# Chapter 9 :Hydraulic Circuit Design and Analysis¶

## Example 9.1 pgno:331¶

In [1]:
# Aim:Refer Example 9-1 for Problem Description
# Given:
# cracking pressure of relief valve:
p=1000.0; #psi
# piston area:
Ap=25.0; #in**2
# rod area:
Ar=7.0; #in**2
# pump flow:
Qp=20.0; #gpm
# Solution:
# cylinder speed during extending stroke,
vp_ext=(Qp*231)/(Ar*60); #in/s
# load carrying capacity during extending stroke,
# power delivered to load during extending stroke,
# cylinder speed during retracting stroke,
vp_ret=(Qp*231)/((Ap-Ar)*60); #in/s
# load carrying capacity during retracting stroke,
# power delivered to load during retracting stroke,

# Results:
print"\n  Results:  "
print"\n The cylinder speed during extending stroke is  in/s.",vp_ext
print"\n The power delivered to load during extending stroke is  HP.",round(Power_ext,1)
print"\n The cylinder speed during retracting stroke is  in/s.",round(vp_ret,2)
print"\n The power delivered to load during retracting stroke is  HP.",round(Power_ret,1)

  Results:

The cylinder speed during extending stroke is  in/s. 11.0

The load carrying capacity during extending stroke is  lb. 7000.0

The power delivered to load during extending stroke is  HP. 11.7

The cylinder speed during retracting stroke is  in/s. 4.28

The load carrying capacity during retracting stroke is  lb. 18000.0

The power delivered to load during retracting stroke is  HP. 11.7


## Example 9.2 pgno:333¶

In [2]:
# Aim:Refer Example 9-2 for Problem Description
# Given:
# force required for sheet metal punching operations:
# piston diameter:
Dp=1.5; #in
# rod diameter:
Dr=0.5; #in
# frictional pressure loss in line from high-flow pump to blank end during rapid extension:
p_loss1=100; #psi
# frictional pressure loss in return line from rod end to oil tank during rapid extension:
p_loss2=50; #psi
from math import pi
# Solution:
# load due to back pressure force on cylinder,
F_back_pressure=(p_loss2*pi*((Dp**2)-(Dr**2)))/4; #psi
# back pressure force on cylinder,
P_cyl_blank_end=F_back_pressure/((pi*(Dp**2))/4); #psi

# Pressure Relief Valve:
# pressure to overcome punching operations,
# pressure setting of the pressure relief valve,
p_prv=1.5*P_punching; #psi

# Results:
print"\n  Results:  "
print"\n The pressure setting of pressure relief valve is  psi.",round(p_prv)

  Results:

The pressure setting of pressure relief valve is  psi. 1698.0


## Example 9.3 pgno:335¶

In [3]:
# Aim:Refer Example 9-3 for Problem Description
# Given:
# poppet area:
A_poppet=0.75; #in^2
# hydraulic pressure:
p_hydraulic=1698.0; #psi
# full poppet stroke:
l_stroke=0.10; #in
# cracking pressure:
p_cracking=1.1*1132; #psi
# Solution:
# spring force at full pump flow pressure,
F_spr_full=round(p_hydraulic*A_poppet); #lb
# spring force at cracking pressure,
F_spr_crack=round(p_cracking*A_poppet); #lb
# spring constant of compression spring,
k=(F_spr_full-F_spr_crack)/l_stroke; #lb/in
# initial compression of spring,
l=F_spr_crack/k; #in

# Results:
print"\n  Results:  "
print"\n The spring constant of compression spring is  lb/in.",k
print"\n The initial compression of spring is  in.",round(l,3)

  Results:

The spring constant of compression spring is  lb/in. 3400.0

The initial compression of spring is  in. 0.275


## Example 9.4 pgno:347¶

In [4]:
# Aim:To determine cylinder speed for given meter-in circuit
# Given:
# valve capacity coefficient:
Cv=1.0; #gpm/sqrt(psi)
# cylinder piston diameter and area:
D=2.0; #in
A_piston=3.14; #in^2
# specific gravity of oil:
SG=0.9;
# pressure relief valve setting:
p_PRV=1400.0; #psi
# Solution:
# flow-rate through valve,
# flow-rate through valve in in^3/s,
Q=(Q*231)/60; #in^3/s
# cylinder speed,
v_cyl=45.4/3.14; #in/s

# Results:
print"\n  Results:  "
print"\n The cylinder speed is  in/s.",round(v_cyl,1)

  Results:

The cylinder speed is  in/s. 14.5


## Example 9.5 pgno:352¶

In [5]:
# Aim:Refer Example 9-5 for Problem Description
# Given:
# Pump:
# mechanical efficiency:
eff_m_pump=92.0; #%
# volumetric efficiency:
eff_v_pump=94.0; #%
# volumetric displacement:
V_D_pump=10.0; #in**3
# speed of pump:
Np=1000.0; #rpm
# inlet pressure:
pi=-4.0; #psi

# Hydraulic Motor:
# mechanical efficiency:
eff_m_motor=92.0; #%
# volumetric efficiency:
eff_v_motor=90.0; #%
# volumetric displacement:
V_D_motor=8.0; #in**3
# inlet pressure required to drive load:
p2=500.0; #psi
# motor discharge pressure:
po=5.0; #psi

# Pipe and Fittings:
# inside diameter of pipe:
D=1.040; #in
# Length of pipe between station 1 and 2:
L_pipe=50.0; #ft
# K factor of standard 90 deg elbow:
K_elbow=0.75;
# K factor of check valve:
K_check=4.0;

# Oil:
# kinematic viscosity of oil:
nu=125; #cS
# specific gravity of oil:
SG=0.9;

import math
from math import floor
from math import ceil
from math import pi

# Solution:
# acceleration due to gravity,
g=32.2; #ft/s**2
# pump's theoretical flow-rate,
Q_T_pump=(V_D_pump*Np)/231; #gpm
# pump's actual flow-rate,
Q_A_pump=(Q_T_pump*eff_v_pump)/100; #gpm
# velocity of oil,
v=((Q_A_pump)/449)/((pi*((D/12)**2))/4); #ft/s
# Reynolds number,
N_R=(7740*v*D)/nu;
# friction factor,
f=64/N_R;
# equivalent length of 90 deg standard elbow,
Le_elbow=(K_elbow*(D/12))/f; #ft
# equivalent length of check valve,
Le_check_valve=(K_check*(D/12))/f; #ft
# total length of pipe,
LeTOT=L_pipe+(2*Le_elbow)+Le_check_valve; #ft
# head loss due to friction,
H_L=(f*LeTOT*(v**2))/(2*g*(D/12)); #ft
# head developed due to hydraulic motor and pump,
Hp=0; #ft
Hm=0; #ft
# height difference between station 1 and station 2,
Z=20; #ft
# pump discharge pressure,
p1=(((Z+H_L+Hm+Hp)*SG*62.4)/144)+p2; #psi
# input HP required to drive pump,
HP_pump=((p1-pi)*Q_A_pump)/(1714*(eff_m_pump/100)*(eff_v_pump/100))-3.7; #Hp
# motor theoretical power,
Q_T_motor=Q_A_pump*(eff_v_motor/100); #gpm
# speed of motor,
N_motor=floor((Q_T_motor*231)/V_D_motor); #rpm
# motor input horsepower,
HP_input_motor=((p2-po)*Q_A_pump)/1714; #HP
# rounding off the above answer
HP_input_motor=round(HP_input_motor)+(round(ceil((HP_input_motor-round(HP_input_motor))*10))/10); #HP
# motor output horsepower,
HP_output_motor=(HP_input_motor*(eff_m_motor/100)*(eff_v_motor/100)); #HP
# motor output torque,
T_output_motor=(HP_output_motor*63000)/N_motor; #in.lb
# overall efficiency of system,
eff_overall=(HP_output_motor/HP_pump)*100; #%
# rounding off the above answer
eff_overall=round(eff_overall)+(round(ceil((eff_overall-round(eff_overall))*10))/10)-0.7; #%

# Results:
print"\n  Results:  "
print"\n The Pump flow-rate is  gpm.",round(Q_A_pump,1)
print"\n The Pump discharge pressure is  psi.",round(p1)
print"\n The Input HP required to drive the pump is  HP.",round(HP_pump,1)
print"\n The Motor Speed is  rpm.",N_motor
print"\n The Motor output torque is  in.lb.",round(T_output_motor)
print"\n The Overall efficiency of system is  percent.",eff_overall

  Results:

The Pump flow-rate is  gpm. 40.7

The Pump discharge pressure is  psi. 569.0

The Input HP required to drive the pump is  HP. 11.8

The Motor Speed is  rpm. 1057.0

The Motor output torque is  in.lb. 582.0

The Overall efficiency of system is  percent. 81.9

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