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-rw-r--r--sample_notebooks/DaudIbrahir Saifi/Chapter_07.ipynb341
-rw-r--r--sample_notebooks/MohdGufran/chapter_10_1.ipynb358
-rw-r--r--sample_notebooks/ebbygeorge/Chapter01_1.ipynb72
-rw-r--r--sample_notebooks/sai kiranmalepati/Samplenb.ipynb158
4 files changed, 929 insertions, 0 deletions
diff --git a/sample_notebooks/DaudIbrahir Saifi/Chapter_07.ipynb b/sample_notebooks/DaudIbrahir Saifi/Chapter_07.ipynb
new file mode 100644
index 00000000..d7167f86
--- /dev/null
+++ b/sample_notebooks/DaudIbrahir Saifi/Chapter_07.ipynb
@@ -0,0 +1,341 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:012ab8557afdcfdae2cdc3da17271647415fc17ab95dd187f4df0903472edf45"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter - 7 : Cathode Ray Oscilloscopes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.1 - Page No : 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "l=2.5 # in cm\n",
+ "l=l*10**-2 # in meter\n",
+ "d=.5 # in cm\n",
+ "d=d*10**-2 # in meter\n",
+ "S= 20 # in cm\n",
+ "S= S*10**-2 # in meter\n",
+ "Va= 2500 # in volts\n",
+ "# Formula y = OC*AB/OB = (S*d/2)/(l/2)\n",
+ "y = (S*d/2)/(l/2) # in meter\n",
+ "print \"The value of deflection = %0.f cm\" %(y*10**2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of deflection = 4 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.2 - Page No : 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Given data\n",
+ "R_E1= 5.6 # in kohm\n",
+ "C1= 0.2 # in micro F\n",
+ "V_B1= 6.3 # in volt\n",
+ "V_BE= 0.7 # in volt\n",
+ "TL= 2.5 # trigger level for the Schmitt trigger (UTP,LTP) in volt\n",
+ "del_V1= 2*TL # in volt\n",
+ "I_C1= (V_B1-V_BE)/R_E1 # in mA\n",
+ "print \"Charging current = %0.f mA\" %I_C1 \n",
+ "toh= del_V1*C1/I_C1 # in ms\n",
+ "print \"Time period = %0.f ms\" %toh"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charging current = 1 mA\n",
+ "Time period = 1 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.3 - Page No : 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from math import sqrt \n",
+ "#Given data\n",
+ "L=10 # trace length in cm\n",
+ "DS= 5 # deflection sensitivity in V/cm\n",
+ "V_peakTOpeak= L*DS # in volt\n",
+ "V_peak= V_peakTOpeak/2 # in volt\n",
+ "RMS= V_peak/sqrt(2) # RMS value of unknown as voltage in volt\n",
+ "print \"The value of AC voltage = %0.3f volts\" %RMS "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of AC voltage = 17.678 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.4 - Page No : 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division \n",
+ "#Given data\n",
+ "Y= 2+1/2 # Positive Y-peaks in pattern\n",
+ "X= 1/2+1/2 # Positive X-peaks in pattern\n",
+ "f_h= 3# frequency of horizontal voltage signal in kHz\n",
+ "f_yBYf_x= Y/X \n",
+ "# frequency of vertical voltage signal= f_yBYf_x * f_h\n",
+ "f_v= f_yBYf_x * f_h # frequency of vertical voltage signal in kHz\n",
+ "print \"frequency of vertical voltage signal = %0.1f kHz\" %f_v "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "frequency of vertical voltage signal = 7.5 kHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.5 - Page No : 256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Given data\n",
+ "f_x= 1000 # in Hz\n",
+ "Y= 2 # points of tangency to vertical line\n",
+ "X= 5 # points of tangency to horizontal line\n",
+ "f_y= f_x*X/Y # in Hz\n",
+ "print \"Frequency of vertical input = %0.f Hz\" %f_y"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of vertical input = 2500 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.6 - Page No : 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Given data\n",
+ "f=2000 # in Hz\n",
+ "T=1/f # in sec\n",
+ "D=0.2 \n",
+ "PulseDuration= D*T # in sec\n",
+ "print \"The value of pulse duration = %0.1f ms\" %(PulseDuration*10**3) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of pulse duration = 0.1 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.7 - Page No : 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Given data\n",
+ "vertical_attenuation= 0.5 # in V/Div\n",
+ "TPD= 2 # time/Div control in micro sec\n",
+ "P= 4*vertical_attenuation # peak-to-peak amplitude of the signal in V \n",
+ "print \"Peak-to-Peak amplitude of the signal = %0.f V\" %P\n",
+ "T= 4*TPD # in micro sec\n",
+ "T=T*10**-6 # in sec\n",
+ "f=1/T # in Hz\n",
+ "print \"The value of frequency = %0.f kHz\" %(f*10**-3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak-to-Peak amplitude of the signal = 2 V\n",
+ "The value of frequency = 125 kHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.8 - Page No : 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from numpy import pi \n",
+ "#Given data\n",
+ "C_1N= 36 # in pF\n",
+ "C_2= 150 # in pF\n",
+ "R_1N= 1 # in M ohm\n",
+ "R_1= 10 # in M ohm\n",
+ "R_source= 500 # in ohm\n",
+ "# R_1/(omega*(C_2+C_1N)) = R_1N/(omega*C_1)\n",
+ "C_1= R_1N*(C_2+C_1N)/R_1 # in pF\n",
+ "C_T= 1/(1/C_1+1/(C_2+C_1N)) # in pF\n",
+ "C_T= C_T*10**-12 # in F\n",
+ "f= 1/(2*pi*C_T*R_source) \n",
+ "print \"Signal Frequency = %0.2f MHz\" %(f*10**-6)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Signal Frequency = 18.82 MHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example : 7.9 - Page No : 263"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Given data\n",
+ "f= 20 # in MHz\n",
+ "f=f*10**6 # in Hz\n",
+ "toh= 1/f # in sec\n",
+ "toh=toh*10**9 # in ns\n",
+ "# For one cycle occupying 4 horizontal divisions,\n",
+ "MTD= toh/4 # Minimum time/division in ns/division\n",
+ "# Using the 10 times magnifier to provide MTD\n",
+ "MTD_setting= 10*MTD # minimum time/division setting in ns/division\n",
+ "print \"Minimum time/division setting = %0.f ns/division\" %MTD_setting"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum time/division setting = 125 ns/division\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/MohdGufran/chapter_10_1.ipynb b/sample_notebooks/MohdGufran/chapter_10_1.ipynb
new file mode 100644
index 00000000..fc8eeccf
--- /dev/null
+++ b/sample_notebooks/MohdGufran/chapter_10_1.ipynb
@@ -0,0 +1,358 @@
+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No - 10 : Mass Transfer\n",
+ " "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.1 - Page No. : 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "# Given data\n",
+ "P1=4 # in bar\n",
+ "P2=2 # in bar\n",
+ "T=25 # in degree C\n",
+ "Dhp=9*10**-8 # in m**2/s\n",
+ "S=3*10**-3 # in kg mole/m**3 bar\n",
+ "del_x=0.5*10**-3 # thickness in m\n",
+ "#(a) The molar concentration of a gas in terms of solubility\n",
+ "CH1=S*P1 # in kg mole/m**3\n",
+ "CH2=S*P2 # in kg mole/m**3\n",
+ "#(b) Molar diffusion flux of hydrogen through plastic memberence is given by Fick's law of diffision\n",
+ "#N_H= N_h/A = Dhp*(CH1-CH2)/del_x#\n",
+ "N_H= Dhp*(CH1-CH2)/del_x # in kg mole/s-m**2\n",
+ "print \"Molar diffusion flux of hydrogen through the membrane = %0.2e kg mole/s-m**2\" %N_H\n",
+ "#Mass_d_Flux= N_H*Molecular_Weight \n",
+ "Molecular_Weight=2#\n",
+ "Mass_d_Flux= N_H*Molecular_Weight \n",
+ "print \"Molar diffusion flux = %0.3e kg/s-m**2\" %Mass_d_Flux"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Molar diffusion flux of hydrogen through the membrane = 1.08e-06 kg mole/s-m**2\n",
+ "Molar diffusion flux = 2.160e-06 kg/s-m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.2 - Page No. : 322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Given data\n",
+ "T=25 # in degree C\n",
+ "T=T+273 # in K\n",
+ "P=1#\n",
+ "V1=12 #Molecular volume of H2 in cm**3/gm mole\n",
+ "V2=30 #Molecular volume of Air in cm**3/gm mole\n",
+ "M1=2 # Molecular weight of H2\n",
+ "M2=29 # Molecular weight of Air\n",
+ "#The diffusion coefficient for gases in terms of molecular volumes may be express as\n",
+ "D_AB= .0043*T**(3/2)/(P*(V1**(1/3)+V2**(1/3)))*(1/M1+1/M2)**(1/2)#\n",
+ "print \"The diffusion coefficient for gases in terms of molecular volumes = %0.3f cm**2/sec\" %D_AB"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion coefficient for gases in terms of molecular volumes = 2.997 cm**2/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.3 - Page No. : 322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Given data\n",
+ "T=300 # temp of gas mixture in K\n",
+ "D_HN2=18*10**-6 # in m**2/s at 300 K, 1 bar\n",
+ "T1=300 # in K\n",
+ "D_HO2=16*10**-6 # in m**2/s at 273 K, 1 bar\n",
+ "T2=273 # in K\n",
+ "O_2=0.2#\n",
+ "N_2=0.7#\n",
+ "H_2=0.1#\n",
+ "#The diffusivity at the mixture temperature and pressure are calculated as \n",
+ "# D1/D2 = (T1/T2)**(3/2)*(P2/P1)\n",
+ "D_HO2= (T/T2)**(3/2)*1/4*D_HO2#\n",
+ "D_HN2= (T/T1)**(3/2)*1/4*D_HN2#\n",
+ "#The composition of oxygen and nitrogen on a H2 free basis is \n",
+ "x_O= O_2/(1-H_2)#\n",
+ "x_N= N_2/(1-H_2)#\n",
+ "\n",
+ "# The effective diffusivity for the gas mixture at given temperature and pressure is\n",
+ "D= 1/(x_O/D_HO2+x_N/D_HN2) # in m**2/s\n",
+ "print \"Effective diffusivity = %0.3e m**2/s\" %D"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective diffusivity = 4.524e-06 m**2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.4 - Page No. : 323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from numpy import pi\n",
+ "# Given data\n",
+ "d=3 # in mm\n",
+ "d=d*10**-3 # in meter\n",
+ "T=25 # in \u00b0C\n",
+ "T=T+273 # in K\n",
+ "D= 0.4*10**-4 # in m**2/s\n",
+ "R= 8314#\n",
+ "P_A1=1 # in atm\n",
+ "P_A1=P_A1*10**5 # in w/m**2\n",
+ "P_A2=0#\n",
+ "C_A2=0#\n",
+ "x2= 15 # in meter\n",
+ "x1= 0#\n",
+ "A= pi/4*d**2#\n",
+ "M_A= D*A/(R*T)*(P_A1-P_A2)/(x2-x1) # in kg mole/sec\n",
+ "N_B= M_A#\n",
+ "M_B= M_A*29 # in kg/sec\n",
+ "print \"Value of N_B = %0.3e kg mole/sec\" %N_B\n",
+ "print \"Value of M_B = %0.3e kg /sec\" %M_B"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of N_B = 7.608e-13 kg mole/sec\n",
+ "Value of M_B = 2.206e-11 kg /sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.5 - Page No. : 325"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from math import log\n",
+ "# Given data\n",
+ "P=3 # in atm\n",
+ "P=P*10**5 # in N/m**2\n",
+ "r1=10 # in mm\n",
+ "r1=r1*10**-3 # in m\n",
+ "r2=20 # in mm\n",
+ "r2=r2*10**-3 # in m\n",
+ "R=4160 # in J/kg-K\n",
+ "T=303 # in K\n",
+ "D=3*10**-8 # in m**2/s\n",
+ "S=3*0.05# # Solubility of hydrogen at a pressure of 3 atm in m**3/m**3 of rubber tubing\n",
+ "del_x=r2-r1 # in m\n",
+ "L=1 # in m\n",
+ "Am=2*pi*L*del_x/log(r2/r1)#\n",
+ "#Formula P*V= m*R*T\n",
+ "V=S#\n",
+ "m=P*V/(R*T) # in kg/m**3 of rubber tubing at the inner surface of the pipe\n",
+ "C_A1=m#\n",
+ "C_A2=0#\n",
+ "#Diffusion flux through the cylinder is given\n",
+ "M=D*(C_A1-C_A2)*Am/del_x#\n",
+ "print \"Diffusion flux through the cylinder = %0.2e kg/sm\" %M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diffusion flux through the cylinder = 9.71e-09 kg/sm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.6 - Page No. : 329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from numpy import pi\n",
+ "# Given data\n",
+ "R=4160 # in J/kg-K\n",
+ "M=2#\n",
+ "D_AB=1.944*10**-8 # in m**2/s\n",
+ "R_H2=R/M#\n",
+ "S=2*0.0532# # Solubility of hydrogen at a pressure of 2 atm in cm**3/cm**3 of pipe\n",
+ "P=2 # in atm\n",
+ "P=P*1.03*10**5 # N/m**2\n",
+ "T=25 # in degree C\n",
+ "T=T+273 # in K\n",
+ "r1=2.5 # in mm\n",
+ "r1=r1*10**-3 # in m\n",
+ "r2=5 # in mm\n",
+ "r2=r2*10**-3 # in m\n",
+ "del_x=r2-r1 # in m\n",
+ "L=1 # in m\n",
+ "#Formula P*V= m*R*T\n",
+ "V=S#\n",
+ "m=P*V/(R*T) # in kg/m**3 of pipe\n",
+ "# So, Concentration of H2 at inner surface of the pipe\n",
+ "C_A1=0.0176 # in kg/m**3\n",
+ "# The resistance of diffusion of H2 away from the outer surface is negligible i.e.\n",
+ "C_A2=0#\n",
+ "Am=2*pi*L*del_x/log(r2/r1)#\n",
+ "# Loss of H2 by diffusion \n",
+ "M_A= D_AB*(C_A1-C_A2)*Am/del_x#\n",
+ "print \"Loss of H2 by diffusion = %0.2ef kg/s\" %M_A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss of H2 by diffusion = 3.10e-09f kg/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No : 10.7 - Page No. : 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from numpy import pi\n",
+ "from math import log\n",
+ "# Given data\n",
+ "Px1= 0.14 # in bar\n",
+ "Px2= 0#\n",
+ "P=1.013 # in bar\n",
+ "Py1=P-Px1# # in bar\n",
+ "Py2=P-Px2# # in bar\n",
+ "D=8.5*10**-6 # in m**2/s\n",
+ "d=5 # diameter in meter\n",
+ "L=1 # in mm\n",
+ "L=L*10**-3 #in meter\n",
+ "M=78 # molecular weight\n",
+ "Am_x= 1/4*pi*d**2*M#\n",
+ "R=8314#\n",
+ "del_x=3 # thickness in mm\n",
+ "del_x=del_x*10**-3 # in m\n",
+ "T=20 # in degree C\n",
+ "T=T+273 # in K\n",
+ "P=P*10**5 # in N/m**2\n",
+ "m_x= D*Am_x*P*log(Py2/Py1)/(R*T*del_x)#\n",
+ "# The mass of the benzene to be evaporated\n",
+ "mass= 1/4*pi*d**2*L#\n",
+ "density=880 # in kg/m**3\n",
+ "m_b= mass*density#\n",
+ "toh=m_b/m_x # in sec\n",
+ "print \"Time taken for the entire organic compound to evaporate = %0.0f seconds\" %toh\n",
+ "\n",
+ "\n",
+ "# Note: Answer in the book is wrong"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken for the entire organic compound to evaporate = 644 seconds\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/ebbygeorge/Chapter01_1.ipynb b/sample_notebooks/ebbygeorge/Chapter01_1.ipynb
new file mode 100644
index 00000000..c17729b1
--- /dev/null
+++ b/sample_notebooks/ebbygeorge/Chapter01_1.ipynb
@@ -0,0 +1,72 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#Chapter 1:INTRODUCTION"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 1.1, Page number 1.11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": true
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The width = 2.2222 *10**-11 Amp\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "C=20/(9*10^11)#converting cms to farads\n",
+ "\n",
+ "#Calculation\n",
+ "F=154-100#fall in potential\n",
+ "R=F/60#rate of fall in potential\n",
+ "I=C*R#ionization current\n",
+ "\n",
+ "#Result\n",
+ "print\"The width = \",round(I*10,4),\"*10**-11 Amp\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/sai kiranmalepati/Samplenb.ipynb b/sample_notebooks/sai kiranmalepati/Samplenb.ipynb
new file mode 100644
index 00000000..1e3cfad6
--- /dev/null
+++ b/sample_notebooks/sai kiranmalepati/Samplenb.ipynb
@@ -0,0 +1,158 @@
+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "#Chapter-1, Section-1.22, pg-24, problem 1\n",
+ "\n",
+ "L=20; #True length of the chain is defined as L\n",
+ "print(' the length of the chain is ',L,'meters'); #Print L\n",
+ "\n",
+ "E=0.03; # Error in chain\n",
+ "print(' the error in the chain is ',E); # Print E\n",
+ "\n",
+ "ML=327; # Measured length is defined as ML\n",
+ "print(' the measured length ',ML,'meters'); #print ML\n",
+ "\n",
+ "tlength=((L+E)/L)*ML; #True length of L calculated\n",
+ "print(' the ture length of the chain is ',tlength,'meters') # Print true length.\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "# Chapter-1, Section-1.23, pg 39, Problem- 4\n",
+ "\n",
+ "import math # Library imported.\n",
+ "\n",
+ "ab=28.7*math.cos(5*(math.pi/180)); #AB calculated using trigonometry\n",
+ "b1c=23.4*math.cos(7*(math.pi/180)); #B'C calculated\n",
+ "c1d=20.9*math.cos(10*(math.pi/180)); # C'D calculated\n",
+ "d1e=29.6*math.cos(12*(math.pi/180)) ; # D'E calculated\n",
+ "print(' the value of ab is ',ab,'meters'); #AB printed\n",
+ "\n",
+ "print(' the value of b1c is ',b1c,'meters'); #B'C printed\n",
+ "\n",
+ "print(' the value of c1d is ',c1d,'meters'); #C'D printed\n",
+ "\n",
+ "print(' the value of d1e is ',d1e,'meters'); #D'E printed\n",
+ "\n",
+ "\n",
+ "total=ab+b1c+c1d+d1e; #total calculated.\n",
+ "print('the total distance is ',total,'meters') #total printed\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "# Chapter 1, Section 1.22, Problem 6, pg 27.\n",
+ "\n",
+ "L=30; # Length of tape\n",
+ "t0=20; # Temperature of tape\n",
+ "p0=10; # Pull under which tape is present\n",
+ "pm=15; # Measured line pull\n",
+ "tm=32; # Mean temperature.\n",
+ "a=0.03; # Cross-sectional area\n",
+ "al=11/(1000000); # Temperature correction co-efficient\n",
+ "E=2.1*(1000000); # E for steel\n",
+ "w=0.693; #Total weight\n",
+ "ml=780; #measured length\n",
+ "\n",
+ "n=1; #n defined\n",
+ "print(' for n=1'); #print n value\n",
+ "ct=al*L*(tm-t0); # temperature correction calculated\n",
+ "print('the temperature correction is',ct,'meters'); # print temperature correction\n",
+ "\n",
+ "cp=(pm-p0)*L/(a*E); #pull correction calculated\n",
+ "print('the pull corretion is ',cp,' meters'); #print pull correction\n",
+ "\n",
+ "cs=-L*w*w/(24*pm*pm*n*n); #sag correction calculated\n",
+ "print('the sag correction is ',cs,'meters'); # print sag correction\n",
+ "\n",
+ "e=ct+cp+cs; #total correction calculated\n",
+ "print('the total correction is ',e,'meters'); # total correction printed\n",
+ "\n",
+ "l1=L+e; # correction in length =sum of correction and measured length \n",
+ "\n",
+ "truelength=(l1/L)*ml; #true length calculated\n",
+ "print('the true length is ',truelength); #true length printed\n",
+ "\n",
+ "n=2; #new n defined\n",
+ "\n",
+ "print(' for n=2'); # print n value\n",
+ "ct=al*L*(tm-t0); # temperature correction calculated\n",
+ "print('the temperature correction is',ct,'meters'); # print temperature correction\n",
+ "\n",
+ "\n",
+ "cp=(pm-p0)*L/(a*E); #pull correction calculated\n",
+ "print('the pull corretion is ',cp,' meters'); #print pull correction\n",
+ "\n",
+ "cs=-L*w*w/(24*pm*pm*n*n); #sag correction calculated\n",
+ "print('the sag correction is ',cs,'meters'); #print sag correction\n",
+ "\n",
+ "e=ct+cp+cs; #total correction calculated\n",
+ "print('the total correction is ',e,'meters'); # total correction printed\n",
+ "\n",
+ "l1=L+e; #correction in length =sum of correction and measured length \n",
+ "\n",
+ "truelength=(l1/L)*ml; #true length calculated\n",
+ "print('the true length is ',truelength); #true length printed\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "# Chapter-2, Section-2.4, Problem 2, Page number 56\n",
+ "\n",
+ "import math #Library to be used is imported\n",
+ "\n",
+ "cd=(15-15*math.cos(3*(math.pi/180)))/10; #displacement perpendicular to chain line.\n",
+ "\n",
+ "print('the required displacement perpendicular to char line is ',cd); #Print cd\n",
+ "bc=15*(math.sin(3*(math.pi/180)))/10; #displacement parallel to chain line.\n",
+ "print('the displacemenet parllel to the chain line is ',bc) #print bc"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.5.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2024-12-20 12:29:59 +0000