diff options
Diffstat (limited to 'Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch5.ipynb')
-rwxr-xr-x | Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch5.ipynb | 1241 |
1 files changed, 0 insertions, 1241 deletions
diff --git a/Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch5.ipynb b/Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch5.ipynb deleted file mode 100755 index 0aec293d..00000000 --- a/Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch5.ipynb +++ /dev/null @@ -1,1241 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 5 - Second law of themodynamics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.1 Page: 150" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.1 - Page: 150\n", - "\n", - "\n", - "The theoretical efficiency of heat engine is 63.5 %\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "print \"Example: 5.1 - Page: 150\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Th = 550 + 273## [K]\n", - "Tl = 27 + 273## [K]\n", - "#************#\n", - "\n", - "# The theoretical efficiency of a heat engine is given by:\n", - "# eta = Net Work Output/Net Work Input\n", - "# eta = Wnet/Qin\n", - "# eta = (Qin - Qout)/Qin = (Th - Tl)/Th\n", - "eta = (Th - Tl)/Th#\n", - "print \"The theoretical efficiency of heat engine is %.1f %%\"%(eta * 100)" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.2 Page: 150" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.2 - Page: 150\n", - "\n", - "\n", - "(a) The efficiency of the heat engine is 63.0 %\n", - "\n", - "(b) The efficiency of the heat engine is 78.0 %\n", - "\n", - "(c) The efficiency of the heat engine is 57.5 %\n", - "\n" - ] - } - ], - "source": [ - "print \"Example: 5.2 - Page: 150\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Th = 810## [K]\n", - "Tl = 300## [K]\n", - "#*************#\n", - "\n", - "# Solution (a)\n", - "eta = (Th - Tl)/Th#\n", - "print \"(a) The efficiency of the heat engine is %.1f %%\\n\"%(eta*100)#\n", - "\n", - "# Solution (b)\n", - "Th = 1366## [K]\n", - "Tl = 300## [K]\n", - "eta = (Th - Tl)/Th#\n", - "print \"(b) The efficiency of the heat engine is %.1f %%\\n\"%(eta*100)#\n", - "\n", - "# Solution (c)\n", - "Th = 810## [K]\n", - "Tl = 344## [K]\n", - "eta = (Th - Tl)/Th#\n", - "print \"(c) The efficiency of the heat engine is %.1f %%\\n\"%(eta*100)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.3 Page: 151" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.3 - Page: 151\n", - "\n", - "\n", - "(a) The efficiency of the Carnot engine is 67.2 %\n", - "\n", - "(b) Heat released to cold reservoir is 192 kJ\n", - "\n" - ] - } - ], - "source": [ - "print \"Example: 5.3 - Page: 151\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Th = 650 + 273## [K]\n", - "Tl = 30 + 273## [K]\n", - "Qh = 585## [kJ/cycle]\n", - "#*************#\n", - "\n", - "# Solution (a)\n", - "# From Eqn. (5.9)\n", - "eta = (Th - Tl)/Th#\n", - "print \"(a) The efficiency of the Carnot engine is %.1f %%\\n\"%(eta*100)#\n", - "\n", - "# Soluton (b)\n", - "# From the knowledge of the ratio of heat and temperature between the two regions:\n", - "Ql = Qh*Tl/Th## [kJ]\n", - "print \"(b) Heat released to cold reservoir is %d kJ\\n\"%(Ql)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.4 Page: 151" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.4 - Page: 151\n", - "\n", - "\n", - "Minimum Work requirement is 27 kJ\n", - "\n", - "Amount of heat released to the surrounding is 362 kJ\n", - "\n" - ] - } - ], - "source": [ - "print \"Example: 5.4 - Page: 151\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "m = 1## [kg]\n", - "Tl = 273## [K]\n", - "Th = 295## [K]\n", - "Ql = 335## [kJ/kg]\n", - "#*************#\n", - "\n", - "# Solution (a)\n", - "# The coeffecient of performance of refrigerating machine is:\n", - "# COP = Ql/Wnet = Tl/(Th - Tl)\n", - "Wnet = Ql*(Th - Tl)/Tl## [kJ]\n", - "print \"Minimum Work requirement is %d kJ\\n\"%(round(Wnet))#\n", - "\n", - "# Solution (b)\n", - "# Amount of heat released:\n", - "# Wnet = Qh - Ql\n", - "Qh = Wnet + Ql## [kJ]\n", - "print \"Amount of heat released to the surrounding is %d kJ\\n\"%(round(Qh))#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.5 Page: 152" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.5 - Page: 152\n", - "\n", - "\n", - "Minimum Work Required is 13.1 kJ\n", - "\n", - "The efficiency of Heat Engine is 0.263\n", - "\n", - "Amount of Heat released is 36.9 kJ\n", - "\n" - ] - } - ], - "source": [ - "print \"Example: 5.5 - Page: 152\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Th = 373## [K]\n", - "Tl = 275## [K]\n", - "Qh = 50## [kJ]\n", - "#*************#\n", - "\n", - "# Solution (a)\n", - "# Theral Efficiency of the engine can be given as:\n", - "# eta_HE = Wnet/Qh#\n", - "# Wnet = Qh*COP = Qh*(Th - Tl)/Th#\n", - "Wnet = Qh*(Th - Tl)/Th## [kJ]\n", - "print \"Minimum Work Required is %.1f kJ\\n\"%(Wnet)#\n", - "\n", - "# Solution (b)\n", - "eta = (Th - Tl)/Th#\n", - "print \"The efficiency of Heat Engine is %.3f\\n\"%(eta)#\n", - "\n", - "# Solution (c)\n", - "# Amount of heat released can be calculated as:\n", - "# eta = Net Work Output/Net Work Input#\n", - "# eta = Wnet/Qin#\n", - "# eta = (Qin - Qout)/Qin#\n", - "Qin = Qh## [kJ]\n", - "Qout = Qin*(1 - eta)#\n", - "print \"Amount of Heat released is %.1f kJ\\n\"%(Qout)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.6 Page: 153" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.6 - Page: 153\n", - "\n", - "\n", - "Claim is not Valid\n" - ] - } - ], - "source": [ - "print \"Example: 5.6 - Page: 153\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "W = 5## [hp]\n", - "Q = 7000## [J/s]\n", - "Th = 400 + 273## [K]\n", - "Tl = 24 + 273## [K]\n", - "#*************#\n", - "\n", - "W = 5*745.7## [W]\n", - "thermal_eta = W/Q#\n", - "theoretical_eta = (Th - Tl)/Th#\n", - "\n", - "if theoretical_eta <= thermal_eta:\n", - " print \"Claim is Valid\"\n", - "else:\n", - " print \"Claim is not Valid\"\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.7 Page: 162" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.7 - Page: 162\n", - "\n", - "\n", - "Change in Entropy is 23.81 J/mol K\n" - ] - } - ], - "source": [ - "print \"Example: 5.7 - Page: 162\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "n = 1## [mol]\n", - "deltaH_fusion = 6500## [J/mol]\n", - "T_Tr = 273## [transition temperature, K]\n", - "P = 1## [atm]\n", - "#************#\n", - "\n", - "# By Eqn. (9.40)\n", - "deltaS_fusion = deltaH_fusion/T_Tr## [J/mol K]\n", - "print \"Change in Entropy is %.2f J/mol K\"%(deltaS_fusion)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.8 Page: 164" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.8 - Page: 164\n", - "\n", - "\n", - "Change in Entropy is 22.876 eu\n" - ] - } - ], - "source": [ - "from math import log\n", - "print \"Example: 5.8 - Page: 164\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "V1 = 5## [L]\n", - "V2 = 50## [L]\n", - "n = 5## [moles]\n", - "R = 1.987## [cal/mol K]\n", - "#**************#\n", - "\n", - "# Change in entropy for an isothermal change for an ideal gas:\n", - "# deltaS = n*R*log(P1/P2) = n*R*log(V2/V1)\n", - "deltaS = n*R*log(V2/V1)## [cal/degree]\n", - "print \"Change in Entropy is %.3f eu\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.9 Page: 164" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.9 - Page: 164\n", - "\n", - "\n", - "deltaS is 115.26 J/K\n", - "Change in Entropy is 115.257 eu\n" - ] - } - ], - "source": [ - "print \"Example: 5.9 - Page: 164\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "n = 8## [mol]\n", - "R = 8.314## [J/mol K]\n", - "T2 = 700## [K]\n", - "T1 = 350## [K]\n", - "Cp = (5/2)*R## [J/mol K]\n", - "#*************#\n", - "\n", - "deltaS = n*Cp*log(T2/T1)## [J/K]\n", - "print \"deltaS is %.2f J/K\"%(deltaS)#\n", - "\n", - "R*log(V2/V1)#// [cal/degree]\n", - "print \"Change in Entropy is %.3f eu\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.10 Page: 164" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.10 - Page: 164\n", - "\n", - "\n", - "Change in Entropy is -19.179548 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.10 - Page: 164\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "n = 5## [moles]\n", - "T1 = 300## [K]\n", - "T2 = 400## [K]\n", - "P1 = 3## [bars]\n", - "P2 = 12## [bars]\n", - "Cp = 26.73## [J/mol K]\n", - "R = 8.314## [K/mol K]\n", - "#*************#\n", - "\n", - "deltaS = n*((Cp*log(T2/T1)) + (R*log(P1/P2)))## [kJ/K]\n", - "print \"Change in Entropy is %f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.11 Page: 166" - ] - }, - { - "cell_type": "code", - "execution_count": 13, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.11 - Page: 166\n", - "\n", - "\n", - "Entropy Change is 4.27 kJ/kmol K\n" - ] - } - ], - "source": [ - "print \"Example: 5.11 - Page: 166\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "N = 1## [kmol]\n", - "xA = 0.21## [for Oxygen]\n", - "xB = 0.79## [for Nitrogen]\n", - "R = 8.314## [kJ/kmol K]\n", - "#*************#\n", - "\n", - "deltaS = - (N*R*(xA*log(xA) + xB*log(xB)))## [kJ/mol K]\n", - "print \"Entropy Change is %.2f kJ/kmol K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.12 Page: 167" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.12 - Page: 167\n", - "\n", - "\n", - "Change in Entropy is 1.117 cal/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.12 - Page: 167\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Vol_O2 = 5.6## [L]\n", - "Vol_H2 = 16.8## [L]\n", - "R = 1.987## [cal/mol K]\n", - "#*************#\n", - "\n", - "xA = Vol_O2/22.4## [mole fraction O2]\n", - "xB = Vol_H2/22.4## [mle fraaction H2]\n", - "N = xA + xB## [total number of moles]\n", - "# From Eqn. 5.21:\n", - "deltaS = - (N*R*(xA*log(xA) + xB*log(xB)))## [cal/K]\n", - "print \"Change in Entropy is %.3f cal/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.13 Page: 168" - ] - }, - { - "cell_type": "code", - "execution_count": 15, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.13 - Page: 168\n", - "\n", - "\n", - "Entropy Change is 0.319 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.13 - Page: 168\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "m = 80## [mass of Argon, g]\n", - "T1 = 300## [K]\n", - "T2 = 500## [K]\n", - "Cv = 0.3122## [kJ/kg K]\n", - "#**************#\n", - "\n", - "Mw = 40## [Molecular Weight of Argon]\n", - "n = m/Mw## [moles]\n", - "deltaS = n*Cv*log(T2/T1)## [kJ/K]\n", - "print \"Entropy Change is %.3f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.14 Page: 168" - ] - }, - { - "cell_type": "code", - "execution_count": 17, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.14 - Page: 168\n", - "\n", - "\n", - "The upper temperature of the process is 864.929 K\n" - ] - } - ], - "source": [ - "from math import exp\n", - "print \"Example: 5.14 - Page: 168\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "deltaS = 1## [kJ/kg K]\n", - "Cv = 0.918## [kJ/kg K]\n", - "T1 = 273 + 18## [K]\n", - "#*************#\n", - "\n", - "# Let the upper temperature be T.\n", - "# deltaS = integrate('Cv/T','T',T1,T)# \n", - "# deltaS = Cv*log(T/T1)\n", - "T = T1*exp(deltaS/Cv)## [K]\n", - "print \"The upper temperature of the process is %.3f K\"%(T)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.15 Page: 169" - ] - }, - { - "cell_type": "code", - "execution_count": 24, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.15 - Page: 169\n", - "\n", - "\n", - "Change in Entropy is -5.450 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.15 - Page: 169\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "m1 = 5## [kg]\n", - "m2 = 20## [kg]\n", - "C = 4.2## [kJ/kg K]\n", - "T1 = 350## [K]\n", - "T2 = 250## [K]\n", - "#**************#\n", - "\n", - "# Suppose the final temperature is T\n", - "#deff('[y] = f(T)','y = m1*C*(T1 - T) - m2*C*(T - T2)')#\n", - "def f(T):\n", - " y = m1*C*(T1 - T) - m2*C*(T - T2)\n", - " return y\n", - "from scipy.optimize import fsolve\n", - "T = fsolve(f, 7)[0]## [K]\n", - "# Change in entropy of Hot Water:\n", - "from sympy.mpmath import quad\n", - "deltaS1 = m1*C*quad(lambda T:(1/T),[T1,T])## [kJ/K]\n", - "# Change in Entopy of Hot Water:\n", - "deltaS2 = m2*C*quad('(1/T)','T',T2,T)## [kJ/K]\n", - "deltaS = deltaS1 + deltaS2## [kJ/K]\n", - "print \"Change in Entropy is %.3f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.16 Page: 169" - ] - }, - { - "cell_type": "code", - "execution_count": 26, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.16 - Page: 169\n", - "\n", - "\n", - "Change in Entropy is 0.540 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.16 - Page: 169\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "m = 12## [g]\n", - "T1 = 294## [K]\n", - "T2 = 574## [K]\n", - "T = 505## [melting point, K]\n", - "H_fusion = 4.5## [cal/K]\n", - "C_solid = 0.052## [cal/g K]\n", - "C_liquid = 0.062## [cal/g K]\n", - "#*************#\n", - "\n", - "# Entropy Change in heating 12 g of metal from T1 to T\n", - "deltaS1 = m*C_solid*quad(lambda T:(1/T),[T1,T])## [kJ/K]\n", - "# Entropy Change in fusion of metal:\n", - "deltaS2 = m*H_fusion/T## [kJ/K]\n", - "# Entropy Change in heating liquid metal from 505 K to 574 K\n", - "deltaS3 = m*C_liquid*quad(lambda T:(1/T),[T,T2])## [kJ/K]\n", - "deltaS = deltaS1 + deltaS2 + deltaS3## [kJ/K]\n", - "print \"Change in Entropy is %.3f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.17 Page: 170" - ] - }, - { - "cell_type": "code", - "execution_count": 31, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.17 - Page: 170\n", - "\n", - "\n", - "Change in Entropy is 9.165 cal/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.17 - Page: 170\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "#deff('[y] = Cp(T)','y = 7.25 + 2.28*10**(-3)*T')#\n", - "def Cp(T):\n", - " y = (7.25 + 2.28*10**(-3)*T)/T\n", - " return y\n", - "T1 = 273 + 137## [K]\n", - "T2 = 273 + 877## [K]\n", - "#************#\n", - "\n", - "deltaS = quad(Cp,[T1,T2])## [cal/K]\n", - "print \"Change in Entropy is %.3f cal/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.18 Page: 170" - ] - }, - { - "cell_type": "code", - "execution_count": 32, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.18 - Page: 170\n", - "\n", - "\n", - "Total Entropy Change is 7.68 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.18 - Page: 170\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "m_iron = 40## [kg]\n", - "T1 = 625## [K]\n", - "m_water = 160## [kg]\n", - "T2 = 276## [K]\n", - "C_iron = 0.45## [kJ/kg K]\n", - "C_water = 4.185## [kJ/kg K]\n", - "#**************#\n", - "\n", - "#deff('[y] = f(T)','y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)')#\n", - "def f(T):\n", - " y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)\n", - " return y\n", - "from scipy.optimize import fsolve\n", - "T = fsolve(f, 7)## [K]\n", - "# Change in Entropy of the iron casting can be estimated as:\n", - "deltaS1 = m_iron*C_iron*log(T/T1)## [kJ/K]\n", - "# Change in Entropy of Water is given by:\n", - "deltaS2 = m_water*C_water*log(T/T2)## [kJ/K]\n", - "deltaS = deltaS1 + deltaS2## [kJ/K]\n", - "print \"Total Entropy Change is %.2f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.19 Page: 172" - ] - }, - { - "cell_type": "code", - "execution_count": 33, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.19 - Page: 172\n", - "\n", - "\n", - "Entropy at 500 K is 160.73 J/kmol\n" - ] - } - ], - "source": [ - "from math import log\n", - "print \"Example: 5.19 - Page: 172\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "Cp = 21## [J/kmol]\n", - "T1 = 300## [K]\n", - "T2 = 500## [K]\n", - "S1 = 150## [Entropy at T1, J/kmol]\n", - "#*************#\n", - "\n", - "# This is a constant Entropy process. Therefore:\n", - "deltaS = Cp*log(T2/T1)## [J/kmol]\n", - "S2 = S1 + deltaS## [J/kmol]\n", - "print \"Entropy at 500 K is %.2f J/kmol\"%(S2)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.20 Page: 173" - ] - }, - { - "cell_type": "code", - "execution_count": 37, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.20 - Page: 173\n", - "\n", - "\n", - "Total Entropy Change is 369.49 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.20 - Page: 173\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "T1_oil = 273 + 150## [K]\n", - "T2_oil = 50 +273## [K]\n", - "m_water = 4000## [kg]\n", - "T1_water = 273 + 20## [K]\n", - "T2_water = 273 + 130## [K]\n", - "C_water = 4.185## [kJ/kg K]\n", - "C_oil = 2.5## [kJ/kg K]\n", - "#***************#\n", - "\n", - "# For oil:\n", - "deltaT_oil = T1_oil - T2_oil## [K]\n", - "# For water:\n", - "deltaT_water = T2_water - T1_water## [K]\n", - "# The mass flow rate of oil can be measured by the enthalpy balance over the process:\n", - "m_oil = m_water*C_water*deltaT_water/(deltaT_oil*C_oil)## [kg]\n", - "# Change in the Entropy of oil:\n", - "deltaS_oil = m_oil*C_oil*log(T2_oil/T1_oil)## [kJ/K]\n", - "# Change in Entropy of water:\n", - "deltaS_water = m_water*C_water*log(T2_water/T1_water)## [kJ/K]\n", - "deltaS = deltaS_oil + deltaS_water## [kJ/K]\n", - "print \"Total Entropy Change is %.2f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.21 Page: 174" - ] - }, - { - "cell_type": "code", - "execution_count": 38, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.21 - Page: 174\n", - "\n", - "\n", - "Change in Entropy is 1.49 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.21 - Page: 174\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "t = 20*60## [s]\n", - "P = 650## [W]\n", - "T = 273 + 250## [K]\n", - "#*************#\n", - "\n", - "Q = P*t/1000## [kJ]\n", - "deltaS = Q/T## [kJ/K]\n", - "print \"Change in Entropy is %.2f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.22 Page: 174" - ] - }, - { - "cell_type": "code", - "execution_count": 39, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.22 - Page: 174\n", - "\n", - "\n", - "Change in Entropy of the gas is 0.5199 kJ/K\n" - ] - } - ], - "source": [ - "print \"Example: 5.22 - Page: 174\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "#*****Data*****#\n", - "T1 = 400## [K]\n", - "P1 = 300## [kPa]\n", - "V1 = 1## [cubic m]\n", - "V2 =2## [cubic m]\n", - "R = 8.314## [kJ/kmol K]\n", - "#**************#\n", - "\n", - "# Since the system is well insulated, there is no scope of transferring heat between system & surrounding.\n", - "deltaQ = 0## [kJ]\n", - "deltaW = 0## [kJ]\n", - "# By first law of thermodynamics:\n", - "deltaU =deltaQ - deltaW## [kJ]\n", - "# As the internal energy of the gas depends only on temperature,\n", - "deltaT = 0## [K]\n", - "T2 = T1 + deltaT## [K]\n", - "P2 = (P1*V1/T1)*(T2/V2)## [kPa]\n", - "n = P1*V1/(R*T1)## [kmol]\n", - "deltaS_system = n*R*log(P1/P2)## [kJ/K]\n", - "# Since process is adiabatic:\n", - "deltaS_surrounding = 0## [kJ/K]\n", - "deltaS = deltaS_system + deltaS_surrounding## [kJ/K]\n", - "print \"Change in Entropy of the gas is %.4f kJ/K\"%(deltaS)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.23 Page: 174" - ] - }, - { - "cell_type": "code", - "execution_count": 41, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.23 - Page: 174\n", - "\n", - "\n", - " This problem involves proving a relation in which no mathematics and no calculations are involved.\n", - "\n", - "\n", - " For prove refer to this example 5.23 on page 174 of the book.\n" - ] - } - ], - "source": [ - "print \"Example: 5.23 - Page: 174\\n\\n\"\n", - "\n", - "# This problem involves proving a relation in which no mathematics and no calculations are involved.\n", - "# For prove refer to this example 5.23 on page number 174 of the book.\n", - "\n", - "print \" This problem involves proving a relation in which no mathematics and no calculations are involved.\\n\\n\"\n", - "print \" For prove refer to this example 5.23 on page 174 of the book.\"" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.24 Page: 182" - ] - }, - { - "cell_type": "code", - "execution_count": 44, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.24 - Page: 182\n", - "\n", - "\n", - "Work lost is 2119.67 kJ\n" - ] - } - ], - "source": [ - "print \"Example: 5.24 - Page: 182\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "# From Example 5.18 (Pg: 170)\n", - "#*****Data*****#\n", - "m_iron = 40## [kg]\n", - "T1 = 625## [K]\n", - "m_water = 160## [kg]\n", - "T2 = 276## [K]\n", - "C_iron = 0.45## [kJ/kg K]\n", - "C_water = 4.185## [kJ/kg K]\n", - "#**************#\n", - "\n", - "#deff('[y] = f(T)','y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)')#\n", - "def f(T):\n", - " y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)\n", - " return y\n", - "T = fsolve(f, 7)[0]## [K]\n", - "# Change in Entropy of the iron casting can be estimated as:\n", - "deltaS1 = m_iron*C_iron*quad(lambda T:(1/T),[T1,T])## [kJ/K]\n", - "# Change in Entropy of Water is given by:\n", - "deltaS2 = m_water*C_water*quad(lambda T:(1/T),[T2,T])## [kJ/K]\n", - "deltaS = deltaS1 + deltaS2## [kJ/K]\n", - "# By Eqn. 5.63:\n", - "W_lost = T2 * deltaS## [kJ]\n", - "print \"Work lost is %.2f kJ\"%(W_lost)#" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example: 5.25 Page: 182" - ] - }, - { - "cell_type": "code", - "execution_count": 45, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Example: 5.25 - Page: 182\n", - "\n", - "\n", - "Total Entropy Change is 1831.74 kJ/K\n", - "\n", - "Since deltaS is a positive quantity, process is irreversible\n", - "\n" - ] - } - ], - "source": [ - "print \"Example: 5.25 - Page: 182\\n\\n\"\n", - "\n", - "# Solution\n", - "\n", - "# *****Data******#\n", - "m_oil = 4750## [kg]\n", - "T1_oil = 515## [K]\n", - "T2_oil = 315## [K]\n", - "m_water = 9500## [kg]\n", - "T1_water = 290## [K]\n", - "Cp_oil = 3.2## [kJ/kg K]\n", - "Cp_water = 4.185## [kJ/kg K]\n", - "#*****************#\n", - "\n", - "# From enthalpy Balance:\n", - "#def('[y] = f(T2_water)','y = m_oil*Cp_oil*(T1_oil - T2_oil) - m_water*Cp_water*(T2_water - T1_water)')#\n", - "def f(T2_water):\n", - " y = m_oil*Cp_oil*(T1_oil - T2_oil) - m_water*Cp_water*(T2_water - T1_water)\n", - " return y\n", - "T2_water = fsolve(f, 7)[0]## [K]\n", - "# Change in the Entropy of oil:\n", - "deltaS_oil = m_oil*Cp_oil*quad(lambda T : (1/T),[T1_oil,T2_oil])## [kJ/K]\n", - "# Change in Entropy of water:\n", - "deltaS_water = m_water*Cp_water*quad(lambda T: (1/T),[T1_water,T2_water])## [kJ/K]\n", - "deltaS = deltaS_oil + deltaS_water## [kJ/K]\n", - "print \"Total Entropy Change is %.2f kJ/K\\n\"%(deltaS)#\n", - "if deltaS > 0:\n", - " print \"Since deltaS is a positive quantity, process is irreversible\\n\"\n", - "else:\n", - " print \"Since deltaS is a negative quantity, process is reversible\\n\"\n" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} |