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+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:b8d3bc6c59d0efdfca4aa426416ba1f24e3078d69f0411d3e3ed7b293bd81c78"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "ELEMENTS OF MECHANICAL ENGINEERING by R.K. RAJPUT"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "CHAPTER NUMBER 2 : FUELS AND COMBUSTION"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 2.1"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " #DATA GIVEN\n",
+      "c=88;                    #% of carbon in coal\n",
+      "h=4.2;                   #% of hydrogen in coal\n",
+      "Wf=0.848;                #weight of coal in g\n",
+      "Wfw=0.027;               #weight of fuse wire in calorimeter in g\n",
+      "W=1950;                  #weight of water in calorimeter in g\n",
+      "We=380;                  #water equivalent of calorimeter\n",
+      "Dt=3.06;                 #observed temperature rise (t2-t1) in deg celsius\n",
+      "tc=0.017;                #cooling correction in deg celsius\n",
+      "cfw=6700;                #calorific value of fuse wire in J/g\n",
+      "\n",
+      " #CALCULATIONS\n",
+      "ctr=(Dt)+tc;             #corrected temp. rise\n",
+      "Hw=(W+We)*4.18*(ctr);    #heat recieved by water in J\n",
+      "Hfw=Wfw*cfw;             #heat given out by fuse wire in J\n",
+      "Hcf=Hw-Hfw;              #heat produced due to combustion of fuel in J\n",
+      "HCV=Hcf/Wf;              #higher calorific value of fuel in kJ/kg\n",
+      "Ms=9*h/100;              #steam produced per kg of coal\n",
+      "LCV=HCV-2465*Ms;         #lower calorific value of fuel in kJ/kg\n",
+      "\n",
+      "print \"The Higher calorific value of fuel, H.C.V. is: \",round(HCV,4),\" kJ/kg.\"\n",
+      "print \"The Lower calorific value of fuel, L.C.V. is: \",round(LCV,4),\" kJ/kg.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Higher calorific value of fuel, H.C.V. is:  35126.455  kJ/kg.\n",
+        "The Lower calorific value of fuel, L.C.V. is:  34194.685  kJ/kg.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 2.2"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "V1=0.08;                  #gas burnt in calorimeter in m^3\n",
+      "Pg=5.2;                   #pressure of gas supply in cm of water\n",
+      "Pb=75.5;                  #barometer reading in cm of Hg\n",
+      "Ww=28;                    #weight of water heated by gas in kg\n",
+      "Tg=13;                    #temperature of gas in deg celsius\n",
+      "Twi=10;                   #temperature of water at inlet in deg celsius\n",
+      "Two=23.5;                 #temperature of water at outlet in deg celsius\n",
+      "Ms=0.06;                  #steam condensed in kg\n",
+      "\n",
+      " #CALCULATIONS\n",
+      " #by using general gas equation, reducing the volume to S.T.P.\n",
+      " #p1*V1/T1=p2*V2/T2\n",
+      "p1=Pb+(Pg/13.6);          #in cm of Hg\n",
+      "T1=Tg+273;                #in K\n",
+      "p2=76;                    #in cm of Hg\n",
+      "T2=15+273;                #in K\n",
+      "V2=p1*V1*T2/T1/p2;        #in m^3\n",
+      "Hw=Ww*4.18*(Two-Twi);     #heat recieved by water in kJ\n",
+      "HCV=Hw/V1;                #higher calorific value of fuel in kJ/m^3\n",
+      "LCV=HCV-2465*Ms/V1;       #lower calorific value of fuel in kJ/m^3\n",
+      "\n",
+      "print \" The Calorific values of fuel per m^3 of gas at 15 deg celsius and 76 cm of Hg pressure are:\"\n",
+      "print \"  The Higher calorific value of fuel, H.C.V. is: \",HCV,\" kJ/m^3.\"\n",
+      "print \"  The Lower calorific value of fuel, L.C.V. is: \",LCV,\" kJ/m^3.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The Calorific values of fuel per m^3 of gas at 15 deg celsius and 76 cm of Hg pressure are:\n",
+        "  The Higher calorific value of fuel, H.C.V. is:  19750.5  kJ/m^3.\n",
+        "  The Lower calorific value of fuel, L.C.V. is:  17901.75  kJ/m^3.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "CHAPTER NUMBER 3 : PROPERTIES OF GASES"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 3.1"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "Q=-50;                           #heat rejected to cooling water in kJ/kg\n",
+      "W=-100;                          #work input in kJ/kg\n",
+      "\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      "Du=Q-W;                          #(u2-u1) change in internal energy in kJ/kg\n",
+      " #since Du is +ve, there is gain in internal energy\n",
+      "\n",
+      "print \"The GAIN in internal energy is: \",Du,\" kJ/kg.\"   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The GAIN in internal energy is:  50  kJ/kg.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 3.2"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "u1=450;                          #internal energy at beginning of the expansion in kJ/kg\n",
+      "u2=220;                          #internal energy after expansion in kJ/kg\n",
+      "W=120;                           #work done by the air during expansion in kJ/kg\n",
+      "\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      "Q=(u2-u1)+W;                     #heat flow in kJ/kg\n",
+      " #since Q is -ve, there is rejection of heat\n",
+      "\n",
+      "print \"The heat REJECTED by air is: \",(-Q),\" kJ/kg.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat REJECTED by air is:  110  kJ/kg.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 3.3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "m=0.3;                           #mass of nitrogen in kg\n",
+      "p1=0.1;                          #pressure in MPa\n",
+      "T1=40+273;                       #temperature before compression in K\n",
+      "p2=1;                            #pressure in MPa\n",
+      "T2=160+273;                      #temperature after compression in K\n",
+      "W=-30;                           #work done during the compression in kJ/kg\n",
+      "Cv=0.75                          #in kJ/kgK\n",
+      "\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      " #(u2-u1)=m*Cv*(T2-T1)\n",
+      "Du=m*Cv*(T2-T1);\n",
+      "Q=Du+W;                          #heat flow in kJ/kg\n",
+      " #since Q is -ve, there is rejection of heat\n",
+      "\n",
+      "print \"The heat REJECTED by air is: \",(-Q),\" kJ. \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat REJECTED by air is:  3.0  kJ. \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER 3.4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      " #initial state\n",
+      "p1=0.105;                        #pressure of gas in MPa\n",
+      "V1=0.4;                          #volume of gas in m^3\n",
+      " #final state\n",
+      "p2=0.105;                        #pressure of gas in MPa\n",
+      "V2=0.20;                         #volume of gas in m^3\n",
+      "\n",
+      "Q=-42.5;                         #heat transferred in kJ\n",
+      "p=p1;\n",
+      "\n",
+      " #process used- ISOBARIC (Constant pressure)\n",
+      "W12=p*(V2-V1)*1000;              #work in kJ\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      "Du=Q-W12;                        #(u2-u1) change in internal energy in kJ\n",
+      " #since Du is -ve, there is decrease in internal energy\n",
+      "\n",
+      "print \"The DECREASE in internal energy is: \",(-Du),\" kJ.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The DECREASE in internal energy is:  21.5  kJ.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      " #part-1\n",
+      " #pressure=p1,temperature=T1\n",
+      " #part-2\n",
+      " #pressure=p2,temperature=T2\n",
+      "\n",
+      " #Acc. First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      " #when partition moved\n",
+      "DQ=0;\n",
+      "DW=0;\n",
+      "DU=DQ-DW;\n",
+      " #DU=0\n",
+      "\n",
+      "print \"     CONCLUSION:   \"\n",
+      "print \"      Acc. to First Law of Thermodynamics,   \"\n",
+      "print \"      When partion moved, there is conservation of internal energy.   \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "     CONCLUSION:   \n",
+        "      Acc. to First Law of Thermodynamics,   \n",
+        "      When partion moved, there is conservation of internal energy.   \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      " #initial state\n",
+      "p1=10**5;                         #initial pressure of air in Pa\n",
+      "v1=1.8;                          #volume of air in m^3/kg\n",
+      "T1=25+273;                       #initial temperature of air in K\n",
+      " #final state\n",
+      "p2=5*10**5;                       #final pressure of air in Pa\n",
+      "T2=25+273;                       #final temperature of air in K\n",
+      "\n",
+      " #process used- ISOTHERMAL (Constant temperature)\n",
+      "W12=(p1*v1*float(math.log(float(p1)/float(p2))/1000));     #work in kJ/kg\n",
+      " #since W is -ve, work is supplied to the air\n",
+      "\n",
+      " #since temperature is constant\n",
+      "Du=0;                            #(u2-u1) change in internal energy in kJ/kg\n",
+      "\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      "Q=Du+W12;\n",
+      " #since Q is -ve, there is rejection of heat from system to surroundings\n",
+      "\n",
+      "print \" (i) The Work done on the air is: \",round(-W12,4),\" kJ/kg. \"\n",
+      "print \" (ii) The change in internal energy is: \",(Du),\" kJ/kg. \"\n",
+      "print \" (iii) The Heat REJECTED is: \",round(-Q,4),\" kJ/kg.  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (i) The Work done on the air is:  289.6988  kJ/kg. \n",
+        " (ii) The change in internal energy is:  0  kJ/kg. \n",
+        " (iii) The Heat REJECTED is:  289.6988  kJ/kg.  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "p1=4*10**5;                       #initial pressure in N/m^2\n",
+      "V1=0.2;                          #initial volume in m^3\n",
+      "T1=130+273;                      #initial temperature in K\n",
+      "p2=1.02*10**5;                    #final pressure after adiabatic expansion in N/m^2\n",
+      "Q23=72.5;                        #increase in enthalpy during constant pressure process in kJ\n",
+      "Cp=1;                            #in kJ/kgK\n",
+      "Cv=0.714;                        #in kJ/khK\n",
+      "\n",
+      " #gamma for air, g\n",
+      "g=Cp/Cv;\n",
+      "R=(Cp-Cv)*1000;\n",
+      "\n",
+      " #for reversible adiabatic process 1-2\n",
+      " #p1*(V1**g)=p2*(V2**g)\n",
+      "V2=V1*(p1/p2)**(1/g);             #final volume in m^3\n",
+      " #(T2/T1)=(p2/p1)**((g-1)/g);\n",
+      "T2=T1*(p2/p1)**((g-1)/g);       #final temp. T2 in K\n",
+      "\n",
+      "m=p1*V1/R/T1;                    #mass in kg\n",
+      "\n",
+      " #for constant pressure process 2-3\n",
+      " #Q23=m*Cp*(T3-T2);\n",
+      "T3=Q23/m/Cp+T2;\n",
+      " #V2/T2=V3/T3\n",
+      "V3=V2/T2*T3;\n",
+      "\n",
+      " #Work done by the path 1-2-3, W123=W12+W23\n",
+      "W12=(p1*V1-p2*V2)/(g-1);\n",
+      "W23=p2*(V3-V2);\n",
+      "W123=W12+W23;\n",
+      "\n",
+      " #if the above processes are replaced by a single reversible polytropic process giving the same work between initial and final states,\n",
+      " #W13=W123=(p1V1-p3V3)/(n-1)\n",
+      "p3=p2;\n",
+      "n=1+(p1*V1-p3*V3)/W123;          #index of expansion, n\n",
+      "\n",
+      "print \" (i) The Total Work done is: \",round(W123,4),\" Nm or J.\"\n",
+      "print \" (ii) The value of index of expansion, n is: \",round(n,4),\".\"\n",
+      "\n",
+      " #NOTE:\n",
+      " #there is slight variation in answers of the book due to rounding off of the values "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (i) The Total Work done is:  85343.6734  Nm or J.\n",
+        " (ii) The value of index of expansion, n is:  1.0603 .\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      " #initial state\n",
+      "p1=10**5;                         #initial pressure of gas in Pa\n",
+      "V1=0.45;                         #initial volume of gas in m^3\n",
+      "T1=80+273;                       #initial temperature of gas in K\n",
+      " #final state\n",
+      "p2=5*10**5;                       #final pressure of gas in Pa\n",
+      "V2=0.13;                         #final volume of gas in m^3\n",
+      "\n",
+      " #gamma for air, g\n",
+      "g=1.4;\n",
+      "R=294.2                          #J/kgK\n",
+      "\n",
+      "m=p1*V1/R/T1;                    #mass in kg\n",
+      "\n",
+      " #p1*(V1^n)=p2*(V2^n)\n",
+      "n=math.log(p2/p1)/math.log(V2/V1);         #index n\n",
+      "\n",
+      " #In a polytropic process\n",
+      " #(T2/T1)=(V1/V2)^(n-1);\n",
+      "T2=T1*(V1/V2)**(n-1);             #temp. T2 in K\n",
+      "\n",
+      "Cv=R/(g-1);\n",
+      "Du=m*Cv*(T2-T1)/1000;            #increase in internal energy in kJ\n",
+      "\n",
+      " #using First Law of Thermodynamics, Q=(u2-u1)+W\n",
+      " #W12=(p1*V1-p2*V2)/(n-1)=mR(T2-T1)/(n-1)\n",
+      "W12=m*R*(T1-T2)/(n-1)/1000;\n",
+      "Q=Du+W12;\n",
+      " #since Q is -ve, there is rejection of heat from system to surroundings\n",
+      "\n",
+      "print \" (i) The Mass of the gas is: \",round(m,4),\" kg.\"\n",
+      "print \" (ii) The index n is: \",round(n,4),\".\"\n",
+      "print \"(iii) The change in internal energy is: \",(Du),\" kJ.\"\n",
+      "print \" (iv) The Heat REJECTED is: \",round(-Q,4),\"kJ.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (i) The Mass of the gas is:  0.4333  kg.\n",
+        " (ii) The index n is:  -1.2961 .\n",
+        "(iii) The change in internal energy is:  -106.0  kJ.\n",
+        " (iv) The Heat REJECTED is:  124.4657 kJ.\n"
+       ]
+      }
+     ],
+     "prompt_number": 70
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      " #initial state\n",
+      "p1=1.02;                         #initial pressure of air in bar\n",
+      "V1=0.015;                        #initial volume of air in m^3\n",
+      "T1=22+273;                       #initial temperature of air in K\n",
+      " #final state\n",
+      "p2=6.8;                          #final pressure of air in bar\n",
+      " #Law of adiabatic compression, pV^g=C\n",
+      "\n",
+      " #gamma for air, g\n",
+      "g=1.4\n",
+      "R=0.287;\n",
+      "\n",
+      " #In a adiabatic process\n",
+      " #(T2/T1)=(p2/p1)**((g-1)/g);\n",
+      "T2=T1*(p2/p1)**((g-1)/g);       #final temp. T2 in K\n",
+      "\n",
+      " #p1*(V1**g)=p2*(V2**g)\n",
+      "V2=V1*(p1/p2)**(1/g);             #final volume in m^3\n",
+      "\n",
+      "m=p1*10**5*V1/10**3/R/T1;          #mass in kg\n",
+      "\n",
+      " #W=(p1*V1-p2*V2)/(g-1)=mR(T2-T1)/(g-1)\n",
+      "W=m*R*(T1-T2)/(g-1);\n",
+      " #since W is -ve, the work is done on the air\n",
+      "\n",
+      "print \" (i) The Final temperature is: \",(T2-273),\" deg. celsius.\"\n",
+      "print \" (ii) The Final Volume is: \",V2,\" m**3. \"\n",
+      "print \"(iii) The Work done on the air is: \",(-W),\" kJ.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (i) The Final temperature is:  234.252870551  deg. celsius.\n",
+        " (ii) The Final Volume is:  0.00386887782624  m**3. \n",
+        "(iii) The Work done on the air is:  2.7520923046  kJ.\n"
+       ]
+      }
+     ],
+     "prompt_number": 66
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EXAMPLE NUMBER: 3.13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      " \n",
+      " #DATA GIVEN\n",
+      "m=1;                             #mass of etahne gas in kg\n",
+      "M=30;                            #molecular weight of ethane\n",
+      "p1=1.1;                          #initial pressure in bar\n",
+      "T1=27+273;                       #initial temperature in K\n",
+      "p2=6.6;                          #final pressure in bar\n",
+      "Cp=1.75;                         #in kJ/kgK\n",
+      "\n",
+      " #Law of compression, pV**1.3=C\n",
+      "n=1.3;\n",
+      "\n",
+      " #Characteristic gas constant, R = Universal gas constant (Ro)/Molecular weight(M)\n",
+      "Ro=8314;\n",
+      "R=Ro/(M);                     #kJ/kgK\n",
+      "R1 = float(R)/1000;\n",
+      " #R=Cp-Cv\n",
+      "Cv=Cp-float(R1);\n",
+      "g=Cp/Cv;                         #gamma g\n",
+      "\n",
+      " #In a polytropic process\n",
+      " #(T2/T1)=(p2/p1)**((n-1)/n);\n",
+      "T2=T1*(p2/p1)**((n-1)/n);        #final temp. T2 in K\n",
+      "\n",
+      " #W=(p1*V1-p2*V2)/(n-1)=mR(T2-T1)/(g-1)\n",
+      "W=m*R*(T1-T2)/(n-1);\n",
+      "\n",
+      "Q=(g-n)*W/(g-1);               #heat flow in kJ/kg\n",
+      "\n",
+      "print \" The Heat SUPPLIED is: \",float(Q),\" kJ/kg.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The Heat SUPPLIED is:  84441.1861346  kJ/kg.\n"
+       ]
+      }
+     ],
+     "prompt_number": 102
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file