{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 - Properties of the pure substance"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - Pg 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the internal energy of the system.\n",
"#Initialization of variables\n",
"T=32. #F\n",
"m=1 #lbm\n",
"J=778.16\n",
"#calculations\n",
"print '%s' %(\"From steam tables,\")\n",
"hf=0 \n",
"p=0.08854 #psia\n",
"vf=0.01602 #ft^3/lbm\n",
"u=hf-p*144*vf/J\n",
"#results\n",
"print '%s %.7f %s' %(\"Internal energy =\",u,\"Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From steam tables,\n",
"Internal energy = -0.0002625 Btu/lbm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - Pg 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in entropy of the process\n",
"#Initialization of variables\n",
"P=40. #psia\n",
"#calculations\n",
"print '%s' %(\"from steam tables,\")\n",
"hf=200.8 #Btu/lbm\n",
"hg=27 #Btu/lbm\n",
"T=495. #R\n",
"ds=(hf-hg)/T\n",
"#results\n",
"print '%s %.3f %s' %(\"Change in entropy =\",ds,\"Btu/lbm R\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from steam tables,\n",
"Change in entropy = 0.351 Btu/lbm R\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - Pg 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the specific enthalpy of the mixture\n",
"#Initialization of variables\n",
"x=0.35\n",
"T=18. #F\n",
"#calculations\n",
"print '%s' %(\"From table B-14,\")\n",
"hf=12.12 #Btu/lbm\n",
"hg=80.27 #Btu.lbm\n",
"hfg=-hf+hg\n",
"h=hf+x*hfg\n",
"#results\n",
"print '%s %.1f %s' %(\"specific enthalpy =\",h,\"Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table B-14,\n",
"specific enthalpy = 36.0 Btu/lbm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - Pg 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat required for the process\n",
"#Initialization of variables\n",
"x=0.35\n",
"T=18. #F\n",
"T2=55.5 #F\n",
"#calculations\n",
"print '%s' %(\"From table B-14,\")\n",
"hf=12.12 #Btu/lbm\n",
"hg=80.27 #Btu.lbm\n",
"hfg=-hf+hg\n",
"h=hf+x*hfg\n",
"h2=85.68 #Btu/lbm\n",
"dh=h2-h\n",
"#results\n",
"print '%s %.2f %s' %(\"Heat required =\",dh,\"Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table B-14,\n",
"Heat required = 49.71 Btu/lbm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5 - Pg 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the enthalpy given out and the quality of the gas\n",
"#Initialization of variables\n",
"P=1460. #psia\n",
"T=135. #F\n",
"P2=700. #psia\n",
"#calculations\n",
"print '%s' %(\"From mollier chart,\")\n",
"h=120 #Btu/lbm\n",
"x=0.83\n",
"#results\n",
"print '%s %d %s' %(\"enthalpy =\",h,\" Btu/lbm\")\n",
"print '%s %.2f' %(\"\\n Qulaity = \",x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From mollier chart,\n",
"enthalpy = 120 Btu/lbm\n",
"\n",
" Qulaity = 0.83\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - Pg 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat transferred in the process\n",
"#Initialization of variables\n",
"m=1 #lbm\n",
"P1=144. #psia\n",
"P2=150. #psia\n",
"T1=360. #F\n",
"J=778.16\n",
"#calculations\n",
"print '%s' %(\"From table 3,\")\n",
"v1=3.160 #ft^3/lbm\n",
"h1=1196.5 #Btu/lbm\n",
"u1=h1-P1*144*v1/J\n",
"h2=1211.4 #Btu/lbm\n",
"u2=h2-P2*144*v1/J\n",
"dq=u2-u1\n",
"#results\n",
"print '%s %.1f %s' %(\"Heat transferred =\",dq,\" Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 3,\n",
"Heat transferred = 11.4 Btu/lbm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7 - Pg 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the work required, reversible work and the work done in compression \n",
"#Initialization of variables\n",
"T1=100 #F\n",
"P2=1000. #psia\n",
"x=0.6\n",
"J=778.16\n",
"tir=2 \n",
"P1=0.9 #psia\n",
"#calculations\n",
"print '%s' %(\"From table 3,\")\n",
"hf=67.97\n",
"htc=2.7\n",
"hpc=0.32\n",
"h1=67.97\n",
"dv=0.000051\n",
"v=0.01613\n",
"h2=hf+htc+hpc\n",
"wrev=h1-h2\n",
"wact=wrev/x\n",
"dt=hpc+tir\n",
"t2act=T1+dt\n",
"wrev2=-v*144*(P2-P1)/J\n",
"dw=(P1+P2)/2. *dv *144/J\n",
"#results\n",
"print '%s %.2f %s' %(\"Work required =\",wact,\" Btu/lbm\")\n",
"print '%s %.2f %s' %(\"\\n reversible work done =\",wrev2,\" Btu/lbm\")\n",
"print '%s %.4f %s' %(\"\\n Work done in compression =\",dw,\"Btu/lbm\")\n",
"print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 3,\n",
"Work required = -5.03 Btu/lbm\n",
"\n",
" reversible work done = -2.98 Btu/lbm\n",
"\n",
" Work done in compression = 0.0047 Btu/lbm\n",
"The answers are a bit different due to rounding off error in textbook\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8 - Pg 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat transferred in the process\n",
"#Initialization of variables\n",
"pa=1000. #atm\n",
"ta=100. #F\n",
"#calculations\n",
"hf=67.97 #Btu/lbm\n",
"w=3 #Btu/lbm\n",
"ha=hf+w\n",
"print '%s' %(\"from steam table 2,\")\n",
"hc=1191.8 #Btu/lbm\n",
"qrev=hc-ha\n",
"#results\n",
"print '%s %.1f %s' %(\"Heat transferred =\",qrev,\"Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from steam table 2,\n",
"Heat transferred = 1120.8 Btu/lbm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10 - Pg 197"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the work done and final pressure\n",
"#Initialization of variables\n",
"P1=144 #psia\n",
"T1=400 #F\n",
"y=0.7\n",
"#calculations\n",
"print '%s' %(\"From steam tables,\")\n",
"h1=1220.4 #Btu/lbm\n",
"s1=1.6050 #Btu/lbm R\n",
"s2=1.6050 #Btu/lbm R\n",
"P2=3 #psia\n",
"sf=0.2008 #Btu/lbm R\n",
"sfg=1.6855 #Btu/lbm R\n",
"x=(s1-sf)/sfg\n",
"hf=109.37 #Btu/lbm\n",
"hfg=1013.2 #Btu/;bm\n",
"h2=hf+x*hfg\n",
"work=h1-h2\n",
"dw=y*work\n",
"h2d=h1-dw\n",
"#results\n",
"print '%s %d %s' %(\"Work done =\",work,\"Btu/lbm\")\n",
"print '%s %.1f %s' %(\"\\n work done in case 2 =\",dw,\"Btu/lbm\")\n",
"print '%s %d %s' %(\"\\n Final state pressure =\",P2,\"psia\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From steam tables,\n",
"Work done = 266 Btu/lbm\n",
"\n",
" work done in case 2 = 186.8 Btu/lbm\n",
"\n",
" Final state pressure = 3 psia\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11 - Pg 200"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the quality of wet steam\n",
"#Initialization of variables\n",
"pb=14.696 #psia\n",
"pa=150 #psia\n",
"tb=300 #F\n",
"#calculations\n",
"print '%s' %(\"From steam tables,\")\n",
"hb=1192.8 #Btu/lbm\n",
"ha=hb\n",
"hf=330.51 #Btu/lbm\n",
"hfg=863.6 #Btu/lbm\n",
"x=(ha-hf)/hfg*100\n",
"#results\n",
"print '%s %.1f %s' %(\"Quality of wet steam =\",x,\"percent\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From steam tables,\n",
"Quality of wet steam = 99.8 percent\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12 - Pg 204"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the thermal and furnace efficiencies\n",
"#Initialization of variables\n",
"p1=600 #psia\n",
"p2=0.2563 #psia\n",
"t1=486.21 #F\n",
"t2=60 #F\n",
"fur=0.75\n",
"#calculations\n",
"print '%s' %(\"from steam tables,\")\n",
"h1=1203.2\n",
"hf1=471.6\n",
"hfg1=731.6\n",
"h2=1088\n",
"hf2=28.06\n",
"hfg2=1059.9\n",
"s1=1.4454\n",
"sf1=0.6720\n",
"sfg1=0.7734\n",
"s2=2.0948\n",
"sf2=0.0555\n",
"sfg2=2.0393\n",
"xd=(s1-sf2)/sfg2\n",
"hd=hf2+xd*hfg2\n",
"xa=0.3023\n",
"ha=hf2+xa*hfg2\n",
"wbc=0\n",
"wda=0\n",
"wcd=h1-hd\n",
"wab=ha-hf1\n",
"W=wab+wcd+wbc+wda\n",
"Wrev=hfg1- (t2+459.7)*sfg1\n",
"etat=(t1-t2)/(t1+459.7)*100\n",
"eta=fur*etat\n",
"#results\n",
"print '%s %d %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
"print '%s %.1f %s' %(\"\\n Furnace efficiency =\",eta,\"percent\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from steam tables,\n",
"Thermal efficiency = 45 percent\n",
"\n",
" Furnace efficiency = 33.8 percent\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13 - Pg 204"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the thermal and overall efficiencies\n",
"#Initialization of variables\n",
"dhab=-123.1\n",
"etac=0.5\n",
"ha=348.5\n",
"etaf=0.75\n",
"eta=0.85\n",
"hf=471.6\n",
"hfg=731.6\n",
"hc=1203.2\n",
"dhcd=452.7\n",
"#calculations\n",
"dwabs=dhab/etac\n",
"hbd=ha-dwabs\n",
"dwcds=dhcd*eta\n",
"dqa=hc-hbd\n",
"etat=(dwcds+dwabs)/dqa*100\n",
"eta=etat*etaf\n",
"#results\n",
"print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
"print '%s %.1f %s' %(\"\\n Overall efficiency =\",eta,\"percent\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency = 22.8 percent\n",
"\n",
" Overall efficiency = 17.1 percent\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14 - Pg 205"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the thermal and overall efficiencies\n",
"#Initialization of variables\n",
"t=60 #F\n",
"J=778.16\n",
"p1=600 #psia\n",
"p2=0.2563 #psia\n",
"etaf=0.85 \n",
"#calculations\n",
"print '%s' %(\"From steam tables,\")\n",
"vf=0.01604 #ft^3/lbm\n",
"dw=-vf*(p1-p2)*144/J\n",
"ha=28.06 #Btu/lbm\n",
"hb=29.84 #Btu/lbm\n",
"hd=1203.2 #Btu/lbm\n",
"he=750.5 #Btu/lbm\n",
"dqa=hd-hb\n",
"dqr=ha-he\n",
"dw=dqa+dqr\n",
"dwturb=hd-he\n",
"dwpump=ha-hb\n",
"etat=dw/dqa*100\n",
"eta=etat*etaf\n",
"#results\n",
"print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
"print '%s %.1f %s' %(\"\\n Overall efficiency =\",eta,\"percent\")\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From steam tables,\n",
"Thermal efficiency = 38.4 percent\n",
"\n",
" Overall efficiency = 32.7 percent\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15 - Pg 206"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the thermal and overall efficiencies\n",
"#Initialization of variables\n",
"dhab=-1.78\n",
"etac=0.5\n",
"ha=28.06\n",
"eta=0.85\n",
"hf=471.6\n",
"hfg=731.6\n",
"hd=1203.2\n",
"dhcd=452.7\n",
"#calculations\n",
"dwabs=dhab/etac\n",
"hbd=ha-dwabs\n",
"dwcds=dhcd*eta\n",
"dqa=hd-hbd\n",
"etat=(dwcds+dwabs)/dqa*100\n",
"eta=etat*eta\n",
"#results\n",
"print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
"print '%s %.1f %s' %(\"\\n Overall efficiency =\",eta,\"percent\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency = 32.5 percent\n",
"\n",
" Overall efficiency = 27.7 percent\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 16 - Pg 207"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the coefficient of performance, hp required, work of compression and expansion\n",
"#Initialization of variables\n",
"Ta=500. #R\n",
"Tr=540. #R\n",
"#calculations\n",
"cop=Ta/(Tr-Ta)\n",
"hp=4.71/cop\n",
"print '%s' %(\"From steam tables,\")\n",
"ha=48.02\n",
"hb=46.6\n",
"hc=824.1\n",
"hd=886.9\n",
"Wc=-(hd-hc)\n",
"We=-(hb-ha)\n",
"#results\n",
"print '%s %.1f' %(\"Coefficient of performance = \",cop)\n",
"print '%s %.3f %s' %(\"\\n horsepower required per ton of refrigeration =\",hp,\"hp/ton refrigeration\")\n",
"print '%s %.1f %s' %(\"\\n Work of compression =\",Wc,\"Btu/lbm\")\n",
"print '%s %.2f %s' %(\"\\n Work of expansion =\",We,\"Btu/lbm\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From steam tables,\n",
"Coefficient of performance = 12.5\n",
"\n",
" horsepower required per ton of refrigeration = 0.377 hp/ton refrigeration\n",
"\n",
" Work of compression = -62.8 Btu/lbm\n",
"\n",
" Work of expansion = 1.42 Btu/lbm\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17 - Pg 209"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the coefficient of performance\n",
"#Initialization of variables\n",
"x=0.8\n",
"he=26.28 #Btu/lbm\n",
"hb=26.28 #Btu/lbm\n",
"pe=98.76 #psia\n",
"pc=51.68 #psia\n",
"hc=82.71 #Btu/lbm\n",
"hf=86.80+0.95\n",
"#calculations\n",
"dwisen=-(hf-hc)\n",
"dwact=dwisen/x\n",
"hd=hc-dwact\n",
"cop=(hc-hb)/(hd-hc)\n",
"#results\n",
"print '%s %.2f' %(\"Coefficient of performance = \",cop)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Coefficient of performance = 8.96\n"
]
}
],
"prompt_number": 16
}
],
"metadata": {}
}
]
}