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diff --git a/Fluid_Mechanics-Fundamentals_&_Applications/Chapter06.ipynb b/Fluid_Mechanics-Fundamentals_&_Applications/Chapter06.ipynb deleted file mode 100755 index 9e5422ae..00000000 --- a/Fluid_Mechanics-Fundamentals_&_Applications/Chapter06.ipynb +++ /dev/null @@ -1,453 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:82135d7c3c860b18d45a6212b67de54362fa2b4454d970c3f68f6f27e472d43e" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 06:Momentum Analysis of Flow Systems" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-1, Page No:248" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "import scipy.integrate\n", - "\n", - "#Variable Decleration\n", - "a=1 #Lower limit of the intergral to be carried out\n", - "b=0 #Upper limit of the intergral to be carried out\n", - "\n", - "#Intergration\n", - "\n", - "func = lambda y:-4*y**2 #Decleration of the variable and the function to be integrated\n", - "X=scipy.integrate.quadrature(func, a,b)\n", - "\n", - "#Result\n", - "\n", - "print \"The Momentum Flux correction factor becomes\",round(X[0],2)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Momentum Flux correction factor becomes 1.33\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-2, Page No:251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "m_dot=14 #Mass flow rate in kg/s\n", - "rho=1000 #Density of water in kg/m^3\n", - "theta=pi/6 #Angle at which the pipe is deflected w.r.t the horizontal\n", - "A1=0.0113 #Cross-sectional Area at the inlet of the elbow in m^2\n", - "A2=7*10**-4 #Cross-sectional Area at the outlet of the elbow in m^2\n", - "C=10**-3 #Conversion factor\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "z2=0.3 #Elevational difference betweem inlet and outlet in m\n", - "z1=0 #Considering Datum in m\n", - "beta=1.03 #Momentum correction factor \n", - "\n", - "#Calculations\n", - "V1=m_dot/(rho*A1) #Velocity at the inlet of the elbow in m/s\n", - "V2=m_dot/(rho*A2) #Velocity at the outlet of the elbow in m/s\n", - "\n", - "#Part(a)\n", - "#Applying the Bernoulli Principle\n", - "#Simplfying the calculations in two steps\n", - "a=(V2**2-V1**2)/(2*g)\n", - "P1_gauge=(a+z2-z1)*g*rho*C #Gauge pressure at inlet in kPa\n", - "\n", - "#Part(b)\n", - "#Applying the momentum equation\n", - "#Anchoring forces required\n", - "F_rx=-(P1_gauge*1000*A1)+(beta*m_dot*((V2*cos(theta))-V1)) #N\n", - "F_rz=beta*m_dot*V2*sin(theta) #N\n", - "\n", - "#Result\n", - "print \"The gauge pressure at the inlet is\",round(P1_gauge,1),\"kPa\"\n", - "print \"The anchoring forces required to hold it in place are\",round(F_rx,),\"N and\",round(F_rz),\"N\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The gauge pressure at the inlet is 202.2 kPa\n", - "The anchoring forces required to hold it in place are -2053.0 N and 144.0 N\n" - ] - } - ], - "prompt_number": 51 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-3, Page No:253" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable decleration\n", - "beta=1.03 #Momentum Correction factor\n", - "m_dot=14 #mass flow rate in kg/s\n", - "V2=20 #Velocity at outlet in m/s\n", - "V1=1.24 #Velocity at inlet in m/s\n", - "P1_gauge=202200 #gauge pressure at inlet in N/m^2\n", - "A1=0.0113 #Area at the inlet in m^2\n", - "\n", - "#Calculations\n", - "#Applying the momentum equation\n", - "F_rx=-beta*m_dot*(V2+V1)-P1_gauge*A1 #Horiznotal force in N\n", - "\n", - "#Result\n", - "print \"The horizontal force is\",round(F_rx),\"N\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The horizontal force is -2591.0 N\n" - ] - } - ], - "prompt_number": 52 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-4, Page No:253" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "beta=1 #Momentum correction factor\n", - "m_dot=10 #Mass flow rate in kg/s\n", - "V1=20 #Velocity of flow of water in m/s\n", - "\n", - "#Calculations\n", - "#Applying the momentum equation\n", - "F_r=beta*m_dot*V1 #The force exerted in N\n", - "\n", - "#Result\n", - "print \"The force exerted is\",round(F_r),\"N\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The force exerted is 200.0 N\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-5, Page No:254" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable decleration\n", - "W_s=11 #Wind speed in km/h\n", - "C=3.6**-1 #Conversion from km/h to m/s\n", - "D=9 #Diameter of the blade in m\n", - "rho1=1.22 #Density of air in kg/m^3\n", - "W_actual=0.4 #Actual power generated in kW\n", - "\n", - "#Calculations\n", - "#Part(a)\n", - "V1=W_s*C #Velocity in m/s\n", - "m_dot=(rho*V1*pi*D**2)/4 #Mass flow rate of air in kg/s\n", - "W_dot_max=0.5*m_dot*V1**2*10**-3 #Work done in kW\n", - "n_windturbine=W_actual/W_dot_max #Efficiency of the turbine-generator \n", - "\n", - "#Part(b)\n", - "V2=V1*((1-n_windturbine)**0.5) #Exit velocity in m/s\n", - "\n", - "#Applying momentun equation\n", - "F_r=m_dot*(V2-V1) #Force exerted in N\n", - "\n", - "#Result\n", - "print \"The efficiency of the turbine is\",round(n_windturbine,3)\n", - "print \"The horizontal force exerted is\",round(F_r,1),\"N\"\n", - "#Answer differs by 0.5 due to floating point accuracy in second part" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The efficiency of the turbine is 0.361\n", - "The horizontal force exerted is -145.5 N\n" - ] - } - ], - "prompt_number": 59 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-6, Page No:256" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "V_gas=3000 #Velocity of the gas exiting in m/s\n", - "m_dot_gas=80 #mass flow rate of gas escaping in kg/s\n", - "m_spacecraft=12000 #mass of the spacecraft in kg\n", - "delta_t=5 #Time in s\n", - "#Calculations\n", - "#Part(a)\n", - "a_spacecraft=-(m_dot_gas*V_gas)/m_spacecraft #Acceleration of the spacecraft in m/s^2\n", - "\n", - "#Part(b)\n", - "dV=a_spacecraft*delta_t #Change in velocity of the spacecraft in m/s\n", - "\n", - "#PArt(c)\n", - "F_thrust=-(m_dot_gas*V_gas)/1000 #Thrust force exerted in kN\n", - "\n", - "#Result\n", - "print \"The acceleration of the spacecraft is\",round(a_spacecraft),\"m/s^2\"\n", - "print \"The change in velocity is\",round(dV),\"m/s\"\n", - "print \"The thrust force exerted is\",round(F_thrust),\"kN\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The acceleration of the spacecraft is -20.0 m/s^2\n", - "The change in velocity is -100.0 m/s\n", - "The thrust force exerted is -240.0 kN\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 6, - "metadata": {}, - "source": [ - "Example 6.6-7, Page No:257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "V_dot=70 #Volumertic Flow rate in L/min\n", - "D=0.02 #Inner diameter of the pipe in m\n", - "C=60*10**3 #Conversion factor\n", - "rho=997 #Density of water in kg/m^3\n", - "P1_gauge=90000 #Pressure at location in Pa\n", - "X=57 #Total weight of faucet in N\n", - "\n", - "#Calculations\n", - "V=((V_dot*4)/(pi*D**2))/C #Velocity of flow in m/s\n", - "m_dot=(rho*V_dot)/C #mass flow rate in kg/s\n", - "\n", - "#Applying the Momentum equation\n", - "F_rx=-(m_dot*V)-((P1_gauge*pi*D**2)/4) #X-component of force in N\n", - "F_rz=-m_dot*V+X #z-Component of force in N\n", - "\n", - "#Result\n", - "print \"The net force exerted on the flange in vector notation is Fr\",round(F_rx,2),\"i+\",round(F_rz,2),\"k N\"\n", - "#NOTE:The answer in the textbook differs due to decimal point accuracy difference in computation" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The net force exerted on the flange in vector notation is Fr -31.99 i+ 53.29 k N\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-9, Page NO:266" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "rho=1000 #Denisty of water in kg/m^3\n", - "D=0.10 #Diameter of the pipe in m\n", - "V=3 #Average velocity of water in m/s\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "m=12 #Mass of horizontal pipe section when filled with water in kg\n", - "r1=0.5 #Moment arm 1 in m\n", - "r2=2 #Moment arm 2 in m\n", - "\n", - "#Calculation\n", - "m_dot=rho*((pi*D**2)/4)*V #Mass flow rate in kg/s\n", - "W=m*g #Weight in N\n", - "\n", - "#Applying the momentum equation\n", - "M_A=r1*W-(r2*m_dot*V) #Momentum about point A in N.m\n", - "\n", - "#Setting M as zero and using the momentum equation\n", - "L=((2*r2*m_dot*V)/W)**0.5 #Length in m\n", - "\n", - "#Result\n", - "print \"The bending Moment at A is\",round(M_A,1),\"N.m and the length required is\",round(L,1),\"m\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The bending Moment at A is -82.5 N.m and the length required is 1.5 m\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6-9, Page No:267" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "V_dot_nozzle=5 #Volumertic flow rate in L/s\n", - "D_jet=0.01 #Diameter of the jet in m\n", - "C=10**-3 #Conversion Factor\n", - "n_dot=300 #R.P.M of the nozzle\n", - "r=0.6 #Radial arm in m\n", - "m_dot=20 #Mass flow rate in kg/s\n", - "s=60**-1 #Conversion factor\n", - "#Calculations\n", - "V_jet_r=(V_dot_nozzle*4)/(pi*D_jet**2)*C #Velocity relative to the rotating nozzle in m/s\n", - "w=(2*pi*n_dot)*s #Angular speed in rad/s\n", - "V_nozzle=r*w #Tangential Velocity in m/s\n", - "\n", - "#Applying thr relative velocity principle\n", - "V_jet=V_jet_r-V_nozzle #Velocity of the jet in m/s\n", - "\n", - "#Applying the momentum Equation and using the torque concept\n", - "T_shaft=r*m_dot*V_jet #Torque transmitted through the shaft in N.m\n", - "W_dot=w*T_shaft*C #Power generated in kW\n", - "\n", - "#Result\n", - "print \"The sprinkler-type turbine has the potential to produce\",round(W_dot,1),\"kW\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The sprinkler-type turbine has the potential to produce 16.9 kW\n" - ] - } - ], - "prompt_number": 33 - } - ], - "metadata": {} - } - ] -}
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