# Chapter No 14 :Troubleshooting Hydraulic systems

### Example 14.1 Page No 386

In [1]:
import math

#initialisation of variables
p=1800      #lbf/in^2 pressure
Vd=1.5      #in^3/rev displacement

#CALCULATIONS
T=0.013*p*Vd

#RESULTS
print('The breakway torque = %.2f lbf-ft' %T)

The breakway torque = 35.10 percent


### Example 14.2, Page No 386

In [2]:
import math

#initialisation of variables
T=80        #lbf-ft torque
Vd=3.5      #in^3/rev displacement

#CALCULATIONS
P=(T/(0.013*Vd))*(1/0.75)

#RESULTS
print('The min pressure = %.2f lbf/in^2' %P)

The min pressure = 2344.32 lbf/in^2


### Example 14.3, Page No 389

In [4]:
import math

#initialisation of variables
N=1200        #rev/min theotrical power
Vd=0.75      #in^3/rev displacement
Vd2=0.57
#CALCULATIONS
# part a
Q=(Vd*N)/231
# part b
Qa=Q-0.50
N2=(231*Qa)/Vd2
ev=(Qa/Q)*100
#RESULTS
print('The flow from the pump = %.2f lbf/in^2' %Qa)
print('The actual speed of the motor = %.2f rev/min' %N2)
print('The Volumetric efficiency of the system = %.2f percent' %ev)

The flow from the pump = 3.40 lbf/in^2
The actual speed of the motor = 1376.32 rev/min
The Volumetric efficiency of the system = 87.17 percent


### Example 14.4, Page No 390

In [4]:
#initialisation of variables
Q=3.9        #gal/min flow rate
Sg=0.87      # density
Ac=4.9       #in^2 area

#CALCULATIONS
Qa=0.872*Q
Vrod=Qa/(0.26*Ac)

#RESULTS
print('The Velocity = %.2f in/sec' %Vrod)

The Velocity = 2.67 in/sec


### Example 14.5, Page No 390

In [5]:
#initialisation of variables
Q=10.0        #gal/min flow rate
P=1500        #lbf/in^2 pressure

#CALCULATIONS
Fhp=((P*Q)/1714)*0.25
A=Fhp/(0.001*75)

#RESULTS
print('The area is = %.2f ft^2' %A)

The area is = 29.17 ft^2


### Example 14.6, Page No 391

In [6]:
#initialisation of variables
A=15.0         #ft^2 area
HP=0.25*5     # power

#CALCULATIONS
T=(HP/(0.001*A))+75

#RESULTS
print('The temparature is = %.2f F' %T)

The temparature is = 158.33 F


### Example 14.7, Page No 392

In [7]:
#initialisation of variables
A=100        #ft^2 area
T=40         #F  temperature

#CALCULATIONS
q=2.545*A*T

#RESULTS
print('The heat that should be dissipated is = %.2f Btu/hr' %q)

The heat that should be dissipated is = 10180.00 Btu/hr