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{
"metadata": {
"name": "",
"signature": "sha256:79711d3b5ce866eb594601c655e2c85bdace3e8f861ba88d25f2b6f7e8d8655f"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter8:ENTROPY"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.1:pg-290"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 1\n",
"#coefficient of performance of refrigerator\n",
"\n",
"Th=60 #temperature at which heat is rejected from R-134a\n",
"Tl=0 #temperature at which heat is absorbed into the R-134a \n",
"s1=1.7262 #specific entropy at 0 Celsius\n",
"s2=s1 #process of state change from 1-2 is isentropic \n",
"s3=1.2857 #specific entropy at 60 celsius\n",
"s4=s3 #process of state change from 3-4 is isentropic\n",
"print\"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\"\n",
"P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) #pressure after compression in kPa\n",
"B=(Th+273.15)/(Th-Tl) #coefficient of performance of refrigerator\n",
"print\" \\n hence,pressure after compression is \",round(P2,1),\"kPa\"\n",
"print\"\\n and coefficient of performance of refrigerator is\",round(B,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore\n",
" \n",
" hence,pressure after compression is 1487.1 kPa\n",
"\n",
" and coefficient of performance of refrigerator is 5.55\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.2:pg-290"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 2 \n",
"#heat transfer in a given process\n",
"\n",
"u1=87.94 #specific internal energy of R-12 at state 1 in kJ/kg\n",
"u2=276.44 #specific internal energy of R-12 at state 2 in kJ/kg\n",
"s1=0.3357 #specific entropy at state 1 in kJ/kg-K\n",
"s2=1.2108 #specific entropy at state 2 in kJ/kg-K\n",
"V=0.001 #volume of saturated liquid in m^3\n",
"v1=0.000923 #specific volume in m^3/kg\n",
"m=V/v1 #mass of saturated liquid in kg\n",
"T=20 #temperature of liquid in celsius\n",
"Q12=m*(T+273.15)*(s2-s1) #heat transfer in kJ to accomplish the process\n",
"W12=m*(u1-u2)+Q12 #work required to accomplish the process\n",
"print\" \\n hence,work required to accomplish the process is\",round(W12,1),\"KJ\"\n",
"print\" \\n and heat transfer is\",round(Q12,1),\"KJ\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" \n",
" hence,work required to accomplish the process is 73.7 KJ\n",
" \n",
" and heat transfer is 277.9 KJ\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.3:pg-293"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3\n",
"#entropy change\n",
"import math\n",
"C=4.184 # specific heat of water in kJ/kg-K\n",
"T1=20 #initial temperature of water in celsius\n",
"T2=90 #final temperature of water in celsius\n",
"dS1=C*math.log((T2+273.2)/(T1+273.2)) #change in entropy in kJ/kg-K\n",
"dS2=1.1925-0.2966 #in kJ/kg-K using steam tables\n",
"print\"\\n hence,change in entropy assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",
"print\"\\n using steam table is\",round(dS2,4),\"KJ/kg-k\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,change in entropy assuming constant specific heat is 0.8958 KJ/kg-k\n",
"\n",
" using steam table is 0.8959 KJ/kg-k\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.4:pg-295"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 4\n",
"#entropy change with different assumptions\n",
"import math\n",
"T1=300 #initial temperature in kelvins\n",
"T2=1500 #final temperature in kelvins\n",
"P1=200.0 #initial pressure in kPa\n",
"P2=150.0 #final pressure in kPa\n",
"R=0.2598 # in kJ/kg-K\n",
"Cp=0.922 #specific heat in kJ/kg-K at constant pressure\n",
"dsT2=8.0649 #in kJ/kg-K\n",
"dsT1=6.4168 #in kJ/kg-K\n",
"dS1=dsT2-dsT1-R*math.log(P2/P1) #entropy change calculated using ideal gas tables\n",
"dS3=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat in kJ/kg-K\n",
"dS4=1.0767*math.log(T2/T1)+0.0747 #entropy change assuming specific heat is constant at its value at 990K\n",
"print\"\\n hence,change in entropy using ideal gas tables is\",round(dS1,4),\"KJ/kg-k\"\n",
"print\"\\n hence,change in entropy using the value of specific heat at 300K is\",round(dS3,4),\"KJ/kg-k\"\n",
"print\"\\n hence,change in entropy assuming specific heat is constant at its value at 900K is \",round(dS4,4),\"KJ/kg-k\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,change in entropy using ideal gas tables is 1.7228 KJ/kg-k\n",
"\n",
" hence,change in entropy using the value of specific heat at 300K is 1.5586 KJ/kg-k\n",
"\n",
" hence,change in entropy assuming specific heat is constant at its value at 900K is 1.8076 KJ/kg-k\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.5:pg-296\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 5\n",
"#entropy change\n",
"import math\n",
"Cp=1.004 #specific heat at constant pressure in kJ/kg-K\n",
"R=0.287 #gas constant in kJ/kg-K\n",
"P1=400.0 #initial pressure in kPa\n",
"P2=300.0 #final pressure in kPa\n",
"T1=300 #initial temperature in K\n",
"T2=600 #final temperature in K\n",
"dS1=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat\n",
"s1=6.8693 #specific entropy at T1\n",
"s2=7.5764 #specific entropy at T2\n",
"dS2=s2-s1-R*math.log(P2/P1) #entropy change assuming variable specific heat\n",
"print\"\\n hence,entropy change assuming constant specific heat is\",round(dS1,4),\"KJ/kg-k\" \n",
"print\"\\n and assuming variable specific heat is\",round(dS2,4),\"KJ/kg-k\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,entropy change assuming constant specific heat is 0.7785 KJ/kg-k\n",
"\n",
" and assuming variable specific heat is 0.7897 KJ/kg-k\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.6:pg-297"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 6\n",
"#work done by air\n",
"import math\n",
"T1=600 #initial temperature of air in K\n",
"P1=400.0 #intial pressure of air in kPa\n",
"P2=150.0 #final pressure in kPa\n",
"u1=435.10 #specific internal energy at temperature T1 in kJ/kg\n",
"sT1=7.5764 #specific entropy at temperature T1 in kJ/kg-K\n",
"R=0.287 #gas constant in kJ/kg-K\n",
"ds=0\n",
"sT2=ds+sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n",
"print\"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\"\n",
"T2=457 #final temperature in K\n",
"u2=328.14 #specific internal energy at temperature T2 in kJ/kg\n",
"w=u1-u2 #work done by air in kJ/kg\n",
"print\"\\n hence,work done by air is\",round(w,4),\"KJ/kg\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table\n",
"\n",
" hence,work done by air is 106.96 KJ/kg\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.7:pg-300"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 7\n",
"#work and heat transfer\n",
"\n",
"P2=500 #final pressure in cylinder in kPa\n",
"P1=100 #initial pressure in cylinder in kPa\n",
"T1=20+273.2 #initial temperature inside cylinder in Kelvins\n",
"n=1.3 \n",
"T2=(T1)*(P2/P1)**((n-1)/n) #final temperature inside cylinder in K\n",
"R=0.2968 #gas constant in kJ/kg-K\n",
"w12=R*(T2-T1)/(1-n) #work in kJ/kg\n",
"Cvo=0.745 #specific heat at constant volume in kJ/kg-K\n",
"q12=Cvo*(T2-T1)+w12 #heat transfer in kJ/kg\n",
"print\" \\n hence,work done is\",round(w12,2),\"KJ/kg\" \n",
"print\"\\n and heat transfer are\",round(q12,1),\"KJ/kg\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" \n",
" hence,work done is -130.47 KJ/kg\n",
"\n",
" and heat transfer are -32.2 KJ/kg\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.8:pg-308"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3\n",
"#calculating increase in entropy\n",
"\n",
"m=1 #mass of saturated water vapour\n",
"sfg=6.0480 #in kJ/K\n",
"T=25 #temperature of surrounding air in celsius\n",
"dScm=-m*sfg #change in entropy of control mass in kJ/K\n",
"hfg=2257.0 #in kJ/kg\n",
"Qtosurroundings=m*hfg #heat transferred to surroundings in kJ\n",
"dSsurroundings=Qtosurroundings/(T+273.15) #in kJ/K\n",
"dSnet=dScm+dSsurroundings #net increase in entropy in kJ/K\n",
"print\" hence,net increase in entropy of water plus surroundings is \",round(dSnet,3),\"KJ/k\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" hence,net increase in entropy of water plus surroundings is 1.522 KJ/k\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.9:pg-309"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 9\n",
"#entropy generation\n",
"\n",
"Qout=1.0 #value of heat flux generated by 1kW of electric power\n",
"T=600.0 #temperature of hot wire surface in K\n",
"Sgen=Qout/T #entropy generation in kW/K\n",
"print\"\\n hence,entropy generation is\",round(Sgen,5),\"KW/k\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,entropy generation is 0.00167 KW/k\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.10:pg-310"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 10\n",
"#Determiining the entropy generated\n",
"\n",
"B=4.0#COP of air conditioner\n",
"W=10.0 #power input of air conditioner in kW\n",
"Qh=B*W #in kW\n",
"Ql=Qh-W #in kW\n",
"Thigh=323 #in Kelvin\n",
"Tlow=263 #in Kelvin\n",
"SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) #in W/K\n",
"Tl=281 # in K\n",
"Th=294 #in K\n",
"SgenCV1=Ql*1000/Tlow-Ql*1000/Tl #in W/K\n",
"SgenCV2=Qh*1000/Th-Qh*1000/Thigh #in W/K\n",
"SgenTOT=SgenCV1+SgenCV2+SgenHP #in W/K\n",
"print\" \\n Hence,Total entropy generated is\",round(SgenTOT,1),\"W/k\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" \n",
" Hence,Total entropy generated is 29.3 W/k\n"
]
}
],
"prompt_number": 16
}
],
"metadata": {}
}
]
}
|
