diff options
Diffstat (limited to 'Fluid_Mechanics-Fundamentals_&_Applications/Chapter05.ipynb')
| -rwxr-xr-x | Fluid_Mechanics-Fundamentals_&_Applications/Chapter05.ipynb | 626 |
1 files changed, 0 insertions, 626 deletions
diff --git a/Fluid_Mechanics-Fundamentals_&_Applications/Chapter05.ipynb b/Fluid_Mechanics-Fundamentals_&_Applications/Chapter05.ipynb deleted file mode 100755 index e6b94086..00000000 --- a/Fluid_Mechanics-Fundamentals_&_Applications/Chapter05.ipynb +++ /dev/null @@ -1,626 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:8bd11344481cd08353414df9b95ff36a77bcdb1a4b70dab675e71546fa1a4453" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 05:Mass, Bernoulli and Energy Equations" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-1, Page No:190" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Vairable Decleration\n", - "V=10#Volume of the bucket in Gal\n", - "r_in=1 #Radius of the hose in cm\n", - "r_e=0.4 #Radius of the hose at the nozzle exit in cm\n", - "t=50 #Time taken to fill the bucket in s\n", - "C_gl=3.7854 #Conversion factor gal to Lit\n", - "rho=1 #Denisty of water in kg/Lit\n", - "C_v=10**-3 #Conersion factor in m^3/lit\n", - "\n", - "#Calculations\n", - "\n", - "#Part (a)\n", - "V_dot=(V*C_gl)/t #Volume flow rate in Lit/s\n", - "m_dot=rho*V_dot #Mass flow rate in kg/s\n", - "\n", - "#Part(b)\n", - "A_e=pi*r_e**2*10**-4 #Cross-Sectional Area of the nozzle at exit in m^2\n", - "V_e=(V_dot*C_v)/A_e #Average Velocity of water at nozzle exit in m/s\n", - "\n", - "#Result\n", - "print \"The Volume Flow rate is\",round(V_dot,3),\"L/s and the mass flow rate is\",round(m_dot,3),\"kg/s\"\n", - "print \"The area of cross section at nozzle exit is\",round(A_e,5),\"m^2\"\n", - "print \"The Average Velocity of water is\",round(V_e,1),\"m/s\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Volume Flow rate is 0.757 L/s and the mass flow rate is 0.757 kg/s\n", - "The area of cross section at nozzle exit is 5e-05 m^2\n", - "The Average Velocity of water is 15.1 m/s\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-2, Page No:191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "h_o=1.2 #Original Height in m\n", - "h_2=0.6 #Water level drop in m\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "D_tank=0.9 #Diameter of the tank in m\n", - "D_jet=0.013 #Diameter at the jet in m\n", - "\n", - "#Calculations\n", - "#After carrying out the theroetical calculations and integration we arrive to obtain\n", - "t_min=((h_o**0.5-h_2**0.5)/((g/2)**0.5))*((D_tank/D_jet)**2) #Time required to reach a level 0.6m in s\n", - "t=t_min/60 #Converting time from sec to min\n", - "\n", - "#Result\n", - "print \"The time it takes to half empty the tank is\",round(t,1),\"min\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The time it takes to half empty the tank is 11.6 min\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-3, Page No:195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "##Note:The symbols in the textbook are cumbersome to code hence a different one has been used in this coding\n", - "\n", - "#Variable Decleration\n", - "h=50 #Elevation difference in m\n", - "m_dot=5000 #Mass flow rate at which the water is to be supplied in kg/s\n", - "W_dot_out=1862 #Electric Power generated in kWh\n", - "n_generator=0.95 #Efficiency of the generator in fraction\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "C=10**-3 #Conversion factor in kJ/kg/m^2/s^2\n", - "\n", - "#Calculations\n", - "\n", - "#Part(a)\n", - "\n", - "#Calling e_mech_in-e_mech_out as del_e for convienence \n", - "del_e=g*h*C #Change in water's mechanical energy per unit mass in kJ/kg\n", - "delta_E_fluid=m_dot*del_e # Change in energy of the fluid in kW\n", - "\n", - "n_overall=W_dot_out/delta_E_fluid #Overall Efficiency in fraction\n", - "\n", - "#Part(b)\n", - "n_turbine_gen=n_overall/n_generator #Mechanical efficiency os the turbine in fraction\n", - "\n", - "#Part(c)\n", - "W_dot_shaft_out=n_turbine_gen*delta_E_fluid #Shaft power output in kW\n", - "\n", - "#Result\n", - "print \"The overall efficiency is\",round(n_overall,2)\n", - "print \"The mechanical efficiency of the turbine is\",round(n_turbine_gen,1)\n", - "print \"The shaft power output is\",round(W_dot_shaft_out,1),\"kW\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The overall efficiency is 0.76\n", - "The mechanical efficiency of the turbine is 0.8\n", - "The shaft power output is 1960.0 kW\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-5, Page No:205" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "P1=400 #Pressure at upstream of the jet in kPa\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "rho=1000 #Density of water in kg/m^3\n", - "C1=1000 #Conversion factor in N/m^2.kPa\n", - "C2=1 #Conversion factor in kg.m/s^2.N\n", - "\n", - "#Calculations\n", - "#Applying the Bernoulli Equation\n", - "z2=(P1*C1*C2)/(rho*g) #maximun height the water jet reaches in m\n", - "\n", - "#Result\n", - "print \"The water jet rises up to\",round(z2,1),\"m\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The water jet rises up to 40.8 m\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-6, Page No:206" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "h=5 #Height at which the water tank is filled in m\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "\n", - "#Calculations\n", - "z1=h #Decleration in terms of datum in m\n", - "#Applying the Bernoulli Equation\n", - "V2=(2*g*z1)**0.5 #Maximum velocity that the water jet can attain in m/s\n", - "\n", - "#Result\n", - "print \"The maximum velocity that the water jet can attain is\",round(V2,1),\"m/s\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum velocity that the water jet can attain is 9.9 m/s\n" - ] - } - ], - "prompt_number": 40 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-7, Page No:207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "P_atm=101.3 #Atmospheric pressure in kPa\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "rho=750 #Denisty of gasoline in kg/m^3\n", - "z1=0.75 #Location of point 2 w.r.t point 1\n", - "D=5*10**-3 #Diameter of the siphon pipe in m\n", - "V=4 #Volume of gasoline to be siphoned in Lit\n", - "z3=2.75 #Height of point 3 w.r.t to point 2 in m\n", - "C1=1 #conversion factor in N.s^2/kg.m\n", - "C2=10**-3 #Conversion factor in kPa.m^2/N\n", - "#Calculations\n", - "\n", - "#Part (a)\n", - "#Applying the Bernoulli Equation\n", - "V2=(2*g*z1)**0.5 #Velocity in m/s\n", - "A=(pi*D**2)/4 #Cross-Sectional Area in m^2\n", - "V_dot=V2*A*1000#Volume flow rate in L/s\n", - "delta_t=V/V_dot #Time required to siphon gasoline in s\n", - "\n", - "#Part(b)\n", - "#Applying Bernoulli Equations\n", - "P3=P_atm-(rho*g*z3*C1*C2) #Pressure at point 3 in kPa\n", - "\n", - "#result\n", - "print \"The time requires to siphon 4L gasoline is\",round(delta_t,1),\"s\"\n", - "print \"The pressure at point 3 is\",round(P3,1),\"kPa\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The time requires to siphon 4L gasoline is 53.1 s\n", - "The pressure at point 3 is 81.1 kPa\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-8, Page No:208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Declerations\n", - "g=9.81 #Acceleration due to Gravity in m/s^2\n", - "h3=0.12 #Difference in level in m\n", - "\n", - "#Calculations\n", - "#Applying Bernoulli Equations\n", - "V1=(2*g*h3)**0.5 #Velocity of Fluid in m/s\n", - "\n", - "#Result\n", - "print \"The velocity of fkuid is\",round(V1,2),\"m/s\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The velocity of fkuid is 1.53 m/s\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-9, Page No:209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "rho_hg=13600 #density of mercury in kg/m^3\n", - "rho_sw=1025 #density of sea-water in kg/m^3\n", - "rho_atm_air=1.2 #Density of air in kg/m^3\n", - "P_atm_air=762 #Atmospheric pressure 320km away from the eye in mm oh Hg\n", - "P_air=560 #Atmospheric pressure at the eye of the strom in mm og Hg\n", - "C=10**-3 #Conversion factor in m/mm\n", - "V_A=250 #Hurricane Wind Velocity in km/hr\n", - "C_k=1/3.6 #Conversion Factor from km/hr to m/s \n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "\n", - "#Calculations\n", - "\n", - "#part(a)\n", - "h3=(rho_hg*(P_atm_air-P_air)*C)/rho_sw #Pressure difference in m\n", - "\n", - "#Part(b)\n", - "#Applying Bernoulli Equations\n", - "h_air=(V_A**2*C_k**2)/(2*g) #Height of air column in m\n", - "rho_air=(P_air*rho_atm_air)/P_atm_air #Density of Air in the hurricane in kg/m^3\n", - "h_dynamic=(rho_air*h_air)/rho_sw #Sea-Water column equivalent to air-column in m\n", - "h2=h3+h_dynamic #Total storm surge at point 2 in m\n", - "\n", - "#Result\n", - "print \"The pressure difference between point's 1 and 3 in terms of sea-water column is\",round(h3,2),\"m\"\n", - "print \"The total Storm Surge at point2 is\",round(h2,2),\"m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The pressure difference between point's 1 and 3 in terms of sea-water column is 2.68 m\n", - "The total Storm Surge at point2 is 2.89 m\n" - ] - } - ], - "prompt_number": 61 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-12, Page No:221" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "V_dot=50 #Volumetric Flow rate in L/s\n", - "rho=1 #Density of water \n", - "n_motor=0.9 #efficiency of the electric motor in fraction\n", - "W_dot_electric=15 #Power of the electric motor in kW\n", - "P2=300 #Absolute pressure at the outlet in kPa\n", - "P1=100 #Absolute pressure at the inlet in kPa\n", - "c=4.18 #Specific heat of water in kJ/kg C\n", - "#Calculations\n", - "\n", - "#Part(a)\n", - "m_dot=rho*V_dot #Mass flow rate in kg/s\n", - "W_dot_pump=n_motor*W_dot_electric #Mechanical shaft power delivered in kW\n", - "delta_E_dot_mech_fluid=(m_dot*((P2-P1)/rho))/1000 #Increase in mechanical energy in kW\n", - "n_pump=delta_E_dot_mech_fluid/W_dot_pump #Efficiency in fraction\n", - "\n", - "#part (b)\n", - "E_dot_loss=W_dot_pump-delta_E_dot_mech_fluid #Lost mechanical energy in kW\n", - "delta_T=(E_dot_loss)/(m_dot*c) #Temperature rise of water due to mechanical inefficiency in degree C\n", - "\n", - "#Result\n", - "print \"The Mechanical efficiency of the pump is\",round(n_pump,3)\n", - "print \"The temperature rise of water due to mechanical inefficiency is\",round(delta_T,3),\"Degree Centigrade\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Mechanical efficiency of the pump is 0.741\n", - "The temperature rise of water due to mechanical inefficiency is 0.017 Degree Centigrade\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-13, Page No:222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "V_dot=100 #Discharge through the power plant in m^3/s\n", - "rho=1000 #Density of water in kg/m^3\n", - "z1=120 #Elevation from which the water flows in m\n", - "h_l=35 #Elevation of point 2 in m\n", - "n_turbine_gen=0.8 #Overall efficiency of the generator in fraction\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "C=10**-3 #Conversion Factor\n", - "\n", - "#Calculations\n", - "m_dot=rho*V_dot #mass flow rate through the turbine in kg/s\n", - "\n", - "#Applying Bernoullis principle and taking point 2 as reference point z2=0\n", - "h_turbine=z1-h_l #extracted turbine head in m\n", - "W_dot_turbine=m_dot*g*h_turbine*C #Turbine Power in kW\n", - "W_dot_electric=C*n_turbine_gen*W_dot_turbine #Electrical Power Generated by the actual Unit in MW\n", - "\n", - "#Result\n", - "print \"The electrical Power generated is\",round(W_dot_electric,1),\"MW\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The electrical Power generated is 66.7 MW\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-14, Page No:223" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "void_fraction=0.5 #Void Fraction\n", - "l=12 #Dimension of the fan in cm\n", - "w=40 #Dimension of the fan in cm\n", - "h=40 #Dimension of the fan in cm\n", - "delta_t=1 #time in s\n", - "rho=1.2 #Ddensity of air in kg/m^3\n", - "D=0.05 #Diameter of opening in the case in m\n", - "alpha2=1.1 #kinetic correction factor\n", - "n_fan=0.3 #Efficiency of the fan-motor\n", - "#Calculations\n", - "#Part(a)\n", - "V=void_fraction*l*w*h #Volume in cm^3\n", - "V_dot=(V/delta_t)*10**-6 #Volumetric flow rate in m^3/s\n", - "m_dot=rho*V_dot #mass flow rate in kg/s\n", - "A=(pi*D**2)/4 #Area of the opening is the case in m^2\n", - "\n", - "#Notation has been changed to avoid conflict\n", - "Vel=V_dot/A #Velocity of the air thorught the opening in m/s\n", - "\n", - "#Applying Bernoullis principle\n", - "W_dot_fan=m_dot*alpha2*Vel**2*0.5 #Work done in W\n", - "W_dot_electric=W_dot_fan/n_fan #Electric Work done in W\n", - "\n", - "#Part(b)\n", - "#Applying Brnoullis principle\n", - "#Notation has been changed here\n", - "delta_P=(rho*W_dot_fan)/m_dot #Pressure rise across fan in Pa\n", - "\n", - "#Result\n", - "print \"Wattage of the fan to be purchased is\",round(W_dot_electric,4),\"W\"\n", - "print \"The pressure difference across the fan is\",round(delta_P,1),\"Pa\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Wattage of the fan to be purchased is 0.5049 W\n", - "The pressure difference across the fan is 15.8 Pa\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5-15, Page No:225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "W_shaft=5 #Shaft Power in kW\n", - "n_pump=0.72 #Efficiency of the pump in fraction\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "h_l=4 #Head loss in m\n", - "z2=25 #Datum in m\n", - "rho=1000 #Density of water in kg/m^3\n", - "\n", - "#Calculations\n", - "W_dot_pump=n_pump*W_shaft #Useful mechanical power returned in kW\n", - "\n", - "#Applying Bernoullis Principle\n", - "m_dot=(W_dot_pump/(g*(z2+h_l)))*1000 #mass floe rate in kg/s\n", - "V_dot=(m_dot/rho) #Volumetric flow rate in m^3/s\n", - "delta_P=W_dot_pump/V_dot #Pressure difference in kPa\n", - "\n", - "#Result\n", - "print \"Discharge of water is\",round(V_dot,4),\"m^3/s\"\n", - "print \"The pressure difference across the pump is\",round(delta_P),\"kPa\"\n", - "#Answer in the coding is off by 1 kPa due to decimal point accuracy" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge of water is 0.0127 m^3/s\n", - "The pressure difference across the pump is 284.0 kPa\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file |