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{
"metadata": {
"name": "",
"signature": "sha256:c4763e5901f11f7dc8699953a49ca33923db1ee352fad2402763eb19eb84fa47"
},
"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": 2,
"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": {}
}
]
}
|