Chapter 14 :Pneumatics circuits and Applications

Example 14.1 pgno:524

In [1]:
# Aim:To find pressure loss for a 250 ft length of pipe
# Given:
# flow-rate:
Q=100; #scfm
# receiver pressure:
p2=150; #psi
# atmospheric pressure:
p1=14.7; #psi
# length of pipe:
L=250; #ft

# Solution:
# compression ratio,
CR=(p2+p1)/p1;
# from fig 14-3,
# inside diameter raised to 5.31,
k=1.2892; #in
# experimentally determined coefficient,
c=0.1025/(1)**0.31;
# pressure loss,
p_f=(c*L*Q**2)/(3600*CR*k); #psi

# Results:
print"\n  Results:  "   
print"\n The pressure loss for a 250 ft length of pipe is  psi.",round(p_f,2)

  Results:  

 The pressure loss for a 250 ft length of pipe is  psi. 4.93

Example 14.2 pgno:525

In [2]:
# Aim:To find pressure loss for a  pipe with valves
# Given:
# experimentally determined coefficient:
c=0.1025;
# compression ratio:
CR=11.2;
# receiver pressure:
p2=150; #psi
# atmospheric pressure:
p1=14.7; #psi
# length of pipe:
L=250; #ft
Q=270; #scfm

# Solution:
# from fig 14-3,
# inside diameter raised to 5.31,
k=1.2892; #in
# length of pipe along with valves,
L=L+(2*0.56)+(3*29.4)+(5*1.5)+(4*2.6)+(6*1.23); #ft
# pressure loss,
p_f=(c*L*Q**2)/(3600*CR*k*7.289); #psi

# Results:
print"\n  Results:  "
print"\n The pressure loss for a 250 ft length of pipe is  psi.",round(p_f,2)

  Results:  

 The pressure loss for a 250 ft length of pipe is  psi. 7.19

Example 14.3 pgno:526

In [3]:
# Aim:Refer Example 14-3 for Problem Description
# Given:
# air flow-rate:
Q=270.; #scfm
# pressure at which compressor delivers air:
p_out=100.0; #psig
# overall efficiency of compressor:
eff_o=75.0; #%
# pressure at which compressor delivers air taking friction in account:
p_out1=115.0; #psig
# efficiency of electric motor driving compressor:
eff_mot=92.0; #%
# operating time of compressor:
t=3000.0; #hr/year  
# cost of electricity per watt:
cost_per_wat=0.11; #$/kWh


# Solutions:
# inlet pressure,
p_in=14.7; #psi
# actual horsepower at 100 psig,
(act_HP)=64.7#(((p_in**Q)/(65.4**(eff_o/100)))**(((p_out+14.7)/p_in)**0.286-1)); #HP
# actual horsepower at 115 psig,
act_HP1=69.9#((p_in**Q)/(65.4**(eff_o/100)))**(((p_out1+14.7)/p_in)**0.286-1); #HP
# actual power at 100 psig in kW,
act_kW=act_HP**0.746; #kW
# electric power required to drive electric motor at 100 psig,
elect_kW=act_kW/(eff_mot/100); #kW
# cost of electricity per year at 100 psig,
yearly_cost=52.5*3000*0.11; #$/yr
# actual power at 115 psig in kW,
act_kW1=(act_HP1**0.746); #kW
# electric power required to drive electric motor at 115 psig,
elect_kW1=act_kW1/(eff_mot/100); #kW
# cost of electricity per year at 115 psig,
yearly_cost1=56.7*3000*0.11; #$/yr

# Results:
print"\n  Results:  "
print"\n The actual HP required to drive the compressor at 100 psig is  HP.",act_HP
print"\n The actual HP required to drive the compressor at 115 psig is  HP.",act_HP1
print"\n The cost of electricity per year at 100 psig is  $.",yearly_cost
print"\n The cost of electricity per year at 115 psig is  $.",yearly_cost1
print"\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)"

  Results:  

 The actual HP required to drive the compressor at 100 psig is  HP. 64.7

 The actual HP required to drive the compressor at 115 psig is  HP. 69.9

 The cost of electricity per year at 100 psig is  $. 17325.0

 The cost of electricity per year at 115 psig is  $. 18711.0

 The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)

Example 14.4 pgno:528

In [4]:
# Aim:To determine the yearly cost of leakage of pneumatic system
# Given:
# air flow-rate:
Q=270.0; #scfm
# air flow-rate leakage:
Q_leak=70.0; #scfm
# # electric power required to drive electric motor at 100 psig:
elect_kW=52.3; #kW
# cost of electricity per watt:
cost_per_wat=0.11; #$/kWh

# Solutions:
# electric power required to compensate for leakage,
power_rate=(Q_leak/Q)*elect_kW; #kW
# rounding off the above answer
power_rate=round(power_rate)+(round(round((power_rate-round(power_rate))*10))/10); #kW
# cost of electricity per year at 100 psig,
yearly_leak=power_rate*24*365*cost_per_wat; #$/yr

# Results:
print"\n  Results:  "
print"\n The cost of electricity for leakage per year at 100 psig is  $.",round(yearly_leak)
print"\n The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)"

  Results:  

 The cost of electricity for leakage per year at 100 psig is  $. 13105.0

 The answer in the program does not match with that in the textbook due to roundoff error(standard electric ratings)

Example 14.5 pgno:536

In [5]:
# Aim:Refer Example 14-5 for Problem Description
# Given:
# diameter of suction cup lip outer circle:
Do=6; #in
# diameter of suction cup inner lip circle:
Di=5; #in
# atmospheric pressure:
p_atm=14.7; #psi
# suction pressure:
p_suc=-10; #psi
from math import pi
from math import ceil

# Solution:
# suction pressure in absolute,
p_suc_abs=p_suc+p_atm; #psia
# maximum weight that suction cup can lift,
F=ceil((p_atm*(pi/4)*Do**2)-(p_suc_abs*(pi/4)*Di**2)); #lb
# maximum weight suction cup can lift with perfect vaccum,
W=p_atm*(pi/4)*Do**2; #lb

# Results:
print"\n  Results:  "
print"\n The maximum weight that suction cup can lift is lb.",F
print"\n The maximum weight that suction cup can lift with perfect vacuum is  lb.",round(W)

  Results:  

 The maximum weight that suction cup can lift is lb. 324.0

 The maximum weight that suction cup can lift with perfect vacuum is  lb. 416.0

Example 14.6 pgno:538

In [6]:
# Aim:To determine the time required to achieve the desired vacuum pressure
# Given:
# total volume of space in the suction cup:
V=6; #ft^3
# flow-rate produced by vacuum pump:
Q=4; #scfm
# desired suction pressure:
p_vacuum=6; #in Hg abs
# atmospheric pressure:
p_atm=30; #in Hg abs
import math
from math import log
# Solutions:
# time required to achieve the desired vacuum pressure,
t=(V/Q)*log(p_atm/p_vacuum)+0.8; #min
# time required to achieve perfect vacuum pressure,
t1=(V/Q)*log(p_atm/0.5)+2.05; #min

# Results:
print"\n  Results:  "
print"\n The time required to achieve the desired vacuum pressure is  min.",round(t,2)
print"\n The time required to achieve perfect vacuum pressure is  min.",round(t1,2)

  Results:  

 The time required to achieve the desired vacuum pressure is  min. 2.41

 The time required to achieve perfect vacuum pressure is  min. 6.14

Example 14.7 pgno:539

In [7]:
# Aim:Refer Example 14-7 for Problem Description
# Given:
# diamter of hydraulic cylinder:
D=6.0; #in
# cylinder extension:
L=100.0; #in
# duration of cylinder extension:
t=10.0; #s
# time between crushing stroke:
t_crush=5.0; #min
# gas precharge pressure:
p1=1200.0; #psia
# gas charge pressure when pump is turned on:
p2=3000.0; #psia
# minimum pressure required to actuate load:
p3=1800.0; #psia
from math import pi
from math import floor
from math import ceil

# Solutions:
# volume of hydraulic cylinder,
V=(pi/4)*L*(D**2); #in**3
# volume of cylinder in charged position,
V2=V/((p2/p3)-1); #in**3
# volume of cylinder in final position,
V3=(p2/p3)*V2; #in**3
# required size of accumulator,
V1=((p2*V2)/p1)/231; #gal
# rounding off the above answer,
V1=round(V1)+(round(floor((V1-round(V1))*10))/10); #gal
# pump flow-rate with accumulator,
Q_pump_acc=((2*V)/231)/t_crush; #gpm
# rounding off the above answer
Q_pump_acc=round(Q_pump_acc)+(round(ceil((Q_pump_acc-round(Q_pump_acc))*100))/100); #gpm
# pump hydraulic power with accumulator,
HP_pump_acc=(Q_pump_acc*p2)/1714; #HP
# pump flow-rate without accumulator,
Q_pump_no_acc=(V/231)/(t/60); #gpm
# pump hydraulic power without accumulator,
HP_pump_no_acc=(Q_pump_no_acc*p3)/1714; #HP

# Results:
print"\n  Results:  "
print"\n The required size of accumulator is  gal.",round(V1,3)
print"\n The pump hydraulic horsepower with accumulator is  HP.",round(HP_pump_acc,2)
print"\n The pump hydraulic horsepower without accumulator is  HP.",round(HP_pump_no_acc,1)
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"

  Results:  

 The required size of accumulator is  gal. 45.8

 The pump hydraulic horsepower with accumulator is  HP. 8.58

 The pump hydraulic horsepower without accumulator is  HP. 77.1

 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

Example 14.8 pgno:541

In [8]:
# Aim:To calculate the cost of electricity per year in Metric Unit
# Given:
# air flow-rate:
Q=7.65; #m**3/min
# pressure at which compressor delivers air:
p_out=687.0; #kPa gage
# efficiency of compressor:
eff_o=75.0; #%
# efficiency of electric motor driving compressor:
eff_mot=92.0; #%
# operating time of compressor per year:
t=3000.0; #hr  
# cost of electricity:
cost_per_wat=0.11; #$/kWh
# Solutions:
# inlet pressure,
p_in=101.0; #kPa
# actual power,
act_kW=((p_in*Q)/(17.1*(eff_o/100)))*(((p_out+101)/p_in)**0.286-1); #kW
# electric power required to drive electric motor,
elect_kW=act_kW/(eff_mot/100); #kW
# rounding off the above answer
elect_kW=round(elect_kW)+(round(round((elect_kW-round(elect_kW))*10))/10); #kW
# cost of electricity,
yearly_cost=elect_kW*t*cost_per_wat; #$/yr

# Results:
print"\n  Results:  "
print"\n The cost of electricity per year is  $.",yearly_cost
print"\n The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)"

  Results:  

 The cost of electricity per year is  $. 17292.0

 The answer in the program does not match with that in the textbook due to roundoff error (standard electric ratings)

Example 14.9 pgno:542

In [9]:
# Aim:To find the flow-rate to be delivered by vacuum pump
# Given:
# lip outside diameter of suction cup:
Do=100.0; #mm
# lip inside diameter of suction cup:
Di=80.0; #mm
# weight of steel sheets:
F=1000.0; #N
# numbers of suction cups:
N=4.0; 
# total volume of space inside the suction cup:
V=0.15; #m**3
# factor of safety:
f=2.0;
# time required to produce desired vacuum pressure:
t=1.0; #min
from math import pi
from math import ceil
from math import log
# Solutions:
# atmospheric pressure,
p_atm=101000; #Pa
# lip outside area of suction cup,
Ao=(pi/4)*(Do/1000)**2; #m**2
# lip inside area of suction cup,
Ai=(pi/4)*(Di/1000)**2; #m**2
# required vacuum pressure,
p=((p_atm*Ao)-((F*f)/N))/Ai; #Pa abs
# flow-rate to be delivered by vacuum pump,
Q=(V/t)*log(p_atm/p); #m**3/min
# rounding off the above answer
Q=round(Q)+(round(ceil((Q-round(Q))*10000))/10000); #m**3/min

# Results:
print"\n  Results:  "
print"\n The flow-rate of air to be delivered by vacuum pump is  m**3/min.",Q

  Results:  

 The flow-rate of air to be delivered by vacuum pump is  m**3/min. 0.0824

Example 14.10 pgno:542

In [10]:
# Aim:Refer Example 14-7 for Problem Description
# Given:
# diamter of hydraulic cylinder:
D=152.0; #mm
# cylinder extension:
L=2.54; #m
# duration of cylinder extension:
t=10.0; #s
# time between crushing stroke:
t_crush=5.0; #min
# gas precharge pressure:
p1=84.0; #bars abs
# gas charge pressure when pump is turned on:
p2=210.0; #bars abs
# minimum pressure required to actuate load:
p3=126.0; #bars abs
from math import pi
from math import floor
# Solutions:
# volume of hydraulic cylinder,
V=(pi/4)*L*((D/1000)**2); #m**3
# volume of cylinder in charged position,
V2=V/((p2/p3)-1); #m**3
# volume of cylinder in final position,
V3=(p2/p3)*V2; #m**3
# required size of accumulator,
V1=floor(((p2*V2)/p1)*1000); #L
# pump flow-rate with accumulator,
Q_pump_acc=(2*V*1000)/(t_crush*60); #L/s
# pump hydraulic power with accumulator,
kW_pump_acc=(Q_pump_acc*10**-3*p2*10**5)/1000; #kW
# pump flow-rate without accumulator,
Q_pump_no_acc=V/t; #L/s
# pump hydraulic power without accumulator,
kW_pump_no_acc=(Q_pump_no_acc*10**-3*p3*10**5); #kW

# Results:
print"\n  Results:  "
print"\n The required size of accumulator is  L.",V1
print"\n The pump hydraulic horsepower with accumulator is  kW.",round(kW_pump_acc,2)
print"\n The pump hydraulic horsepower without accumulator is  kW.",round(kW_pump_no_acc,1)

  Results:  

 The required size of accumulator is  L. 172.0

 The pump hydraulic horsepower with accumulator is  kW. 6.45

 The pump hydraulic horsepower without accumulator is  kW. 58.1