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{
"metadata": {
"name": "",
"signature": "sha256:eb1b6239947f111d6ea8f1bcaafe0bb84b121b2561bbb9f5f8a8bdec77169fd2"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter2:SOME CONCEPTS AND DEFINITIONS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.1:pg-19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 1\n",
"#weight of a person\n",
"\n",
"\n",
"m=1 #kg\n",
"g=9.75 #acc.due to gravity in m/s^2\n",
"F=m*g #weight of 1 kg mass in N\n",
"print\"\\n hence,weight of person is\",round(F,2),\" N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,weight of person is 9.75 N\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.2:pg-24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 2\n",
"#average volume and density\n",
"\n",
"Vliq=0.2 #volume of liquid in m^3\n",
"dliq=997 #density of liquid in kg/m^3\n",
"Vstone=0.12 #volume of stone in m^3\n",
"Vsand=0.15 #volume of sand in m^3\n",
"Vair=0.53 #vo;ume of air in m^3\n",
"mliq=Vliq*dliq #mass of liquid in kg\n",
"dstone=2750 #density of stone in kg/m^3\n",
"dsand=1500 #density of sand in kg/m^3\n",
"mstone=Vstone*dstone #volume of stone in m^3\n",
"msand=Vsand*dsand #volume of sand in m^3\n",
"Vtot=1 #total volume in m^3\n",
"dair=1.1 #density of air in kg/m^3\n",
"mair=Vair*dair #mass of air\n",
"mtot=mair+msand+mliq+mstone #total mass in kg\n",
"v=Vtot/mtot #specific volume in m^3/kg\n",
"d=1/v #overall density in kg/m^3\n",
"print\"\\n hence,average specific volume is\",round(v,6),\"m^3/kg\" \n",
"print\"\\n and overall density is\", round(d),\"kg/m^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,average specific volume is 0.001325 m^3/kg\n",
"\n",
" and overall density is 755.0 kg/m^3\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.3:pg-26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3\n",
"#calculating the required force\n",
"import math\n",
"\n",
"Dcyl=0.1 #cylinder diameter in m\n",
"Drod=0.01 #rod diameter in m\n",
"Acyl=math.pi*Dcyl**2/4 #cross sectional area of cylinder in m^2\n",
"Arod=math.pi*Drod**2/4 #cross sectional area of rod in m^2\n",
"Pcyl=250000 #inside hydaulic pressure in Pa\n",
"Po=101000 #outside atmospheric pressure in kPa\n",
"g=9.81 #acc. due to gravity in m/s^2\n",
"mp=25 #mass of (rod+piston) in kg\n",
"F=Pcyl*Acyl-Po*(Acyl-Arod)-mp*g #the force that rod can push within the upward direction in N\n",
"print\"\\n hence,the force that rod can push within the upward direction is\",round(F,1),\" N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence,the force that rod can push within the upward direction is 932.9 N\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.4:pg-28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 4\n",
"#Calculating atmospheric pressure \n",
"\n",
"dm=13534 #density of mercury in kg/m^3\n",
"H=0.750 #height difference between two columns in metres\n",
"g=9.80665 #acc. due to gravity in m/s^2\n",
"Patm=dm*H*g/1000 #atmospheric pressure in kPa\n",
"print\"\\n hence, atmospheric pressure is\",round(Patm,2),\"kPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence, atmospheric pressure is 99.54 kPa\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.5:pg-28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 5\n",
"#pressure inside vessel\n",
"\n",
"dm=13590 #density of mercury in kg/m^3\n",
"H=0.24 #height difference between two columns in metres\n",
"g=9.80665 #acc. due to gravity in m/s^2\n",
"dP=dm*H*g #pressure difference in Pa\n",
"Patm=13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",
"Pvessel=dP+Patm #Absolute Pressure inside vessel in Pa\n",
"Pvessel=Pvessel/101325 #Absolute Pressure inside vessel in atm\n",
"print\"\\n hence, the absolute pressure inside vessel is\",round(Pvessel,3),\"atm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence, the absolute pressure inside vessel is 1.302 atm\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.6:pg-29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 6\n",
"#calculating pressure\n",
"\n",
"dg=750 #density of gaasoline in kg/m^3\n",
"dR=1206 #density of R-134a in kg/m^3\n",
"H=7.5 #height of storage tank in metres\n",
"g=9.807 #acc. due to gravity in m/s^2\n",
"dP1=dg*g*H/1000 #in kPa\n",
"Ptop1=101 #atmospheric pressure in kPa\n",
"P1=dP1+Ptop1\n",
"print\"hence,pressure at the bottom of storage tank if fluid is gasoline is\",round(P1,1),\"kPa\" \n",
"dP2=dR*g*H/1000 #in kPa\n",
"Ptop2=1000 #top surface pressure in kPa\n",
"P2=dP2+Ptop2\n",
"print\"\\n hence, pressure at the bottom of storage tank if liquid is R-134a is\",round(P2),\"kPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"hence,pressure at the bottom of storage tank if fluid is gasoline is 156.2 kPa\n",
"\n",
" hence, pressure at the bottom of storage tank if liquid is R-134a is 1089.0 kPa\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2.7:pg-29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 7\n",
"#calculating balancing force\n",
"\n",
"Po=100#Outside atmospheric pressure in kPa\n",
"F1=25 #net force on the smallest piston in kN\n",
"A1=0.01 #cross sectional area of lower piston in m^2\n",
"P1=Po+F1/A1 #fluid pressure in kPa\n",
"d=900 #density of fluid in kg/m^3\n",
"g=9.81 #acc. due to gravity in m/s^2\n",
"H=6 #height of second piston in comparison to first one in m\n",
"P2=P1-d*g*H/1000 #pressure at higher elevation on piston 2 in kPa\n",
"A2=0.05 #cross sectional area of higher piston in m^3\n",
"F2=(P2-Po)*A2 #balancing force on second piston in kN\n",
"print\"\\n hence, balancing force on second larger piston is\",round(F2,1),\"N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" hence, balancing force on second larger piston is 122.4 N\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
