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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2 - First law of thermodynamics"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.1 Page: 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.1 - Page: 39\n",
+ "\n",
+ "\n",
+ "The work done during the process is -9486.67 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "from __future__ import division\n",
+ "\n",
+ "print \"Example: 2.1 - Page: 39\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "#deff('[E] = f1(T)','E = 50 + 25*T + 0.05*T**2')## [J]\n",
+ "#deff('[Q] = f2(T)','Q = 4000 + 10*T')## [J]\n",
+ "\n",
+ "def f1(T):\n",
+ " E = 50 + 25*T + 0.05*T**2\n",
+ " return E\n",
+ "\n",
+ "def f2(T):\n",
+ " Q = 4000 + 10*T\n",
+ " return Q\n",
+ "Ti = 400## [K]\n",
+ "Tf = 800## [K]\n",
+ "#*************#\n",
+ "\n",
+ "# From the first law of thermodynamics:\n",
+ "# W = Q - delta_E\n",
+ "# W = f2 -f1\n",
+ "from sympy.mpmath import quad\n",
+ "W = quad(lambda T:(4000 + 10*T) - (50 + (25*T) + (0.05*T**2)),[Ti,Tf])#\n",
+ "print \"The work done during the process is %.2f kJ\\n\"%(W/1000)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.2 Page: 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.2 - Page: 40\n",
+ "\n",
+ "\n",
+ "The final internal energy of the fluid is 500 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.2 - Page: 40\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "U1 = 1000## [kJ]\n",
+ "Q = -600# # [kJ]\n",
+ "W = -100## [kJ]\n",
+ "#************#\n",
+ "\n",
+ "# The system is considered to be a closed system. No mass transfer takes place across the system. The tank is rigid. \n",
+ "# So, the kinetic and the potential energies is zero.\n",
+ "# Therefore:\n",
+ "# delta_E = delta_U + delta_PE + delta_KE\n",
+ "# delta_E = delta_U\n",
+ "# From the first law of thermodynamics:\n",
+ "# Q = delta_U + W\n",
+ "# delta_U = Q - W\n",
+ "# U2 - U1 = Q - W\n",
+ "U2 = U1 + Q - W## [kJ]\n",
+ "print \"The final internal energy of the fluid is %d kJ\\n\"%(U2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.3 Page: 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.3 - Page: 40\n",
+ "\n",
+ "\n",
+ "The change in the internal energy of the system would be 1125.99 J/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.3 - Page: 40\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "W = -3## [hp]\n",
+ "Q = -4000## [kJ/h]\n",
+ "#**************#\n",
+ "\n",
+ "# The work done by the stirrer on the system is given by\n",
+ "W = W*745.7## [W]\n",
+ "# The amount of heat transferred to the suroundings can be expressed in terms of J/s:\n",
+ "Q = Q*1000/3600## [J/s]\n",
+ "# From the first law of thermodynamics:\n",
+ "# Q = delta_U - W\n",
+ "delta_U = Q - W## [J/s]\n",
+ "print \"The change in the internal energy of the system would be %.2f J/s\\n\"%(delta_U)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.4 Page: 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.4 - Page: 41\n",
+ "\n",
+ "\n",
+ "The amount of heat transferred from the system to the surroundings during process B-3-A is 95 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.4 - Page: 41\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "# From Fig. 2.4 (Page: 41)\n",
+ "# For process A-1-B:\n",
+ "Q1 = 60## [kJ]\n",
+ "W1 = 35## [kJ]\n",
+ "# For process A-2-B:\n",
+ "W2 = 50## [kJ]\n",
+ "# For process B-3-A:\n",
+ "W3 = -70## [kJ]\n",
+ "#************#\n",
+ "\n",
+ "# For process A-1-B:\n",
+ "# The internal energy of the process A-1-B can be estimated as:\n",
+ "# Q = delta_U + W\n",
+ "delta_U = Q1 - W1## [kJ]\n",
+ "# For process A-2-B:\n",
+ "Q2 = delta_U + W2## [kJ]\n",
+ "# For process B-3-A:\n",
+ "Q3 = -delta_U + W3## [kJ]\n",
+ "print \"The amount of heat transferred from the system to the surroundings during process B-3-A is %d kJ\\n\"%(-Q3)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.5 Page: 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.5 - Page: 41\n",
+ "\n",
+ "\n",
+ "Change in Internal Energy for process 1-2 is 50 kJ\n",
+ "\n",
+ "Change in Internal Energy for process 2-3 is 80 kJ\n",
+ "\n",
+ "Change in Internal Energy for process 3-1 is -130 kJ\n",
+ "\n",
+ "The work done during the adiabatic process is 130 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.5 - Page: 41\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "# For constant pressure process 1-2:\n",
+ "W12 = -100## [kJ]\n",
+ "Q12 = -50## [kJ]\n",
+ "# For constant volume process 2-3:\n",
+ "Q23 = 80## [kJ]\n",
+ "# process 3-1: Adiabatic process\n",
+ "#**************#\n",
+ "\n",
+ "# The internal energy of process 1-2 can be calculated as:\n",
+ "delta_U12 = Q12 - W12## [kJ]\n",
+ "print \"Change in Internal Energy for process 1-2 is %d kJ\\n\"%(delta_U12)#\n",
+ "# For the process 2-3:\n",
+ "# As the process is constant volume process:\n",
+ "W23 = 0## [kJ]\n",
+ "delta_U23 = Q23 - W23## [kJ]\n",
+ "print \"Change in Internal Energy for process 2-3 is %d kJ\\n\"%(delta_U23)#\n",
+ "\n",
+ "# For process 3-1:\n",
+ "# Since the process is adiabatic, ther is no heat transfer between the system and the surrounding.\n",
+ "Q31 = 0## [kJ]\n",
+ "# For a cyclic process, the internal energy change is zero.\n",
+ "# delta_U12 + delta_U23 + delta_U31 = 0\n",
+ "delta_U31 = -(delta_U12 + delta_U23)## [kJ]\n",
+ "# Putting the value of delta_U31:\n",
+ "W31 = Q31 - delta_U31## [kJ]\n",
+ "print \"Change in Internal Energy for process 3-1 is %d kJ\\n\"%(delta_U31)#\n",
+ "print \"The work done during the adiabatic process is %d kJ\\n\"%(W31)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.6 Page: 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.6 - Page: 44\n",
+ "\n",
+ "\n",
+ "Change in Internal Energy is 2087.59 kJ\n",
+ "\n",
+ "Change in enthalpy is 2257 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.6 - Page: 44\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "m = 1## [kg]\n",
+ "Temp = 373## [K]\n",
+ "P = 101325## [N/square m]\n",
+ "V_Liquid = 1.04*10**(-3)## [cubic m/kg]\n",
+ "V_Vapour = 1.673## [cubic m/kg]\n",
+ "Q = 2257## [kJ]\n",
+ "#**************#\n",
+ "\n",
+ "# Work done due to expansion:\n",
+ "Wexpansion = P*(V_Vapour - V_Liquid)## [N-m]\n",
+ "deltaU = Q - Wexpansion/1000## [kJ]\n",
+ "deltaH = deltaU + Wexpansion/1000## [kJ]\n",
+ "\n",
+ "print \"Change in Internal Energy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Change in enthalpy is %d kJ\\n\"%(deltaH)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.7 Page: 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.7 - Page: 45\n",
+ "\n",
+ "\n",
+ "Change in Internal Energy is 26705.16 J\n",
+ "\n",
+ "Change in Enthalpy is 29640 J\n",
+ "\n",
+ "Amount of Heat supplied is 29640 J\n",
+ "\n",
+ "Work done is 2934.84 J\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.7 - Page: 45\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "n = 1## [mol]\n",
+ "Temp = 353## [K]\n",
+ "P = 1## [atm]\n",
+ "Hv = 380## [J/g]\n",
+ "Mwt = 78## [g/mol]\n",
+ "R = 8.314## [J/K mol]\n",
+ "#*************#\n",
+ "\n",
+ "Q = Hv*Mwt## [J/mol]\n",
+ "# Since Vv >> Vl:\n",
+ "# P*(Vv - Vl) = P*Vv =n*R*Temp\n",
+ "Wexpansion = n*R*Temp## [J]\n",
+ "# By first law of thermodynamics:\n",
+ "deltaU = Q - Wexpansion## [J]\n",
+ "deltaH = deltaU + Wexpansion## [J]\n",
+ "\n",
+ "print \"Change in Internal Energy is %.2f J\\n\"%(deltaU)#\n",
+ "print \"Change in Enthalpy is %d J\\n\"%(deltaH)#\n",
+ "print \"Amount of Heat supplied is %d J\\n\"%(Q)#\n",
+ "print \"Work done is %.2f J\\n\"%(Wexpansion)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.8 Page: 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.8 - Page: 45\n",
+ "\n",
+ "\n",
+ "Change in enthalpy is 589.48 cal\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.8 - Page: 45\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "deltaU = 200## [cal]\n",
+ "Vinit = 10## [L]\n",
+ "Vfinal = 50## [L]\n",
+ "#deff('[P] = f(V)','P = 10/V')#\n",
+ "def f(V):\n",
+ " P=10/V\n",
+ " return P\n",
+ "#**************#\n",
+ "\n",
+ "# By definition of enthalpy:\n",
+ "# deltaQ = deltaU + PdV\n",
+ "deltaQ = deltaU + quad(f,[Vinit,Vfinal])*24.2## [cal]\n",
+ "print \"Change in enthalpy is %.2f cal\\n\"%(deltaQ)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.9 Page: 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.9 - Page: 48\n",
+ "\n",
+ "\n",
+ "Rise in Temperature is 58.67 K\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.9 - Page: 48\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "m_water = 1## [kg]\n",
+ "Cv = 4.18## [kJ/kg K]\n",
+ "m_stirrer = 40## [kg]\n",
+ "h = 25## [m]\n",
+ "g = 9.81## [m/square s]\n",
+ "#***************#\n",
+ "\n",
+ "# Since the system is thermally insulated:\n",
+ "# Q = 0\n",
+ "# From the first law of thermodynamics:\n",
+ "# dQ = dE + dW\n",
+ "# As E = U + Ek +Ep and Ek = Ep = 0\n",
+ "# dQ = dU + dW\n",
+ "dT = g*h/Cv## [K]\n",
+ "print \"Rise in Temperature is %.2f K\\n\"%(dT)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.10 Page: 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.10 - Page: 53\n",
+ "\n",
+ "\n",
+ "The final temperature is 753.6 K\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.10 - Page: 53\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "T1 = 300## [K]\n",
+ "V1 = 30## [L]\n",
+ "V2 = 3## [L]\n",
+ "Cv = 5## [cal/mol]\n",
+ "R = 2## [cal/K mol]\n",
+ "#*************#\n",
+ "\n",
+ "Cp = Cv + R## [cal/mol]\n",
+ "gama = Cp/Cv#\n",
+ "# The relation between temperature and volume of ideal gas undergoing adiabatic change is given by:\n",
+ "# (T2/T1) = (V1/V2)**(gama - 1)\n",
+ "T2 = T1 * (V1/V2)**(gama - 1)## [K]\n",
+ "print \"The final temperature is %.1f K\\n\"%(T2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.11 Page: 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.11 - Page: 53\n",
+ "\n",
+ "\n",
+ "The final volme is 7.44 L\n",
+ "\n",
+ "The final temperature is 226.80 K\n",
+ "\n",
+ "Adiabatic work done is 35.72 L-atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.11 - Page: 53\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "n = 2## [mol]\n",
+ "T1 = 293## [K]\n",
+ "P1 = 15##[atm]\n",
+ "P2 = 5## [atm]\n",
+ "Cp = 8.58## [cal/degree mol]\n",
+ "#**************#\n",
+ "\n",
+ "R = 2## [cal/degree mol]\n",
+ "Cv = Cp - R## [cal /degree mol]\n",
+ "gama = Cp/Cv#\n",
+ "R = 0.082## [L atm/degree K]\n",
+ "# Since the gas is ideal:\n",
+ "V1 = n*R*T1/P1## [L]\n",
+ "# Under adiabatic conditions:\n",
+ "# (V2/V1) = (P1/P2)**(1/gama)\n",
+ "V2 = V1*(P1/P2)**(1/gama)## [L]\n",
+ "print \"The final volme is %.2f L\\n\"%(V2)#\n",
+ "\n",
+ "# To determine the final temperature:\n",
+ "# (T2/T1) = (V1/V2)**(gama - 1)#\n",
+ "T2 = T1*(V1/V2)**(gama - 1)## [K]\n",
+ "print \"The final temperature is %.2f K\\n\"%(T2)#\n",
+ "\n",
+ "# Adiabatic Work done can be calculated as:\n",
+ "W = (P1*V1 - P2*V2)/(gama - 1)#\n",
+ "print \"Adiabatic work done is %.2f L-atm\\n\"%(W)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.12 Page: 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.12 - Page: 57\n",
+ "\n",
+ "\n",
+ "Adiabatic Process \n",
+ "\n",
+ "Final Pressure is 0.84 atm\n",
+ "\n",
+ "Final Temperature is 169.673696 K\n",
+ "\n",
+ "Adiabatic Work done is 31.432 L-atm\n",
+ "\n",
+ "\n",
+ "\n",
+ "Isothermal Process\n",
+ "\n",
+ "Final temperature is 323 K\n",
+ "\n",
+ "Final pressure is 1.6 atm\n",
+ "\n",
+ "Work done during the isothermal process is 42.63 L-atm\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "print \"Example: 2.12 - Page: 57\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "m = 1## [kg]\n",
+ "P1 = 8## [atm]\n",
+ "T1 = 50 + 273## [K]\n",
+ "# V1 = V## [L]\n",
+ "# V2 = 5V## [L]\n",
+ "V1_by_V2 = 1/5#\n",
+ "gama = 1.4#\n",
+ "R = 0.082## [L-atm]\n",
+ "#***************#\n",
+ "\n",
+ "# Adiabatic process:\n",
+ "print \"Adiabatic Process \\n\"\n",
+ "P2 = P1*V1_by_V2**gama## [atm]\n",
+ "print \"Final Pressure is %.2f atm\\n\"%(P2)#\n",
+ "T2 = T1*V1_by_V2**(gama - 1)## [K]\n",
+ "print \"Final Temperature is %f K\\n\"%(T2)#\n",
+ "Wad = R*(T2 - T1)/(1 - gama)## [L-atm]\n",
+ "print \"Adiabatic Work done is %.3f L-atm\\n\"%(Wad)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Isothermal Process:\n",
+ "print \"Isothermal Process\\n\"\n",
+ "# In an isothermal Process, the temperature remans constant:\n",
+ "T2 = T1## [K]\n",
+ "print \"Final temperature is %d K\\n\"%(T2)#\n",
+ "# From the ideal gas:\n",
+ "# (P2*V2/T2) = (P1*V1/T1)\n",
+ "# Since T2 = T1\n",
+ "# P2*V2 = P1*V1\n",
+ "P2 = P1*V1_by_V2## [atm]\n",
+ "print \"Final pressure is %.1f atm\\n\"%(P2)#\n",
+ "W = R*T1*log(1/V1_by_V2)## [L-atm]\n",
+ "print \"Work done during the isothermal process is %.2f L-atm\\n\"%(W)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.13 Page: 58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.13 - Page: 58\n",
+ "\n",
+ "\n",
+ "Isochoric Process\n",
+ "\n",
+ "Change in Internal Energy is 716.55 kJ\n",
+ "\n",
+ "Heat Supplied is 716.55 kJ\n",
+ "\n",
+ "Work done is 0 kJ\n",
+ "\n",
+ "Change in Enthalpy is 1003.24 kJ\n",
+ "\n",
+ "\n",
+ "\n",
+ "Isobaric Process\n",
+ "\n",
+ "Change in Internal Energy is 716.76 kJ\n",
+ "\n",
+ "Heat Supplied is 1003.45 kJ\n",
+ "\n",
+ "Work done is 286.69 kJ\n",
+ "\n",
+ "Change in Enthalpy is 1003.45 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.13 - Page: 58\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "m = 5## [kg]\n",
+ "M = 29## [kg/mol]\n",
+ "T1 = 37 + 273## [K]\n",
+ "P1 = 101.33## [kPa]\n",
+ "T2 = 237 + 273## [K]\n",
+ "Cp = 29.1## [J/mol K]\n",
+ "Cv = 20.78## [J/mol K]\n",
+ "R = 8.314## [J/K mol]\n",
+ "#*****************#\n",
+ "\n",
+ "n = m/M#\n",
+ "# From ideal gas equation:\n",
+ "V1 = n*R*T1/P1## [cubic m]\n",
+ "\n",
+ "# Isochoric Process:\n",
+ "print \"Isochoric Process\\n\"\n",
+ "# Volume = constant\n",
+ "V2 = V1## [cubic m]\n",
+ "deltaU = n*Cv*(T2 - T1)## [kJ]\n",
+ "# Since Volume is constant\n",
+ "W = 0#\n",
+ "Q = deltaU + W## [kJ]\n",
+ "# deltaH = deltaU + P*deltaV\n",
+ "# deltaH = deltaU + n*R*deltaT\n",
+ "deltaH = deltaU + n*R*(T2 - T1)## [kJ]\n",
+ "print \"Change in Internal Energy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Heat Supplied is %.2f kJ\\n\"%(Q)#\n",
+ "print \"Work done is %d kJ\\n\"%(W)#\n",
+ "print \"Change in Enthalpy is %.2f kJ\\n\"%(deltaH)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Isobaric Process\n",
+ "print \"Isobaric Process\\n\"\n",
+ "# Since Pressure is constant.\n",
+ "P2 = P1## [kPa]\n",
+ "deltaH = n*Cp*(T2 - T1)## [kJ]\n",
+ "Qp = deltaH## [kJ]\n",
+ "# deltaU = deltaH - P*deltaV\n",
+ "# From ideal gas equation:\n",
+ "deltaU = deltaH - n*R*(T2 - T1)## [kJ]\n",
+ "W = Qp - deltaU## [kJ]\n",
+ "print \"Change in Internal Energy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Heat Supplied is %.2f kJ\\n\"%(Qp)#\n",
+ "print \"Work done is %.2f kJ\\n\"%(W)#\n",
+ "print \"Change in Enthalpy is %.2f kJ\\n\"%(deltaH)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.14 Page: 60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.14 - Page: 60\n",
+ "\n",
+ "\n",
+ "Step 1: Isothermal Expansion Of Ideal Gas\n",
+ "\n",
+ "delta_E for Step 1 is 0 J/mol\n",
+ "\n",
+ "Q for step 1 is 11679.76 J/mol\n",
+ "\n",
+ "W for step 1 is 11679.76 J/mol\n",
+ "\n",
+ "\n",
+ "\n",
+ "Step 2: Adiabatic Expansion of ideal gas\n",
+ "\n",
+ "delta_E for Step 2 is -6234 J/mol\n",
+ "\n",
+ "Q for step 2 is 0.00 J/mol\n",
+ "\n",
+ "W for step 2 is 6234.00 J/mol\n",
+ "\n",
+ "\n",
+ "\n",
+ "Step 3: Isothermal Compression Of Ideal Gas\n",
+ "\n",
+ "delta_E for Step 3 is 0 J/mol\n",
+ "\n",
+ "Q for step 3 is -5935.61 J/mol\n",
+ "\n",
+ "W for step 3 is -5935.61 J/mol\n",
+ "\n",
+ "\n",
+ "\n",
+ "Step 4: Adiabatic Compression of ideal gas\n",
+ "\n",
+ "delta_E for Step 4 is 6234 J/mol\n",
+ "\n",
+ "Q for step 4 is 0.00 J/mol\n",
+ "\n",
+ "W for step 4 is -6234.00 J/mol\n",
+ "\n",
+ "\n",
+ "\n",
+ "Net Work done for the complete cycle is 5744.14 J/mol\n",
+ "\n",
+ "The efficiency of the cycle is 0.49\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log10\n",
+ "print \"Example: 2.14 - Page: 60\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "n = 1## [mol]\n",
+ "T1 = 610## [K]\n",
+ "P1 = 10**6## [N/square m]\n",
+ "T2 = 310## [K]\n",
+ "P2 = 10**5## [N/square m]\n",
+ "Cv = 20.78## [J/mol K]\n",
+ "#*************#\n",
+ "\n",
+ "R = 8.314## [J/K mol]\n",
+ "\n",
+ "# Step 1: Isothermal Expansion Of Ideal Gas:\n",
+ "print \"Step 1: Isothermal Expansion Of Ideal Gas\\n\"\n",
+ "T1 = 610## [K]\n",
+ "P1 = 10**6## [N/square m]\n",
+ "P2 = 10**5## [N/square m]\n",
+ "# Work done:\n",
+ "W1 = 2.303*n*R*T1*log10(P1/P2)## [J/mol]\n",
+ "# For isothermal expansion:\n",
+ "delta_E1 = 0## [J/mol]\n",
+ "# From first law of thermodynamics:\n",
+ "Q1 = delta_E1 + W1## [J/mol]\n",
+ "print \"delta_E for Step 1 is %d J/mol\\n\"%(delta_E1)#\n",
+ "print \"Q for step 1 is %.2f J/mol\\n\"%(Q1)#\n",
+ "print \"W for step 1 is %.2f J/mol\\n\"%(W1)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Step 2: Adiabatic Expansion of ideal gas:\n",
+ "print \"Step 2: Adiabatic Expansion of ideal gas\\n\"\n",
+ "Q2 = 0## [J/mol]\n",
+ "delta_E2 = Cv*(T2 - T1)## [J/mol]\n",
+ "# From first law of thermodynamics:\n",
+ "W2 = Q2 - delta_E2## [J/mol]\n",
+ "print \"delta_E for Step 2 is %d J/mol\\n\"%(delta_E2)#\n",
+ "print \"Q for step 2 is %.2f J/mol\\n\"%(Q2)#\n",
+ "print \"W for step 2 is %.2f J/mol\\n\"%(W2)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Step 3: Isothermal Compression Of Ideal Gas:\n",
+ "print \"Step 3: Isothermal Compression Of Ideal Gas\\n\"\n",
+ "T2 = 310## [K]\n",
+ "P1 = 10**5## [N/square m]\n",
+ "P2 = 10**6## [N/square m]\n",
+ "# Work done:\n",
+ "W3 = 2.303*n*R*T2*log10(P1/P2)## [J/mol]\n",
+ "# For isothermal expansion:\n",
+ "delta_E3 = 0## [J/mol]\n",
+ "# From first law of thermodynamics:\n",
+ "Q3 = delta_E3 + W3## [J/mol]\n",
+ "print \"delta_E for Step 3 is %d J/mol\\n\"%(delta_E3)#\n",
+ "print \"Q for step 3 is %.2f J/mol\\n\"%(Q3)#\n",
+ "print \"W for step 3 is %.2f J/mol\\n\"%(W3)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Step 4: Adiabatic Compression of ideal gas:\n",
+ "print \"Step 4: Adiabatic Compression of ideal gas\\n\"\n",
+ "T1 = 310## [K]\n",
+ "T2 = 610## [K]\n",
+ "Q4 = 0## [J/mol]\n",
+ "delta_E4 = Cv*(T2 - T1)## [J/mol]\n",
+ "# From first law of thermodynamics:\n",
+ "W4 = Q4 - delta_E4## [J/mol]\n",
+ "print \"delta_E for Step 4 is %d J/mol\\n\"%(delta_E4)#\n",
+ "print \"Q for step 4 is %.2f J/mol\\n\"%(Q4)#\n",
+ "print \"W for step 4 is %.2f J/mol\\n\"%(W4)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Net work done for the complete cycle:\n",
+ "W = W1 + W2 + W3 + W4## [J/mol]\n",
+ "print \"Net Work done for the complete cycle is %.2f J/mol\\n\"%(W)#\n",
+ "\n",
+ "# The efficiency of the cycle is given by:\n",
+ "eta = 1- T1/T2#\n",
+ "print \"The efficiency of the cycle is %.2f\\n\"%(eta)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.15 Page: 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.15 - Page: 61\n",
+ "\n",
+ "\n",
+ " Mathematics is involved in proving but just that no numerical computations are involved.\n",
+ "\n",
+ "\n",
+ " For prove refer to this example 2.15 on page 61 of the book.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.15 - Page: 61\\n\\n\"\n",
+ "\n",
+ "# Mathematics is involved in proving but just that no numerical computations are involved.\n",
+ "# For prove refer to this example 2.15 on page number 61 of the book.\n",
+ "\n",
+ "print \" Mathematics is involved in proving but just that no numerical computations are involved.\\n\\n\"\n",
+ "print \" For prove refer to this example 2.15 on page 61 of the book.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.16 Page: 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.16 - Page: 62\n",
+ "\n",
+ "\n",
+ "First Process\n",
+ "\n",
+ "Change in Internal Energy is -2223.97 kJ\n",
+ "\n",
+ "Change in Enthalpy is -5612.22 kJ\n",
+ "\n",
+ "Heat Requirement is -2246.40 kJ\n",
+ "\n",
+ "\n",
+ "\n",
+ "Second Process\n",
+ "\n",
+ "Change in Internal Energy is -22441.57 kJ\n",
+ "\n",
+ "Change in Enthalpy is -56122.20 kJ\n",
+ "\n",
+ "Heat Requirement is -22464.00 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.16 - Page: 62\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "P1 = 1## [bar]\n",
+ "T1 = 300##[K]\n",
+ "V1 = 24.92## [cubic m/kmol]\n",
+ "P2 = 10## [bar]\n",
+ "T2 = 300## [K]\n",
+ "Cp = 29.10## [kJ/kmol K]\n",
+ "Cv = 20.78## [kJ/kmol K]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#**************#\n",
+ "\n",
+ "# Basis: 1 kmol of ideal gas:\n",
+ "n = 1#\n",
+ "V2 = P1*V1/P2## [cubic m]\n",
+ "\n",
+ "# First Process:\n",
+ "print \"First Process\\n\"\n",
+ "# In the first step of the first process, the cooling of ga takes place at constant pressure.\n",
+ "# Here the volume is reduced appreciably and consequently the temperature decreases.\n",
+ "T_prime = T1*V2/V1## [K]\n",
+ "# Heat Requirement:\n",
+ "Q1 = n*Cp*(T_prime - T1)## [kJ]\n",
+ "deltaH1 = Q1## [kJ]\n",
+ "deltaU1 = deltaH1 - P1*(V2 - V1)## [kJ]\n",
+ "# In the second step, the gas is heated at constant Volume:\n",
+ "# V = constant\n",
+ "Q2 = n*Cv*(T2 - T_prime)## [kJ]\n",
+ "deltaU2 = Q2## [kJ]\n",
+ "deltaH2 = n*R*(T2 - T_prime)## [kJ]\n",
+ "deltaU = deltaU1 + deltaU2## [kJ]\n",
+ "deltaH = deltaH1 + deltaH2## [kJ]\n",
+ "Q = Q1 + Q2## [kJ]\n",
+ "print \"Change in Internal Energy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Change in Enthalpy is %.2f kJ\\n\"%(deltaH)#\n",
+ "print \"Heat Requirement is %.2f kJ\\n\"%(Q)#\n",
+ "print \"\\n\"\n",
+ "\n",
+ "# Second Process:\n",
+ "print \"Second Process\\n\"\n",
+ "# In the first step of the second process, the gas is heated at constant volume.\n",
+ "T_prime = T1*P2/P1## [K]\n",
+ "# Heat Requirement:\n",
+ "Q1 = n*Cv*(T_prime - T1)## [kJ]\n",
+ "deltaU1 = Q1## [kJ]\n",
+ "deltaH1 = n*R*(T_prime - T1)## [kJ]\n",
+ "# In the second step, the gas is cooled at constant presure:\n",
+ "# V = constant\n",
+ "Q2 = n*Cp*(T2 - T_prime)## [kJ]\n",
+ "deltaH2 = Q2## [kJ]\n",
+ "deltaU2 = deltaH2 - P1*(V2 - V1)## [kJ]\n",
+ "deltaU = deltaU1 + deltaU2## [kJ]\n",
+ "deltaH = deltaH1 + deltaH2## [kJ]\n",
+ "Q = Q1 + Q2## [kJ]\n",
+ "print \"Change in Internal Energy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Change in Enthalpy is %.2f kJ\\n\"%(deltaH)#\n",
+ "print \"Heat Requirement is %.2f kJ\\n\"%(Q)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.17 Page: 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.17 - Page: 64\n",
+ "\n",
+ "\n",
+ "Workdone by the gas is 251.33 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi\n",
+ "print \"Example: 2.17 - Page: 64\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "D1 = 1## [m]\n",
+ "P1 = 120## [kPa]\n",
+ "P2 = 360## [kPa]\n",
+ "# P = k*D**3\n",
+ "#***************#\n",
+ "\n",
+ "k = P1/D1**3## [proportionality constant]\n",
+ "D2 = (P2/k)**(1/3)## [m]\n",
+ "# Work done by the gas inside the balloon can be estimated as:\n",
+ "# W = integral(P*dV)#\n",
+ "# W = integral((k*D**3)*d((4/3)*pi*r**3)#\n",
+ "# W = (pi*k/6)*integral((D**3)*d(D**3))#\n",
+ "# W = (pi*k/12)*(D2**6 - D1**6)#\n",
+ "W = (pi*k/12)*(D2**6 - D1**6)## [kJ]\n",
+ "print \"Workdone by the gas is %.2f kJ\\n\"%(W)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.18 Page: 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.18 - Page: 65\n",
+ "\n",
+ "\n",
+ "Change in Inernal Enrgy is 1040.65 kJ\n",
+ "\n",
+ "Change in Enthalpy is 1456.35 kJ\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.18 - Page: 65\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "P1 = 10*100## [kPa]\n",
+ "T1 = 250## [K]\n",
+ "P2 = 1*100## [kPa]\n",
+ "T2 = 300## [K]\n",
+ "R = 8.314## [J/mol K]\n",
+ "Cv = 20.78## [kJ/kmol K]\n",
+ "Cp = 29.10## [kJ/kmol K]\n",
+ "#**********#\n",
+ "\n",
+ "V1 = R*T1/P1## [cubic m]\n",
+ "V2 = R*T2/P2## [cubic m]\n",
+ "\n",
+ "# Calculation based on First Process:\n",
+ "# In this constant-volume process, the initial pressure of 10 bar is reduced to a final pressure of 1 bar and consequently the temperature decreases.\n",
+ "T_prime = P2*V1/R## [K]\n",
+ "deltaU1 = Cv*(T_prime - T1)## [kJ]\n",
+ "deltaH1 = deltaU1 + V1*(P2 -P1)## [kJ]\n",
+ "# Since V = constant\n",
+ "W1 = 0##[kJ]\n",
+ "# By first law of thermodynamics:\n",
+ "Q = W1 + deltaU1## [kJ]\n",
+ "\n",
+ "# Calculation based on second process:\n",
+ "# In this process, the gas is heated at constant pressure to the final temperature of T2.\n",
+ "deltaH2 = Cp*(T2 - T_prime)## [kJ]\n",
+ "deltaU2 = deltaH2 - P2*(V2 - V1)## [kJ]\n",
+ "Q = deltaH2## [kJ]\n",
+ "W2 = Q - deltaU2## [kJ]\n",
+ "\n",
+ "deltaU = deltaU1 + deltaU2## [kJ]\n",
+ "deltaH = deltaH1 + deltaH2## [kJ]\n",
+ "print \"Change in Inernal Enrgy is %.2f kJ\\n\"%(deltaU)#\n",
+ "print \"Change in Enthalpy is %.2f kJ\\n\"%(deltaH)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 2.19 Page: 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 2.19 - Page: 69\n",
+ "\n",
+ "\n",
+ "Tempertaure of water in the second tank is 312.86 K\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 2.19 - Page: 69\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data*****#\n",
+ "T1 = 273## [K]\n",
+ "T2 = 273 + 67## [K]\n",
+ "m_dot = 20000## [kg/h]\n",
+ "Ws = -1.5## [hp]\n",
+ "Q = -38000## [kJ/min]\n",
+ "Z = 20## [m]\n",
+ "Cp = 4.2## [kJ/kg K]\n",
+ "g = 9.81## [m/second square]\n",
+ "#***************#\n",
+ "\n",
+ "Q = Q*60/m_dot## [kJ/kg]\n",
+ "Ws = Ws*0.7457*3600/m_dot## [kJ/kg]\n",
+ "PE = g*Z*10**(-3)## [kJ/kg]\n",
+ "# KE is assumed to be negligible.\n",
+ "# For Steady Flow process: dE/dt = 0\n",
+ "# From Eqn. 2.47:\n",
+ "deltaH = Q - Ws - PE## [kJ/kg]\n",
+ "H1 = Cp*(T2 - T1)## [kJ/kg]\n",
+ "H2 = H1 + deltaH## [kJ/kg]\n",
+ "\n",
+ "# Now, the temperature of the tank can be determined as:\n",
+ "T = (H2/Cp) + T1## [K]\n",
+ "print \"Tempertaure of water in the second tank is %.2f K\\n\"%(T)#"
+ ]
+ }
+ ],
+ "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
+}
generated by cgit (git 2.34.1) at 2025-01-09 21:31:05 +0000