summaryrefslogtreecommitdiff
path: root/Electrical_Power_System_by_C.L._Wadhwa/Chapter20.ipynb
blob: c307f665630bd025623ee52346f8fe0f20bc7c8d (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 20 : Load Frequency Control"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20.1, Page No 676"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "#let x MW flows from A to B\n",
      "#Load on station A=75+x\n",
      "#%drop in speed =5*(75+x)/200\n",
      "#load on station B =(30-x)\n",
      "#%drp in speed=(30-x)*4/75\n",
      "x=(1.6-1.875)/(.025+.12+.0533) #by manipulating equation : 5*(75+x)/200 + 3*x/25 =(30-x)*4/75 \n",
      "\n",
      "#Results\n",
      "print(\"x=%.2f MW\\n\" %x)\n",
      "print(\"which means power of magnitude %.2f MW will be from B to A\" %abs(x))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "x=-1.39 MW\n",
        "\n",
        "which means power of magnitude 1.39 MW will be from B to A\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20.2, Page No 676"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "x=(250*11.0)/(21.0+11)\t\t# by manipulating equation : 5x/110=5x(250-x)/210\n",
      "P=250-x\t\t\t\t\t\t#Power shared by 210 MW unit \n",
      "\n",
      "#Results\n",
      "print(\"Power supplied by 210 MW unit = %.2f MW \\n\" %P)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Power supplied by 210 MW unit = 164.06 MW \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20.3, Page No 677"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#initialisation of variables\n",
      "E=4.5*100\t\t#Energy stored at no load(MJ)\n",
      "E1=25*.6\t\t#Energy lost by rotor(MJ)\n",
      "\n",
      "#Calculations\n",
      "fnew=math.sqrt((E-E1)/E)*50\n",
      "\n",
      "#Results\n",
      "print(\"New frequency will be %.2f Hz\" %fnew)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "New frequency will be 49.16 Hz\n"
       ]
      }
     ],
     "prompt_number": 3
    }
   ],
   "metadata": {}
  }
 ]
}