{ "metadata": { "name": "Chapter_6" }, "nbformat": 2, "worksheets": [ { "cells": [ { "cell_type": "markdown", "source": [ "

Chapter 6: Level

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

Example 6.1,Page Number:370

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

Example 6.2, Page Number:371

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

Example 6.3, Page Number: 372

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

Example 6.4, Page Number: 373

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

Example 6.5, Page Number: 374

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

Example 6.6, Page Number: 374

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

Example 6.7, Page Number: 374

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

Example 6.8, Page Number: 374

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

Example 6.9, Page Number:374

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

Example 6.10, Page Number:375

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