{
"metadata": {
"name": "",
"signature": "sha256:ef26cf09debc2accb827957d575ac9576db35c1e13d184b4161dad1098add2f0"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2 : Fluid statics"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.1 page no : 40\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"g=32.2; #ft/s^2\n",
"rho_water=62.3; #lbm/ft^3\n",
"\n",
"# calculation\n",
"#specific weoight=(density)*(acceleration due to gravity)\n",
"specific_wt=rho_water*g; #lbm.ft/ft^3.s^2\n",
"\n",
"#1 lbf=32.2 lbm.ft/s^2\n",
"specific_wt=specific_wt/32.2; #lbf/ft^3\n",
"\n",
"# result\n",
"print \"Specific weight of water is\" ,specific_wt , \"lbf/ft^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Specific weight of water is 62.3 lbf/ft^3\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.2 page no : 40\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"d=304.9; #m\n",
"rho_water=1024.; #Kg/m^3\n",
"g=9.81; #m/s^2\n",
"p_atm=101.3; #KPa\n",
"\n",
"# calculation\n",
"#gauge pressure=(desity)*(acc. due to gravity)*(depth)\n",
"p_depth=p_atm+rho_water*g*d/1000.0; #KPa\n",
"\n",
"# result\n",
"print \"pressure at the depth is\" , (p_depth) , \"KPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure at the depth is 3164.154656 KPa\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.3 page no : 41\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"rho_oil=55.; #lbm/ft^3\n",
"g=32.2; #ft/s^2\n",
"d=60.; #ft (depth of oil cylinder)\n",
"\n",
"# calculation and result\n",
"gauge_pressure=rho_oil*g*d/32.2; #lbf/ft^2\n",
"print \"Gauge pressure is\",\n",
"print gauge_pressure,\n",
"print \"lbf/ft^2\"\n",
"\n",
"#1 ft=12 in\n",
"gauge_pressure=gauge_pressure/144.0; #lbf/in^2\n",
"print \"Gauge pressure is\",\n",
"print gauge_pressure,\n",
"print \"lbf/in^2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gauge pressure is 3300.0 lbf/ft^2\n",
"Gauge pressure is 22.9166666667 lbf/in^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.4 page no : 42\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# varirbles\n",
"#calc of density of air at a certain height\n",
"p_atm=14.7; #psia\n",
"T=289.; #K\n",
"\n",
"#P2=P1*exp^(-(acc. due to gravity)*(mass of air)*(height)/(universal gas const.)/(temp.))\n",
"g=9.81; #m/s^2\n",
"R=8314; #N.m^2/Kmol/K\n",
"\n",
"#for height of 1000 ft=304.8m\n",
"h=304.8; #m\n",
"p_1000=14.7*math.exp(-g*29*h/R/289);\n",
"print \"pressure at 1000ft is\",\n",
"print p_1000,\n",
"print \"psia\"\n",
"\n",
"#for height of 10000 ft=3048m\n",
"h=3048.; #m\n",
"p_10000=p_atm*math.exp(-g*29.*h/R/289.);\n",
"print \"pressure at 10000ft is\",\n",
"print p_10000,\n",
"print \"psia\"\n",
"\n",
"#for height of 100000 ft=30480m\n",
"h=30480.; #m\n",
"p_100000=14.7*math.exp(-g*29.*h/R/289.);\n",
"print \"pressure at 100000ft is\",\n",
"print p_100000,\n",
"print \"psia\","
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure at 1000ft is 14.1789512072 psia\n",
"pressure at 10000ft is 10.2467246829 psia\n",
"pressure at 100000ft is 0.398102276652 psia\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.5 page no : 42\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"p_atm=14.7; #psia\n",
"g=9.81; #m/s^2\n",
"\n",
"#P2=P1*[1-(acc. due to gravity)*(mass of air)*(height)/(univ. gas const.)/(temp.)]\n",
"T=289.; #K\n",
"R=8314. #N.m^2/Kmol/K\n",
"\n",
"\n",
"# calculation and result\n",
"#for height of 1000ft=304.8m\n",
"h=304.8 #m\n",
"p_1000=p_atm*(1-g*29*h/R/T)\n",
"print \"pressure at 1000ft is\",\n",
"print p_1000,\n",
"print \"psia\"\n",
"\n",
"#for height of 10000ft=3048m\n",
"h=3048. #m\n",
"p_10000=p_atm*(1-g*29*h/R/T)\n",
"print \"pressure at 10000ft is\",\n",
"print p_10000,\n",
"print \"psia\"\n",
"\n",
"#for height of 100000ft=30480m\n",
"h=30480. #m\n",
"p_100000=p_atm*(1-g*29*h/R/T)\n",
"print \"pressure at 100000ft is\",\n",
"print p_100000,\n",
"print \"psia\"\n",
"\n",
"#NOTE that the pressure comes out to be negative at 100000ft justifying that density of air changes with altitude"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure at 1000ft is 14.1694926079 psia\n",
"pressure at 10000ft is 9.39492607874 psia\n",
"pressure at 100000ft is -38.3507392126 psia\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.6 page no : 45\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# variables\n",
"#calc atm pressure on a storage tank roof\n",
"p_atm=14.7; #psia\n",
"\n",
"#diameter of roof is 120ft\n",
"d_roof=120.; #ft\n",
"\n",
"# calculation\n",
"#force=(pressure)*(area)\n",
"f_roof=p_atm*(math.pi)*d_roof**2/4.*144; #lbf ;144 because 1ft=12inch\n",
"\n",
"# result\n",
"print \"Force exerted by atmosphere on the roof is\",\n",
"print f_roof,\n",
"print \"lbf\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Force exerted by atmosphere on the roof is 23940443.9848 lbf\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.7 page no : 45\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# variables\n",
"#calc atm pressure on a storage tank roof\n",
"p_atm=14.7; #psia\n",
"\n",
"#diameter of roof is 120ft\n",
"d_roof=120.; #ft\n",
"#force=(atm. pressure + gauge pressure)*(area)\n",
"#gauge pressure=(desity)*(acc. due to gravity)*(depth)\n",
"rho_water=62.3 #lbm/ft^3\n",
"g=32.2; #ft/s^2\n",
"\n",
"# calculation\n",
"#depth of water on roof=8 inch=o.667 ft\n",
"h=0.667; #ft\n",
"gauge_pressure=rho_water*g*h/32.2*(math.pi)*d_roof**2/4.; #lbf\n",
"\n",
"# result\n",
"print gauge_pressure"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"469965.799032\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.8 page no : 46\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"#lock gate has water on one side and air on the other at atm. pressure\n",
"w=20.; #m (width of the lock gate)\n",
"h=10.; #m (height of the lock gate)\n",
"p_atm=1.; #atm\n",
"rho_water=1000.; #Kg/m^3\n",
"g=9.81 #m/s^2\n",
"\n",
"# calculation\n",
"#for a small strip of dx height at the depth of x on the lock gate\n",
"#net pressure on strip = (p_atm+(rho_water)*g*x) - p_atm\n",
"#thus, net pressure on strip = (rho_water)*g*x\n",
"#force on strip = (rho_water*g*x)*w.dx = (rho_water)*g*w*(x.dx)\n",
"#force on lock gate = integration of force on strip fromm h=0 to h=10\n",
"#integration(x.dx) = x^2/2\n",
"#for h=0 to h=10; integration (x.dx) = h^2/2\n",
"force_lockgate=(rho_water)*g*w*h**2/2;\n",
"\n",
"# result\n",
"print \"The net force on the lock gate is\",force_lockgate/10**6,\"MN\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net force on the lock gate is 9.81 MN\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 2.9 page no : 49\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"sigma_tensile=20000. #lbf/in^2 (tensile stress is normally 1/4 rupture stress)\n",
"\n",
"#max pressure is observed at the bottom of the storage\n",
"p_max=22.9; #lbf/in^2\n",
"\n",
"#diameter of storaeg tank = 120ft =1440in\n",
"d=1440.; #in\n",
"\n",
"# calculation\n",
"t=(p_max)*d/sigma_tensile/2; #in\n",
"\n",
"# result\n",
"print \"Thichness of the storage tank is\",\n",
"print t,\n",
"print \"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thichness of the storage tank is 0.8244 in\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.10 page no : 50\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"p_working=250.0; #lbf/in^2\n",
"\n",
"#diameter of the cylinder = 10ft = 120in\n",
"d=120.0; #in\n",
"sigma_tensile=20000.; #lbf/in^2\n",
"\n",
"# calculation\n",
"t=p_working*d/sigma_tensile/2; #in\n",
"\n",
"# result\n",
"print \"Thichness of the storage tank is\",\n",
"print t,\n",
"print \"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thichness of the storage tank is 0.75 in\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.11 page no : 53\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# variables\n",
"p_atm=1.; #atm\n",
"T=293.; #K\n",
"d=3.; #m (diameter of the balloon)\n",
"\n",
"# calculation\n",
"#buoyant force=(density of air)*g*(volume of balloon)\n",
"#weight of balloon = (density of helium)*g*(volume of balloon)\n",
"#density for gases = PM/RT\n",
"#payload of balloon = buoyant force - weight\n",
"V_balloon=(math.pi)*d**3/6.; #m^3\n",
"R=8.2*10**(-2); #m^3.atm/mol/K\n",
"M_air=29.; #Kg/Kmol\n",
"M_he=4.; #Kg/Kmol\n",
"g=9.81; #m/s^2\n",
"payload=(V_balloon)*g*p_atm*(M_air-M_he)/R/T; #N\n",
"\n",
"# result\n",
"print \"Payload of the balloon is\",\n",
"print payload,\n",
"print \"N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Payload of the balloon is 144.307841185 N\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.12 page no : 54\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"#calc fraction of block in water\n",
"SG_wood=0.96; #Specific gravity\n",
"SG_gasoline=0.72;\n",
"\n",
"# calculation\n",
"#Let r be the ratio - V_water/V_wood\n",
"r=(SG_wood-SG_gasoline)/(1-SG_gasoline);\n",
"\n",
"# result\n",
"print \"Fraction of wood in water\",\n",
"print r"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fraction of wood in water 0.857142857143\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.13 page no : 54\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"#height of water above pt.C = 2.5ft\n",
"rho_water=62.3; #lbm/ft^3;\n",
"h1=2.5; #ft\n",
"rho_gas=0.1; #lbm/ft^3\n",
"h2=0.5; #ft (height of gas)\n",
"g=32.2; #ft/s^2\n",
"\n",
"# calculation\n",
"gauge_pressure=((rho_water)*g*h1+(rho_gas)*g*h2)/144/32.2 #lbf/in^2\n",
"\n",
"# result\n",
"print \"Gauge pressure is\",\n",
"print gauge_pressure,\n",
"print \"lbf/in^2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gauge pressure is 1.08194444444 lbf/in^2\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.14 page no : 56\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"rho_water=62.3; #lbm/ft^3\n",
"SG_oil=1.1;\n",
"rho_oil=SG_oil*(rho_water);\n",
"g=32.2; #ft/s^2\n",
"h1_1=1.; #ft\n",
"h1_2=2.; #ft\n",
"h2_1=2.; #ft\n",
"h2_2=1.; #ft\n",
"\n",
"# calculation\n",
"p_diff=((rho_water)*g*(h1_1-h1_2)+(rho_oil)*g*(h2_1-h2_2))/32.2/144.0; #lbf/in^2\n",
"\n",
"# result\n",
"print \"The pressure difference is\",\n",
"print p_diff,\n",
"print \"lbf/in^2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure difference is 0.0432638888889 lbf/in^2\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.15 page no : 57\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"k=10000.; #N/m (spring constant)\n",
"x=0.025; #m (displacement in spring)\n",
"A=0.01; #m^2 (area of piston)\n",
"\n",
"# calculation\n",
"gauge_pressure=k*x/A/1000.; #KPa\n",
"\n",
"# result\n",
"print \"The gauge pressure is\",\n",
"print gauge_pressure,\n",
"print \"KPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gauge pressure is 25.0 KPa\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.16 page no : 60\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# variables\n",
"g=32.2; #ft/s^2\n",
"h=20.; #ft (height of fireplace)\n",
"rho_air=0.075; #lbm/ft^3\n",
"T_air=293.0; #K (surrounding temperature)\n",
"T_fluegas=422.0; #K\n",
"\n",
"# calculation\n",
"p_diff=g*h*(rho_air)*(1-(T_air/T_fluegas))/32.2/144; #lbf/in^2\n",
"\n",
"# result\n",
"print \"The pressure difference is\",\n",
"print p_diff,\n",
"print \"lbf/in^2\","
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The pressure difference is 0.00318424170616 lbf/in^2\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.17 page no : 64\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables\n",
"rho_water=1000. #Kg/m^3\n",
"g=9.81; #m/s^2\n",
"h=5.; #m (depth of water)\n",
"\n",
"# calculation and result\n",
"#for elevator not accelerated\n",
"p_gauge=(rho_water)*g*h/1000.0; #KPa\n",
"print \"THe gauge pressure is\",\n",
"print p_gauge,\n",
"print \"KPa\"\n",
"\n",
"#for elevator accelerated at 5m/s^2 in upward direction\n",
"a=5.; #m/s^2\n",
"p_gauge=(rho_water)*(g+a)*h/1000.0; #KPa\n",
"print \"THe gauge pressure is\",\n",
"print p_gauge,\n",
"print \"KPa\"\n",
"\n",
"#for elevator accelerated at 5m/s^2 in downward direction\n",
"a=5.; #m/s^2\n",
"p_gauge=(rho_water)*(g-a)*h/1000.0; #KPa\n",
"print \"THe gauge pressure is\",\n",
"print p_gauge,\n",
"print \"KPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"THe gauge pressure is 49.05 KPa\n",
"THe gauge pressure is 74.05 KPa\n",
"THe gauge pressure is 24.05 KPa\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 2.18 page no : 65\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# variables\n",
"#angle free surface makes with the horizontal in an accelerated body\n",
"a=1.; #ft/s^2\n",
"g=32.2; #ft/s^2\n",
"\n",
"# calculation\n",
"theta=math.atan(a/g); #radians\n",
"theta=theta*180./math.pi; #degrees\n",
"\n",
"# result\n",
"print \"The angle made by free surface with the horizontal is\",\n",
"print theta,\n",
"print \"degrees\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The angle made by free surface with the horizontal is 1.77880031567 degrees\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.19 page no : 66\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"# variables\n",
"f=78/60.0; #rps\n",
"r=0.15; #m\n",
"g=9.81; #m/s^2\n",
"\n",
"# calculation\n",
"#omega=2*(%pi)*f\n",
"z=((2*(math.pi)*f)**2)*r**2/2/g; #m\n",
"\n",
"# result\n",
"print \"The liquid in the cylinder rises to a height of\",\n",
"print z,\n",
"print \"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The liquid in the cylinder rises to a height of 0.0765120708158 m\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" example 2.20 page no : 67\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"# variables\n",
"#Let difference between heights at bottom and top be d\n",
"d=20.; #in\n",
"r_a=14.; #in\n",
"f=1000/60.; #rps\n",
"g=32.2; #ft/s^2\n",
"\n",
"# calculation\n",
"r_b=((r_a)**2-2*(d)*g*12/(2*(math.pi)*f)**2)**0.5; #in\n",
"\n",
"# result\n",
"print \"The thickness of water strip at bottom of industrial centrifuge\",\n",
"print r_b,\n",
"print \"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thickness of water strip at bottom of industrial centrifuge 13.9495728181 in\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}