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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 10 : Pumps compressors and turbines"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.1 page no : 362"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The efficiency of the pump is 0.729167\n"
]
}
],
"source": [
"#Calculate the efficiency of a pump\n",
"\n",
"# variables\n",
"Q=50. #gal/min\n",
"P1=30. #psia 0r lbf/in**2\n",
"P2=100. #psia 0r lbf/in**2\n",
"dP=P2-P1 #psia 0r lbf/in**2\n",
"power=2.8 #hp\n",
"\n",
"# calculation\n",
"#1 ft = 12 in\n",
"#1 hp.min = 33000 lbf.ft\n",
"#1 gal = 231 in**3\n",
"eta=(Q*dP/power)*(1/33000.0)*231*(1/12.0) #dimentionless\n",
"\n",
"# result\n",
"print \"The efficiency of the pump is %f\"%eta"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.2 page no : 366"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the maximum elevation above the lowest water level in sump at which pump inlet can be placed is 19.770533 ft\n"
]
}
],
"source": [
"#Calculate the maximum elevation above the lowest water level in sump at which pump inlet can be placed\n",
"\n",
"# variables\n",
"P1=3.72 #psia 0r lbf/in**2\n",
"P2=14.5 #psia 0r lbf/in**2\n",
"dP=P2-P1 #psia 0r lbf/in**2\n",
"rho=61.3 #lbm/ft**3\n",
"g=32.2 #ft/s**2\n",
"\n",
"# calculation\n",
"#1 ft = 12 in\n",
"#1 lbf.s**2 = 32.2 lbm.ft\n",
"h_loss=4 #ft\n",
"v=10 #ft/s\n",
"h_max=(dP/rho/g)*144*32.2-(v**2/2.0/g)-h_loss #ft\n",
"\n",
"# result\n",
"print \"the maximum elevation above the lowest water level in sump at which pump inlet can be placed is %f ft\"%h_max"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.3 page no : 370"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The work required per pound mole for a 100 percent efficient isothermal compressor is 2415.724914 Btu/lbmol\n",
"\n",
"The work required per pound mole for a 100 percent efficient adiabatic compressor is 3417.499724 Btu/lbmol\n",
"\n"
]
}
],
"source": [
"#Calculate the work requird per pound mole for a 100% efficient isothermal and adiabatic compressor\n",
"import math\n",
"\n",
"# variables\n",
"R=1.987 #Btu/lbmol/R (universal gas constant)\n",
"T=528. #R (Rankine temperature scale)\n",
"ratio_P=10. #dimentionless\n",
"\n",
"# calculations and results\n",
"#for isothermal compressor\n",
"W1=R*T*math.log(ratio_P) #Btu/lbmol\n",
"print \"The work required per pound mole for a 100 percent efficient isothermal compressor is %f \"%W1,\"Btu/lbmol\\n\"\n",
"\n",
"#for adiabatic compressor\n",
"gama=1.4#dimentionless\n",
"W2=(gama/(gama-1))*R*T*(ratio_P**((gama-1)/gama)-1)#Btu/lbmol\n",
"print \"The work required per pound mole for a 100 percent efficient adiabatic compressor is %f \"%W2,\"Btu/lbmol\\n\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.4 page no : 370"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The work required per pound mole for a 100 percent efficient adiabatic compressor is 2861.592127 Btu/lbmol\n"
]
}
],
"source": [
"#Calculate the work requird per pound mole for a 100% efficient 2 stage adiabatic compressor\n",
"\n",
"# variables\n",
"R=1.987 #Btu/lbmol/R (universal gas constant)\n",
"T=528.0 #R (Rankine temperature scale)\n",
"ratio_P1=3 #dimentionless\n",
"ratio_P2=10/3.0 #dimentionless\n",
"gama=1.4 #dimentionless\n",
"\n",
"# calculation\n",
"W=(gama/(gama-1))*R*T*((ratio_P1**((gama-1)/gama)-1)+(ratio_P2**((gama-1)/gama)-1)) #Btu/lbmol\n",
"\n",
"# result\n",
"print \"The work required per pound mole for a 100 percent efficient adiabatic compressor is %f \"%W,\n",
"print \"Btu/lbmol\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.5 page no : 373"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The pump head is 112.471747 ft\n"
]
}
],
"source": [
"#Calculate the pump head\n",
"import math\n",
"\n",
"# variables\n",
"N=1750. #rev/min\n",
"#1 min 60 sec\n",
"omega=2*(math.pi)*N/60 #radians/sec\n",
"Q=100. #gal/min\n",
"#1 gallon = 231 in**3\n",
"#1 ft =12 in\n",
"#1 min = 60 sec\n",
"d_inlet = 2.067 #ft\n",
"\n",
"# calculations\n",
"A_inlet=(math.pi)/4.*(d_inlet**2) #ft**2\n",
"V1=(Q/A_inlet)*231/60.0/12.0 #ft/s\n",
"d_outlet = 1.61 #ft\n",
"A_outlet=(math.pi)/4*(d_outlet**2) #ft**2\n",
"V2=(Q/A_outlet)*231/60.0/12.0 #ft/s\n",
"g=32.2 #ft/s**2\n",
"d_inner=0.086 #ft\n",
"d_outer=0.336 #ft\n",
"h=(omega)**2/g*((d_outer**2)-(d_inner)**2)+(V2**2-V1**2)/2./g #ft\n",
"\n",
"#result\n",
"print \"The pump head is %f ft\"%h"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.6 page no : 378"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The pump head is 2.734925 ft\n"
]
}
],
"source": [
"#Calculate the pump head\n",
"\n",
"# variables\n",
"rho=62.3 #lbm/ft**3\n",
"g=32.2 #ft/s**2\n",
"v=18.46 #ft/s\n",
"\n",
"# calculation\n",
"#1 lbf/s**2 = 32.2 lbm.ft\n",
"h=1/(rho*g) * (v**2/2)*32.2 #ft\n",
"\n",
"# result\n",
"print \"The pump head is %f ft\"%h"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.7 page no : 381"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the estimated pressure rise in the first stage of mutisatge centrifugal compressor is 8.865562 psia\n"
]
}
],
"source": [
"#Calculate the estimated pressure rise in the first stage of mutisatge centrifugal compressor\n",
"\n",
"# variables\n",
"rho=0.075 #lbm/ft**3\n",
"omega=1047. #rad/sec\n",
"d=2. #ft\n",
"\n",
"# calculation\n",
"dP=(1/2.0)*(rho)*(omega*d/2.0)**2/32.2/144.0 #psia\n",
"#1 lbf.s**2 = 32.2 lbm into feed\n",
"#1 ft = 144 in**2\n",
"\n",
"# result\n",
"print \"the estimated pressure rise in the first stage of mutisatge centrifugal compressor is %f psia\"%dP"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 10.8 page no : 383"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The efficiency of the compressor is 0.749243\n",
"dT_real = 190 K\n",
"dT_isentropic = 142 K\n"
]
}
],
"source": [
"#Calculate the efficiency of a compressor and the change respective change in temperature\n",
"\n",
"# variables\n",
"m=100.0 #Kg/hr\n",
"M=29. #gm/mol\n",
"gama=1.4 #dimentionless\n",
"R=8.314 #J/mol/K\n",
"T=293.15 #K\n",
"ratio_P=4. #dimentionless\n",
"\n",
"# calculations\n",
"Po=(m/M)*R*T*(gama/(gama-1))*((ratio_P)**((gama-1)/gama)-1)/3600.0 #kW\n",
"P_real=5.3 #kW\n",
"eta=Po/P_real #dimentionless\n",
"\n",
"# result\n",
"print \"The efficiency of the compressor is %f\"%eta\n",
"Cp=29.1 #J/mol/K\n",
"dT_real=P_real*(M/m)*3600/Cp #K\n",
"print \"dT_real = %d K\"%dT_real\n",
"dT_isentropic=Po*(M/m)*3600/Cp #K\n",
"print \"dT_isentropic = %d K\"%dT_isentropic"
]
}
],
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"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
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},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
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|
