{ "metadata": { "name": "", "signature": "sha256:6148d2be2796832e7d2e9cefcc5c361a4c9dc07b22c355b627a74bd914199046" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "

Chapter 6: Level

" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.1,Page Number:370

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#(a)\n", "\n", "# variable declaration\n", "p=1.5 # pressure applied\n", "a=4.0 # mA corresponds to 0 kg/cm^2\n", "b=20.0 # mA corresponds to 2 kg/cm^2\n", "\n", "#calculation\n", "wh=(((b-a)/2)*p)+a\n", "\n", "#result\n", "print('(a)just at the bottom level of the tank')\n", "print('Water head applied to the transmitter =%d mA'%wh)\n", "\n", "#(b)\n", "\n", "#calculation\n", "wh2=(((b-a)/2)*p)+2*a\n", "\n", "#result\n", "print('\\n\\n(b)5m below the bottom of the tank')\n", "print('Water head applied to the transmitter =%d mA' %wh2)\n", "\n", "#(c)\n", "\n", "#calculation\n", "wh3=(((b-a)/2)*p)\n", "\n", "#result\n", "print('\\n\\n(c)5m above the bottom of the tank')\n", "print('Water head applied to the transmitter =%d mA'%wh3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)just at the bottom level of the tank\n", "Water head applied to the transmitter =16 mA\n", "\n", "\n", "(b)5m below the bottom of the tank\n", "Water head applied to the transmitter =20 mA\n", "\n", "\n", "(c)5m above the bottom of the tank\n", "Water head applied to the transmitter =12 mA" ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.2, Page Number:371

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#(a)\n", "\n", "#variable declaration\n", "b=20.0 # Maximum output\n", "a=4.0 # minimum output \n", "op=16.0 # output in mA\n", "\n", "#calculation\n", "p=(op-a)*2/(b-a)\n", "p_h=p*10.0\n", "h=p_h-2-5\n", "\n", "#result\n", "print('(a)\\nh = %dm'%h)\n", "\n", "#(b)\n", "\n", "#variable declaration\n", "p1=1 # pressure applied\n", "\n", "#calculation\n", "t_op=((b-a)/2)*p1+4\n", "\n", "#result\n", "print('\\n(b)\\nTransmitter output =%d mA'%t_op)\n", "\n", "#(c)\n", "\n", "#variable declaration\n", "p2=0.5 # applied pressure\n", "\n", "#calculation\n", "t_op1=((b-a)/2)*p2+4\n", "\n", "#result\n", "print('\\n(c)\\nTransmitter output =%d mA'%t_op1)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)\n", "h = 8m\n", "\n", "(b)\n", "Transmitter output =12 mA\n", "\n", "(c)\n", "Transmitter output =8 mA" ] } ], "prompt_number": 2 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.3, Page Number: 372

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#(a)\n", "\n", "#variable declaration\n", "b=20.0 # Maximum output\n", "a=4.0 # minimum output\n", "op=16.0 # actual output \n", "wt_l1=25.0 # water level (i)\n", "\n", "#calculation\n", "t_op=((b-a)/100)*(100-75)+4\n", "\n", "#result\n", "print('(a)\\nWater level=+25cm\\nTransmitter output = %d mA' %t_op)\n", "\n", "#(b)\n", "\n", "#calculation\n", "wt_l2=-25.0 # water level (ii)\n", "t_op2=((b-a)/100)*(100-25)+4\n", "\n", "#result\n", "print('\\n(b)\\nWater level=-25cm\\nTransmitter output = %d mA' %t_op2)\n", "\n", "#(c)\n", "\n", "#Variable declaration\n", "t_op3=12.0 # Transmitter output \n", "\n", "#calculation\n", "H=(100.0/(b-a))*(12-4) \n", "\n", "#result\n", "print('\\n(c)\\nHead Applied = %d cm\\nLevel corresponding to 50 cm head =0 cm' %H)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)\n", "Water level=+25cm\n", "Transmitter output = 8 mA\n", "\n", "(b)\n", "Water level=-25cm\n", "Transmitter output = 16 mA\n", "\n", "(c)\n", "Head Applied = 50 cm\n", "Level corresponding to 50 cm head =0 cm" ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.4, Page Number: 373

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#(a)\n", "\n", "#variable declaration\n", "a=5.0*10**-4 #area\n", "l=8.0 #length\n", "dens=6.0*1000.0 #density\n", "\n", "#calculation\n", "w=a*l*dens\n", "\n", "#result\n", "print('(a)\\nWeight of the displacer if weighed in air = %d kg'%w)\n", "\n", "\n", "#(i)\n", "\n", "#variable declaration\n", "sbr1=23.0 # spring balance reading\n", "\n", "#calculation\n", "wloss1=w-sbr1\n", "L1=wloss1/(1000.0*a)\n", "\n", "#result\n", "print('\\n(i)\\tL1=%dm'%L1)\n", "\n", "\n", "#(ii)\n", "\n", "#variable declaration\n", "sbr2=22.0 # spring balance reading\n", "\n", "#calculation\n", "wloss2=w-sbr2\n", "L2=wloss2/(1000.0*a)\n", "\n", "#result\n", "print('\\n(ii)\\tL2=%dm'%L2)\n", "\n", "#(iii)\n", "\n", "#variable declaration\n", "sbr3=21.0 # spring balance reading\n", "\n", "#calculation\n", "wloss3=w-sbr3\n", "L3=wloss3/(1000.0*a)\n", "\n", "#result\n", "print('\\n(iii)\\tL3=%dm'%L3)\n", "\n", "#(b)\n", "\n", "#variable declaration\n", "level=8.0 # level wen tank is full \n", "\n", "#calculation\n", "wt=a*level*1000.0\n", "spring=w-wt\n", "\n", "#result\n", "print('\\n(b):when the tank is full\\nSpring Balance reading = %d kg'%spring)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)\n", "Weight of the displacer if weighed in air = 24 kg\n", "\n", "(i)\tL1=2m\n", "\n", "(ii)\tL2=4m\n", "\n", "(iii)\tL3=6m\n", "\n", "(b):when the tank is full\n", "Spring Balance reading = 20 kg" ] } ], "prompt_number": 4 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.5, Page Number: 374

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "rho=1000.0 # density of water \n", "v=3.0 # displaced volume of water \n", "\n", "#calculation\n", "Bw=rho*v\n", "\n", "#Result\n", "print('Buoyance Force(Bw) = %d kg'%Bw)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Buoyance Force(Bw) = 3000 kg" ] } ], "prompt_number": 5 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.6, Page Number: 374

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "rho=1000.0 # density of water\n", "Bw=5000.0 # Buoyancy Force\n", "\n", "#calculation\n", "v=Bw/rho\n", "\n", "#result\n", "print('V = %d m^3' %v)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "V = 5 m^3" ] } ], "prompt_number": 6 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.7, Page Number: 374

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "rho=1000.0 # density of water\n", "h=10.0 # height of liquid\n", "\n", "#calculation\n", "P=rho*h\n", "\n", "#result\n", "print('P = %d kg/m^2 = %d kg/cm^2 '%(P,P/10000))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "P = 10000 kg/m^2 = 1 kg/cm^2 " ] } ], "prompt_number": 7 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.8, Page Number: 374

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "rho=1000.0 # density of water\n", "h=15.0 # height of liquid \n", "ex_p=1.0 # External pressure on liquid\n", "\n", "#calculation\n", "P=(rho*h/10000.0)+ex_p\n", "\n", "#result\n", "print('P = %.1f kg/cm^2' %P)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "P = 2.5 kg/cm^2" ] } ], "prompt_number": 8 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.9, Page Number:374

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "rho=1000.0 # density of water\n", "ex_p=0.5*10**4 # External pressure on liquid \n", "P=1.6*10**4 #(rho*h/10000)+ex_p\n", "\n", "#calculation\n", "h=(P-ex_p)/1000.0\n", "\n", "#result\n", "print('h = %d m' %h)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "h = 11 m" ] } ], "prompt_number": 9 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6.10, Page Number:375

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "c2=100.0*10**-6 # capacitance in capacitance probe\n", "r1=10.0*10**3 # value of resistor in bride\n", "r2=100.0*10**3 # value of resistor in bride\n", "r3=50.0*10**3 # value of resistor in bride\n", "\n", "#calculation\n", "Cx=r1*c2/r3\n", "Cx=Cx*10**6\n", "\n", "#result\n", "print('Cx = %d microFarad'%Cx)\n", "c=5.0\n", "\n", "#calculation\n", "l=Cx/c\n", "\n", "#result\n", "print('\\nLevel on the probe = %dm'%l)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Cx = 20 microFarad\n", "\n", "Level on the probe = 4m" ] } ], "prompt_number": 10 } ], "metadata": {} } ] }