# Chpater No 2: Basics of Hydraulics

### Example 2.1, Page No 23

In [1]:
#initialisation of variables
W=1000.0        #weight lbf
D=1.25          #distance ft

#CALCULATIONS
Work=W*D
#IN SI unit
W=4448          #weight N
D=0.381         #distance m
Work1=W*D

#RESULTS
print('The work from the system would be = %.2f ft-lbf' %Work)
print('The work from the system in SI unit would be = %.2f m-N' %Work1)

The work from the system would be = 1250.00 ft-lbf
The work from the system in SI unit would be = 1694.69 m-N


### Example 2.2, Page No 24

In [2]:
#initialisation of variables
F1=100.0        #Force lbf
F2=2000.0       #Force lbf
L2=12            #in

#CALCULATIONS
L11=L2*(F2/F1)
#IN SI unit
F1=444.8        #Force N
F2=8896.0       #Force N
L2=0.3048       #Force m
L12=L2*(F2/F1)

#RESULTS
print('The pump piston have to travel = %.2f in' %L11)
print('The pump piston have to travel = %.2f m' %L12)

The pump piston have to travel = 240.00 in
The pump piston have to travel = 6.10 m


### Example 2.3, Page No 25

In [3]:
#initialisation of variables
F1=150.0        #lbf Force
A1=0.049        #in^2 area
A2=0.785        #in^2 area

#CALCULATIONS
F21=F1*(A2/A1)
#IN SI unit
F1=667.2        #N(Force)
A1=31.7        #mm^2(Area)
A2=506.7      #mm^2(Area)
F22=F1*(A2/A1)

#RESULTS
print('The weight lifted by a downward force = %.2f lbf' %F21)
print('weight lifted by a downward force in SI unit = %.2f N' %F22)

The weight lifted by a downward force = 2403.06 lbf
weight lifted by a downward force in SI unit = 10664.68 N


### Example 2.4, Page No 27

In [4]:
#initialisation of variables
F =800.0        #lbf Force
D=20            #ft distance
t=20            #sec time
C=550           #ft-lbf/sec

#CALCULATIONS
HP=(F*D)/(t*C)

#RESULTS
print('The Horse powe required is = %.2f HP' %HP)

The Horse powe required is = 1.45 HP


### Example 2.5, Page No 27

In [5]:
#initialisation of variables
V1=2000.0        #watts voltage
V2=746.0        #watts voltage

#CALCULATIONS
HP=V1/V2

#RESULTS
print('The Horse powe required is = %.2f HP' %HP)

The Horse powe required is = 2.68 HP


### Example 2.6, Page No 28

In [6]:
#initialisation of variables
N=1200              #rev/min speed
A=(3.14*((1.125*1.125)*5))/4 #Cross section of the bore in in^2
L=1 # cylinder stroke (inch)
#CALCULATIONS
V=A*L
Q=V*N/231

#RESULTS
print('The displacement would be = %.2f in^3/rev' %V)
print('The flow rate would be = %.2f gal/min' %Q)

The displacement would be = 4.97 in^3/rev
The flow rate would be = 25.81 gal/min


### Example 2.7, Page No 31

In [14]:
#initialisation of variables
P1=2000.0          #lbf/in^2 pressure
Q1=5              #gal/min flow rate
P3=13793.0        #kPa   pressure
Q3=18.9          #liters/min flow rate

#CALCULATIONS
Fhp1=P1*Q1/1714  #In English units
Fhp2=(P1*4.448*1550*Q1*3.7851*0.001)/44760 #In SI unit
Fhp3=P3*Q3/44760 #with technique 3

#RESULTS
print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp1)
print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp2)
print('Using technique 1 the horsepower transmitted by system would be = %.2f hp' %Fhp3)

Using technique 1 the horsepower transmitted by system would be = 5.83 hp
Using technique 1 the horsepower transmitted by system would be = 5.83 hp
Using technique 1 the horsepower transmitted by system would be = 5.82 hp


### Example 2.8, Page No 35

In [8]:
#initialisation of variables
THP=5.83     # Horse power
pi=3.14     #Pi value
N=500       #rev/min speed

#CALCULATIONS
T=(THP*33000)/(2*pi*N)

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

The torque would be = 61.27 lbf-ft


### Example 2.9, Page No 35

In [17]:
#initialisation of variables
T=2000.0          #lbf-ft    torque
N=20.0            #rev/min   speed
P=2000          #lbf/in^2    pressure

#CALCULATIONS
Q1=(1714*N*T)/(5252*P)
Fhp1=(P1*Q1)/1714
Q2=(44760*N*2712)/(7122*13793)
Fhp2=(13793*Q2)/44760

#RESULTS
print('In English units the flow rate would be = %.2f gal/min' %Q1)
print('In English units the Horsepower would be = %.2f hp' %Fhp1)
print('In SI units the flow rate would be = %.2f lit/min' %Q2)
print('In SI units the Horsepower would be = %.2f hp' %Fhp2)

The flow rate would be = 6.53 gal/min
The Horsepower would be = 7.62 hp
The flow rate would be = 24.71 gal/min
The Horsepower would be = 7.62 hp


### Example 2.10, Page No 36

In [10]:
#initialisation of variables
N=750.0          #rev/min   speed
Q=5             #gal/min    flow rate
P=1000          #lbf/in^2    pressure

#CALCULATIONS
T=(5252*P*Q)/(1714*N)

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

The expected torque would be = 20.43 lbf-ft