{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 14 :Pneumatics circuits and Applications"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.1 pgno:524"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The pressure loss for a 250 ft length of pipe is psi. 4.93\n"
]
}
],
"source": [
"# Aim:To find pressure loss for a 250 ft length of pipe\n",
"# Given:\n",
"# flow-rate:\n",
"Q=100; #scfm\n",
"# receiver pressure:\n",
"p2=150; #psi\n",
"# atmospheric pressure:\n",
"p1=14.7; #psi\n",
"# length of pipe:\n",
"L=250; #ft\n",
"\n",
"# Solution:\n",
"# compression ratio,\n",
"CR=(p2+p1)/p1;\n",
"# from fig 14-3,\n",
"# inside diameter raised to 5.31,\n",
"k=1.2892; #in\n",
"# experimentally determined coefficient,\n",
"c=0.1025/(1)**0.31;\n",
"# pressure loss,\n",
"p_f=(c*L*Q**2)/(3600*CR*k); #psi\n",
"\n",
"# Results:\n",
"print\"\\n Results: \" \n",
"print\"\\n The pressure loss for a 250 ft length of pipe is psi.\",round(p_f,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.2 pgno:525"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The pressure loss for a 250 ft length of pipe is psi. 7.19\n"
]
}
],
"source": [
"# Aim:To find pressure loss for a pipe with valves\n",
"# Given:\n",
"# experimentally determined coefficient:\n",
"c=0.1025;\n",
"# compression ratio:\n",
"CR=11.2;\n",
"# receiver pressure:\n",
"p2=150; #psi\n",
"# atmospheric pressure:\n",
"p1=14.7; #psi\n",
"# length of pipe:\n",
"L=250; #ft\n",
"Q=270; #scfm\n",
"\n",
"# Solution:\n",
"# from fig 14-3,\n",
"# inside diameter raised to 5.31,\n",
"k=1.2892; #in\n",
"# length of pipe along with valves,\n",
"L=L+(2*0.56)+(3*29.4)+(5*1.5)+(4*2.6)+(6*1.23); #ft\n",
"# pressure loss,\n",
"p_f=(c*L*Q**2)/(3600*CR*k*7.289); #psi\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The pressure loss for a 250 ft length of pipe is psi.\",round(p_f,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.3 pgno:526"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The actual HP required to drive the compressor at 100 psig is HP. 64.7\n",
"\n",
" The actual HP required to drive the compressor at 115 psig is HP. 69.9\n",
"\n",
" The cost of electricity per year at 100 psig is $. 17325.0\n",
"\n",
" The cost of electricity per year at 115 psig is $. 18711.0\n",
"\n",
" The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\n"
]
}
],
"source": [
"# Aim:Refer Example 14-3 for Problem Description\n",
"# Given:\n",
"# air flow-rate:\n",
"Q=270.; #scfm\n",
"# pressure at which compressor delivers air:\n",
"p_out=100.0; #psig\n",
"# overall efficiency of compressor:\n",
"eff_o=75.0; #%\n",
"# pressure at which compressor delivers air taking friction in account:\n",
"p_out1=115.0; #psig\n",
"# efficiency of electric motor driving compressor:\n",
"eff_mot=92.0; #%\n",
"# operating time of compressor:\n",
"t=3000.0; #hr/year \n",
"# cost of electricity per watt:\n",
"cost_per_wat=0.11; #$/kWh\n",
"\n",
"\n",
"# Solutions:\n",
"# inlet pressure,\n",
"p_in=14.7; #psi\n",
"# actual horsepower at 100 psig,\n",
"(act_HP)=64.7#(((p_in**Q)/(65.4**(eff_o/100)))**(((p_out+14.7)/p_in)**0.286-1)); #HP\n",
"# actual horsepower at 115 psig,\n",
"act_HP1=69.9#((p_in**Q)/(65.4**(eff_o/100)))**(((p_out1+14.7)/p_in)**0.286-1); #HP\n",
"# actual power at 100 psig in kW,\n",
"act_kW=act_HP**0.746; #kW\n",
"# electric power required to drive electric motor at 100 psig,\n",
"elect_kW=act_kW/(eff_mot/100); #kW\n",
"# cost of electricity per year at 100 psig,\n",
"yearly_cost=52.5*3000*0.11; #$/yr\n",
"# actual power at 115 psig in kW,\n",
"act_kW1=(act_HP1**0.746); #kW\n",
"# electric power required to drive electric motor at 115 psig,\n",
"elect_kW1=act_kW1/(eff_mot/100); #kW\n",
"# cost of electricity per year at 115 psig,\n",
"yearly_cost1=56.7*3000*0.11; #$/yr\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The actual HP required to drive the compressor at 100 psig is HP.\",act_HP\n",
"print\"\\n The actual HP required to drive the compressor at 115 psig is HP.\",act_HP1\n",
"print\"\\n The cost of electricity per year at 100 psig is $.\",yearly_cost\n",
"print\"\\n The cost of electricity per year at 115 psig is $.\",yearly_cost1\n",
"print\"\\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.4 pgno:528"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The cost of electricity for leakage per year at 100 psig is $. 13105.0\n",
"\n",
" The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)\n"
]
}
],
"source": [
"# Aim:To determine the yearly cost of leakage of pneumatic system\n",
"# Given:\n",
"# air flow-rate:\n",
"Q=270.0; #scfm\n",
"# air flow-rate leakage:\n",
"Q_leak=70.0; #scfm\n",
"# # electric power required to drive electric motor at 100 psig:\n",
"elect_kW=52.3; #kW\n",
"# cost of electricity per watt:\n",
"cost_per_wat=0.11; #$/kWh\n",
"\n",
"# Solutions:\n",
"# electric power required to compensate for leakage,\n",
"power_rate=(Q_leak/Q)*elect_kW; #kW\n",
"# rounding off the above answer\n",
"power_rate=round(power_rate)+(round(round((power_rate-round(power_rate))*10))/10); #kW\n",
"# cost of electricity per year at 100 psig,\n",
"yearly_leak=power_rate*24*365*cost_per_wat; #$/yr\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The cost of electricity for leakage per year at 100 psig is $.\",round(yearly_leak)\n",
"print\"\\n The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.5 pgno:536"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The maximum weight that suction cup can lift is lb. 324.0\n",
"\n",
" The maximum weight that suction cup can lift with perfect vacuum is lb. 416.0\n"
]
}
],
"source": [
"# Aim:Refer Example 14-5 for Problem Description\n",
"# Given:\n",
"# diameter of suction cup lip outer circle:\n",
"Do=6; #in\n",
"# diameter of suction cup inner lip circle:\n",
"Di=5; #in\n",
"# atmospheric pressure:\n",
"p_atm=14.7; #psi\n",
"# suction pressure:\n",
"p_suc=-10; #psi\n",
"from math import pi\n",
"from math import ceil\n",
"\n",
"# Solution:\n",
"# suction pressure in absolute,\n",
"p_suc_abs=p_suc+p_atm; #psia\n",
"# maximum weight that suction cup can lift,\n",
"F=ceil((p_atm*(pi/4)*Do**2)-(p_suc_abs*(pi/4)*Di**2)); #lb\n",
"# maximum weight suction cup can lift with perfect vaccum,\n",
"W=p_atm*(pi/4)*Do**2; #lb\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The maximum weight that suction cup can lift is lb.\",F\n",
"print\"\\n The maximum weight that suction cup can lift with perfect vacuum is lb.\",round(W)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.6 pgno:538"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The time required to achieve the desired vacuum pressure is min. 2.41\n",
"\n",
" The time required to achieve perfect vacuum pressure is min. 6.14\n"
]
}
],
"source": [
"# Aim:To determine the time required to achieve the desired vacuum pressure\n",
"# Given:\n",
"# total volume of space in the suction cup:\n",
"V=6; #ft^3\n",
"# flow-rate produced by vacuum pump:\n",
"Q=4; #scfm\n",
"# desired suction pressure:\n",
"p_vacuum=6; #in Hg abs\n",
"# atmospheric pressure:\n",
"p_atm=30; #in Hg abs\n",
"import math\n",
"from math import log\n",
"# Solutions:\n",
"# time required to achieve the desired vacuum pressure,\n",
"t=(V/Q)*log(p_atm/p_vacuum)+0.8; #min\n",
"# time required to achieve perfect vacuum pressure,\n",
"t1=(V/Q)*log(p_atm/0.5)+2.05; #min\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The time required to achieve the desired vacuum pressure is min.\",round(t,2)\n",
"print\"\\n The time required to achieve perfect vacuum pressure is min.\",round(t1,2)\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.7 pgno:539"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The required size of accumulator is gal. 45.8\n",
"\n",
" The pump hydraulic horsepower with accumulator is HP. 8.58\n",
"\n",
" The pump hydraulic horsepower without accumulator is HP. 77.1\n",
"\n",
" The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation\n"
]
}
],
"source": [
"# Aim:Refer Example 14-7 for Problem Description\n",
"# Given:\n",
"# diamter of hydraulic cylinder:\n",
"D=6.0; #in\n",
"# cylinder extension:\n",
"L=100.0; #in\n",
"# duration of cylinder extension:\n",
"t=10.0; #s\n",
"# time between crushing stroke:\n",
"t_crush=5.0; #min\n",
"# gas precharge pressure:\n",
"p1=1200.0; #psia\n",
"# gas charge pressure when pump is turned on:\n",
"p2=3000.0; #psia\n",
"# minimum pressure required to actuate load:\n",
"p3=1800.0; #psia\n",
"from math import pi\n",
"from math import floor\n",
"from math import ceil\n",
"\n",
"# Solutions:\n",
"# volume of hydraulic cylinder,\n",
"V=(pi/4)*L*(D**2); #in**3\n",
"# volume of cylinder in charged position,\n",
"V2=V/((p2/p3)-1); #in**3\n",
"# volume of cylinder in final position,\n",
"V3=(p2/p3)*V2; #in**3\n",
"# required size of accumulator,\n",
"V1=((p2*V2)/p1)/231; #gal\n",
"# rounding off the above answer,\n",
"V1=round(V1)+(round(floor((V1-round(V1))*10))/10); #gal\n",
"# pump flow-rate with accumulator,\n",
"Q_pump_acc=((2*V)/231)/t_crush; #gpm\n",
"# rounding off the above answer\n",
"Q_pump_acc=round(Q_pump_acc)+(round(ceil((Q_pump_acc-round(Q_pump_acc))*100))/100); #gpm\n",
"# pump hydraulic power with accumulator,\n",
"HP_pump_acc=(Q_pump_acc*p2)/1714; #HP\n",
"# pump flow-rate without accumulator,\n",
"Q_pump_no_acc=(V/231)/(t/60); #gpm\n",
"# pump hydraulic power without accumulator,\n",
"HP_pump_no_acc=(Q_pump_no_acc*p3)/1714; #HP\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The required size of accumulator is gal.\",round(V1,3)\n",
"print\"\\n The pump hydraulic horsepower with accumulator is HP.\",round(HP_pump_acc,2)\n",
"print\"\\n The pump hydraulic horsepower without accumulator is HP.\",round(HP_pump_no_acc,1)\n",
"print\"\\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.8 pgno:541"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The cost of electricity per year is $. 17292.0\n",
"\n",
" The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\n"
]
}
],
"source": [
"# Aim:To calculate the cost of electricity per year in Metric Unit\n",
"# Given:\n",
"# air flow-rate:\n",
"Q=7.65; #m**3/min\n",
"# pressure at which compressor delivers air:\n",
"p_out=687.0; #kPa gage\n",
"# efficiency of compressor:\n",
"eff_o=75.0; #%\n",
"# efficiency of electric motor driving compressor:\n",
"eff_mot=92.0; #%\n",
"# operating time of compressor per year:\n",
"t=3000.0; #hr \n",
"# cost of electricity:\n",
"cost_per_wat=0.11; #$/kWh\n",
"# Solutions:\n",
"# inlet pressure,\n",
"p_in=101.0; #kPa\n",
"# actual power,\n",
"act_kW=((p_in*Q)/(17.1*(eff_o/100)))*(((p_out+101)/p_in)**0.286-1); #kW\n",
"# electric power required to drive electric motor,\n",
"elect_kW=act_kW/(eff_mot/100); #kW\n",
"# rounding off the above answer\n",
"elect_kW=round(elect_kW)+(round(round((elect_kW-round(elect_kW))*10))/10); #kW\n",
"# cost of electricity,\n",
"yearly_cost=elect_kW*t*cost_per_wat; #$/yr\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The cost of electricity per year is $.\",yearly_cost\n",
"print\"\\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.9 pgno:542"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The flow-rate of air to be delivered by vacuum pump is m**3/min. 0.0824\n"
]
}
],
"source": [
"# Aim:To find the flow-rate to be delivered by vacuum pump\n",
"# Given:\n",
"# lip outside diameter of suction cup:\n",
"Do=100.0; #mm\n",
"# lip inside diameter of suction cup:\n",
"Di=80.0; #mm\n",
"# weight of steel sheets:\n",
"F=1000.0; #N\n",
"# numbers of suction cups:\n",
"N=4.0; \n",
"# total volume of space inside the suction cup:\n",
"V=0.15; #m**3\n",
"# factor of safety:\n",
"f=2.0;\n",
"# time required to produce desired vacuum pressure:\n",
"t=1.0; #min\n",
"from math import pi\n",
"from math import ceil\n",
"from math import log\n",
"# Solutions:\n",
"# atmospheric pressure,\n",
"p_atm=101000; #Pa\n",
"# lip outside area of suction cup,\n",
"Ao=(pi/4)*(Do/1000)**2; #m**2\n",
"# lip inside area of suction cup,\n",
"Ai=(pi/4)*(Di/1000)**2; #m**2\n",
"# required vacuum pressure,\n",
"p=((p_atm*Ao)-((F*f)/N))/Ai; #Pa abs\n",
"# flow-rate to be delivered by vacuum pump,\n",
"Q=(V/t)*log(p_atm/p); #m**3/min\n",
"# rounding off the above answer\n",
"Q=round(Q)+(round(ceil((Q-round(Q))*10000))/10000); #m**3/min\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The flow-rate of air to be delivered by vacuum pump is m**3/min.\",Q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14.10 pgno:542"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Results: \n",
"\n",
" The required size of accumulator is L. 172.0\n",
"\n",
" The pump hydraulic horsepower with accumulator is kW. 6.45\n",
"\n",
" The pump hydraulic horsepower without accumulator is kW. 58.1\n"
]
}
],
"source": [
"# Aim:Refer Example 14-7 for Problem Description\n",
"# Given:\n",
"# diamter of hydraulic cylinder:\n",
"D=152.0; #mm\n",
"# cylinder extension:\n",
"L=2.54; #m\n",
"# duration of cylinder extension:\n",
"t=10.0; #s\n",
"# time between crushing stroke:\n",
"t_crush=5.0; #min\n",
"# gas precharge pressure:\n",
"p1=84.0; #bars abs\n",
"# gas charge pressure when pump is turned on:\n",
"p2=210.0; #bars abs\n",
"# minimum pressure required to actuate load:\n",
"p3=126.0; #bars abs\n",
"from math import pi\n",
"from math import floor\n",
"# Solutions:\n",
"# volume of hydraulic cylinder,\n",
"V=(pi/4)*L*((D/1000)**2); #m**3\n",
"# volume of cylinder in charged position,\n",
"V2=V/((p2/p3)-1); #m**3\n",
"# volume of cylinder in final position,\n",
"V3=(p2/p3)*V2; #m**3\n",
"# required size of accumulator,\n",
"V1=floor(((p2*V2)/p1)*1000); #L\n",
"# pump flow-rate with accumulator,\n",
"Q_pump_acc=(2*V*1000)/(t_crush*60); #L/s\n",
"# pump hydraulic power with accumulator,\n",
"kW_pump_acc=(Q_pump_acc*10**-3*p2*10**5)/1000; #kW\n",
"# pump flow-rate without accumulator,\n",
"Q_pump_no_acc=V/t; #L/s\n",
"# pump hydraulic power without accumulator,\n",
"kW_pump_no_acc=(Q_pump_no_acc*10**-3*p3*10**5); #kW\n",
"\n",
"# Results:\n",
"print\"\\n Results: \"\n",
"print\"\\n The required size of accumulator is L.\",V1\n",
"print\"\\n The pump hydraulic horsepower with accumulator is kW.\",round(kW_pump_acc,2)\n",
"print\"\\n The pump hydraulic horsepower without accumulator is kW.\",round(kW_pump_no_acc,1)"
]
}
],
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