Chapter 8 :Hydraulic Valves

Example 8.1 pgno:293

In [1]:
# Aim:Refer Example 8-1 for Problem Description 
# Given:
# area of relief valve:
A=0.75; #in**2
# spring constant:
k=2500.0; #lb/in
# initial compressed length of spring:
S=0.20; #in
# poppet displacement to pass full pump flow:
L=0.10; #in






# Solution:
# spring force excerted on poppet when it is fully closed,
F=k*S; #lb
# Cracking pressure,
p_crack=F/A; #psi
# spring force when poppet moves 0.10 in from its fully closed position,
F_new=k*(L+S); #lb
# Full pump flow pressure,
p_ful_pump_flow=F_new/A; #psi
  
# Results:
print"\n  Results:  "   
print"\n The Cracking pressure is  psi.",round(p_crack)
print"\n The Full pump flow pressure  is  psi.",p_ful_pump_flow

  Results:  

 The Cracking pressure is  psi. 667.0

 The Full pump flow pressure  is  psi. 1000.0

Example 8.2 pgno:299

In [2]:
# Aim:To compute horsepower across the pressure relief valve
# Given:
# pressure relief valve setting:
p=1000.0; #psi
# pump flow to the tank:
Q=20.0; #gpm



# Solution:
# Horsepower across the valve,
HP=((p*Q)/1714); #HP
 
# Results:
print"\n  Results:  "  
print"\n The Horsepower across the pressure relief valve is  HP.",round(HP,1)

  Results:  

 The Horsepower across the pressure relief valve is  HP. 11.7

Example 8.3 pgno:299

In [3]:
# Aim:To compute horsepower across the unloading valve
# Given:
# pump pressure during unloading:
p=25.0; #psi
# pump flow to the tank:
Q=20.0; #gpm




# Solution:
# Horsepower across the valve,
HP=((p*Q)/1714); #HP
 
# Results:
print"\n  Results:  "   
print"\n The Horsepower across the unloading valve is  HP.",round(HP,2)

  Results:  

 The Horsepower across the unloading valve is  HP. 0.29

Example 8.4 pgno:302

In [4]:
# Aim:To find flow-rate through given orifice
# Given:
# pressure drop across orifice:
del_p=100.0; #psi
# orifice diameter:
D=1.0; #in
# specific gravity of oil:
SG_oil=0.9;
# flow coefficient for sharp edge orifice:
C=0.80;
import math 
from math import pi

# Solution:
# flow-rate through orifice,
Q=38.1*C*((pi*(D**2))/4)*(del_p/SG_oil)**0.5; #gpm

# Results:
print"\n  Results:  "  
print"\n The flow-rate through orifice is gpm.",round(Q)

  Results:  

 The flow-rate through orifice is gpm. 252.0

Example 8.5 pgno:304

In [5]:
# Aim:To determine the capacity coefficient of flow control valve 
# Given:
# pressure drop across flow control valve:
del_p=100.0; #psi
del_p1=687.0; #kPa
# flow-rate across valve:
Q=25.0; #gpm
Q1=94.8; #Lpm
# specific gravity of oil:
SG_oil=0.9; 





# Solution:
# capacity coefficient in English Units,
Cv=Q/((del_p/SG_oil)**0.5); #gpm/sqrt(psi)
# capacity coefficient in Metric Units,
Cv1=Q1/((del_p1/SG_oil)**0.5); #Lpm/sqrt(kPA)

# Results:
print"\n  Results:  "   
print"\n The capacity coefficient in English unit is  gpm/sqrt(psi).",round(Cv,2)
print"\n The capacity coefficient in Metric unit is  Lpm/sqrt(kPa).",round(Cv1,2)

  Results:  

 The capacity coefficient in English unit is  gpm/sqrt(psi). 2.37

 The capacity coefficient in Metric unit is  Lpm/sqrt(kPa). 3.43

Example 8.6 pgno:304

In [6]:
# Aim:To determine the capacity coefficient of needle valve 
# Given:
# Desired cylinder speed:
v2=10.0; #in/s
# Cylinder piston area:
A1=3.14; #in^2
# Cylinder rod area:
Ar=0.79; #in^2
# Cylinder load:
F_load=1000.0; #lb
# Specific gravity of oil:
SG_oil=0.9;
# Pressure relief valve setting:
p1=500.0; #psi





# Solution:
# annular area of cylinder,
A2=A1-Ar; #in^2
# back pressure in the rod end,
p2=((p1*A1)-F_load)/A2; #psi
# flow rate through needle valve based on desired cylinder speed,
Q=(A2*v2*60)/231; #gpm
# capacity coefficient of needle valve,
Cv=Q/((p2/SG_oil)**0.5); #gpm/sqrt(psi)

# Results:
print"\n  Results:  "
print"\n The capacity coefficient of needle valve is  gpm/sqrt(psi).",round(Cv,2)

  Results:  

 The capacity coefficient of needle valve is  gpm/sqrt(psi). 0.37
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