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diff --git a/Fluid_Power_With_Applications_by_A._Esposito/Chapter2PhysicalPropertiesofHydraulicFluids.ipynb b/Fluid_Power_With_Applications_by_A._Esposito/Chapter2PhysicalPropertiesofHydraulicFluids.ipynb new file mode 100755 index 00000000..587bcf9a --- /dev/null +++ b/Fluid_Power_With_Applications_by_A._Esposito/Chapter2PhysicalPropertiesofHydraulicFluids.ipynb @@ -0,0 +1,851 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2:Physical Properties of Hydraulic Fluids "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.1 pgno:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the weight is 129.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Find Weight of Body\n",
+ "# Given:\n",
+ "# Mass of the Body:\n",
+ "m=4; #slugs\n",
+ "\n",
+ "# Solutions:\n",
+ "# we know acceleration due to gravity,\n",
+ "g=32.2; #ft/s**2\n",
+ "W=(m*g);\n",
+ "\n",
+ "# Results:\n",
+ "print \"the weight is\",round(W)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.2 pgno:43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The specific weight of Body is lb/ft**3. 71.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the specific weight of a body\n",
+ "# Given:\n",
+ "# Weigth of the Body:\n",
+ "W=129; #lb\n",
+ "# Volume of the Body:\n",
+ "V=1.8; #ft**3\n",
+ "\n",
+ "# Solution:\n",
+ "# we know specific weight,\n",
+ "# gamma=(Weigth of the Body/Volume of the Body)\n",
+ "gamma1=(W/V); #lb/ft^3\n",
+ "# rounding off the above answer\n",
+ "#gamma1=round(gamma1)+(round((gamma1-round(gamma1))*10)/10); #lb/ft^3\n",
+ " \n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The specific weight of Body is lb/ft**3.\",round(gamma1,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.3 pgno:44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Results: \n",
+ "The specific gravity of air 0.00121\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find the specific gravity of air at 68 degF\n",
+ "# Given:\n",
+ "# specific weight of air at 68 degF:\n",
+ "gamma_air=0.0752; #lb/ft**3\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# we know,\n",
+ "# specific gravity of air=(specific weight of air/specific weight of water)\n",
+ "# also we know,specific weight of water at 68 degF,\n",
+ "gamma_water=62.4; #lb/ft**3\n",
+ "SG_air=gamma_air/gamma_water;\n",
+ "\n",
+ "# Results:\n",
+ "print \"Results: \"\n",
+ "print \"The specific gravity of air \",round(SG_air,5) \n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.4 pgno:45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The Density of Body is slugs/ft**3. 2.22\n",
+ " The Density of Body is slugs/ft**3. 2.22\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find Density of body of Example 2-1 and 2-2\n",
+ "# Given:\n",
+ "# mass of the Body:\n",
+ "m=4; #slugs\n",
+ "# Volume of the Body:\n",
+ "V=1.8; #ft**3\n",
+ "\n",
+ "# Solution:\n",
+ "# we know density,\n",
+ "# rho1=(mass of the Body/Volume of the Body)\n",
+ "rho1=(m/V); #slugs/ft**3\n",
+ "# also density,rho2=(specific weight/acceleration due to gravity)\n",
+ "g=32.2; #ft/s**2\n",
+ "gamma1=71.6; #lb/ft**3\n",
+ "rho2=(gamma1/g); #slugs/ft**3\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The Density of Body is slugs/ft**3.\",round(rho1,2)\n",
+ "print \" The Density of Body is slugs/ft**3.\",round(rho2,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.5 pgno:48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The pressure on skin diver is psi. 26.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find pressure on the skin diver\n",
+ "# Given:\n",
+ "# Depth of Water Body:\n",
+ "H=60; #ft\n",
+ "\n",
+ "# Solution:\n",
+ "# specific Weight of water,\n",
+ "gamma1=0.0361; #lb/in**3 \n",
+ "# Conversion: \n",
+ "# 1 feet = 12 inches\n",
+ "# 1 lb/in**2 = 1 psi \n",
+ "# we know pressure,\n",
+ "# p=(specific weight of liquid * liquid column height)\n",
+ "p=(gamma1*H*12); #psi\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The pressure on skin diver is psi.\",round(p)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.6 pgno:50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Height of water column is ft. 34.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find tube height of a Barometer\n",
+ "# Given:\n",
+ "# liquid used is Water instead of Mercury.\n",
+ "\n",
+ "# Solution:\n",
+ "# specific Weight of water,\n",
+ "gamma1=0.0361; #lb/in**3 \n",
+ "# We also knows Atmospheric Pressure,\n",
+ "p=14.7; #psi\n",
+ "# Conversion: \n",
+ "# 1 feet = 12 inches\n",
+ "# 1 lb/in**2 = 1 psi \n",
+ "# we know pressure,\n",
+ "# p=(specific weight of liquid * liquid column height)\n",
+ "# Therefore,\n",
+ "H=(p/gamma1); #in\n",
+ "# He=Height in Feet.\n",
+ "He=H*0.083; #ft\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Height of water column is ft.\",round(He,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": []
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.7 pgno:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Absolute Pressure is psi. 9.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To convert given pressure into absolute pressure\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=-5; #psi\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=14.7; #psi \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po;\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Absolute Pressure is psi.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## example 2.8 pgno:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The Absolute Pressure is psi. 40.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find absolute pressure on skin diver of Example 2-5\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=26; #psi\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=14.7; #psi \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po; #psi\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The Absolute Pressure is psi.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.9 pgno:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ "The specific weights is N/m**3. 8792\n",
+ " The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To Determine specific weights in N/m**3\n",
+ "# Given:\n",
+ "# specific weight:\n",
+ "gamma1=56; #lb/ft**3\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# We know,\n",
+ "# 1 N/m**3 = 157 lb/ft**3\n",
+ "gamma2=157*gamma1; #N/m**3\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \"The specific weights is N/m**3.\",gamma2\n",
+ "print \" The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.10 pgno:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The temp at which Fahrenheit and Celsius values are equal is deg. -40.0\n"
+ ]
+ }
+ ],
+ "source": [
+ " # Aim:To find Temperature at which Fahrenheit and Celsius values are equal \n",
+ "# Given:\n",
+ "# T(degF) = T(degC) #Eqn - 1\n",
+ "\n",
+ "# Solution:\n",
+ "# We know that,\n",
+ "# T(degF)=((1.8*T(degC))+32) #Eqn - 2 \n",
+ "# From Eqn 1 and 2\n",
+ "# ((1.8*T(degC))+32)= T(degC)\n",
+ "# (1-1.8)*T(degC)=32\n",
+ "# -0.8*T(degC)=32\n",
+ "TdegC=-32/0.8;\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The temp at which Fahrenheit and Celsius values are equal is deg.\",TdegC\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.11 pgno:57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The change in volume of oil is in^3. -0.076\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find change in volume of the oil\n",
+ "# Given:\n",
+ "# Volume of original oil:\n",
+ "V=10.0; #in^3\n",
+ "# Initial Pressure:\n",
+ "P1=100.0; #psi\n",
+ "# Final pressure:\n",
+ "P2=2000.0; #psi\n",
+ "# Bulk Modullus:\n",
+ "betaa=250000.0; #psi\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Change in pressure,\n",
+ "delP=P2-P1; #psi\n",
+ "# Change in volume,\n",
+ "delV=-((V*delP)/betaa); #in^3 ,- sign indicates oil is being compressed\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The change in volume of oil is in^3.\",delV\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.12 pgno:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil in centistokes is cS. 50.0\n",
+ " The viscosity of oil in centipoise is cP. 45.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find viscosity of oil in centistokes and centipoise\n",
+ "# Given:\n",
+ "# viscosity of oil:\n",
+ "nu=230; #SUS at\n",
+ "t=150; #degF.\n",
+ "# specific gravity of oil:\n",
+ "gamma1=0.9;\n",
+ "# Solution:\n",
+ "# kinematic viscosity of oil in centistokes,\n",
+ "nu_cs=((0.220*nu)-(135/230)); #centistokes\n",
+ "# absolute viscosity of oil in centipoise,\n",
+ "mu_cp=(gamma1*nu_cs); #centipoise\n",
+ "from math import floor\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil in centistokes is cS.\",floor(nu_cs)\n",
+ "print \" The viscosity of oil in centipoise is cP.\",floor(mu_cp)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.13 pgno:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil in centistokes is cS. 25.0\n",
+ " The viscosity of oil in centipoise is cP. 22.3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find kinematic and absolute viscosity of oil in cS and cP respectively\n",
+ "# Given:\n",
+ "# Density of oil:\n",
+ "Den=0.89; #g/cm^3\n",
+ "# Time flow:\n",
+ "t=250; #s\n",
+ "# Calibration constant:\n",
+ "cc=0.100;\n",
+ "\n",
+ "# Solution:\n",
+ "# kinematic viscosity of oil in centistokes,\n",
+ "nu_cs=(t*cc); #centistokes\n",
+ "# absolute viscosity of oil in centipoise,\n",
+ "SG=Den;\n",
+ "mu_cp=(SG*nu_cs); #centipoise\n",
+ "# rounding off the above answer\n",
+ "mu_cp=round(mu_cp)+(round(round((mu_cp-round(mu_cp))*10))/10); #centipoise\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil in centistokes is cS.\",nu_cs\n",
+ "print \" The viscosity of oil in centipoise is cP.\",mu_cp\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.14 pgno:65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of sample oil at 100 degF is SUS. 200\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find viscosity of oil at 100 degF in SUS\n",
+ "# Given:\n",
+ "# Viscosity Index:\n",
+ "VI=80;\n",
+ "# viscosity of O-VI oil at 100 degF:\n",
+ "L=400; #SUS\n",
+ "# viscosity of 100-VI oil at 100 degF:\n",
+ "H=150; #SUS\n",
+ "\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# viscosity of sample oil at 100 degF,\n",
+ "U=L-(((L-H)*VI)/100); #SUS\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of sample oil at 100 degF is SUS.\",U\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.15 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The pressure on skin diver is kPa. 179.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find pressure on the skin diver in SI units\n",
+ "# Given:\n",
+ "# Depth of Water Body:\n",
+ "H=18.3; #m\n",
+ "\n",
+ "# Solution:\n",
+ "# specific Weight of water,\n",
+ "gamma1=9800; #N/m^3 \n",
+ "# we know pressure,\n",
+ "# p=(specific weight of liquid * liquid column height)\n",
+ "p=(gamma1*H); #Pa\n",
+ "pK=p/1000; #kPa\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The pressure on skin diver is kPa.\",round(pK)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.16 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Absolute Pressure is Pa. 67000\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To convert given pressure into absolute pressure\n",
+ "# Given:\n",
+ "# Gage Pressure:\n",
+ "Pg=-34000; #Pa\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Atmospheric Pressure,\n",
+ "Po=101000; #Pa \n",
+ "# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure\n",
+ "Pa=Pg+Po; #Pa\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Absolute Pressure is Pa.\",Pa\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.17 pgno:67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The Percentage change in volume of oil is -0.76\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find % change in volume of the oil\n",
+ "# Given:\n",
+ "# Volume of original oil:V=164 #cm^3\n",
+ "# Initial Pressure:\n",
+ "P1=687.0; #kPa\n",
+ "# Final pressure:\n",
+ "P2=13740.0; #kPa\n",
+ "# Bulk Modullus:\n",
+ "betaa=1718.0; #MPa\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# Change in pressure,\n",
+ "delP=P2-P1; #kPa \n",
+ "betaa=betaa*1000; #kPA\n",
+ "# % Change in volume,\n",
+ "delV=-(delP/betaa)*100; #% ,- sign indicates oil is being compressed\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The Percentage change in volume of oil is \",round(delV,2)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.18pgno:68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Results: \n",
+ " The viscosity of oil is Ns/m**2. 0.05\n",
+ " The viscosity of oil is cP. 50.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Aim:To find absolute viscosity of oil in Ns/m**2 and cP\n",
+ "# Given:\n",
+ "# Area of moving plate surface in contact with oil:\n",
+ "A=1; #m**2\n",
+ "# Force applied to the moving plate:\n",
+ "F=10; #N \n",
+ "# velocity of the moving plate:\n",
+ "v=1; #m/s\n",
+ "# oil film thickness:\n",
+ "y=5; #mm\n",
+ "y=5*0.001; #m\n",
+ "\n",
+ "\n",
+ "# Solution:\n",
+ "# absolute viscosity of oil,\n",
+ "mu=(F/A)/(v/y); #Ns/m**2\n",
+ "# absolute viscosity of oil in cP,\n",
+ "mu_P=(F*100000*y*100)/(v*100*A*10000); #poise\n",
+ "mu_cP=mu_P*100; #centipoise\n",
+ "\n",
+ "# Results:\n",
+ "print \" Results: \"\n",
+ "print \" The viscosity of oil is Ns/m**2.\",mu\n",
+ "print \" The viscosity of oil is cP.\",mu_cP\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
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+ "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",
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