Chapter 11: Ancillary Hydraulic Devices

Example 11.1 pgno:409

In [1]:
# Aim:To find the discharge flow and pressure 
# Given:
# high inlet flow-rate:
Q_high_inlet=20.0; #gpm
# low inlet pressure:
p_low_inlet=500.0; #psi
# Ratio of piston area to rod area:
Ratio=5/1;




# Solution:
# high discharge pressure,
p_high_discharge=Ratio*p_low_inlet; #psi
# low discharge flow-rate,
Q_low_discharge=Q_high_inlet/Ratio; #gpm

# Results:
print"\n  Results:  " 
print"\n The high discharge pressure is  psi.",p_high_discharge
print"\n The low discharge flow-rate is  gpm.",Q_low_discharge

  Results:  

 The high discharge pressure is  psi. 2500.0

 The low discharge flow-rate is  gpm. 4.0

Example 11.2 pgno:424

In [2]:
# Aim:To find the downstream oil temperature
# Given:
# temperature of oil flowing through pressure relief valve:
T_oil=120.0; #deg F
# pressure of oil flowing through pressure relief valve:
p=1000.0; #psi
# oil flow through pressure relief valve:
Q_gpm=10.0; #gpm
# Solution:
# heat generation rate,
HP=(p*Q_gpm)/1714; #HP
# heat generation rate in Btu/min,
HP_btu=HP*42.4; #Btu/min
# oil flow-rate in lb/min,
Q_lb=7.42*Q_gpm; #lb/min
# temperature increase,
T_increase=HP_btu/(0.42*Q_lb); #deg F
# downward oil temperature,
T_downward=T_oil+T_increase; #deg F

# Results:
print "\n  Results:  " 
print "\n The downstream oil temperature is  deg F.",round(T_downward,1)

  Results:  

 The downstream oil temperature is  deg F. 127.9

Example 11.3 pgno:424

In [3]:
# Aim:To find the downstream oil temperature in SI Unit
# Given:
# temperature of oil flowing through pressure relief valve:
T_oil=50.0; #deg C
# pressure of oil flowing through pressure relief valve:
p=70.0; #bar
# oil flow through pressure relief valve:
Q=0.000632; #m**3/s
# Solution:
# heat generation rate,
kW=((p*10**5)*Q)/1000; #kW
# oil flow-rate,
Q_kg_s=895*Q; #kg/s
# temperature increase,
T_increase=kW/(1.8*Q_kg_s); #deg C
# downward oil temperature,
T_downward=T_oil+T_increase; #deg C

# Results:
print "\n  Results:  "   
print "\n The downstream oil temperature is  deg C.",round(T_downward,1)

  Results:  

 The downstream oil temperature is  deg C. 54.3

Example 11.4 pgno:425

In [4]:
# Aim:To find the rating of heat exchanger required to dissipate generated heat
# Given:
# oil flow-rate:
Q=20.0; #gpm
# operating pressure:
p=1000.0; #psi
# overall efficiency of pump:
eff_overall=85.0; #%
# power lost due to friction:
HP_frict=10.0; #%

# Solution:
# pump power loss,
pump_HP_loss=((1/(eff_overall/100))-1)*((p*Q)/1714); #HP
# PRV average HP loss,
PRV_loss=0.5*((p*Q)/1714); #HP
# line average HP loss,
line_loss=(HP_frict/100)*PRV_loss; #HP
# total average loss,
total_loss=pump_HP_loss+PRV_loss+line_loss; #HP
# heat exchanger rating,
HEx_rating=total_loss*2544; #Btu/hr

# Results:
print"\n  Results:  "   
print"\n The heat exchanger rating is  Btu/hr.",round(HEx_rating)
print"\n The answer in the program does not match with that in the textbook due to roundoff error (standard ratings) in textbook"

  Results:  

 The heat exchanger rating is  Btu/hr. 21565.0

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

Example 11.5 pgno:426

In [5]:
# Aim:To find the rating of heat exchanger required to dissipate generated heat in SI unit
# Given:
# oil flow-rate:
Q=0.00126; #m**3/s
# operating pressure:
p=70.0; #bar
# overall efficiency of pump:
eff_overall=85.0; #%
# power lost due to friction:
HP_frict=10.0; #%

# Solution:
# pump power loss,
pump_loss=((1/(eff_overall/100))-1)*((p*10**5*Q)/1000); #kW
# PRV average HP loss,
PRV_loss=0.5*((p*10**5*Q)/1000); #kW
# line average HP loss,
line_loss=(HP_frict/100)*PRV_loss; #kW
# total average loss,
HEx_rating=pump_loss+PRV_loss+line_loss; #kW

# Results:
print"\n  Results:  "  
print"\n The heat exchanger rating is  kW.",round(HEx_rating,2)

  Results:  

 The heat exchanger rating is  kW. 6.41
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