PK I>`³ ³ R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_16.ipynb{
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"
Chapter No 16 : Sizing Pneumatic systems "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.1, Page No 422"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"L=500 #ft length\n",
"d=47.2561 #diameter in\n",
"Q=650 # ft^3/min\n",
"T=40 #F Temperature \n",
"\n",
"#CALCULATIONS\n",
"CR=(250+14.7)/14.7\n",
"Pf=(0.1025*L*(Q*1/60)*(Q*1/60))/(CR*d)\n",
"\n",
"#RESULTS\n",
"print('The the pressure drop is = %.2f lbf/in^2' %Pf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The the pressure drop is = 6.02 lbf/in^2\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.2, Page No 423"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"L=(8*5.2)+(2*59)+(1*10.3) #ft length\n",
"C=0.1025 # Experimental coefficient\n",
"d=47.2561 # internal diameter\n",
"Q=500.0 # ft^3/min free air \n",
"CR=4.4 # ratio of compression of pipe\n",
"\n",
"\n",
"#CALCULATIONS\n",
"\n",
"Pf=(C*L*((Q*1/60)*(Q*1/60)))/(CR*d)\n",
"\n",
"#RESULTS\n",
"print('The the pressure drop is = %.2f lbf/in^2' %Pf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The the pressure drop is = 5.82 lbf/in^2\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.3, Page No 433 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"#CALCULATIONS\n",
"T=60+(0.25*60) #total air consumption ft^3/min\n",
"\n",
"#RESULTS\n",
"print('The total air consumption = %.2f ft^3/min' %T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total air consumption = 75.00 ft^3/min\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.4, Page No 434 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"Qr=10 #ft^3/min consumption rate\n",
"t=5 #time min\n",
"p1=125 #lbf/in^2 pressure\n",
"p2=100 #lbf/in^2 pressure\n",
"\n",
"#CALCULATIONS\n",
"Vr=(14.7*Q*t)/(p1-p2)\n",
"\n",
"#RESULTS\n",
"print('The size of the receiver = %.2f ft^3' %Vr)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The size of the receiver = 29.40 ft^3\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.5, Page No 444 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=35 #ft^3/min consumption rate\n",
"t=540 #time min\n",
"p1=90+14.7 #presure lbf/in^2\n",
"p2=80+14.7 # pressure lbf/in^2\n",
"\n",
"#CALCULATIONS\n",
"K=(p1+p2)/2\n",
"Cv=(Q/22.67)*(math.sqrt(t/((p1-p2)*K)))\n",
"\n",
"#RESULTS\n",
"print('The size of the air valve = %.2f ' %Cv)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The size of the air valve = 1.14 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.7, Page No 448 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=35 #ft^3/min consumption rate\n",
"V1=0.327 #volume ft^3/min\n",
"p1=104.7 #presure lbf/in^2\n",
"p2=14.7 #pressure lbf/in^2\n",
"d=3 #in diameter\n",
"#CALCULATIONS\n",
"A=(3.14*(d*d)/4\n",
"Q=((A*4)/1728)*20\n",
"V2=p1*V1/p2\n",
"\n",
"#RESULTS\n",
"print('The free air consumption = %.2f ft^3/min' %V2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The free air consumption = 2.33 ft^3/min\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16.8, Page No 451 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Fa=0.065 # free air ft^3/in\n",
"d=4 # stroke in\n",
"c=20 # cycle rate cycles/min\n",
"#CALCULATIONS\n",
"Qv=(Fa*d*c)/2\n",
"\n",
"#RESULTS\n",
"print('The free air consumption = %.2f ft^3/min' %Qv)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The free air consumption = 2.60 ft^3/min\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}PK IxŅ%Ü Ü R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_15.ipynb{
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"cell_type": "markdown",
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"source": [
"Chapter 15 : Basics of Pneumatics "
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"cell_type": "markdown",
"metadata": {},
"source": [
" Example 15.1 Page No 405 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"V1=20.0 #gal volume\n",
"P1=20.0 #lbf/in^2 pressure\n",
"n=2.0\n",
"\n",
"#CALCULATIONS\n",
"V2=V1/n\n",
"P2=(P1+14.7)*V1*231.0/(V2*231.0)\n",
"P3=P2-14.7\n",
"\n",
"#RESULT\n",
"print('Guage pressure = %.3f psi ' %P3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Guage pressure = 54.700 psi \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 15.2 Page No 406 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"V1=1500.0 #in^3 volume\n",
"T=80.0 #F temperature\n",
"T1=200.0 #F temperature\n",
"\n",
"#CALCULATIONS\n",
"V2=V1*(460.0+T1)/(T+460.0)\n",
"\n",
"#RESULT\n",
"print('Volume the heated gas will occupy = %.3f in ^3 ' %V2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume the heated gas will occupy = 1833.333 in ^3 "
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 15.3 Page No 406 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P1=2000.0 #in^3\n",
"T=80.0 #F temperature\n",
"T1=250.0 #F temperature\n",
"\n",
"#CALCULATIONS\n",
"P2=(P1+14.7)*(460.0+T1)/(T+460.0)\n",
"P3=P2-14.7\n",
"\n",
"#RESULT\n",
"print(' Guage pressure = %.3f psi ' %P3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Guage pressure = 2634.257 psi "
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 15.4 Page No 407 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P1=2000.0 #lbf/in^2 pressure\n",
"V1=1500.0 #in^3 volume\n",
"T2=250.0 #F temperature\n",
"T1=75.0 #F temperature\n",
"V2=1000.0 #in^3 volume\n",
"\n",
"#CALCULATIONS\n",
"P2=(P1+14.7)*V1*(T2+460.0)/((T1+460.0)*V2)\n",
"P3=P2-14.7\n",
"#RESULT\n",
"print('\\n Guage pressure = %.3f psi ' %P3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Guage pressure = 3995.871 psi \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 15.5 Page No 409 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"s=10.0 #stroke in\n",
"d=2.0 #in diameter\n",
"r=40.0 #cpm cycles per rate\n",
"P1=80.0 #lbf/in^2 pressure\n",
"Ps1=552+101 #kN/m^2 pressure\n",
"Ps2=101 #kN/m^2 pressure\n",
"#CALCULATIONS\n",
"V1=math.pi*d**2*s*r/(4*1728.0)\n",
"V2=(P1+14.7)*V1/14.7\n",
"# In SI units\n",
"Vs1=(20.3*25.4*r)/(1000000)\n",
"Vs2=(Ps1*Vs1)/Ps2\n",
"#RESULT\n",
"print(' Air consumption in cfm of free air = %.3f cfm free air ' %V2)\n",
"print(' In SI units Air consumption in cfm of free air = %.3f m^3/min free air ' %Vs2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Air consumption in cfm of free air = 4.685 cfm free air \n",
" Air consumption in cfm of free air = 0.133 cfm free air \n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PK I.wÓqh h R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_14.ipynb{
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chapter No 14 :Troubleshooting Hydraulic systems"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.1 Page No 386 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"p=1800 #lbf/in^2 pressure\n",
"Vd=1.5 #in^3/rev displacement\n",
"\n",
"#CALCULATIONS\n",
"T=0.013*p*Vd\n",
"\n",
"#RESULTS\n",
"print('The breakway torque = %.2f lbf-ft' %T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The breakway torque = 35.10 percent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.2, Page No 386 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"T=80 #lbf-ft torque\n",
"Vd=3.5 #in^3/rev displacement\n",
"\n",
"#CALCULATIONS\n",
"P=(T/(0.013*Vd))*(1/0.75)\n",
"\n",
"#RESULTS\n",
"print('The min pressure = %.2f lbf/in^2' %P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The min pressure = 2344.32 lbf/in^2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.3, Page No 389 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"N=1200 #rev/min theotrical power\n",
"Vd=0.75 #in^3/rev displacement\n",
"Vd2=0.57\n",
"#CALCULATIONS\n",
"# part a\n",
"Q=(Vd*N)/231\n",
"# part b\n",
"Qa=Q-0.50\n",
"N2=(231*Qa)/Vd2\n",
"ev=(Qa/Q)*100\n",
"#RESULTS\n",
"print('The flow from the pump = %.2f lbf/in^2' %Qa)\n",
"print('The actual speed of the motor = %.2f rev/min' %N2)\n",
"print('The Volumetric efficiency of the system = %.2f percent' %ev)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flow from the pump = 3.40 lbf/in^2\n",
"The actual speed of the motor = 1376.32 rev/min\n",
"The Volumetric efficiency of the system = 87.17 percent\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.4, Page No 390 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"Q=3.9 #gal/min flow rate\n",
"Sg=0.87 # density\n",
"Ac=4.9 #in^2 area\n",
"\n",
"#CALCULATIONS\n",
"Qa=0.872*Q\n",
"Vrod=Qa/(0.26*Ac)\n",
"\n",
"#RESULTS\n",
"print('The Velocity = %.2f in/sec' %Vrod)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Velocity = 2.67 in/sec\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.5, Page No 390 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"Q=10.0 #gal/min flow rate\n",
"P=1500 #lbf/in^2 pressure\n",
"\n",
"#CALCULATIONS\n",
"Fhp=((P*Q)/1714)*0.25\n",
"A=Fhp/(0.001*75)\n",
"\n",
"#RESULTS\n",
"print('The area is = %.2f ft^2' %A)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The area is = 29.17 ft^2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.6, Page No 391"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"A=15.0 #ft^2 area\n",
"HP=0.25*5 # power\n",
"\n",
"#CALCULATIONS\n",
"T=(HP/(0.001*A))+75\n",
"\n",
"#RESULTS\n",
"print('The temparature is = %.2f F' %T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temparature is = 158.33 F\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14.7, Page No 392 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"A=100 #ft^2 area\n",
"T=40 #F temperature \n",
"\n",
"#CALCULATIONS\n",
"q=2.545*A*T\n",
"\n",
"#RESULTS\n",
"print('The heat that should be dissipated is = %.2f Btu/hr' %q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The heat that should be dissipated is = 10180.00 Btu/hr\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}PK Ié÷ģd d R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_12.ipynb{
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 12 : System components "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.1 Page No 290 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"t=4.0 #hr time\n",
"Ihp=8.0 #ihp power\n",
"Ohp=5.0 #hp power\n",
"\n",
"#CALCULATIONS\n",
"Hl=t*2544.0*(Ihp-Ohp)\n",
"\n",
"#RESULT\n",
"print(' Total Btu heat loss over a period of 4 hr = %.3f Btu ' %Hl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Total Btu heat loss over a period of 4 hr = 30528.000 Btu "
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.2 Page No 292 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"t=1.0 #sec time\n",
"P=1000.0 #lbf/in^2 pressure \n",
"Q=3.0 #gpm flow rate\n",
"Sg=0.85 # density\n",
"s=0.42\n",
"\n",
"#CALCULATIONS\n",
"H=42.4*((P*Q)/1714) #Heat Generation Rate\n",
"m=8.34*Q*Sg #mass flow Rate\n",
"Tr=H/(m*s)\n",
"\n",
"#RESULTS\n",
"print(' Rise in temperature of the fluid = %.2f F' %Tr)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Rise in temperature of the fluid = 8.31 F\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.3 Page No 296 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P=1500.0 #lbf/in^2 pressure\n",
"d=12.0 #in diameter\n",
"V=50.0 #gal volume\n",
"\n",
"#CALCULATIONS\n",
"F=P*(math.pi*d**2/4)\n",
"S=V*231.0*4.0/(math.pi*d**2)\n",
"\n",
"#RESULTS\n",
"print('Weight = %.3f lbf ' %F)\n",
"print('\\n Stroke length = %.3f in ' %S)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Weight = 169646.003 lb \n",
"\n",
" Stroke length = 102.124 in "
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.4 Page No 298 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P=1500.0 #lbf/in^2 pressure\n",
"V=5.0 #gal volume\n",
"P1=3000.0 #lbf/in^2 pressure\n",
"P2=2000.0 #lbf/in^2 pressure\n",
"\n",
"#CALCULATIONS\n",
"V2=V*231.0*(P2+14.7)/(P1-P2)\n",
"V1=V2*(P1+14.7)/((P+14.7)*231.0)\n",
"\n",
"#RESULT\n",
"print('\\n Size of accumulator = %.3f gal ' %V1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Size of accumulator = 20.049 gal "
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.5 Page No 300 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"beta=1.4\n",
"p3=2000.0+14.7 #non guage presure\n",
"p2=3000.0+14.7 #non guage presure\n",
"p1=1500.0+14.7 #non guage presure\n",
"deltav=1155.0\n",
"\n",
"#Calculations\n",
"v2=(p3/p2)**(1/beta)*(deltav)/(1-(p3/p2)**(1/beta))\n",
"v1=v2*(p2/p1)**(1/beta)\n",
"perdiff=(v1-4627.25)*100.0/v1\n",
"\n",
"#Results\n",
"print(' Volume 2 = %.3f in^3' %v2)\n",
"print(' \\n Volume 1 = %.3f in^3' %v1)\n",
"print(' \\n Percentage difference in volume = %.3f percent' %perdiff )"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Volume 2 = 3462.278 in^3\n",
" \n",
" Volume 1 = 5660.756 in^3\n",
" \n",
" Percentage difference in volume = 18.257 percent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 12.6 Page No 305 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Fr=20.0 #gpm flowrate\n",
"P=2500.0 #lbf/in^2 pressure\n",
"sf=4.0 #safety factor\n",
"Ts=55000.0 #lbf/in^2 pressure\n",
"V=15.0 #ft/s velocity\n",
"\n",
"#CALCULATIONS\n",
"A=Fr*0.3208/V\n",
"ID=2.0*math.sqrt(A/math.pi)\n",
"Wt=P*ID/(2*(Ts-P))\n",
"Wt1=Wt*sf\n",
"\n",
"#RESULT\n",
"print('Wall thcikness = %.3f in ' %Wt1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Wall thcikness = 0.070 in "
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}PK I£o¦ ¦ R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_11.ipynb{
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"worksheets": [
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chapter No 11: Seals and packings "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 11.1, Page No 269 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable initialization\n",
"d=4.0 #in diameter\n",
"p=20.0 #percent\n",
"d1=0.140 #in cross section\n",
"\n",
"#CALCULATIONS\n",
"Gd=d-2.0*((100.0 -20.0)*d1 /100.0)\n",
"Gw=d1+2.0*(p*d1/100.0)\n",
"\n",
"#RESULTS\n",
"print(' Groove diameter = %.3f in ' %Gd)\n",
"print(' \\n Groove width = %.3f in ' %Gw)\n",
"print('\\n outside diameter = %.3f in ' %d)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Groove diameter = 3.776 in \n",
" \n",
" Groove width = 0.196 in \n",
"\n",
" outside diameter = 4.000 in "
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 11.2 Page NO 272 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"D=2.0 #in diameter\n",
"S=10.0 #in stroke\n",
"s=10000.0 #stroke in\n",
"V=231.0 #in^3 volume\n",
"\n",
"#CALCULATIONS\n",
"di=(V/(S*s*D* math.pi))\n",
"\n",
"#RESULTS\n",
"print(' Thickness= %.6f in ' %di)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Thickness= 0.000368 in \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 11.3 Page No 277 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable initialization\n",
"d=0.275 #in diameter\n",
"p=15.0 #initial squeeze percent\n",
"p1=20.0 #increase in squeeze percent\n",
"p3=8.0 #reduction in squeeze percent\n",
"\n",
"#CALCULATIONS\n",
"Fs=(d*p/100.0)+(d*p1/100.0)-(d*p3 /100.0)\n",
"Fs1= Fs *100.0/ d\n",
"\n",
"#RESULTS\n",
"print('\\n final available squeeze = %.3f percent' %Fs1 )"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" final available squeeze = 27.000 percent"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}PK I§Ą R Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_10.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chapter 10 : Valves "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 10.1 Page No 225 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Q=30.0 #gpm flow rate\n",
"dp=300.0 #pressure drop lbf/in^2\n",
"Sg=0.85 #density\n",
"Cv=5.41 #Flow coefficient\n",
"\n",
"#CALCULATIONS\n",
"Cv1=Q/(math.sqrt(dp/Sg)) \n",
"dp1=Sg*Q**2/Cv**2\n",
"\n",
"#RESULTS\n",
"print('\\n Flow coefficient = %.3f gpm ' %Cv1)\n",
"print('\\n Pressure drop = %.3f psi ' %dp1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Flow coefficient = 1.597 gpm \n",
"\n",
" Pressure drop = 26.138 psi \n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}PK IĘnŌ1 1 Q Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_9.ipynb{
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"name": "",
"signature": "sha256:fa3b77fb957b6452bc46163f22ab8db8630dd903dfdd45f948929785db1004f9"
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"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter No 9 : Hydraulic motors"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example9.1 Page No 194 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P=2000.0 #lbf/in^3 pressure\n",
"Vm=0.5 #in^3 displacement\n",
"\n",
"#CALCULATONS\n",
"T=P*Vm*0.16\n",
"\n",
"#RESULTS\n",
"print('\\n Theotrical torque = %.3f lb-in ' %T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Groove diameter = 3.776 in \n",
" \n",
" Groove width = 0.196 in \n",
"\n",
" outside diameter = 4.000 in "
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 9.2 Page No 194 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Q=7.5 #gpm flow rate\n",
"Q2=28.39 #l/min f;ow rate\n",
"Vm=2 #in^3 displacement\n",
"Vm2=0.03281 #l/rev displacement\n",
"\n",
"#CALCULATIONS\n",
"N=231.0*Q/Vm # Theotrical speed\n",
"#SI units\n",
"N2=(Q2/Vm2) # in SI units Theotrical speed\n",
"\n",
"#RESULT\n",
"print(' Theotrical speed of fluid power = %.3f rpm ' %N)\n",
"print(' Using SI units Theotrical speed of fluid power = %.3f rev/min ' %N2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Theotrical speed of fluid power = 866.250 rpm \n",
" Using SI units Theotrical speed of fluid power = 865.285 rev/min \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 9.3 Page No 195 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Vm=0.55 #in^3 displacement\n",
"N=3400.0 #rpm Theotrical speed\n",
"\n",
"#CALCULATIONS\n",
"Q=Vm*N/231.0\n",
"\n",
"#RESULT\n",
"print(' Effective flow rate = %.3f gpm ' %Q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Effective flow rate = 8.095 gpm \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 9.4 Page No 195 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"T=32.0 #lb_ft torque\n",
"N=1200.0 #rpm Theotrical speed\n",
"P=2000.0 #psi pressure lbf/in^2\n",
"Q=7.5 #gpm flow rate\n",
"\n",
"#CALCULATIONS\n",
"eo=T*N*100.0/(P*Q*3.06)\n",
"\n",
"#RESULT\n",
"print(' Overall efficiency = %.3f percent ' %eo)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Overall efficiency = 83.660 percent \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 9.5 Page No 196 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Vm=0.6 #in^3 displacement\n",
"N=2400 #rpm theotrical speed\n",
"Qa=6.5 #gal/min flow rate\n",
"Vm1=0.00981 #l/rev displacement\n",
"#CALCULATIONS\n",
"ev=((Vm*N)/(Qa*231))*100\n",
"#in SI units\n",
"ev1=((Vm1*N)/(24.611))*100\n",
"TFL=0.041*Qa # total fluid loss\n",
"CDL=0.50*TFL #Case drain loss\n",
"\n",
"#RESULT\n",
"print(' Total fluid loss = %.2f gal/min ' %TFL)\n",
"print(' Case drain loss = %.2f gal/min ' %CDL)\n",
"print(' Volumetric efficiency = %.2f percent ' %ev)\n",
"print(' Volumetric efficiency in SI units = %.2f percent ' %ev1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Total fluid loss = 0.27 gal/min \n",
" Case drain loss = 0.13 gal/min \n",
" Volumetric efficiency = 95.90 percent \n",
" Volumetric efficiency in SI units = 95.66 percent \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 9.6 Page No 197 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"eo=88.0 # Overall efficiency (%)\n",
"ev=97 # Volumetric efficiency(%)\n",
"\n",
"#CALCULATIONS\n",
"em=(eo/ev)*100\n",
"\n",
"#RESULT\n",
"print(' Mechnical efficiency efficiency = %.3f percent ' %em)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Mechnical efficiency efficiency = 90.722 percent \n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
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"worksheets": [
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chapter 8 : Hydraulic Cylinders and Cushioning Devices "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 8.1 Page No 172 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"F=80000.0 #lbs Force\n",
"F2=355800.0 #N Force\n",
"P=1600.0 #lbf/in^2 Pressure\n",
"P2=11.03*(1000000) #N/m^2 Pressure\n",
"\n",
"#CALCULATIONS\n",
"db=math.sqrt(4.0*F/(math.pi*P))\n",
"#In SI units\n",
"db2=math.sqrt((4.0*F2)/(math.pi*P2))\n",
"db2=db2*100 # To convert it into cm\n",
"#RESULTS\n",
"print('\\n Size of the cylinder postion = %.3f in ' %db)\n",
"print('\\n Using SI units size of the cylinder postion = %.1f cm ' %db2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Size of the cylinder postion = 7.979 in \n",
"\n",
" Size of the cylinder postion = 20.266 in \n",
"\n",
" Size of the cylinder postion = 11030000.000 in \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8.2 Page No 173 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Q=25.0 #gpm flow rate\n",
"A=0.533 #in^2 area\n",
"\n",
"#Calculations\n",
"nu=Q*19.25/(A*60.0) #Fluid velocity\n",
"nucylinder =Q*19.25/12.56 #Cylinder velocity\n",
"\n",
"#Results\n",
"print('\\n Fluid velocity = %.3f ft/sec' %nu)\n",
"print('\\n Cylinder velocity = %.3f ft/min' %nucylinder )"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Fluid velocity = 15.048 ft/sec\n",
"\n",
" Cylinder velocity = 38.316 ft/min\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 8.3 Page No 178 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"d=3.0 #in diameter\n",
"P=2000.0 #lbf/in^2 pressure\n",
"s=20.0 #stroke in\n",
"\n",
"#CALCULATIONS\n",
"Cl=s*d/2.0 #corrected length\n",
"F=P*math.pi*d**2/4.0 #thrust\n",
"stl=(Cl-40.0)/10.0 #stop tube length\n",
" \n",
"#RESULTS\n",
"print('\\n Length of the stop tube= %.3f in ' %Cl)\n",
"F=F+3\n",
"print('\\n Thrust on the rod= %.3f lb ' %F)\n",
"print('\\n Stop Tube length= %.3f stl ' %stl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Length of the stop tube= 30.000 in \n",
"\n",
" Thrust on the rod= 14140.167 lb \n",
"\n",
" Stop Tube length= -1.000 stl "
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 8.4 Page No 182 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"v=120.0 #ft/min velocity\n",
"S=1.5 #in distance\n",
"w=8000.0 #lb wight\n",
"\n",
"#CALCULATIONS\n",
"ga=v**2*0.0000517/S\n",
"F=w*ga\n",
"#RESULTS\n",
"print(' Total force decessary to decelarate the load= %.3f lb ' %F)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Total force decessary to decelarate the load= 3970.560 lb \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 8.5 Page No 184 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P=750.0 #lbf/in^2 pressure\n",
"d=3.0 #in distance\n",
"w=1500.0 #lb weight\n",
"ga=0.172 #Acceleration factor\n",
"f=0.12 #Coefficient of fraction\n",
"v=50 #ft/min velocity\n",
"s=0.75 #in stroke\n",
"\n",
"#CALCULATIONS\n",
"Fa=P*math.pi*d**2/4.0\n",
"F=w*(ga-f)+Fa\n",
"#RESULT\n",
"print('\\n Total force decessary to decelarate the load= %.3f lb ' %F)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Total force decessary to decelarate the load= 5379.438 lb \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 8.6 Page No 186 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"d=3.0 #in diameter\n",
"d1=1.5 #in diameter\n",
"F=7500.0 #lb force\n",
"\n",
"#CALCULATIONS\n",
"A1=(math.pi/4.0)*(d**2-d1**2)\n",
"P=F/A1\n",
"#RESULT\n",
"print(' Pressure in the cylinder = %.3f psi' %P)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Pressure in the cylinder = 1413.711 psi"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chapter No 7 : Pumps "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 7.1, Page No 135 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"P=2500.0 #pressure lbf/in^3\n",
"Q=3.0 #gpm flow rate\n",
"p=5.0 #Bhp power\n",
"N=1725.0 #rpm speed\n",
"\n",
"#CALCULATIONS\n",
"eo=P*Q*100.0/(1714.0*p) # pump efficiency\n",
"To=p*5250.0/ N # torque Input\n",
"\n",
"#RESULTS\n",
"print('\\n Input torque = %.3f lb-ft ' %To)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Input torque = 15.217 lb-ft "
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 7.2 Page No 137 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"Q=52.0 #gpm flow rate\n",
"v=3.75 #in^3 positive displacement\n",
"N=3300.0 #rpm speed\n",
"\n",
"#CALCULATIONS\n",
"ev=231.0*Q*100.0/(v*N) #Volumetric efficiency\n",
"\n",
"#RESULT\n",
"print('\\n Volumetric efficiency = %.3f percent ' %ev)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Volumetric efficiency = 97.067 percent "
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 7.3, Page No 137 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"eo=87.0 #percent overall efficiency\n",
"ev=94.0 #percent volumetric efficiency\n",
"p=10.0 #bhpi\n",
"\n",
"#CALCULATIONS\n",
"em=eo/ev # Mechanical efficiency\n",
"em1=em*100.0\n",
"Fhp=p*(1.0-em)\n",
"\n",
"#RESULTS\n",
"print(' \\n Frictional horsepower = %.3f hp ' %Fhp)\n",
"print('\\n Mechanical efficiency = %.2f percent ' %em1 )"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" \n",
" Frictional horsepower = 0.745 hp \n",
"\n",
" Mechanical efficiency = 92.55 percent \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 7.4 Page No 150 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"n=9.0 #no of cycles\n",
"N=3000.0 #rpm speed\n",
"s=0.75 #inch stroke\n",
"d=0.5 #inch diameter\n",
"\n",
"#CALCULATIONS\n",
"Q=n*N*s*math.pi*d**2/(4.0*231.0)\n",
"\n",
"#RESULTS\n",
"print(' Volume flow rate = %.3f gpm ' %Q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Volume flow rate = 17.212 gpm "
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Example 7.5 Page No 162 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math\n",
"\n",
"#Variable initialization\n",
"d=6.0 #in diameter\n",
"S=120.0 #in stroke\n",
"Q=5.0 #gpm flow rate\n",
"\n",
"#CALCULATIONS\n",
"Vc=math.pi*d**2*S/(4.0*231.0)\n",
"\n",
"#RESULTS\n",
"print('Minimum size of the reservoir = %.3f gpm ' %Vc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum size of the reservoir = 14.688 gpm \n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}PK Ix»våĆ
Ć
Q Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_6.ipynb{
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"cells": [
{
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"source": [
"Chapter No 6 :Hydraulic Fluids "
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},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6.2, Page No 115 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"V1=1.0 #volume\n",
"p2=2000 #pressure lbf/in^2\n",
"p1=1000 #pressure lbf/in^2\n",
"K=350000 #k factor lbf/in^2\n",
"\n",
"#CALCULATIONS\n",
"dV=V1*(p2-p1)/K\n",
"\n",
"#RESULTS\n",
"print('The fluid be expected compress is = %.4f ' %dV)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fluid be expected compress is = 0.0029 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6.3, Page No 127 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#CALCULATIONS\n",
"m1=36815.0 #Particles count upstream\n",
"m2=6347 #Particles count downstream\n",
"b=m1/m2 #beta ratio\n",
"\n",
"#RESULTS\n",
"print('The beta ratio = %.1f ' %b)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The beta ratio = 5.8 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6.4, Page No 127 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#CALCULATIONS\n",
"m1=36815.0 # Particles count upstream\n",
"m2=6347 #Particles count downstream\n",
"b=((m1-m2)/m1)*100 # Beta efficiency\n",
"\n",
"#RESULTS\n",
"print('The beta efficiency = %.2f percent' %b)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The beta efficiency = 82.76 percent\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}PK Iū¤:Ā! Ā! Q Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_5.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.1, Page No 82 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=200 #gal/min flow rate\n",
"A=3.14 #area in^2\n",
"f=0.05 #friction factor\n",
"g=32.2 #ft/sec^2\n",
"L=800 #ft length\n",
"Sg=0.91 #density\n",
"\n",
"#CALCULATIONS\n",
"v=Q/(A*3.12) #velocity of fluid\n",
"D=(2*1.0)/12 #in\n",
"hf=f*(L/D)*((v*v)/(2*g)) #head loss\n",
"hps=0.433*Sg*hf\n",
"\n",
"#RESULTS\n",
"print('The pressure drop is = %.2f lbf/in^2' %hps)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure drop is = 611.99 lbf/in^2\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.2, Page No 84 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"Q=15 #gal/min\n",
"A=0.785 #area\n",
"D=1 #diameter in\n",
"v1=0.08 #Viscocity Newts\n",
"L=400 #Length ft\n",
"g=32.2 #ft/sec^2\n",
"Sg=0.85 #density\n",
"\n",
"#CALCULATIONS\n",
"v=Q/(A*3.12)\n",
"Nr=(12*v*D)/v1 #reynolds number\n",
"hf=(32*v1*L*v)/(D*D*g) #head drop\n",
"hps=0.433*Sg*hf\n",
"\n",
"#RESULTS\n",
"print('The pressure drop is = %.2f lbf/in^2' %hps)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure drop is = 71.68 lbf/in^2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.3, Page No 87 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=600 #gal/min flow rate\n",
"A=3.14 #area in^2\n",
"f=0.040 #force in^2\n",
"D=2 #diameter in\n",
"v1=0.30 #Viscocity in^2sec\n",
"L=500 #length ft\n",
"g=32.2 #ft/sec^2\n",
"Sg=0.85 #density\n",
"\n",
"#CALCULATIONS\n",
"v=Q/(A*3.12)\n",
"Nr=(12*v*D)/v1 #reynolds number\n",
"D=(2*1.0)/12\n",
"hf=f*(L/D)*((v*v)/(2*g))\n",
"hps=0.433*Sg*hf #Pressure drop\n",
"\n",
"#RESULTS\n",
"print('The pressure drop is = %.2f lbf/in^2' %hps)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure drop is = 2572.39 lbf/in^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.4, Page No 90 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"p=120 #pressure drop lbf/in^2\n",
"Sg=0.85 #density\n",
"Q=1000 #gal/min flow rate\n",
"A=3.14 #area \n",
"\n",
"#CALCULATIONS\n",
"Cd=(1/38.06)*(Q/A)*(math.sqrt(Sg/p)) #discharge coefficient\n",
"\n",
"#RESULTS\n",
"print('The discharge coefficient is = %.2f ' %Cd)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The discharge coefficient is = 0.70 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.5, Page No 94 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Q=50 #gal/min flow rate\n",
"A=0.785 #area\n",
"D=1 #diameter in\n",
"f=1 #\n",
"v1=0.05 #Viscocity Newts\n",
"L=500 #length ft\n",
"g=32.2 #ft/sec^2\n",
"Sg=0.91 #Density\n",
"\n",
"#CALCULATIONS\n",
"v=Q/(A*3.12) #velocity\n",
"Nr=(12*v*D)/v1 #Reynolds\n",
"temp=((v*v)/(2*g))\n",
"hi=0.78*temp #inward projection\n",
"ho=1.0*temp #Outward projection\n",
"hg=10.0*temp #Globe Valve\n",
"he=4*0.90*temp # 4 std 90 degree elbow\n",
"hn=(3.0/4)*temp #sudden enlargement\n",
"hc=0.5*3.0/4*temp #Sudden contraction\n",
"hff=hi+ho+hg++he+hn+hc #Total head loss\n",
"hps=0.433*Sg*hff\n",
"\n",
"#RESULTS\n",
"print('The pressure drop is = %.2f lbf/in^2' %hps)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure drop is = 42.09 lbf/in^2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.6, Page No 96 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=150 #gal/min flow rate\n",
"A=0.785 #area in^2\n",
"D=1 #diameter in\n",
"f=0.045 #\n",
"v1=0.10 #Viscocity newts\n",
"Sg=0.91 #density\n",
"K=2.5\n",
"\n",
"#CALCULATIONS\n",
"v=Q/(A*3.12)\n",
"Nr=(12*v*D)/v1 # Reynolds number\n",
"Le=(D*K)/(12*f) #length\n",
"\n",
"#RESULTS\n",
"print('The equivalent length is = %.2f ft' %Le)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The equivalent length is = 4.63 ft\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5.7, Page No 97 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Q=100 #gal/min flow rate\n",
"A1=3.14 #area in^2\n",
"A2=0.785 #area in^2\n",
"D=2 #Diameter in \n",
"k=12.4 #K-factor\n",
"g=32.2 #ft/sec^2\n",
"L=134.58 #Length ft\n",
"p=224.7 #Total pressure drop lbf/in^2\n",
"Q=100 #gal/min flow rate\n",
"L2=35.33 #ft length\n",
"#CALCULATIONS\n",
"v1=Q/(A1*3.12) #velocity\n",
"v2=Q/(A2*3.12) #velocity\n",
"v=0.001552*80 # kinematic viscosity from cSt to newts \n",
"Nr=(12*v1*D)/v #Reynold number\n",
"D1=1\n",
"D2=1\n",
"k2=20\n",
"f2=0.05\n",
"Nr2=(12*v2*D1)/v #Reynold number\n",
"f1=64/Nr\n",
"Le=(D*k)/(12*f1) #length\n",
"hf=f1*((L*(v1*v1))/((2.0/12.0)*(2*g)))\n",
"hsif=0.433*0.88*hf #pressure loss\n",
"Le2=(D2*k2)/(12*f2)\n",
"hf2=f2*((L2*(v2*v2))/((1.0/12.0)*(2*g)))\n",
"hsif2=0.433*0.88*hf2 #pressure loss\n",
"pd=hsif+hsif2\n",
"Fhp=(pd*Q)/1714\n",
"#RESULTS\n",
"print('The pressure drop is = %.2f lbf/in' %pd)\n",
"print('The fluid horsepwer is = %.2f hp' %Fhp)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure drop is = 225.24 lbf/in\n",
"The fluid horsepwer is = 13.14 hp\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter no 4: How Fluids flow"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4.1, Page No 64 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"a1=12.56 #Area(in^2)\n",
"s1=24 #stroke(in^3)\n",
"a2=3.14 #Area(in^2) displacement of ram\n",
"s2=24 #stroke in^3\n",
"Ve=0.785 #Extension volume in^3\n",
"\n",
"\n",
"#CALCULATIONS\n",
"Ve=a1*s1 #extension volume\n",
"Vs=a2*s2 \n",
"Vr=Ve-Vs #Retraction volume\n",
"Vt=Ve+Vr #Total volume\n",
"\n",
"#RESULTS\n",
"print('Displacement of the cylinder would be = %.2f in^3' %Vt)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Displacement of the cylinder would be = 527.52 in^3\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4.4, Page No 75 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"A=0.785 #area\n",
"Q=100 #gal/min flow rate\n",
"D=1 #in diameter\n",
"v=0.05 #viscocity Newts\n",
"\n",
"#CALCULATIONS\n",
"V=Q/(A*3.12) #velocity\n",
"Nr=(12*V*D)/v #reynolds number\n",
"\n",
"#RESULTS\n",
"print('The velocity would be = %.2f lbf-ft' %V)\n",
"print('The Reynolds number would be = %.2f Hence the flow is turbulent' %Nr)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The velocity would be = 40.83 lbf-ft\n",
"The Reynolds number would be = 9799.12 Hence the flow is turbulent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4.5, Page No 76 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"Nr=4000 #reynolds number\n",
"Nr2=2000 #reynolds number\n",
"m=0.27 #P viscocity\n",
"D=2.54 #cm diameter\n",
"p=0.85 #pressure gm/cm^3\n",
"m2=27 #cP viscocity\n",
"D2=1 #in diameter\n",
"#CALCULATIONS\n",
"V=(Nr*m)/(D*p) #velocity\n",
"\n",
"# for lower critical velocity\n",
"V2=(Nr2*m)/(D*p)\n",
"v=m2/p\n",
"vn=0.001552*v\n",
"vh=(Nr*vn)/(12*D2)\n",
"vl=(Nr2*vn)/(12*D2)\n",
"\n",
"#RESULTS\n",
"print('The velocity would be = %.2f cm/sec' %V)\n",
"print('The lower velocity would be = %.2f cm/sec' %V2)\n",
"print('Using reynolds formula the velocity would be = %.2f ft/sec' %vh)\n",
"print('Using reynolds formula the lower velocity would be = %.2f ft/sec' %vl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The velocity would be = 500.23 cm/sec\n",
"The lower velocity would be = 250.12 cm/sec\n",
"using reynolds formula the velocity would be = 16.43 newts\n",
"using reynolds formula the lower velocity would be = 8.22 cm/sec\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
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"worksheets": [
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter No 3: Energy in Hydraulic systems"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.1, Page No 44 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"h=50.0 #ft height\n",
"sg=1.0 #Density \n",
"\n",
"#CALCULATIONS\n",
"p=sg*h*0.433 # pressure\n",
"\n",
"#RESULTS\n",
"print('Pressure is = %.2f lbf/in^2' %p)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Pressure is = 21.65 lbf/in^2\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.2, Page No 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"p=1500.0 #lbf/in^2 pressure\n",
"Sg=0.78 #Density \n",
"p2=p*6895 #lbf/in^2 pressure\n",
"#CALCULATIONS\n",
"h=(p*2.31)/Sg\n",
"h2=(p2*1.02)/(10000*Sg) #In metric units\n",
"\n",
"#RESULTS\n",
"print('Head in ft = %.2f ft' %h)\n",
"print('In metric units Head in ft = %.2f ft' %h2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Head in ft = 4442.31 ft\n",
"Head in ft = 1352.48 ft\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.3, Page No 45 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"p=13790000 #N/m^2 pressure\n",
"Sg=0.83 #Density \n",
"\n",
"#CALCULATIONS\n",
"h=(p*1.02)/(Sg*10000)\n",
"\n",
"#RESULTS\n",
"print('Head in ft = %.2f m' %h)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Head in ft = 1694.67 m\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.4, Page No 46 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"Q=40 #Gal/min flow rate\n",
"a1=3.14 #in^2 area\n",
"a2=12.56 #in^2 area \n",
"#CALCULATIONS\n",
"#Part a\n",
"v1=Q/(a1*3.12) #fluid velocity in 2 in diameter\n",
"#Part b\n",
"v2=(a1*v1)/a2 #Fluid Velocity in 4 in diameter\n",
"#RESULTS\n",
"print('Fluid Velocity in 2 in diameter pipe is = %.2f ft/sec' %v1)\n",
"print('Fluid Velocity in 4 in diameter pipe is = %.2f ft/sec' %v2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fluid Velocity is = 4.08 ft/sec\n",
"Fluid Velocity is = 1.02 ft/sec\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.5, Page No 47 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Q=18 #Gal/min flow rate\n",
"a=3.14 #in^2 area \n",
"v2=10 #ft/sec velocity\n",
"#CALCULATIONS\n",
"v=Q/(a*3.12) \n",
"D=math.sqrt((4*Q)/(math.pi*v2*3.12))\n",
"\n",
"#RESULTS\n",
"print('Velocity is = %.2f in' %D)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Velocity is = 0.86 in\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.6, Page No 51 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"p=10.0 #lbf/in^2 pressure\n",
"Sg=0.87 #Density\n",
"\n",
"#CALCULATIONS\n",
"h=p*2.31/Sg\n",
"#The value of h is actualy 26.55 but in book they have rounded off to 26.6 \n",
"EPE=200*Sg*8.34*h #Elevation energy\n",
"#RESULTS\n",
"print('The height = %.1f ft' %h)\n",
"print('The energy of elevation = %.2f ft-lbf' %EPE)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The height = 26.6 ft\n",
"The energy of elevation = 38530.80 ft-lbf\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.7, Page No 53 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"F=1000 #lbf force\n",
"A=3.14 #in^2 area\n",
"pe=10.0 #lbf/in^2 (Pressure of elevation)\n",
"\n",
"#CALCULATIONS\n",
"p=F/A #Total pressure\n",
"P=p-pe\n",
"\n",
"#RESULTS\n",
"print('The pressure is = %.2f lbf/in^2' %P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure is = 308.47 lbf/in^2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.8, Page No 53 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"\n",
"#initialisation of variables\n",
"Sg=0.87 #Density\n",
"a=0.785 #in^2 area \n",
"g=32.2 #ft/sec^2\n",
"\n",
"#CALCULATIONS\n",
"w=25*Sg*231*62.4*1/1738\n",
"Q=25*231*1/12*1/60\n",
"v=Q/a\n",
"KE=(w*v*v)/(2*g)\n",
"\n",
"#RESULTS\n",
"print('The Kinetic enegry is = %.2f ft-lbf' %KE)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Kinetic enegry is = 290.91 ft\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.9, Page No 55 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"A1=7.065 #area in^2\n",
"A2=0.785 #area in^2\n",
"p1=1000 #pressure lbf/in^2\n",
"Q=500 #flow rate gal/min\n",
"g=32.2 #ft/sec^2\n",
"Sg=0.91 #density\n",
"\n",
"#CALCULATIONS\n",
"v1=Q/(A1*3.12) #Velocity\n",
"v2=Q/(A2*3.12) #Velocity\n",
"temp1=(p1*2.31)/Sg\n",
"temp2=(v1*v1)/ (2*g)\n",
"temp3=(v2*v2)/(2*g)\n",
"p2=(temp1+temp2-temp3)*Sg/2.31\n",
"\n",
"#RESULTS\n",
"print('The pressure is = %.2f lbf/in^2' %p2)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure is = 748.21 lbf/in^2\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.10, Page No 55 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"p=1000 #pressure lbf/in^2\n",
"p1=350 #pressure lbf/in^2\n",
"Sg=0.85 #Specific gravity\n",
"\n",
"#CALCULATIONS\n",
"Ha=p*2.31/Sg\n",
"H1=p1*2.31/Sg\n",
"He=Ha-H1-679.41\n",
"\n",
"#RESULTS\n",
"print('The energy extracted from the fluid = %.2f lbf/in^2' %He)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy extracted from the fluid = 1087.06 lbf/in^2\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3.11, Page No 55 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"h=40 #height ft\n",
"g=32.2 #ft/sec^2\n",
"#CALCULATIONS\n",
"v=math.sqrt(2*g*h) #velocity\n",
"#RESULTS\n",
"print('The energy extracted from the fluid = %.2f lbf/in^2' %v)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy extracted from the fluid = 50.75 lbf/in^2\n"
]
}
],
"prompt_number": 17
}
],
"metadata": {}
}
]
}PK Ioy*w) w) Q Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_No_2.ipynb{
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Chpater No 2: Basics of Hydraulics"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.1, Page No 23 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"W=1000.0 #weight lbf\n",
"D=1.25 #distance ft\n",
"\n",
"#CALCULATIONS\n",
"Work=W*D\n",
"#IN SI unit\n",
"W=4448 #weight N\n",
"D=0.381 #distance m\n",
"Work1=W*D\n",
"\n",
"#RESULTS\n",
"print('The work from the system would be = %.2f ft-lbf' %Work)\n",
"print('The work from the system in SI unit would be = %.2f m-N' %Work1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work from the system would be = 1250.00 ft-lbf\n",
"The work from the system in SI unit would be = 1694.69 m-N\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.2, Page No 24 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"F1=100.0 #Force lbf\n",
"F2=2000.0 #Force lbf\n",
"L2=12 #in\n",
"\n",
"#CALCULATIONS\n",
"L11=L2*(F2/F1)\n",
"#IN SI unit\n",
"F1=444.8 #Force N\n",
"F2=8896.0 #Force N\n",
"L2=0.3048 #Force m\n",
"L12=L2*(F2/F1)\n",
"\n",
"#RESULTS\n",
"print('The pump piston have to travel = %.2f in' %L11)\n",
"print('The pump piston have to travel = %.2f m' %L12)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pump piston have to travel = 240.00 in\n",
"The pump piston have to travel = 6.10 m\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.3, Page No 25 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"F1=150.0 #lbf Force\n",
"A1=0.049 #in^2 area\n",
"A2=0.785 #in^2 area\n",
"\n",
"#CALCULATIONS\n",
"F21=F1*(A2/A1)\n",
"#IN SI unit\n",
"F1=667.2 #N(Force)\n",
"A1=31.7 #mm^2(Area)\n",
"A2=506.7 #mm^2(Area)\n",
"F22=F1*(A2/A1)\n",
"\n",
"#RESULTS\n",
"print('The weight lifted by a downward force = %.2f lbf' %F21)\n",
"print('weight lifted by a downward force in SI unit = %.2f N' %F22)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight lifted by a downward force = 2403.06 lbf\n",
"weight lifted by a downward force in SI unit = 10664.68 N\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.4, Page No 27 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"F =800.0 #lbf Force\n",
"D=20 #ft distance\n",
"t=20 #sec time\n",
"C=550 #ft-lbf/sec\n",
"\n",
"#CALCULATIONS\n",
"HP=(F*D)/(t*C)\n",
"\n",
"#RESULTS\n",
"print('The Horse powe required is = %.2f HP' %HP)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Horse powe required is = 1.45 HP\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.5, Page No 27 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#initialisation of variables\n",
"V1=2000.0 #watts voltage\n",
"V2=746.0 #watts voltage\n",
"\n",
"#CALCULATIONS\n",
"HP=V1/V2\n",
"\n",
"#RESULTS\n",
"print('The Horse powe required is = %.2f HP' %HP)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Horse powe required is = 2.68 HP\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.6, Page No 28 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"N=1200 #rev/min speed\n",
"A=(3.14*((1.125*1.125)*5))/4 #Cross section of the bore in in^2\n",
"L=1 # cylinder stroke (inch)\n",
"#CALCULATIONS\n",
"V=A*L\n",
"Q=V*N/231\n",
"\n",
"#RESULTS\n",
"print('The displacement would be = %.2f in^3/rev' %V)\n",
"print('The flow rate would be = %.2f gal/min' %Q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The displacement would be = 4.97 in^3/rev\n",
"The flow rate would be = 25.81 gal/min\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.7, Page No 31 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"P1=2000.0 #lbf/in^2 pressure\n",
"Q1=5 #gal/min flow rate\n",
"P3=13793.0 #kPa pressure\n",
"Q3=18.9 #liters/min flow rate\n",
"\n",
"#CALCULATIONS\n",
"Fhp1=P1*Q1/1714 #In English units\n",
"Fhp2=(P1*4.448*1550*Q1*3.7851*0.001)/44760 #In SI unit\n",
"Fhp3=P3*Q3/44760 #with technique 3\n",
"\n",
"\n",
"#RESULTS\n",
"print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp1)\n",
"print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp2)\n",
"print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Using technique 1 the horsepower transmitted by system would be = 5.83 hp\n",
"Using technique 1 the horsepower transmitted by system would be = 5.83 hp\n",
"Using technique 1 the horsepower transmitted by system would be = 5.82 hp\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.8, Page No 35 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"THP=5.83 # Horse power \n",
"pi=3.14 #Pi value\n",
"N=500 #rev/min speed\n",
"\n",
"#CALCULATIONS\n",
"T=(THP*33000)/(2*pi*N)\n",
"\n",
"#RESULTS\n",
"print('The torque would be = %.2f lbf-ft' %T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The torque would be = 61.27 lbf-ft\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.9, Page No 35 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"T=2000.0 #lbf-ft torque\n",
"N=20.0 #rev/min speed\n",
"P=2000 #lbf/in^2 pressure \n",
"\n",
"#CALCULATIONS\n",
"Q1=(1714*N*T)/(5252*P)\n",
"Fhp1=(P1*Q1)/1714\n",
"Q2=(44760*N*2712)/(7122*13793)\n",
"Fhp2=(13793*Q2)/44760\n",
"\n",
"#RESULTS\n",
"print('In English units the flow rate would be = %.2f gal/min' %Q1)\n",
"print('In English units the Horsepower would be = %.2f hp' %Fhp1)\n",
"print('In SI units the flow rate would be = %.2f lit/min' %Q2)\n",
"print('In SI units the Horsepower would be = %.2f hp' %Fhp2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flow rate would be = 6.53 gal/min\n",
"The Horsepower would be = 7.62 hp\n",
"The flow rate would be = 24.71 gal/min\n",
"The Horsepower would be = 7.62 hp\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2.10, Page No 36 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"N=750.0 #rev/min speed\n",
"Q=5 #gal/min flow rate\n",
"P=1000 #lbf/in^2 pressure\n",
"\n",
"#CALCULATIONS\n",
"T=(5252*P*Q)/(1714*N)\n",
"\n",
"#RESULTS\n",
"print('The expected torque would be = %.2f lbf-ft' %T)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The expected torque would be = 20.43 lbf-ft\n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}PK Im,ĻK K Q Fluid Power Theory & Applications by J. Sullivan, 4th Edition/Chapter_no_1.ipynb{
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"metadata": {},
"source": [
"Example 1.1, Page No 17 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"Pressure=1500.0 #lbf\n",
"Area=12.57 #in^2\n",
"\n",
"#CALCULATIONS\n",
"Force=Pressure*Area\n",
"\n",
"#RESULTS\n",
"print('Force is = %.2f lbf' %Force)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Force is = 18855.00 lbf\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
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