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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 24 : Unit Commitment"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 24.3, Page No 803"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Fc1=1.1\t\t\t#Fuel cost(1)=Rs 1.1/MBtu\n",
"Fc2=1\t\t\t#Fuel cost(2)=1/MBtu\n",
"Fc3=1.2\t\t\t#Fuel cost(3)=1.2/MBtu\n",
"P1max=600.0\n",
"P1=P1max\n",
"\n",
"#Calculations\n",
"F1=600+7.1*P1+0.00141*(P1**2)\t#For P1= Pm1ax\n",
"Favg1=F1*Fc1/600.0\t\t\t\t#Full load average production cost\n",
"P2max=450.0\n",
"P2=P2max\n",
"F2=350+7.8*P2+0.00195*(P2**2)\t#For P2= P2max\n",
"Favg2=F2*Fc2/450.0\t\t\t\t#Full load average production cost\n",
"P3max=250.0\n",
"P3=P3max\n",
"F3=80+8*P3+0.0049*(P3**2)\t\t#For P3= P3max\n",
"Favg3=F3*Fc3/250.0\t\t\t\t#Full load average production cost\n",
"\n",
"#Results\n",
"print(\"Priority List is as follows\")\n",
"print(\"Unit Rs/MWhr MinMW Max MW\")\n",
"print(\" 2 %.3f 100 %.0f \" %(Favg2,P2max))\n",
"print(\" 1 %.4f 60 %.0f \" %(Favg1,P1max))\n",
"print(\" 3 %.2f 50 %.0f \" %(Favg3,P3max))\n",
"Fmax1=P1max+P2max+P3max\n",
"Fmax2=P2max+P1max\n",
"Fmax3=P2max\n",
"print(\"Unit Commitment Scheme is follows\")\n",
"print(\"Combination Min.MW from Combination Max.MW from Combination\")\n",
"print(\"2+1+3 310 %.0f \" %Fmax1)\n",
"print(\"2+1 260 %.0f \" %Fmax2)\n",
"print(\"2 100 %.0f \" %Fmax3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Priority List is as follows\n",
"Unit Rs/MWhr MinMW Max MW\n",
" 2 9.455 100 450 \n",
" 1 9.8406 60 600 \n",
" 3 11.45 50 250 \n",
"Unit Commitment Scheme is follows\n",
"Combination Min.MW from Combination Max.MW from Combination\n",
"2+1+3 310 1300 \n",
"2+1 260 1050 \n",
"2 100 450 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 24.4, Page No 805"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"#Calculations\n",
"\n",
"def F1(P1):\n",
" F1=7.1*P1+.00141*(P1^2)\n",
" print(\"F1(%.0f)=%.1f\" %(P1,F1))\n",
"\n",
"def F2(P2):\n",
" f2=7.8*P2+.00195*(P2^2)\n",
" print(\"f2(%.0f)=%.0f\" %(P2,f2))\n",
"\n",
"def F(P1,P2):\n",
" F1=7.1*P1+.00141*(P1**2)\n",
" F2=7.8*P2+.00195*(P2**2)\n",
" F=F1+F2\n",
" print(\"F1(%.0f)+f2(%.0f)=%.0f\" %(P1,P2,F))\n",
"\n",
"\n",
" \n",
"#Results\n",
"P1max=600\n",
"P2max=450\n",
"print(\"Unit Commitment using Load 500MW\")\n",
"F1(500)\n",
"print(\"\\n Since min. Power of second unit is 100MW , we find\")\n",
"F(400,100)\n",
"F(380,120)\n",
"F(360,140)\n",
"print(\"\\n Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\")\n",
"print(\"Next we increase the load by 50 MW and loading unit 1 we get, \\n\")\n",
"F1(550)\n",
"print(\"Also if we distribute a part of load to unit 2 we get ,\")\n",
"F(450,100)\n",
"F(400,150)\n",
"F(350,200)\n",
"print(\"\\n Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Unit Commitment using Load 500MW\n",
"F1(500)=3550.7\n",
"\n",
" Since min. Power of second unit is 100MW , we find\n",
"F1(400)+f2(100)=3865\n",
"F1(380)+f2(120)=3866\n",
"F1(360)+f2(140)=3869\n",
"\n",
" Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\n",
"Next we increase the load by 50 MW and loading unit 1 we get, \n",
"\n",
"F1(550)=3905.8\n",
"Also if we distribute a part of load to unit 2 we get ,\n",
"F1(450)+f2(100)=4280\n",
"F1(400)+f2(150)=4279\n",
"F1(350)+f2(200)=4296\n",
"\n",
" Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
