{
"metadata": {
"name": "",
"signature": "sha256:c5a74c16b8ffa890dee9e16252d9e0c3674909f92103f0b1c670f62652469cd7"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7:Power and Refrigeration Gas Cycles"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.1:PG-175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"r=12 # compression ratio\n",
"k=1.4 # polytropic index for air\n",
"p1=200.0 # pressure at state 1 in kPa\n",
"p3=10000.0 # pressure at state 3 in kPa\n",
"\n",
"c=100/(r-1) # clearance in percentage\n",
"print \"The percent clearance is\",round(c,2),\"% \\n\"\n",
"v3=100.0 # let us assume v3=100 m^3 for calculations\n",
"p2=p1*(r**k) # polytopic process pressure relation\n",
"p4=p3*(1/(r**k))# polytropic process pressure relation\n",
"w34=v3*(r*p4-p3)/(1-k) # polytropic work done in process 3 to 4\n",
"v2=v3 # constant volume process\n",
"w12=v2*(p2-r*p1)/(1-k)\n",
"wcycle=w12+w34 # total work in cycle\n",
" # now equating the polytropic work calculated to work by MEP\n",
"MEP=wcycle/(r*v2-v2) # as work = pressure*change in volume\n",
"print \"The MEP is\",round(MEP),\" kPa\" \n",
"# The solution is wrong in textbook as calculation for P2 is wrong \n",
"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The percent clearance is 9.0 % \n",
"\n",
"The MEP is 503.0 kPa\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.2:PG-179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"r=10.0 # compression ratio\n",
"k=1.4 # polytropic index for air\n",
"R=0.287 # specific gas constant for air\n",
"Cv=0.717 # specific heat at constant volume\n",
"Wnet=1000 # net work output in kJ/kg\n",
"T1=227+273.0 # low air temperaure in kelvin\n",
"p1=200.0 # low pressure in kPa\n",
"\n",
"effi=1-(1/r**(k-1)) # thermal efficeiency \n",
"print \"The maximum possible thermal efficiency is\",round(effi*100,1),\"% \\n\"\n",
"\n",
"T2=T1*(r)**(k-1) # isentropic process temperature relation\n",
"\n",
"T4=((Wnet/Cv)+T2-T1)/((r**(k-1))-1) # using expression for work\n",
"\n",
"T3=T4*(r)**(k-1)\n",
"\n",
"efficarnot=1-T1/T3\n",
"print \"The carnot efficiency is\",round(efficarnot*100),\"%\"\n",
"\n",
"v1=R*T1/p1 # initial volume \n",
"v2=v1/r # from compression ratio\n",
"\n",
"MEP=Wnet/(v1-v2) # mean effective pressure equation\n",
"\n",
"print \"The MEP is\",round(MEP),\" kPa\"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum possible thermal efficiency is 60.2 % \n",
"\n",
"The carnot efficiency is 86.0 %\n",
"The MEP is 1549.0 kPa\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.3:PG-182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"\n",
"r=18 # compression ratio\n",
"k=1.4 # polytropic index for air\n",
"R=0.287 # specific gas constant for air\n",
"Cv=0.717 # specific heat at constant volume\n",
"Cp=1.0 # specific heat at constant pressure\n",
"T1=200+273 # lower temperaure in kelvin\n",
"P1=200.0 # low pressure in kPa\n",
"T3=2000 # higher temperature of cycle in kelvin\n",
"\n",
"v1=R*T1/P1 # specific volume at state 1 in m^3\n",
"v2=v1/r # specific volume after compression in m^3\n",
"\n",
"T2=T1*(v1/v2)**(k-1) # temperature after compression\n",
"P2=P1*(v1/v2)**k # pressure after compression\n",
"P3=P2 # diesel cycle\n",
"v3=R*T3/P3 # volume at state 3\n",
"\n",
"rc=v3/v2 # cutoff ratio\n",
"\n",
"effi=1-((rc**k)-1)/(r**(k-1)*k*(rc-1))\n",
"\n",
"\n",
"print \"The thermal efficiency is\",round(effi*100,1),\"%\"\n",
"\n",
"v4=v1 # diesel cycle\n",
"T4=T3*(v3/v4)**(k-1) # adiabatic process\n",
"\n",
"qin=Cp*(T3-T2) # using first law \n",
"qout=Cv*(T4-T1) # heat rejected \n",
"\n",
"Wnet=qin-qout # net work\n",
"MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",
"\n",
"print \"The MEP is\",round(MEP),\" kPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thermal efficiency is 66.6 %\n",
"The MEP is 515.0 kPa\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.4:PG-183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"r=18 # compression ratio\n",
"k=1.4 # polytropic index for air\n",
"R=0.287 # specific gas constant for air\n",
"T1=200+273 # lower temperaure in kelvin\n",
"P1=200.0 # low pressure in kPa\n",
"T3=2000.0 # higher temperature of cycle in kelvin \n",
"\n",
"v1=R*T1/P1 # specific volume at state 1 in m^3\n",
"#using table E.1\n",
"u1=340.0 # specific internal energy in kJ/kg\n",
"vr1=198.1 # in m^3/kg\n",
"\n",
"vr2=vr1*(1/r) # as r=v1/v2\n",
"\n",
"# now finding corresponding values from table E.1\n",
"T2=1310.0 # temperature in kelvin\n",
"Pr2=34.0 # pressure in kPa\n",
"h2=1408.0 # specific entropy in kJ/kg\n",
"v2=v1/18 # volume at state 2\n",
"P2=R*T2/v2 # pressure at state 2\n",
"\n",
"h3=2252.1 # specific enthalpy in kJ/kg from table E.1\n",
"vr3=2.776 \n",
"P3=P2 # diesel cycle\n",
"v3=R*T3/P3 # after compression volume\n",
"v4=v1 # isochoric process\n",
"vr4=vr3*v4/v3 # isentropic process\n",
"# now using Vr4 we read corresponding value from table E.1\n",
"T4=915 # final temperature in kelvin\n",
"u4=687.5 # specific internal energy at state 4\n",
"\n",
"qin=h3-h2 # using first law \n",
"qout=u4-u1 # heat rejected \n",
"\n",
"Wnet=qin-qout # net work\n",
"effi=100*Wnet/qin # thermal efficiency\n",
"print\" The thermal efficiency is\",round(effi,1),\"%\"\n",
"\n",
"MEP=Wnet/(v1-v2) # expression of mean effective pressure in terms of work\n",
"\n",
"print \" The MEP is\",round(MEP),\" kPa\"\n",
"\n",
"erroreffi=(66.6-effi)*100/effi # error in efficiency\n",
"errorMEP=(515-MEP)*100/MEP # error in MEP\n",
"\n",
"print \" The % error in efficiency is\",round(erroreffi,1),\"%\"\n",
"print \" The % error MEP is\",round(errorMEP,1),\"% \\n\"\n",
"\n",
"# the answers are slight different due to approximation in textbook ... here answers are precise\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The thermal efficiency is 58.8 %\n",
" The MEP is 775.0 kPa\n",
" The % error in efficiency is 13.2 %\n",
" The % error MEP is -33.5 % \n",
"\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.5:PG-186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"Cp=1.0 # specific heat at constant pressure\n",
"k=1.4 # polytropic index for air\n",
"T1=25+273.0 # temperature at compressor inlet\n",
"T3=850+273.0 # maximum temperature in kelvin\n",
"\n",
"r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
"\n",
"T2=T1*(r)**((k-1)/k) # temperature after compression\n",
"\n",
"T4=T3*(1/r)**((k-1.0)/k) # final temperature\n",
"\n",
"Wcomp=Cp*(T2-T1) # compressor work\n",
"Wturb=Cp*(T3-T4) # turbine work\n",
"\n",
"BWR=Wcomp/Wturb # back work ratio\n",
"\n",
"print \" The BWR is\",round(BWR*100,1),\"%\\n\" \n",
"\n",
"Effi=1-r**((1-k)/k) # thermal efficiency\n",
"\n",
"print\" The thermal efficiency is\",round(Effi*100,1),\"%\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The BWR is 42.0 %\n",
" The thermal efficiency is 36.9 %\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.6:PG-186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"Cp=1.0 # specific heat at constant pressure\n",
"k=1.4 # polytropic index for air\n",
"T1=25+273.0 # temperature at compressor inlet\n",
"T3=850+273.0 # maximum temperature in kelvin\n",
"\n",
"r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
"efficomp=0.75 # efficiency of compressor\n",
"effiturb=0.75 # efficiency of turbine\n",
"\n",
"T2dash=T1*(r)**((k-1)/k) # temperature after compression\n",
"Wcomp=Cp*(T2dash-T1)/efficomp # compressor work\n",
"\n",
"T4dash=T3*(1/r)**((k-1)/k) # final temperature\n",
"Wturb=Cp*(T3-T4dash)*effiturb # turbine work\n",
"\n",
"BWR=100*Wcomp/Wturb # back work ratio\n",
"\n",
"print \" The BWR is\",round(BWR,1),\"%\\n\"\n",
"\n",
"T2=(Wcomp/Cp)+T1 # actual temperature of state 2\n",
"\n",
"qin=Cp*(T3-T2) # using first law \n",
"\n",
"Wnet=(Wturb-Wcomp) # net work\n",
"\n",
"effi=100*Wnet/qin # thermal efficiency\n",
"print\" The thermal efficiency is\",round(effi,1),\"%\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The BWR is 74.7 %\n",
" The thermal efficiency is 13.2 %\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.7:PG-191"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"Cp=1.0 # specific heat at constant pressure\n",
"k=1.4 # polytropic index for air\n",
"T1=25+273.0 # temperature at compressor inlet\n",
"T3=850+273.0 # maximum temperature in kelvin\n",
"\n",
"r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
"\n",
"T2=T1*(r)**((k-1)/k) # temperature after compression\n",
"\n",
"T4=T3*(1/r)**((k-1)/k) # final temperature\n",
"\n",
"Wcomp=Cp*(T2-T1) # compressor work\n",
"Wturb=Cp*(T3-T4) # turbine work\n",
"\n",
"BWR=Wcomp/Wturb # back work ratio\n",
"\n",
"print \" The BWR is\",round(BWR,2),\"\\n\"\n",
"\n",
"effi=(1-((T1/T4)*(r**((k-1)/k))))# efficiency\n",
"print\" The thermal efficiency is\",round(effi*100,1),\"%\"\n",
"# The solution in textbook is incorrect due to wrong value of T4 (temperature at state 4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The BWR is 0.42 \n",
"\n",
" The thermal efficiency is 33.4 %\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.8:PG-193"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"#REFER TO FIG.:7.8\n",
"\n",
"Cp=1 # specific constant at constant pressure\n",
"k=1.4 # polytropic constant for air\n",
"T5=25+273.0 # temperature at state 5 in kelvin\n",
"T7=850+273.0 # temperature at state 4 in kelvin\n",
"T9=350 # exit temperature of water from bolier in kelvin\n",
"WdotST=100000.0 # power from steam turbine in Watt\n",
"r=5.0 # pressure ratio=P2/P1 & P4/P3\n",
"\n",
"h1=192.0 # specific enthalpy at 10 Kpa from steam table\n",
"h2=h1 # isenthalpic process\n",
"h3=3214.0 # specific enthalpy at 4 Mpa and 400 degree celsius from steam table\n",
"s3=6.769 # specific entropy at 4 Mpa and 400 degree celsius from steam table\n",
"\n",
"s4=s3 # isentropic process\n",
"sf=0.6491 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",
"sg=8.1510 # specific entropy of saturated liquid at 10 kPa and 45 degree celsiusfrom table C.2\n",
"x4=(s4-sf)/(sg-sf) # quality of steam\n",
"\n",
"hf=h1 # specific enthalpy of saturated liquid @ 10 Kpa \n",
"hg=2584.6\n",
"h4=hf+x4*(hg-hf) # specific entropy at state 4\n",
"\n",
"mdots=WdotST/(h3-h4) # steam mass flow rate from turbine output\n",
"\n",
"T6=T5*(r**((k-1)/k)) # adiabatic process relation\n",
"T8=T7*(1/r**((k-1)/k)) # adiabatic process relation\n",
"\n",
"# Now using energy balance in boiler\n",
"mdota=mdots*(h3-h2)/(Cp*(T8-T9)) # mass flow rate of water\n",
"\n",
"Wdotturb=mdota*Cp*(T7-T8) # power produced by turbine\n",
"\n",
"Wdotcomp=mdota*Cp*(T6-T5) # energy needed by compressor\n",
"\n",
"WdotGT=Wdotturb-Wdotcomp # net turbine work\n",
"\n",
"Qdotin=mdota*Cp*(T7-T6) # energy input by combustor\n",
"\n",
"effi=100*(WdotST+WdotGT)/Qdotin # combined efficiency\n",
"\n",
"print \"The thermal efficiency of the combined cycle is\",round(effi,1),\"%\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thermal efficiency of the combined cycle is 56.4 %\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.9:PG-196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# initialization of variables\n",
"\n",
"Cp=1 # specific constant at constant pressure\n",
"k=1.4 # polytropic constant for air\n",
"r=10.0\n",
"T2=-10+273 # temperature at entry of compressor\n",
"T4=30+273 # temperature at entry of turbine\n",
"\n",
"T3=T2*(r**((k-1)/k)) # temperature at state 3 in kelvin\n",
"T1=T4*(1/r**((k-1)/k)) # temperature at state 1 in degree celsius\n",
"print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",
"\n",
"qin=Cp*(T2-T1) # heat input\n",
"Wcomp=Cp*(T3-T2)# compressor work\n",
"Wturb=Cp*(T4-T1) # turbine work\n",
"\n",
"COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",
"print\" The COP is\",round(COP,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum temperature is -116.0 degree celsius \n",
"\n",
" The COP is 1.07\n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7.10:PG-196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#solution\n",
"# initialization of variables\n",
"\n",
"Cp=1.0 # specific constant at constant pressure\n",
"k=1.4 # polytropic constant for air\n",
"r=10.0\n",
"T3=-10+273 # temperature at entry of compressor\n",
"T6=-40+273 # temperature at entry of turbine\n",
"\n",
"T5=T3 # heat exchanger\n",
"T2=T6 # heat exchanger\n",
"\n",
"T4=T3*(r**((k-1)/k)) # temperature after compression\n",
"T1=T6*(1/r**((k-1)/k)) # temperature after exit from turbine\n",
"\n",
"print \"The minimum temperature is\",round(T1-273),\"degree celsius \\n\"\n",
"\n",
"qin=Cp*(T2-T1) # heat input\n",
"Wcomp=Cp*(T4-T3)# compressor work\n",
"Wturb=Cp*(T6-T1) # turbine work\n",
"\n",
"COP=qin/(Wcomp-Wturb) # COP of refrigeration\n",
"\n",
"print\" The COP is\",round(COP,3)\n",
"\n",
"# the answer is correct within given limits\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum temperature is -152.0 degree celsius \n",
"\n",
" The COP is 0.848\n"
]
}
],
"prompt_number": 51
}
],
"metadata": {}
}
]
}