{
"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": {}
}
]
}