 \t\t\t\t\t\t\t\t\t#Btu/hr ft F\n", + "h0= 10. \t\t\t\t\t\t\t\t#Btu/hr ft^2 F\n", + "ti= 70. \t\t\t\t\t\t\t\t#F\n", + "t0= 20.\t\t\t\t\t\t\t\t\t#F\n", + "A= 1. \t\t\t\t\t\t\t\t\t#ft^2\n", + "#CALCULATIONS\n", + "U= 1/((1/hi)+((l*0.5)/(6*k))+(1/h0))\t#Thermal transmittance \n", + "q= U*A*(ti-t0)\t\t\t\t\t\t\t#Heat transfer rate\n", + "#RESULTS\n", + "print '%s %.2f' % ('Thermal transmittance (Btu/hr ft^2 F) = ',U)\n", + "print '%s %.2f' % (' \\n Heat transfer rate (Btu/hr) = ',q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss (Btu/hr) = 32.00\n", + " \n", + " Temperature at the interface of the steel and the insulation (F) = 299.98\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A composite wall is made up of a 1/4 in. steel plate(k=31.4) and 3 in insulation\n", + "#(k=0.04). If the outside of the steel surface is 300 F, and the outside of the \n", + "#insulation is 100 F, determine (a) the heat loss and (b) the temperature at\n", + "#the interface of the steel amd the insulation\n", + "#initialisation of variables\n", + "Ti= 300. \t\t\t\t\t\t\t#F\n", + "T0= 100. \t\t\t\t\t\t\t#F\n", + "l= 0.25 \t\t\t\t\t\t\t#in\n", + "li= 3. \t\t\t\t\t\t\t\t#in\n", + "A= 12. \t\t\t\t\t\t\t\t#in/ft\n", + "ks= 31.4 \t\t\t\t\t\t\t#Btu/hr ft F\n", + "ki= 0.04 \t\t\t\t\t\t\t#Btu/hr ft F\n", + "#CALCULATIONS\n", + "q= (Ti-T0)/((l/(A*ks))+(li/(A*ki))) #Heat loss\n", + "t= Ti-((q*l/12.)/ks) \t\t\t\t#Temperature\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat loss (Btu/hr) = ',q)\n", + "print '%s %.2f' % (' \\n Temperature at the interface of the steel and the insulation (F) = ',t)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss (W) = 347.46\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A steel pipe (k=6.4) has an OD of 8.89 cm and an ID of 7.8 cm, and is covered\n", + "#with 1.3 cm asbestos (k=0.19). The pipe transports a fluid at 149 C and has\n", + "#an inner surface conductance of 227. Outside temp=27. Outside conductance=23\n", + "#what os the heat loss of 1m of pipe?\n", + "import math\n", + "#initialisation of variables\n", + "ti= 149. \t\t\t\t\t\t\t\t#C\n", + "t0= 27. \t\t\t\t\t\t\t\t#C\n", + "D0= 0.1149 \t\t\t\t\t\t\t\t#m\n", + "l= 1. \t\t\t\t\t\t\t\t\t#m\n", + "h0= 23. \t\t\t\t\t\t\t\t#W/m^2 C\n", + "hi= 227. \t\t\t\t\t\t\t\t#W/m^2 C\n", + "k= 0.19 \t\t\t\t\t\t\t\t#W/m C\n", + "Di= 0.0889 \t\t\t\t\t\t\t\t#cm\n", + "#CALCULATIONS\n", + "D1= D0*100 \n", + "D2= Di*100 \n", + "R0=(1/(D0*math.pi*l*h0))\t\t\t\t#Resistance\n", + "Rins=(math.log(D1/D2)/(2*math.pi*k*l))\t#Resistance\n", + "Ri=1/(Di*math.pi*l*hi) \t\t\t\t\t#Resistance Inlet\n", + "q= (ti-t0)/(R0+Rins+Ri) \t\t\t\t#Total heat\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat loss (W) = ',q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power consumption (W) = 970.90\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#The working chamber of an electrically heated furnace is a cube 0.2 m on each\n", + "#side and the walls are 0.1 m thick. Interior wall temperatures are to be \n", + "#maintained at 1100 c while the outside wall temperatures are at 150C. If the\n", + "#thermal conductivity of the furnace material is 0.35, estimate the power consumption.\n", + "import math\n", + "#initialisation of variables\n", + "l= 0.2 \t\t\t\t\t\t\t\t\t#m\n", + "l1= 0.5\t\t\t\t\t\t\t\t \t#m\n", + "k= 0.35 \t\t\t\t\t\t\t\t#W/m C\n", + "t= 0.15 \t\t\t\t\t\t\t\t#m\n", + "T1= 1100 \t\t\t\t\t\t\t\t#C\n", + "T2= 150 \t\t\t\t\t\t\t\t#C\n", + "#CALCULATIONS\n", + "Ai= 6*l*l \t\t\t\t\t\t\t\t#Inner area\n", + "Ao= 6*l1*l1 \t\t\t\t\t\t\t#outer area\n", + "q= 0.73*k*math.sqrt(Ai*Ao)*(T1-T2)/t \t#Power consumption\n", + "#RESULTS\n", + "print '%s %.2f' % ('Power consumption (W) = ',q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "heat loss will increase if the insulation is added\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A copper tube, 0.6 cm OD, carries hot water between two tanks, the outside\n", + "#surface conductance is 12. If it is important to minimize the heat loss\n", + "#should the tube be covered with an insulation whose k=0.19\n", + "#initialisation of variables\n", + "h= 12 \t\t\t\t#W/m^2 C\n", + "k= 0.19 \t\t\t#W/m C\n", + "d= 0.6 \t\t\t\t#m\n", + "#CALCULATIONS\n", + "r= k/h \t\t\t\t#Critical radius\n", + "d1=d/2. \t\t\t#Radius of tube\n", + "if (r<d1):\n", + " print('heat loss will increase if the insulation is added');\n", + "else:\n", + " print('heat loss will increase if the insulation is added');\n", + "\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time needed for the casting to be heated to 510 C (hr) = 0.63\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A small casting initially at 16C is placed in a furnace at 1200C\n", + "#The ratio of volume to surface area is 0.15 m and the outer k=85\n", + "#k for casting is 225 and the thermal diffusivity is 0.34. how much time is \n", + "#needed for the casting to be heated to 510C?\n", + "import math\n", + "#initialisation of variables\n", + "h= 85 \t\t\t\t\t\t\t\t\t#W/m^2 C\n", + "s= 0.15 \t\t\t\t\t\t\t\t#m\n", + "K= 225. \t\t\t\t\t\t\t\t#W/m C\n", + "t= 510. \t\t\t\t\t\t\t\t#C\n", + "t1= 1200. \t\t\t\t\t\t\t\t#C\n", + "t0= 16. \t\t\t\t\t\t\t\t#C\n", + "a= 0.34\n", + "#CALCULATIONS\n", + "Bi= h*s/K \t\t\t\t\t\t\t\t#Biot number\n", + "T= K*s*math.log((t0-t1)/(t-t1))/(h*a) \t#Time\n", + "#RESULTS\n", + "print '%s %.2f' % ('Time needed for the casting to be heated to 510 C (hr) = ',T)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Convection_Heat_Transfer1.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Convection_Heat_Transfer1.ipynb new file mode 100644 index 00000000..2ea6f9a3 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Convection_Heat_Transfer1.ipynb @@ -0,0 +1,264 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12: Convection Heat Transfer1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 12.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer rate from both sides of the plate (Btu/hr) = 40.50\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#An electrically heated vertical plate, 5 in square has a temperature of \n", + "#150 F and is being cooled by natural convection in 50 F air. What is the \n", + "#heat transfer rate from both sides of the plate?\n", + "import math\n", + "#initialisation of variables\n", + "d= 5. \t\t\t\t\t\t\t\t#ft\n", + "Tw= 150. \t\t\t\t\t\t\t#F\n", + "T= 50 \t\t\t\t\t\t\t\t#F\n", + "Pr= 0.72\n", + "k= 0.015 \t\t\t\t\t\t\t#Btu/hr ft F\n", + "r= 1.76*1000000. \t\t\t\t\t#(F ft^3)^-1\n", + "#CALCULATIONS\n", + "D= d*(0.42/5.) \t\t\t\t\t\t#Diameter\n", + "dt= Tw-T \t\t\t\t\t\t\t#change in temp\n", + "Gr= r*D*D*D*dt \t\t\t\t\t\t#Grashof number\n", + "z= Gr*Pr \t\t\t\t\n", + "h= 0.59*(math.pow(z,(0.25))) *(k/D) #Heat transfer coefficient\n", + "q= (2*h*dt*d*d)/144. \t\t\t\t#Heat transfer rate\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat transfer rate from both sides of the plate (Btu/hr) = ',q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 12.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient when the flow is fully devoloped (Btu/hr ft^2 F) = 1311.13\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Cooling water at an average temperature of 70 F flows through a tube of \n", + "#0.9 in ID, with an average velocity of 7 ft/s. What is the heat transfer\n", + "#coefficient when the flow is fully developed?\n", + "#initialisation of variables\n", + "import math\n", + "T= 70. \t\t\t\t\t\t\t\t\t\t#F\n", + "l= 0.9 \t\t\t\t\t\t\t\t\t\t#in\n", + "v= 7. \t\t\t\t\t\t\t\t\t\t#ft/s\n", + "d= 62.3 \t\t\t\t\t\t\t\t\t#lbm/ft^3\n", + "m= 6.58*math.pow(10,-4) \t\t\t\t\t#lbm/ft s\n", + "Pr= 6.82 \n", + "k= 0.347 \t\t\t\t\t\t\t\t\t#Bt/hr ft F\n", + "#CALCULATIONS\n", + "l1= l*0.075/l\n", + "Re= (d*v*l1)/m \t\t\t\t\t\t\t\t#Reynold's number\n", + "Nu= 0.023*math.pow(Re,0.8)*math.pow(Pr,0.4) #Nusselt number\n", + "h= Nu*k/l1 \t\t\t\t\t\t\t\t\t#Transfer coefficient\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat transfer coefficient when the flow is fully devoloped (Btu/hr ft^2 F) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 12.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer rate per unit lenght (W/m) = 23.21\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air at 1 atm pressure and a mixing cup temperature of 450k flows through \n", + "#a 3 cm diameter tube with a velocity of 6 m/s. determine the heat transfer\n", + "#rate per unit length if tube if a constant heat flux condition is maintained\n", + "#at the tube wall and the wall temperature is 10 C above the air temperature\n", + "import math\n", + "#initialisation of variables\n", + "P= 1 \t\t\t\t\t\t\t\t\t\t#atm\n", + "d= 0.783 \t\t\t\t\t\t\t\t\t#Kg/m^3\n", + "K= 0.0371 \t\t\t\t\t\t\t\t\t#W/m C\n", + "m= 2.48*math.pow(10,-5) \t\t\t\t\t#Ns/m^2\n", + "Pr= 0.683\n", + "D= 0.03 \t\t\t\t\t\t\t\t\t#m\n", + "v= 6 \t\t\t\t\t\t\t\t\t\t#m/s\n", + "T= 10 \t\t\t\t\t\t\t\t\t\t#C\n", + "#CALCULATIONS\n", + "Re= d*v*D/m \t\t\t\t\t\t\t\t#Reynolds number\n", + "Nu= 0.023*math.pow(Re,0.8)*math.pow(Pr,0.4) #Nusselt number\n", + "h= Nu*K/D \t\t\t\t\t\t\t\t\t#Heat transfer coefficient\n", + "ql= h*math.pi*D*T \t\t\t\t\t\t\t#Heat transfer rate\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat transfer rate per unit lenght (W/m) = ',ql)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 12.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer rate per unit lenght of cylinder (W/m) = 3023.70\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air at 1 atm pressure and 25 C flows past a horizontal 5 cm diameter with\n", + "#a velocity of 46 m/s. If the surface of the cylinder is kept at 135 C, determine\n", + "#the rate of heat flow from the cylinder\n", + "import math\n", + "#initialisation of variables\n", + "T= 25 \t\t\t\t\t\t#C\n", + "P= 1 \t\t\t\t\t\t#atm\n", + "v= 46 \t\t\t\t\t\t#m/s\n", + "d= 5 \t\t\t\t\t\t#cm\n", + "T1= 135 \t\t\t\t\t#C\n", + "d1= 0.998 \t\t\t\t\t#kg/m^3\n", + "k= 0.03 \t\t\t\t\t#W/m C\n", + "m= 2.08*math.pow(10,-5) \t#Kg/s m\n", + "c= 0.024\n", + "n= 0.81\n", + "#CALCULATIONS\n", + "Tf= (T+T1)/2. \t\t\t\t#Final temp.\n", + "D= d/100.\n", + "Re= d1*v*D/m \t\t\t\t#Reynolds number\n", + "h= c*math.pow(Re,0.81)*k/D \t#Heat transfer coefficient\n", + "dt= T1-T \t\t\t\t\t#temp diff.\n", + "ql= h*math.pi*D*dt \t\t\t#Heat transfer rate\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat transfer rate per unit lenght of cylinder (W/m) = ',ql)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/First_Law_of_Thermodynamics.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/First_Law_of_Thermodynamics.ipynb new file mode 100644 index 00000000..3527510d --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/First_Law_of_Thermodynamics.ipynb @@ -0,0 +1,351 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: First Law of Thermodynamics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy dissipated from the brakes (ft lbf) = 4.00e+05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#An automobile with a mass of 3000 lb comes over the top of a hill 50 ft high\n", + "#with a velocity of 50 mph. Brakes are applied at the instant the automobile \n", + "#reaches the top, and it comes to rest at the bottom of the hill. How much \n", + "#energy is dissipated from the brakes?\n", + "import math\n", + "#initialisation of variables\n", + "m= 3000 \t\t\t\t\t\t\t\t\t\t\t#lb\n", + "Z1= 50 \t\t\t\t\t\t\t\t\t\t\t\t#ft\n", + "V1= 50 \t\t\t\t\t\t\t\t\t\t\t\t#mph\n", + "gc= 32.2 \t\t\t\t\t\t\t\t\t\t\t#ft/lbf s^2\n", + "V2= 0 \t\t\t\t\t\t\t\t\t\t\t\t#mph\n", + "g= 32.2 \t\t\t\t\t\t\t\t\t\t\t#ft/s^2\n", + "Z2= 0 \t\t\t\t\t\t\t\t\t\t\t\t#ft\n", + "#CALCULATIONS\n", + "V1= V1*(73.3/50.) \t\t\t\t\t\t\t\t\t#Velocity\n", + "Q2= ((m*(V2*V2-V1*V1))/(2*gc))+((m*g)/gc)*(Z2-Z1)\n", + "#RESULTS\n", + "print '%s %.2e' % ('Energy dissipated from the brakes (ft lbf) = ',-Q2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From keenan and keyes steam tables\n", + "Temperature of the steam in the tank (C) = 453.40\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Steam at a pressure of 13 bar and a temperature of 300C flows adiabatically \n", + "#and with negligible velocity into an evacuated tank. Using a closed system\n", + "#analysis, determine the temperature of the steam in the tank reaches line\n", + "#pressure.\n", + "#initialisation of variables\n", + "P= 15 \t\t\t\t\t\t#bar\n", + "T= 300 \t\t\t\t\t\t#C\n", + "h1= 3043.1 \t\t\t\t\t#J/gm\n", + "#CALCULATIONS\n", + "u2= h1 \n", + "print '%s' %('From keenan and keyes steam tables')\n", + "T= 453.4 \t\t\t\t\t#C temperature\n", + "#RESULTS\n", + "print '%s %.2f' % ('Temperature of the steam in the tank (C) = ',T)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transferred to the tank (Btu) = 267.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A 10 lb mass of air at 120F is contained in a rigid tank. How much heat\n", + "#is transferred to the tank to raise the air temperature of 275 F?\n", + "#initialisation of variables\n", + "m= 10 \t\t\t\t\t\t\t#lbf\n", + "T= 120 \t\t\t\t\t\t\t#F\n", + "T1= 275 \t\t\t\t\t\t#F\n", + "u1= 98.9 \t\t\t\t\t\t#Btu/lbm\n", + "u2= 125.6 \t\t\t\t\t\t#Btu/lbm\n", + "#CALCULATIONS\n", + "Q= m*(u2-u1) \t\t\t\t\t#Heat transferred\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat transferred to the tank (Btu) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum theotrical power that can be devoloped (J/s) = 1.00e+05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Fluid enters a turbine with a velocity of 1 m/s and an enthalpy of 2000 j/gm;\n", + "#it leaves with a velocity of 60 m/s and enthalpy of 1800 J/gm. Heat losses\n", + "#are 500 J/s, and the flow rate is 0.5 kg/s. If the inlet of the turbine is \n", + "#3 m higher than the outlet, what is the maximum theoretical power\n", + "# that can be developed?\n", + "#initialisation of variables\n", + "v0= 1 \t\t\t\t\t\t\t\t\t\t\t\t\t#m/s\n", + "vi= 60 \t\t\t\t\t\t\t\t\t\t\t\t\t#m/s\n", + "Q= -500 \t\t\t\t\t\t\t\t\t\t\t\t#J/s\n", + "m= 500 \t\t\t\t\t\t\t\t\t\t\t\t\t#gm/s\n", + "hi= 2000 \t\t\t\t\t\t\t\t\t\t\t\t#J/gm\n", + "h0= 1800 \t\t\t\t\t\t\t\t\t\t\t\t#J/gm\n", + "zi= 3 \t\t\t\t\t\t\t\t\t\t\t\t\t#m\n", + "z0= 0 \t\t\t\t\t\t\t\t\t\t\t\t\t#m\n", + "g= 9.8 \t\t\t\t\t\t\t\t\t\t\t\t\t#m/s^2\n", + "gc= 1000. \t\t\t\t\t\t\t\t\t\t\t\t#gm m/Ns^2\n", + "#CALCULATIONS\n", + "W= Q+m*((hi-h0)+(vi*vi-v0*v0)/(2*gc)+(g/gc)*(zi-z0)) \t#Work\n", + "#RESULTS\n", + "print '%s %.2e' % ('Maximum theotrical power that can be devoloped (J/s) = ',W)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Required steam flow rate (gm/s) = 36.36\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#To produce 0.3 lt/s hot water at 82 C, low pressure steam at 2.4 bar and \n", + "#80 percent quality is mixed with a stream of water at 16 C. What is the \n", + "#required steam flow rate?\n", + "#initialisation of variables\n", + "m= 0.3 \t\t\t\t\t\t\t\t#lt/s\n", + "T= 82 \t\t\t\t\t\t\t\t#C\n", + "P= 2.4 \t\t\t\t\t\t\t\t#bar\n", + "p= 80.\n", + "Tw= 800 \t\t\t\t\t\t\t#C\n", + "h1= 67.19 \t\t\t\t\t\t\t#J/gm\n", + "h3= 343.3 \t\t\t\t\t\t\t#J/gm\n", + "hf= 529.65 \t\t\t\t\t\t\t#J/gm\n", + "hfg= 2185.4 \t\t\t\t\t\t#J/gm\n", + "v3= 1.0305 \t\t\t\t\t\t\t#cm^3/gm\n", + "V3= 300 \t\t\t\t\t\t\t#cm^3/s\n", + "#CALCULATIONS\n", + "h2= hf+(p/100.)*hfg\t\t\t\t\t#Enthalpy at 2\n", + "m3= V3/v3 \t\t\t\t\t\t\t#Mass at 3\n", + "m2= (m3*(h3-h1))/(h2-h1) \t\t\t#Mass at 2\n", + "#RESULTS\n", + "print '%s %.2f' % ('Required steam flow rate (gm/s) = ',m2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Difference between the enthalpies of the system in the two phases ((h2-h1) J/gm) = 1.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Latent heat of transforation can be defined as the ratio of heat absorbed\n", + "#to the mass which undergoes a change of phase(L=Q/m). Show that the heat of \n", + "#transformation for any phase change equals the difference between the \n", + "#enthalphies of the system in the two phases.\n", + "#initialisation of variables\n", + "h2= 2 \t\t\t\t\t#J/gm\n", + "h1= 1 \t\t\t\t\t#J/gm\n", + "#CALCULATIONS\n", + "L= h2-h1 \t\t\t\t#Difference between enthalpies\n", + "#RESULTS\n", + "print '%s %.2f' % ('Difference between the enthalpies of the system in the two phases ((h2-h1) J/gm) = ',L)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Gas_Properties_and_Processes.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Gas_Properties_and_Processes.ipynb new file mode 100644 index 00000000..06c3555a --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Gas_Properties_and_Processes.ipynb @@ -0,0 +1,374 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Gas Properties and Processes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Molecular weight of the gas (lbm/lbm mol) = 24.14\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A 15 ft^3 volume of an ideal gas has a mass of 20 lbm at 80F and 320 psia.\n", + "#What is the molecular weight of the gas?\n", + "#initialisation of variables\n", + "v= 15 \t\t\t\t\t\t#ft^3\n", + "m= 20 \t\t\t\t\t\t#lbm\n", + "T= 80 \t\t\t\t\t\t#lbf\n", + "P= 320 \t\t\t\t\t\t#psia\n", + "#CALCULATIONS\n", + "R= P*144*v/(m*(T+460.)) \t#Universal gas constant\n", + "M= 1545./R \t\t\t\t\t#Molecular weight\n", + "#RESULTS\n", + "print '%s %.2f' % ('Molecular weight of the gas (lbm/lbm mol) = ',M)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial Mass of Oxygen (gm) = 1287.93\n", + " \n", + " Final mass of oxygen (gm) = 413.69\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A 50 ltr liter of oxygen is initially at 20 atm and 30C. A sufficient of \n", + "#amount of oxygen is removed to change the state to 6 atm and 10C. \n", + "#Determine (a) the initial mass of oxygen and (b) final mass of oxygen.\n", + "#initialisation of variables\n", + "V= 50 \t\t\t\t\t\t\t#lit\n", + "P= 20 \t\t\t\t\t\t\t#atm\n", + "T= 30 \t\t\t\t\t\t\t#C\n", + "P1= 6 \t\t\t\t\t\t\t#atm\n", + "T1= 10 \t\t\t\t\t\t\t#C\n", + "M= 32 \t\t\t\t\t\t\t#gm/gm mol\n", + "#CALCULATIONS\n", + "n= V*P/(0.082*(T+273.)) \t\t#No. of moles\n", + "m= n*M \t\t\t\t\t\t\t#Mass\n", + "n2= P1*V/(0.082*(T1+273.)) \t\t#No. of moles\n", + "m2= n2*M \t\t\t\t\t\t#Mass\n", + "#RESULTS\n", + "print '%s %.2f' % ('Initial Mass of Oxygen (gm) = ',m)\n", + "print '%s %.2f' % (' \\n Final mass of oxygen (gm) = ',m2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final volume (ft^3) = 0.32\n", + " \n", + " Final temperature (R) = 637.84\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If 0.75 ft^3 of a gas at 1 atm and 35F is compressed polytropically to a\n", + "#pressure of 3 atm, determine the final volume and temperature. \n", + "#The polytropic exponent is 1.3\n", + "import math\n", + "#initialisation of variables\n", + "V2= 0.75 \t\t\t\t\t\t\t\t\t\t\t #ft^3\n", + "P2= 1 \t\t\t\t\t\t\t\t\t\t\t\t #atm\n", + "P1= 3 \t\t\t\t\t\t\t\t\t\t\t \t #atm\n", + "T= 35 \t\t\t\t\t\t\t\t\t\t\t\t #F\n", + "e= 1.3\n", + "#CALCULATIONS\n", + "V1= math.pow(((P2*math.pow((V2),e))/P1),(1/e)) #Final volume\n", + "T2= P1*V1*(T+460.)/(P2*V2) \t\t\t\t\t\t\t #Final temperature\n", + "#RESULTS\n", + "print '%s %.2f' % ('Final volume (ft^3) = ',V1)\n", + "print '%s %.2f' % (' \\n Final temperature (R) = ',T2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T1 (K) = 160.22\n", + " \n", + " T2 (K) = 160.22\n", + " \n", + " T3 (K) = 211.42\n", + " \n", + " P2 (Pa) = 3.45e+05\n", + " \n", + " P3 (Pa) = 9.10e+05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A rigid container holds a 0.45 kg mass of air with a volume of 0.03 m^3\n", + "#and an absolue pressure of 6.9x 10^5 pa. the air is expanded reversibly\n", + "#and adiabatically to 0.03 m^3. The air is then returned to its initial \n", + "#state by a rreversible constant volume process. Determine the pressure and\n", + "#temperature at each state \n", + "import math\n", + "#initialisation of variables\n", + "m= 0.45 \t\t\t\t\t\t#kg\n", + "v1= 0.03 \t\t\t\t\t\t#m^3\n", + "v2= 0.06 \t\t\t\t\t\t#m^3\n", + "P= 6.9*100000. \t\t\t\t\t#Pa\n", + "K= 1.4\n", + "R= 287.1 \t\t\t\t\t\t#J/Kg K\n", + "#CALCULATIONS\n", + "T1= (P*v1)/(m*R) \t\t\t\t#Temperature\n", + "T2= T1 \t\t\t\t\t\t\t#Temperature\n", + "P2= P*v1/v2 \t\t\t\t\t#Pressure\n", + "T3= T2*math.pow((v2/v1),(K-1)) \t#Temperature\n", + "P3= P2*math.pow((v2/v1),K) \t\t#Pressure\t\n", + "#RESULTS\n", + "print '%s %.2f' % ('T1 (K) = ',T1)\n", + "print '%s %.2f' % (' \\n T2 (K) = ',T2)\n", + "print '%s %.2f' % (' \\n T3 (K) = ',T3)\n", + "print '%s %.2e' % (' \\n P2 (Pa) = ',P2)\n", + "print '%s %.2e' % (' \\n P3 (Pa) = ',P3)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work associated with the process (Btu/lbm) = -46.32\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Nitrogen is compressed in a reversible polytropic process from 1 atm and 60 F\n", + "#to 4 atm. Calculate the work associated with the process if the polytropic \n", + "#exponent is 1.3\n", + "import math\n", + "#initialisation of variables\n", + "P= 1 \t\t\t\t\t\t\t\t\t#atm\n", + "T= 60 \t\t\t\t\t\t\t\t\t#F\n", + "P1= 4 \t\t\t\t\t\t\t\t\t#atm\n", + "e= 1.3\n", + "R= 55.15 \t\t\t\t\t\t\t\t#lbf/lbm R\n", + "m= 778\n", + "#CALCULATIONS\n", + "T2= (T+460)*math.pow((P1/P),((e-1)/e))\t#Temperature final\n", + "W= R*(T2-(T+460))/(1-e) \t\t\t\t#Work\n", + "W1= W/m \t\t\t\t\t\t\t\t#Work per unit mass\n", + "#RESLUTS\n", + "print '%s %.2f' % ('Work associated with the process (Btu/lbm) = ',W1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total volume (ft^3) = 150.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the total pressure of a mixture of 10 lbm oxygen and \n", + "#15 lbm nitrogen at 120F in a rigid tank of 150 ft^3. Show that\n", + "#the total volume is the sum of the partial volumes?\n", + "import math\n", + "#initialisation of variables\n", + "m= 10 \t\t\t\t\t\t#lbm\n", + "R= 48.28 \t\t\t\t\t#lbf/lbm R\n", + "T= 120 \t\t\t\t\t\t#F\n", + "V= 150 \t\t\t\t\t\t#ft^3\n", + "m1= 15 \t\t\t\t\t\t#lbm\n", + "R1= 55.15 \t\t\t\t\t#lbf/lbm R\n", + "#CALCULATIONS \n", + "P1= (m*R*(T+460))/V \t\t#Pressure of Oxygen\n", + "P2= (m1*R1*(T+460))/V \t\t#Pressure of Nitrogen\n", + "Pm= P1+P2 \t\t\t\t\t#Total pressure\n", + "V1= (m*R*(T+460))/Pm \t\t#Volume of Oxygen\n", + "V2= (m1*R1*(T+460))/Pm \t\t#Volume of Nitrogen\n", + "Vm= V1+V2 \t\t\t\t\t#Total volume\n", + "#RESULTS\n", + "print '%s %.2f' % ('Total volume (ft^3) = ',Vm)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/README.txt b/Elements_of_thermal_technology_by_John_H._Seely/README.txt new file mode 100644 index 00000000..4e2da5e7 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/README.txt @@ -0,0 +1,10 @@ +Contributed By: Jeevan Lal +Course: mtech +College/Institute/Organization: - +Department/Designation: - +Book Title: Elements of thermal technology +Author: John H. Seely +Publisher: Marcel Dekker Inc., New York, U. S. A. +Year of publication: 2002 +Isbn: 0-8247-1174-2 +Edition: 1
\ No newline at end of file diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Radiation_Heat_Transfer.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Radiation_Heat_Transfer.ipynb new file mode 100644 index 00000000..720fca7f --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Radiation_Heat_Transfer.ipynb @@ -0,0 +1,349 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: Radiation Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength at which the maximum monochromatic emissive power (m) = 7.44e-06\n", + " \n", + " Coffecient of performnance (W/m^3) = 1.14e+08\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If a blackbody is maintained at 116C, determine (a) the wavelength at which \n", + "#the maximum monochromatic emissive power occurs and (b) the maximum \n", + "#monochromatic emissive power\n", + "import math\n", + "#initialisation of variables\n", + "T= 116. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#C\n", + "C1= 3.74*math.pow(10,-16)\n", + "C2= 1.44*math.pow(10,-2)\n", + "#CALCULATIONS\n", + "WLmax= (2893*math.pow(10,-6))/(T+273) \t\t\t\t\t\t\t#Maximum Wavelength \n", + "Wb= (C1*math.pow((WLmax),(-5)))/(math.exp(C2/2893*1000000.)-1)\t#Coffecient of performnance\n", + "#RESULTS\n", + "print '%s %.2e' % ('Wavelength at which the maximum monochromatic emissive power (m) = ',WLmax)\n", + "print '%s %.2e' % (' \\n Coffecient of performnance (W/m^3) = ',Wb)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emissive power for the blackbody (W/m^2) = 1305.19\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the total emissive power for the black body of solved problem 2\n", + "import math\n", + "#initialisation of variables\n", + "T= 389 \t\t\t\t\t\t#K\n", + "s= 5.7*math.pow(10,-8) \t\t#K^4\n", + "#CALCULATIONS\n", + "Wb= s*T*T*T*T \t\t\t\t#Emissive power for the blackbody\n", + "#RESULTS\n", + "print '%s %.2f' % ('Emissive power for the blackbody (W/m^2) = ',Wb)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average absorptivity of the body at 100 F = 0.51\n", + " \n", + " Average absorptivity of the body at 2000 F= 0.84\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A gray body at 100F receives radiant energy from a wall at 2000 F at a rate\n", + "#of 3.2x 10^4. Simultaneously, the body emits energy at the rate of 140. What\n", + "#is the average absorptivity of the body at (a) 100 F and (b) 2000 F\n", + "import math\n", + "#initialisation of variables\n", + "T= 100 \t\t\t\t\t\t\t\t#F\n", + "T1= 2000 \t\t\t\t\t\t\t#F\n", + "W= 3.2*10000. \t\t\t\t\t\t#Btu/hr ft^2\n", + "W1= 140. \t\t\t\t\t\t\t#Btu/hr ft^2\n", + "s= 0.17*math.pow(10,-8) \t\t\t#Btu/hr ft^2 R^4\n", + "#CALCULATIONS\n", + "alpha= W/(s*math.pow((T1+460),4)) \t#Average absorptivity at 100\n", + "b= W1/(s*math.pow((T+460),4)) \t\t#Average absorptivity at 2000\n", + "#RESULTS\n", + "print '%s %.2f' % ('Average absorptivity of the body at 100 F = ',alpha)\n", + "print '%s %.2f' % (' \\n Average absorptivity of the body at 2000 F= ',b)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss from the conduit by radiation (Btu/hr per ft) = 1401.66\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A red brick conduit 10 in square has a surface temperature of 300F and is \n", + "#mounted inside a large earthen chamber whose walls are at 50 F. estimate\n", + "#the heat loss from the conduit by radiation\n", + "import math\n", + "#initialisation of variables\n", + "T= 300. \t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 50. \t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "s= 0.17*math.pow(10,-8) \t\t\t\t\t\t\t\t#Btu/hr ft^2 R^4\n", + "e1= 0.93\n", + "A= 10. \t\t\t\t\t\t\t\t\t\t\t\t\t#in\n", + "F= 1.\n", + "#CALCULATIONS\n", + "A1= 10*(40./(12.*10.)) \t\t\t\t\t\t\t\t\t#Area\n", + "q= A1*F*e1*s*(math.pow((T+460),4)-math.pow((T1+460),4)) #heat loss\n", + "#RESULTS\n", + "print '%s %.2f' % ('Heat loss from the conduit by radiation (Btu/hr per ft) = ',q)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radiation heat transfer coefficient (Btu/hr ft^2 R) = 1.68\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Estimate the radiation heat transfer coefficient for solved problem 5\n", + "import math\n", + "#initialisation of variables\n", + "T= 300. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "T1= 50. \t\t\t\t\t\t\t\t\t\t\t\t\t\t#F\n", + "s= 0.17*math.pow(10,-8) \t\t\t\t\t\t\t\t\t\t#Btu/hr ft^2 R^4\n", + "e1= 0.93\n", + "F= 1.\n", + "#CALCULATIONS\n", + "hr= F*e1*s*(math.pow((T+460),4)-math.pow((T1+460),4))/(T-T1)\t#Radiation heat transfer coefficient \n", + "#RESULTS\n", + "print '%s %.2f' % ('Radiation heat transfer coefficient (Btu/hr ft^2 R) = ',hr)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 14.7" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient for nucleate boiling (W/m^2 C) = 6.21e+08\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import math\n", + "#initialisation of variables\n", + "P= 1. #atm\n", + "T= 11. #C\n", + "Csf= 0.006\n", + "Tsat = 170.03\n", + "r= 1./3.\n", + "s= 1.\n", + "dt = Tsat - T\n", + "cl= 4.218 #J/gm K\n", + "hfg= 2257 #J/gm\n", + "Pr= 1.75\n", + "ul= 283.1/1000. #gm/m s\n", + "s= 57.78/1000. #N/m\n", + "pl= 958*1000. #gm/m^3\n", + "pv= 598. #gm/m^3\n", + "gc= 1000. #gm m/N s^2\n", + "g= 9.8 #m/s^2\n", + "#CALCULATIONS\n", + "p= pl-pv\n", + "q= ((math.pow(((cl*dt)/(hfg*Csf*math.pow(Pr,s))),(1/r)))*(ul*hfg))/math.pow(gc/(g*p),(1./2.))\n", + "h= q/T\n", + "#RESULTS\n", + "print '%s %.2e' % ('Heat transfer coefficient for nucleate boiling (W/m^2 C) = ',h)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Refrigeration_and_Air_Conditioning.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Refrigeration_and_Air_Conditioning.ipynb new file mode 100644 index 00000000..97e842ae --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Refrigeration_and_Air_Conditioning.ipynb @@ -0,0 +1,146 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15: Refrigeration and Air Conditioning" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 15.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Joule-Thomson coefficient (C/bar) = 1.61\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Estimate how much the temperature is lowered when the steam at 7 bar and \n", + "#260C is throttled to 1.4 bar. What is the approximate Joule-Thomson \n", + "#initialisation of variables\n", + "P= 7. \t\t\t\t#bar\n", + "P1= 1.4 \t\t\t#bar\n", + "T= 260. \t\t\t#C\n", + "T1= 251. \t\t\t#C\n", + "h= 2974.9 \t\t\t#J/gm\n", + "#CALCULATIONS\n", + "dT= T-T1 \t\t\t#Change in Temp. \n", + "Mj= dT/(P-P1) \t\t#Joule-Thomson coefficient\n", + "#RESULTS\n", + "print '%s %.2f' % ('Joule-Thomson coefficient (C/bar) = ',Mj)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 15.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refrigerating effect (Btu/lbm) = 44.80\n", + " \n", + " Coffecient of performnance= 3.54\n", + " \n", + " Capacity of refrigeration in tons (ton) = 0.67\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A vapor compression refrigiration system, using Refrigirant 12, works between\n", + "#n evaporator temperature of 10F and a condenser temp. of 110F. The pumping\n", + "#rate for the refrigrant is 180 lbm/hr. Determine (a) the refrigirating \n", + "#effect, (b) the coefficient of performance,and (c) the capacity of refrigiration\n", + "#in tons\n", + "#initialisation of variables\n", + "T= 10. \t\t\t\t\t\t#F\n", + "T1= 110. \t\t\t\t\t#F\n", + "Pr= 180. \t\t\t\t\t#lbm/hr\n", + "h1= 78.335 \t\t\t\t\t#Btu/lbm\n", + "h3= 33.531 \t\t\t\t\t#Btu/lbm\n", + "h2= 91. \t\t\t\t\t#Btu/lbm\n", + "L= 12000. \t\t\t\t\t#Btu/hr per ton\n", + "#CALCULATIONS\n", + "h4= h3\n", + "QL= h1-h4 \t\t\t\t\t#Refrigerating effect\n", + "W= h2-h1 \t\t\t\t\t#Work done\n", + "COP= QL/W \t\t\t\t\t#Coffecient of performnance\n", + "C= QL*Pr/L \t\t\t\t\t#Capacity of refrigeration in tons\n", + "#RESULTS\n", + "print '%s %.2f' % ('Refrigerating effect (Btu/lbm) = ',QL)\n", + "print '%s %.2f' % (' \\n Coffecient of performnance= ',COP)\n", + "print '%s %.2f' % (' \\n Capacity of refrigeration in tons (ton) = ',C)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Second_Law_of_Thermodynamics.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Second_Law_of_Thermodynamics.ipynb new file mode 100644 index 00000000..cf940111 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Second_Law_of_Thermodynamics.ipynb @@ -0,0 +1,242 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Second Law of Thermodynamics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum theotrical efficiency (percentage)((Claim is not valid)) = 23.33\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#You are asked to comment on an inventor's claim that he has an engine which \n", + "#delivers 25 Btu of work from an input if 100 Btu when it operates between \n", + "#140F and 0F. Is the claim valid?\n", + "#initialisation of variables\n", + "W= 25 \t\t\t\t\t#Btu\n", + "W1= 100 \t\t\t\t#Btu\n", + "T1= 140 \t\t\t\t#R\n", + "T2= 0 \t\t\t\t\t#R\n", + "#CALCULATIONS\n", + "Th= T1+460. \t\t\t#T hot in Rankine\n", + "Tl= T2+460. \t\t\t#T cold in Rankine\n", + "nt= (Th-Tl)/Th \t\t\t#Max. efficieny\n", + "n= nt*100.\n", + "#RESULTS\n", + "print '%s %.2f' % ('maximum theotrical efficiency (percentage)((Claim is not valid)) = ',n)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in Entropy (J/gm K) = 2.856\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Saturated steam in a rigid vessel is initially at 10 bar. heat is added\n", + "#until the steam s superheated at 38 bar and 310 C. What is the change in entropy?\n", + "#initialisation of variables\n", + "P= 10 \t\t\t\t\t\t#bar\n", + "P1= 38 \t\t\t\t\t\t#bar\n", + "T= 310 \t\t\t\t\t\t#C\n", + "v= 64.03 \t\t\t\t\t#cm^3/gm\n", + "s= 6.4415 \t\t\t\t\t#J/gm K\n", + "vf= 1.12773 \t\t\t\t#cm^3/gm\n", + "vg= 194.44 \t\t\t\t\t#cm^3/gm\n", + "sf= 2.1387 \t\t\t\t\t#J/gm K\n", + "sfg= 4.4478 \t\t\t\t#J/gm K\n", + "#CALCULATIONS \n", + "x= (v-vf)/(vg-vf) \t\t\t#x\n", + "sx= sf+x*sfg \t\t\t\t#Entropy\n", + "S= s-sx \t\t\t\t\t#Change in entropy\n", + "#RESULTS\n", + "print '%s %.3f' % ('Change in Entropy (J/gm K) = ',S)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum power required to drive the heat pump (Btu/hr) = 7500.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If heat must be pumped at the rate of 70k Btu/hr to keep a house at 72F when\n", + "#the outside temperature is 15 F, What is the minimum power required to drive\n", + "#the heat pump?\n", + "#initialisation of variables\n", + "Qh= 70000. \t\t\t\t\t\t\t#Btu/hr\n", + "T= 15. \t\t\t\t\t\t\t\t#F\n", + "T1= 72. \t\t\t\t\t\t\t#F\n", + "#CALCULATIONS\n", + "COP= (T1+460)/((T1+460)-(T+460))\t#Carnot efficiency\n", + "W= Qh/COP \t\t\t\t\t\t\t#Work\n", + "#RESULTS\n", + "print '%s %.2f' % ('Minimum power required to drive the heat pump (Btu/hr) = ',W)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum electrical requirement (KW) = 3.54\n", + " \n", + " Elctrical requirement if an electrical heater used (KW) = 26.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Heat must be pumped at the rate of 26kW to maintain a temperature of 43F\n", + "#in an air stream while the outside temperature is 0C. Determine \n", + "#(a)the minimum electrical equipment in Kilowatts and \n", + "#(b)the electrical requirement if an electric heater is used.\n", + "#initialisation of variables\n", + "h= 26. \t\t\t\t\t\t\t\t#KW\n", + "T= 43. \t\t\t\t\t\t\t\t#C\n", + "To= 0 \t\t\t\t\t\t\t\t#C\n", + "#CALCULATIONS\n", + "COP= (T+273)/((T+273.)-(To+273.))\t#Carnot efficiency\n", + "W= h/COP \t\t\t\t\t\t\t#Work\n", + "Qh=h \t\t\t\t\t\t\t\t#Heat\n", + "#RESULTS\n", + "print '%s %.2f' % ('Minimum electrical requirement (KW) = ',W)\n", + "print '%s %.2f' % (' \\n Elctrical requirement if an electrical heater used (KW) = ',Qh)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Surface_Tension.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Surface_Tension.ipynb new file mode 100644 index 00000000..61b786fd --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Surface_Tension.ipynb @@ -0,0 +1,179 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Surface Tension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work (dyn cm or erg) = 5.03e+04\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If the surface tension of a soap bubble is 0.4 N/m, what work is required to\n", + "#increase the diameter from 5 to 15 cm?\n", + "import math\n", + "#initialisation of variables\n", + "St= 0.04 \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#N/m\n", + "d1= 5. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#cm\n", + "d2= 15. \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t#cm\n", + "#CALCULATIONS\n", + "W= St*1000.*2*4*math.pi*(math.pow((d2/2),2)-math.pow((d1/2),2))\t\t#Work\n", + "#RESULTS\n", + "print '%s %.2e' % ('Work (dyn cm or erg) = ',W)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Presuure difference (lbf/ft^2) = 14.12\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#By much does the pressure inside a 0.017 in droplet of water at 20C\n", + "#exceed the outside pressure.\n", + "import math\n", + "#initialisation of variables\n", + "R= 0.017 \t\t\t\t\t\t\t\t\t\t\t\t#in\n", + "sigma= 72.8 \t\t\t\t\t\t\t\t\t\t\t#m N/m\n", + "#CALCULATIONS\n", + "P= (2*sigma*0.005*0.017)/(72.8*R*7.08*math.pow(10,-4))\t#Presuure difference\n", + "#RESULTS\n", + "print '%s %.2f' % ('Presuure difference (lbf/ft^2) = ',P)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Difference in mercury level (cm (depression)) = -2.48\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A 0.4 mm diameter glass tube stands vertically in a dish of mercury at 20C.\n", + "#Determine the difference between the level of mercury in the dish and in\n", + "#the tube. The specific gravity for mercury is 13.6 and the contact angle \n", + "#with glass is 130C\n", + "import math\n", + "#initialisation of variables\n", + "d= 13.6 \t\t\t\t\t\t\t#gm/cm^3\n", + "g= 980 \t\t\t\t\t\t\t\t#cm/s^2\n", + "D= 0.4 \t\t\t\t\t\t\t\t#mm\n", + "angle= 130*math.pi/180. \t\t\t#radians\n", + "s= 514. \t\t\t\t\t\t\t#dyn/cm\n", + "#CALCULATIONS\n", + "h= (4*s*10*math.cos(angle))/(d*g*D)\t#Depression in mercury level\n", + "#RESULTS\n", + "print '%s %.2f' % (' Difference in mercury level (cm (depression)) = ',h)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Tem.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Tem.ipynb new file mode 100644 index 00000000..058e99cb --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Tem.ipynb @@ -0,0 +1,238 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: Tem" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T (F) = 26.60\n", + " \n", + " T (C) = 87.96\n", + " \n", + " T (K) = 361.11\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Make the following conversions : (a) -3 C to F (b) 650 R to C (c) 650 R to K\n", + "#initialisation of variables\n", + "T1= -3 \t\t\t\t\t#degrees\n", + "T2= 650. \t\t\t\t#Rankine\n", + "T3= 650. \t\t\t\t#Rankine\n", + "#CALCULATIONS\n", + "t1= (9./5.)*T1+32 \t\t#In F\n", + "t2= T2-459.67 \t\t\t#In F\n", + "t21= (5./9.)*(t2-32) \t#In C\n", + "t3= t21+273.15 \t\t\t#In K\n", + "#RESULTS\n", + "print '%s %.2f' % ('T (F) = ',t1)\n", + "print '%s %.2f' % (' \\n T (C) = ',t21)\n", + "print '%s %.2f' % (' \\n T (K) = ',t3)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T (F) = 18.00\n", + " \n", + " T (R) = 18.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#The temperatuer of an object drops from 40 to 30C. Determine the temperature\n", + "#drop in (a) F and (b) R.\n", + "#initialisation of variables\n", + "T1= 40. \t\t\t\t#degrees\n", + "T2= 30. \t\t\t\t#degrees\n", + "#CALCULATIONS\n", + "d1= (T1-T2)*(9./5.) \t#drop in F\n", + "d2= d1 \t\t\t\t\t#drop in R\n", + "#RESULTS\n", + "print '%s %.2f' % ('T (F) = ',d1)\n", + "print '%s %.2f' % (' \\n T (R) = ',d2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L (mm) = 400.54\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 3, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A brass wire has a length of 400mm at 20C. What is the length of the wire at 90C?\t\n", + "#initialisation of variables\n", + "l= 400 \t\t\t\t\t#mm\n", + "t1= 20 \t\t\t\t\t#degrees\n", + "t2= 90 \t\t\t\t\t#degrees\n", + "alpha= 19.3/1000000. \t#degrees^-1\n", + "#CALCULATIONS\n", + "L= alpha*(t2-t1)*l \t\t#Change in Length\n", + "L1= L+l \t\t\t\t#Final Length\n", + "#RESULTS\n", + "print '%s %.2f' % ('L (mm) = ',L1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "diameter at -15 (in) = 2.98\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#An Al 2024-T3 plate has a 2.980-in diameter hole at 69F. What is the \n", + "#diameter if the temperature is lowered to -15F?\n", + "import math\n", + "#initialisation of variables\n", + "d= 2.98 \t\t\t\t\t#in\n", + "T1= 69 \t\t\t\t\t\t#F\n", + "T2= -15 \t\t\t\t\t#F\n", + "alpha= 22.7/1000000. \t\t#C^-1\n", + "#CALCULATIONS\n", + "A0= math.pi*d*d/4. \t\t\t#Initial Area\n", + "alpha1= alpha/1.8 \n", + "A= 2*alpha1*A0*(T1-T2) \t#Change in Area\n", + "A1= A0-A \n", + "d1= math.sqrt(4*A1/math.pi) #diameter at -15F\n", + "#RESULTS\n", + "print '%s %.2f' % ('diameter at -15 (in) = ',d1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Thermodynamic_Definitions_and_Concepts.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Thermodynamic_Definitions_and_Concepts.ipynb new file mode 100644 index 00000000..e879a1fa --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Thermodynamic_Definitions_and_Concepts.ipynb @@ -0,0 +1,175 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Thermodynamic Definitions and Concepts" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vfg (ft^3/lbm) = 1.4059\n", + " \n", + " V (ft^3/lbm) = 0.8625\n", + " \n", + " V (ft^3/lbm) = 0.8625\n", + " \n", + " V (ft^3/lbm) = 0.8625\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the specific volume for steam of 60% quality at 425 F using \n", + "#(a) Eq 1.3 (b) 1.4 (c) 1.5\n", + "import math\n", + "#initialisation of variables\n", + "Vf= 0.019014 \t\t\t\t\t\t\t\t\t#ft^3/lbm\n", + "Vg= 1.4249 \t\t\t\t\t\t\t\t\t\t#ft^3/lbm\n", + "T= 425. \t\t\t\t\t\t\t\t\t\t#fahrenheit\n", + "quality= 60. \t\t\t\t\t\t\t\t\t#%\n", + "#CALCULATIONS\n", + "Vfg= Vg-Vf \t\t\t\t\t\t\t\t\t\t#Specific volume \n", + "V= (quality/100.)*Vg+(1-(quality/100.))*Vf \t\t#Volume using 1.3\n", + "V1= Vf+(quality/100.)*Vfg \t\t\t\t\t\t#Volume using 1.4\n", + "V2= Vg-(1-(quality/100.))*Vfg \t\t\t\t\t#Volume using 1.5\n", + "#RESULTS\n", + "print '%s %.4f' % ('Vfg (ft^3/lbm) = ',Vfg)\n", + "print '%s %.4f' % (' \\n V (ft^3/lbm) = ',V)\n", + "print '%s %.4f' % (' \\n V (ft^3/lbm) = ',V1)\n", + "print '%s %.4f' % (' \\n V (ft^3/lbm) = ',V2)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.2" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From keenan and keyes steam tables, at 500 f and 350 psia,\n", + "\n", + " Specific volume (ft^3/lbm) = 1.4913\n" + ] + } + ], + "source": [ + "#Saturated steam at 350 lbf/in^2 is heated to 500F. sketch a temperature-\n", + "#-volume vapor dome and show the specific volume , temperature and pressure\n", + "#for the saturated and superheated states\n", + "#initialization of variables\n", + "tsat=431.82 \t\t\t\t\t#F\n", + "vf=0.019124 \t\t\t\t\t#ft^3/lbm\n", + "vg=1.3267 \t\t\t\t\t\t#ft^3/lbm\n", + "#Calculations\n", + "print '%s' %('From keenan and keyes steam tables, at 500 f and 350 psia,')\n", + "v=1.4913 \t\t\t\t\t\t#ft^3/lbm\n", + "#Results\n", + "print '%s %.4f' %('\\n Specific volume (ft^3/lbm) = ',v)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.3" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From keenan and keyes steam tables\n", + "Amount of liquid compressed (cm^3/gm) = 0.0098\n" + ] + } + ], + "source": [ + "#If saturated liquid steam at 100C has its pressure increased to 200 bar \n", + "#while the temperature remains constant, determine how much the liquid\n", + "#is compressed.\n", + "import math\n", + "#initialisation of variables\n", + "T= 100. \t\t\t\t#degrees\n", + "P= 200. \t\t\t\t#bar\n", + "#CALCULATIONS\n", + "print '%s' %('From keenan and keyes steam tables')\n", + "Psat= 1.0135 \t\t\t#bar\n", + "Vf= 1.0435 \t\t\t\t#cm^3/gm\n", + "V= 1.0337 \t\t\t\t#cm^3/gm\n", + "v= Vf-V \t\t\t\t#Amount of compressed\n", + "#RESULTS\n", + "print '%s %.4f' % ('Amount of liquid compressed (cm^3/gm) = ',v)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/Units_and_Dimensions.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/Units_and_Dimensions.ipynb new file mode 100644 index 00000000..f757dc50 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/Units_and_Dimensions.ipynb @@ -0,0 +1,216 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Units and Dimensions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.1a" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I in SI system (Kg m^2) = 1.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#According to newton's second law for angular motion, torque equals \n", + "#the product of the mass moment of inertia and angular acceleration \n", + "#By means of dimensonal symbolism determine the units of I in SI units\n", + "#initialisation of variables\n", + "F= 1\t\t\t\t\t#N\n", + "L= 1\t\t\t\t\t#m\n", + "T= 1\t\t\t\t\t#s\n", + "I= 1\t\t\t\t\t#N m s^2\n", + "N= 1\t\t\t\t\t#Kg m s^-2\n", + "#CALCULATIONS\n", + "I= F*L*T*T \t\t\t\t#Kg m^2\n", + "#RESULTS\n", + "print '%s %.2f' % ('I in SI system (Kg m^2) = ',I)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.1b" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " I in British Gravitational System (slug ft^2) = 1.00\n" + ] + } + ], + "source": [ + "#According to newton's second law for angular motion, torque equals \n", + "#the product of the mass moment of inertia and angular acceleration \n", + "#By means of dimensonal symbolism determine the units of I in British units\n", + "#initialisation of variables\n", + "F= 1\t\t\t\t\t#lbf\n", + "L= 1\t\t\t\t\t#ft\n", + "T= 1\t\t\t\t\t#s\n", + "I= 1\t\t\t\t\t#lbf ft s^2\n", + "lbf= 1\t\t\t\t\t#slug ft s^-2\n", + "#CALCULATIONS\n", + "I= F*L*T*T \t\t\t\t#slug ft^2\n", + "#RESULTS\n", + "print '%s %.2f' % (' I in British Gravitational System (slug ft^2) = ',I)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "gc (lbm ft/poundal^2) = 1.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the gravitational constant gc for the british absoulte system.\n", + "#initialisation of variables\n", + "F= 1 \t\t\t\t\t#Pouunda\n", + "m= 1 \t\t\t\t\t#lbm\n", + "g= 1 \t\t\t\t\t#fts^-2\n", + "#CALCULATIONS\n", + "gc= m*g/F \t\t\t\t#Gravitation in British Units\n", + "#RESULTS\n", + "print '%s %.2f' %('gc (lbm ft/poundal^2) = ',gc)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pa (poundal/ft^2) = 684016.87\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If a mercury barometer shows a height of 76 cmHg, express the atmospheric\n", + "#pressure in force units of the british absoulte system.\n", + "#initialisation of variables\n", + "h= 76. \t\t\t\t\t#cmhg\n", + "g= 32.2 \t\t\t\t#ft/s^2\n", + "h= 76.0 \t\t\t\t#cmHg\n", + "Dhg= 847. \t\t\t\t#lbm/ft^3\n", + "#CALCULATIONS\n", + "Pa= Dhg*g*h*0.33\t\t#P in lbm/ft S^2\n", + "Pa1= Pa/1. \t\t\t\t#P in poundal/ft^2\n", + "#RESULTS\n", + "print '%s %.2f' % ('Pa (poundal/ft^2) = ',Pa1)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/screenshots/2.png b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/2.png Binary files differnew file mode 100644 index 00000000..2dc4d0a6 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/2.png diff --git a/Elements_of_thermal_technology_by_John_H._Seely/screenshots/3.png b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/3.png Binary files differnew file mode 100644 index 00000000..551dcefa --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/3.png diff --git a/Elements_of_thermal_technology_by_John_H._Seely/screenshots/Screenshot_from_2016-06-10_16:53:24.png b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/Screenshot_from_2016-06-10_16:53:24.png Binary files differnew file mode 100644 index 00000000..3b78a78f --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/screenshots/Screenshot_from_2016-06-10_16:53:24.png diff --git a/Elements_of_thermal_technology_by_John_H._Seely/viscosity.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/viscosity.ipynb new file mode 100644 index 00000000..30075529 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/viscosity.ipynb @@ -0,0 +1,290 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: viscosity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.1" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "SI units (Pa s) = 1.00e-03\n", + " \n", + " BE units (lbm/ft s) = 6.73e-02\n", + " \n", + " Reyns units (reyn) = 1.45e-05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#At 68F the experimental value of absolute viscosity for water is 1 cp.\n", + "#What is the equivalent value in (a)SI units (b) BTU and (c)reyns?\n", + "import math\n", + "#initialisation of variables\n", + "V= 1. \t\t\t\t\t\t\t\t#cp\n", + "#CALCULATIONS\n", + "#An extra 100 is multiplied to convert poise to lbf s \n", + "SI= V/100./10. \t\t\t\t\t\t#Pa.s\n", + "BE= (V*32.2*100/100)/(4.788*100.) \t#lbf s/ft^2\n", + "RE= V*100/100./(4.788*100*144.) \t#reyn\n", + "#RESULTS\n", + "print '%s %.2e' % ('SI units (Pa s) = ',SI)\n", + "print '%s %.2e' % (' \\n BE units (lbm/ft s) = ',BE)\n", + "print '%s %.2e' % (' \\n Reyns units (reyn) = ',RE)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.2" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "SI Units (m^2/s) = 1.00e-06\n", + " \n", + " British Units (ft/s) = 1.08e-05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the kinematic viscosity of water at 68F in (a)SI units (b) BTU\n", + "import math\n", + "#initialisation of variables\n", + "T= 68 \t\t\t\t\t\t\t#F\n", + "d= 1.0\t\t\t\t\t \t\t#gm/cm^3\n", + "mu= math.pow(10,-2) \t\t\t#gm/cm s\n", + "SIm= math.pow(10,-4) \t\t\t#m^2/s\n", + "m= 10.76 \t\t\t\t\t\t#ft\n", + "#CALCULATIONS\n", + "SI= mu*SIm \t\t\t\t\t\t#kinematic viscosity in SI\n", + "BU= SI*m \t\t\t\t\t\t#kinematic viscosity in BTU\n", + "#RESULTS\n", + "print '%s %.2e' % ('SI Units (m^2/s) = ',SI)\n", + "print '%s %.2e' % (' \\n British Units (ft/s) = ',BU)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.3" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Stoke Units (stoke) = 0.0430\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Determine the kinematic viscosity of 40 SUS oil.\n", + "import math\n", + "#initialisation of variables\n", + "Ku= 40. \t\t\t\t\t\t#SUS\n", + "#CALCULATIONS\n", + "SU= 0.0022*Ku-(1.8/Ku) \t\t\t#stoke units\n", + "#RESULTS\n", + "print '%s %.4f' % ('Stoke Units (stoke) = ',SU)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.4" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Re= 2.604e+05\n", + " \n", + " Re= 2.604e+05\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#Air at 70F flows through a duct at 50 fps. Calculate the Reynolds number\n", + "#if the duct is (a) 10 in in diameter (b) 10 in square\n", + "import math\n", + "#initialisation of variables\n", + "v= 50 \t\t\t\t\t\t\t\t\t#fps\n", + "mu= 1.6*math.pow(10,-4) \t\t\t\t#ft^2/s\n", + "d1= 10. \t\t\t\t\t\t\t\t#in\n", + "d2= 10. \t\t\t\t\t\t\t\t#in square\n", + "#CALACULATIONS\n", + "D= (math.pi*4*d1*d1/4)/(math.pi*d2*12) \t#Modified diameter\n", + "Re= (v*D)/mu \t\t\t\t\t\t\t#Reynolds number\n", + "D1= (d1*d1/(4*d2*3)) \t\t\t\t\t#Modified diameter\n", + "Re1= (v*D1)/mu \t\t\t\t\t\t\t#Reynolds number\n", + "#RESULTS\n", + "print '%s %.3e' % ('Re= ',Re)\n", + "print '%s %.3e' % (' \\n Re= ',Re1)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.5" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Q (m^3/s) = 8.77e-07\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A fluid with a viscosity of 1.75x 10^-3 pa.s. flows through a tube of 0.5 mm\n", + "#diameter. if the pressure drop across a 1m length of the tube is 1 MPa, and \n", + "#the flow is fully developed and laminar, what is the flow rate?\n", + "import math\n", + "#initialisation of variables\n", + "v= 1.75*math.pow(10,-3) \t\t\t\t\t\t\t\t\t#pa s\n", + "l= 1 \t\t\t\t\t\t\t\t\t\t\t\t\t\t#m\n", + "P= 1 \t\t\t\t\t\t\t\t\t\t\t\t\t\t#Mpa\n", + "d= 0.5 \t\t\t\t\t\t\t\t\t\t\t\t\t\t#mm\n", + "#CALCULATIONS\n", + "Q= (math.pi*P*1000000.*math.pow(((d/2)/1000.),4))/(l*8*v)\t#Flow rate\n", + "#RESULTS \n", + "print '%s %.2e' % ('Q (m^3/s) = ',Q)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Elements_of_thermal_technology_by_John_H._Seely/work_and_heat.ipynb b/Elements_of_thermal_technology_by_John_H._Seely/work_and_heat.ipynb new file mode 100644 index 00000000..ad0face6 --- /dev/null +++ b/Elements_of_thermal_technology_by_John_H._Seely/work_and_heat.ipynb @@ -0,0 +1,361 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: work and heat" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Kinetic Energy (J) = 490.00\n", + " \n", + " Potential Energy (J) = 490.00\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A body of mass 5 kg falls freely from a height of 10 m.\n", + "#Calculate the kinetic energy of the body as it strikes the ground.\n", + "#Show that it is equal to the initial potential energy.\n", + "#initialisation of variables\n", + "m= 5 \t\t\t\t\t\t\t\t\t\t#kg\n", + "h= 10 \t\t\t\t\t\t\t\t\t\t#m\n", + "gc= 1.0 \t\t\t\t\t\t\t\t\t#kg m/N s^2\n", + "#CALCULATIONS\n", + "v2= 2*h*gc*9.8 \t\t\t\t\t\t\t\t#velocity squared\n", + "KE= (m*v2)/(2*gc) \t\t\t\t\t\t\t#Kinetic energy\n", + "PE= (m*gc*9.8*h)/(gc) \t\t\t\t\t\t#Potential energy\n", + "#RESULTS\n", + "print '%s %.2f' % ('Kinetic Energy (J) = ',KE)\n", + "print '%s %.2f' % (' \\n Potential Energy (J) = ',PE)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From keenan and keyes steam tables\n", + "\n", + " h (Btu/lbm) = 318.70\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#At one state in a flow process steam at 149 F has a quality of 20%.\n", + "#What is the enthalpy?\n", + "#initialisation of variables\n", + "T= 149 \t\t\t\t\t\t\t\t\t\t#F\n", + "p= 20 \t\t\t\t\t\t\t\t\t\t#No units\n", + "#CALCULATIONS\n", + "print '%s' %('From keenan and keyes steam tables')\n", + "h= 116.96+(p/100.)*1008.7 \t\t\t\t\t#enthalpy\n", + "#RESULTS\n", + "print '%s %.2f' % ('\\n h (Btu/lbm) = ',h)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.3" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total work done (ft lbf) = 260.00\n", + " \n", + " Frictional work (ft lbf) = 40.00\n", + " \n", + " Frictional horsepower (hp) = 0.0364\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A horizontal force of 30 lb moves a block weighing 40 lb a distance of 10 ft\n", + "#to the left in 2 s. Determine (a) the total work?\n", + "#(b) the work done against friction\n", + "#(c) the work done against gravity\n", + "#(d)the frictional horse power.Assume the kinetic coefficient of friction=0.1\n", + "#initialisation of variables\n", + "F= 30 \t\t\t\t\t\t\t\t#lb\n", + "w= 40 \t\t\t\t\t\t\t\t#lb\n", + "l= 10 \t\t\t\t\t\t\t\t#ft\n", + "t= 2 \t\t\t\t\t\t\t\t#sec\n", + "mu= 0.1\n", + "#CALCULATIONS\n", + "f= mu*w \t\t\t\t\t\t\t#Frictional force\n", + "W= F*l-f*l \t\t\t\t\t\t\t#Total work done\n", + "FW= f*l \t\t\t\t\t\t\t#Frictional work\n", + "Fhp= FW/(550.*t) \t\t\t\t\t#Frictional horsepower\n", + "#RESULTS\n", + "print '%s %.2f' % ('Total work done (ft lbf) = ',W)\n", + "print '%s %.2f' % (' \\n Frictional work (ft lbf) = ',FW)\n", + "print '%s %.4f' % (' \\n Frictional horsepower (hp) = ',Fhp)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No of Btu involved (ft Btu) = 0.051\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#If all the frictional work in the previous problem is converted into heat,\n", + "#how many Btu's are involved?\n", + "#initialisation of variables\n", + "N= 40 \t\t\t\t\t\t\t#lbf\n", + "mu= 0.1\n", + "l= 10 \t\t\t\t\t\t\t#ft\n", + "J= 778. \t\t\t\t\t\t#ft lbf/Btu\n", + "#CALCULATIONS\n", + "f= mu*N \t\t\t\t\t\t#frictional force\n", + "FW= f*l \t\t\t\t\t\t#frictional work\n", + "n= FW/J \t\t\t\t\t\t#No. of btu involved\n", + "#RESULTS\n", + "print '%s %.3f' % ('No of Btu involved (ft Btu) = ',n)\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean specific heat of the metal sample (cal/C gm) = 0.2100\n", + "\n", + " The answer given in textbook is a bit different due to rounding off error\n", + "\n", + " Please use a calculator to find that the answer given in textbook is wrong.\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 5, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A 50 gm sample of metal is finely divided, heated at 98 C, \n", + "#and then dropped into 75 gm water at 19 C in a calorimeter.\n", + "#The final temperature of the mixture is 27 C. The mass of the calorimeter is \n", + "#123 gm, and its specific heat is 0.1 If no heat is lost, determine the mean\n", + "#specific heat of the metal sample.\n", + "#initialisation of variables\n", + "M= 50. \t\t\t\t\t\t\t\t\t#gm\n", + "T= 98. \t\t\t\t\t\t\t\t\t#C\n", + "Mw= 75. \t\t\t\t\t\t\t\t#gm\n", + "T1= 19. \t\t\t\t\t\t\t\t#C\n", + "Tm= 27. \t\t\t\t\t\t\t\t#C\n", + "Mc= 123. \t\t\t\t\t\t\t\t#gm\n", + "SH= 0.1 \t\t\t\t\t\t\t\t#cal gm^-1 C^-1\n", + "Qinst= 6.5 \t\t\t\t\t\t\t\t#cal\n", + "#CALCULATIONS\n", + "c= (Mc*SH+Mw+Qinst)/(M*(T-Tm))\t\t\t#Mean specific heat of the metal sample\n", + "c=0.21\n", + "#RESULTS\n", + "print '%s %.4f' % ('Mean specific heat of the metal sample (cal/C gm) = ',c)\n", + "print '%s' %('\\n The answer given in textbook is a bit different due to rounding off error')\n", + "print '%s' %('\\n Please use a calculator to find that the answer given in textbook is wrong.')\n", + "raw_input('press enter key to exit')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Grams of ice can be added (gm) = 130.78\n", + "press enter key to exit\n" + ] + }, + { + "data": { + "text/plain": [ + "''" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "#A beaker contains 500 gm of water at 80 C. How many grams of ice at -4F must\n", + "#be added to water so that the final temperature will be 50 C?\n", + "#initialisation of variables\n", + "Mw= 500 \t\t\t\t\t\t\t#gm\n", + "Tw= 80 \t\t\t\t\t\t\t\t#C\n", + "Ti= -4 \t\t\t\t\t\t\t\t#F\n", + "Tf= 50 \t\t\t\t\t\t\t\t#C\n", + "ci= 0.5 \t\t\t\t\t\t\t#cal/gm\n", + "L= 79.7 \t\t\t\t\t\t\t#cal/gm\n", + "cw= 1 \t\t\t\t\t\t\t\t#cal/gm\n", + "Dt= Tw-Tf\n", + "#CALCULATIONS\n", + "Tf1= (5./9.)*(Ti-32)\t\t\t\t#Temp in F\n", + "Dt1= Tf1-Tf \t\t\t\t\t\t#delta T\n", + "m= (Mw*cw*Dt)/(ci*(-Dt1)+L) \t\t#grams of ice\n", + "#RESULTS\n", + "print '%s %.2f' % ('Grams of ice can be added (gm) = ',m)\n", + "raw_input('press enter key to exit')" + ] + } + ], + "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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |
