summaryrefslogtreecommitdiff
path: root/Non_Conventional_Energy_Resources/Chapter11.ipynb
blob: ef0f355fba4bdb644dc1e4f25dd0162950e8b682 (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
{

 "metadata": {

  "name": "",

  "signature": "sha256:131e5807eda8c30572547a9b8e6ec8831945c814f15b621bc05038f1dbaad1ff"

 },

 "nbformat": 3,

 "nbformat_minor": 0,

 "worksheets": [

  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 1,

     "metadata": {},

     "source": [

      "Chapter11:Small Hydro Resources"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex11.1:pg-355"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "# given data\n",

      "rho=996 # density in kg/m^3\n",

      "effi=0.55 # oveall efficiency\n",

      "Q=100/1000.0 # discharge in m^3/sec\n",

      "h=30 # gross head in m\n",

      "g=9.81 # gravitational acceleration in m/sec^2\n",

      "\n",

      "\n",

      "Pnet=effi*rho*Q*g*h/1000 # net power in watts\n",

      "print \"Power available is\",round(Pnet,3),\"kilowatts\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "Power available is 16.122 kilowatts\n"

       ]

      }

     ],

     "prompt_number": 7

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex11.2:pg-356"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "# given data\n",

      "pf=0.8 # power factor\n",

      "Load=3 # load in kW\n",

      "V=230 # voltage of kettle in V\n",

      "P=500 # power of kettle in W\n",

      "VA=pf*Load # VA load in kVA\n",

      "C=4200 # specific heat of water in j/kg-K\n",

      "T=45-20 # temperature difference in degree celsius\n",

      "\n",

      "VAR=VA*1.6 # net required VAR rating thus 60% extra capacity\n",

      "\n",

      "R=(V**2)/P # resistance by kettle in ohms\n",

      "\n",

      "Po=7*(V**2)/R # power dissipation in W for 7 elements\n",

      "\n",

      "print \"7 elements are connected in parallel\"\n",

      "\n",

      "Q=Load*1000.0/(C*T) # flow rate in kg/sec\n",

      "\n",

      "print\"The required flow rate is \",round(Q,5),\"litre/sec\"\n",

      "\n",

      "\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "7 elements are connected in parallel\n",

        "The required flow rate is  0.02857 litre/sec\n"

       ]

      }

     ],

     "prompt_number": 11

    }

   ],

   "metadata": {}

  }

 ]

}