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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 : The balance equation and mass balance"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 3.4 page no : 88\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate vol. flow rate, mass flow rate and average vel of gasoline through pipe\n",
"import math\n",
"\n",
"# variables\n",
"V=15.; #gal volume of gasoline\n",
"t=2.; #min\n",
"rho_water=62.3; #lbm/ft^3\n",
"sg=0.72; #specific gravity\n",
"\n",
"# calculation and Result\n",
"q=(15/2.0)*(0.1336/60) #ft^3/s vol. flow rate\n",
"print \"volumetric flow rate is %f ft^3/s\"%q\n",
"m=q*sg*rho_water #lbm/s\n",
"print \"Mass flow rate is %f lbm/s\"%m\n",
"d=1.; #in diameter of pipe\n",
"a=((math.pi)*d**2/4.0)/144.0; #ft^2 area of pipe\n",
"v_avg=q/a #ft/s\n",
"print \"The average velocity is %f ft/s\"%v_avg"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"volumetric flow rate is 0.016700 ft^3/s\n",
"Mass flow rate is 0.749095 lbm/s\n",
"The average velocity is 3.061886 ft/s\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.5 page no : 90\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate velocity and mass flow rate of natural in a pipe\n",
"import math\n",
"\n",
"# variables\n",
"d1=2.; #ft diameter of pipe at position 1\n",
"a1=(math.pi)/4*d1**2; #ft^2\n",
"v1=50.; #ft/s vel of gas at position 1\n",
"rho1=2.58; #lbm/ft^3 density of gas at position 1\n",
"d2=3.; #ft diameter of pipe at position 2\n",
"\n",
"# calculation\n",
"a2=(math.pi)/4*d2**2;\n",
"rho2=1.54; #lbm/ft^3 density at position 2\n",
"v2=(rho1/rho2)*(a1/a2)*v1 #ft/s\n",
"\n",
"# result\n",
"print \"Velocity is %f ft/s\"%v2\n",
"m=rho1*v1*a1 #lbm/s mass flow rate\n",
"print \"The mass flow rate is %f lbm/s\"%m"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Velocity is 37.229437 ft/s\n",
"The mass flow rate is 405.265452 lbm/s\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.6 page no : 91\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the mass flow rate, volumetric flow rate and velocity of waterin a pipe\n",
"import math\n",
"\n",
"# variables\n",
"d1=0.25; #m diameter of pipe at position 1\n",
"v1=2.; #m/s velocity\n",
"rho=998.2; #kg/m^3 density of water\n",
"a1=(math.pi)/4*d1**2; #m^2\n",
"d2=0.125 #m diameter of pipe at position 2\n",
"\n",
"# calculation\n",
"a2=(math.pi)/4*d2**2; #m^2\n",
"m=rho*a1*v1 #kg/s mass flow rate\n",
"\n",
"# result\n",
"print \"Mass flow rate is %f kg/s\"%m\n",
"q=m/rho #m^3/s volumetric flow rate\n",
"print \"The volumetric flow rate is %f m^3/s\"%q\n",
"v2=(a1/a2)*v1 #m/s velocity\n",
"print \"Velocity of water is %f m/s\"%v2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mass flow rate is 97.998056 kg/s\n",
"The volumetric flow rate is 0.098175 m^3/s\n",
"Velocity of water is 8.000000 m/s\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.7 page no : 92\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calulate the time required\n",
"import math\n",
"\n",
"# variables\n",
"p_initial=1.; #atm pressure initially\n",
"p_final=0.0001; #atm pressure finally\n",
"V=10.; #ft^3 volume of system\n",
"q=1.; #ft^3/min vol. flow rate\n",
"\n",
"# calculation\n",
"t=(V/q)*math.log(p_initial/p_final) #min\n",
"\n",
"# result\n",
"print \"The time required is %f min\"%t"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The time required is 92.103404 min\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.8 page no : 93\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the final or steady state pressure in tank\n",
"\n",
"# variables\n",
"m_in=0.0001; #lbm/min\n",
"q_out=1.; #ft^3/min\n",
"rho_sys=m_in/q_out #lbm/ft^3\n",
"rho_air=0.075; #lbm/ft^3\n",
"p_initial=1.; #atm\n",
"\n",
"# calculation\n",
"p_steady=p_initial*(rho_sys/rho_air) #atm\n",
"\n",
"# result\n",
"print \"The steady state pressure is %f atm\"%p_steady"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The steady state pressure is 0.001333 atm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.9 page no : 94\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate how fast the level of water is rising or falling in a cylindrical tank\n",
"import math\n",
"\n",
"# variables\n",
"d=3.; #m diameter of tank\n",
"a=(math.pi)*d**2/4; #m^2\n",
"d_in=0.1; #m inner diameter of inflow pipe\n",
"d_out=0.2; #m\n",
"v_in=2.0; #m/s\n",
"v_out=1.0; #m/s\n",
"\n",
"# calculation\n",
"q_in=((math.pi)*d_in**2/4.0)*v_in; #m^3/s\n",
"q_out=((math.pi)*d_out**2/4.0)*v_out; #m^3/s\n",
"\n",
"#let D represent d/dt\n",
"DV=q_in-q_out; #m^3/s\n",
"\n",
"# result\n",
"if DV>1:\n",
" print \"The water level in tank is rising\"\n",
"elif DV<1:\n",
" print \"The water level in tank is falling\"\n",
"else:\n",
" print \"No accumulation\"\n",
"#let h be the height of water in tank\n",
"Dh=DV/a #m/s \n",
"print \"The rate of level of water is rising or falling in a cylindrical tank is %f m/s\"%Dh"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The water level in tank is falling\n",
"The rate of level of water is rising or falling in a cylindrical tank is -0.002222 m/s\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.11 page no : 97\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate flow rate of ventilation air supply\n",
"\n",
"# variables\n",
"q=5/8.0; #kg/hr mass evaporation rate of benzene\n",
"c=1.3*10**(-6); #kg/m^3 concentration of benzene\n",
"\n",
"# calculation\n",
"Q=q/c/3600.0 #m^3/s\n",
"\n",
"# result\n",
"print \"The flow rate of ventilation air supply is %f m^3/s\"%Q"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flow rate of ventilation air supply is 133.547009 m^3/s\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
