# Chapter 2:Physical Properties of Hydraulic Fluids¶

## Example 2.1 pgno:42¶

In :
# Aim:To Find Weight of Body
# Given:
# Mass of the Body:
m=4; #slugs

# Solutions:
# we know acceleration due to gravity,
g=32.2; #ft/s**2
W=(m*g);

# Results:
print  "the weight is",round(W)

the weight is 129.0


## Example 2.2 pgno:43¶

In :
# Aim:To find the specific weight of a body
# Given:
# Weigth of the Body:
W=129; #lb
# Volume of the Body:
V=1.8; #ft**3

# Solution:
# we know specific weight,
# gamma=(Weigth of the Body/Volume of the Body)
gamma1=(W/V); #lb/ft^3
# rounding off the above answer
#gamma1=round(gamma1)+(round((gamma1-round(gamma1))*10)/10); #lb/ft^3

# Results:
print " Results:  "
print " The specific weight of Body is lb/ft**3.",round(gamma1,1)

 Results:
The specific weight of Body is lb/ft**3. 71.7


## Example 2.3 pgno:44¶

In :
# Aim:To find the specific gravity of air at 68 degF
# Given:
# specific weight of air at 68 degF:
gamma_air=0.0752; #lb/ft**3

# Solution:
# we know,
# specific gravity of air=(specific weight of air/specific weight of water)
# also we know,specific weight of water at 68 degF,
gamma_water=62.4; #lb/ft**3
SG_air=gamma_air/gamma_water;

# Results:
print  "Results:  "
print  "The specific gravity of air ",round(SG_air,5)

Results:
The specific gravity of air  0.00121


## Example 2.4 pgno:45¶

In :
# Aim:To find Density of body of Example 2-1 and 2-2
# Given:
# mass of the Body:
m=4; #slugs
# Volume of the Body:
V=1.8; #ft**3

# Solution:
# we know density,
# rho1=(mass of the Body/Volume of the Body)
rho1=(m/V); #slugs/ft**3
# also density,rho2=(specific weight/acceleration due to gravity)
g=32.2; #ft/s**2
gamma1=71.6; #lb/ft**3
rho2=(gamma1/g); #slugs/ft**3

# Results:
print "  Results:  "
print  "The Density of Body is  slugs/ft**3.",round(rho1,2)
print " The Density of Body is  slugs/ft**3.",round(rho2,2)

  Results:
The Density of Body is  slugs/ft**3. 2.22
The Density of Body is  slugs/ft**3. 2.22


## Example 2.5 pgno:48¶

In :
# Aim:To find pressure on the skin diver
# Given:
# Depth of Water Body:
H=60; #ft

# Solution:
# specific Weight of water,
gamma1=0.0361; #lb/in**3
# Conversion:
# 1 feet = 12 inches
# 1 lb/in**2 = 1 psi
# we know pressure,
# p=(specific weight of liquid * liquid column height)
p=(gamma1*H*12); #psi

# Results:
print "  Results:  "
print " The pressure on skin diver is  psi.",round(p)

  Results:
The pressure on skin diver is  psi. 26.0


## Example 2.6 pgno:50¶

In :
# Aim:To find tube height of a Barometer
# Given:
# liquid used is Water instead of Mercury.

# Solution:
# specific Weight of water,
gamma1=0.0361; #lb/in**3
# We also knows Atmospheric Pressure,
p=14.7; #psi
# Conversion:
# 1 feet = 12 inches
# 1 lb/in**2 = 1 psi
# we know pressure,
# p=(specific weight of liquid * liquid column height)
# Therefore,
H=(p/gamma1); #in
# He=Height in Feet.
He=H*0.083; #ft

# Results:
print " Results:  "
print " The Height of water column is  ft.",round(He,0)

 Results:
The Height of water column is  ft. 34.0


## Example 2.7 pgno:51¶

In :
# Aim:To convert given pressure into absolute pressure
# Given:
# Gage Pressure:
Pg=-5; #psi

# Solution:
# Atmospheric Pressure,
Po=14.7; #psi
# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure
Pa=Pg+Po;

# Results:
print "  Results:  "
print " The Absolute Pressure is  psi.",Pa

  Results:
The Absolute Pressure is  psi. 9.7


## example 2.8 pgno:51¶

In :
# Aim:To find absolute pressure on skin diver of Example 2-5
# Given:
# Gage Pressure:
Pg=26; #psi

# Solution:
# Atmospheric Pressure,
Po=14.7; #psi
# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure
Pa=Pg+Po; #psi

# Results:
print "  Results:  "
print "The Absolute Pressure is  psi.",Pa

  Results:
The Absolute Pressure is  psi. 40.7


## Example 2.9 pgno:56¶

In :
# Aim:To Determine specific weights in N/m**3
# Given:
# specific weight:
gamma1=56; #lb/ft**3

# Solution:
# We know,
# 1 N/m**3 = 157 lb/ft**3
gamma2=157*gamma1; #N/m**3

# Results:
print "  Results:  "
print "The specific weights is  N/m**3.",gamma2
print " The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation"

  Results:
The specific weights is  N/m**3. 8792
The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation


## Example 2.10 pgno:56¶

In :
 # Aim:To find Temperature at which Fahrenheit and Celsius values are equal
# Given:
# T(degF) = T(degC)  #Eqn - 1

# Solution:
# We know that,
# T(degF)=((1.8*T(degC))+32)  #Eqn - 2
# From Eqn 1 and 2
# ((1.8*T(degC))+32)= T(degC)
# (1-1.8)*T(degC)=32
# -0.8*T(degC)=32
TdegC=-32/0.8;

# Results:
print " Results:  "
print " The temp at which Fahrenheit and Celsius values are equal is  deg.",TdegC

 Results:
The temp at which Fahrenheit and Celsius values are equal is  deg. -40.0


## Example 2.11 pgno:57¶

In :
# Aim:To find change in volume of the oil
# Given:
# Volume of original oil:
V=10.0; #in^3
# Initial Pressure:
P1=100.0; #psi
# Final pressure:
P2=2000.0; #psi
# Bulk Modullus:
betaa=250000.0; #psi

# Solution:
# Change in pressure,
delP=P2-P1; #psi
# Change in volume,
delV=-((V*delP)/betaa); #in^3 ,- sign indicates oil is being compressed

# Results:
print " Results:  "
print " The change in volume of oil is  in^3.",delV

 Results:
The change in volume of oil is  in^3. -0.076


## Example2.12 pgno:62¶

In :
# Aim:To find viscosity of oil in centistokes and centipoise
# Given:
# viscosity of oil:
nu=230; #SUS at
t=150; #degF.
# specific gravity of oil:
gamma1=0.9;
# Solution:
# kinematic viscosity of oil in centistokes,
nu_cs=((0.220*nu)-(135/230)); #centistokes
# absolute viscosity of oil in centipoise,
mu_cp=(gamma1*nu_cs); #centipoise
from math import floor
# Results:
print "  Results:  "
print " The viscosity of oil in centistokes is  cS.",floor(nu_cs)
print " The viscosity of oil in centipoise is  cP.",floor(mu_cp)

  Results:
The viscosity of oil in centistokes is  cS. 50.0
The viscosity of oil in centipoise is  cP. 45.0


## Example 2.13 pgno:62¶

In :
# Aim:To find kinematic and absolute viscosity of oil in cS and cP respectively
# Given:
# Density of oil:
Den=0.89; #g/cm^3
# Time flow:
t=250; #s
# Calibration constant:
cc=0.100;

# Solution:
# kinematic viscosity of oil in centistokes,
nu_cs=(t*cc); #centistokes
# absolute viscosity of oil in centipoise,
SG=Den;
mu_cp=(SG*nu_cs); #centipoise
# rounding off the above answer
mu_cp=round(mu_cp)+(round(round((mu_cp-round(mu_cp))*10))/10); #centipoise

# Results:
print " Results:  "
print " The viscosity of oil in centistokes is  cS.",nu_cs
print " The viscosity of oil in centipoise is cP.",mu_cp

 Results:
The viscosity of oil in centistokes is  cS. 25.0
The viscosity of oil in centipoise is cP. 22.3


## Example 2.14 pgno:65¶

In :
# Aim:To find viscosity of oil at 100 degF in SUS
# Given:
# Viscosity Index:
VI=80;
# viscosity of O-VI oil at 100 degF:
L=400; #SUS
# viscosity of 100-VI oil at 100 degF:
H=150; #SUS

# Solution:
# viscosity of sample oil at 100 degF,
U=L-(((L-H)*VI)/100); #SUS

# Results:
print "  Results:  "
print " The viscosity of sample oil at 100 degF is  SUS.",U

  Results:
The viscosity of sample oil at 100 degF is  SUS. 200


## Example 2.15 pgno:67¶

In :
# Aim:To find pressure on the skin diver in SI units
# Given:
# Depth of Water Body:
H=18.3; #m

# Solution:
# specific Weight of water,
gamma1=9800; #N/m^3
# we know pressure,
# p=(specific weight of liquid * liquid column height)
p=(gamma1*H); #Pa
pK=p/1000; #kPa

# Results:
print "  Results:  "
print " The pressure on skin diver is  kPa.",round(pK)

  Results:
The pressure on skin diver is  kPa. 179.0


## Example 2.16 pgno:67¶

In :
# Aim:To convert given pressure into absolute pressure
# Given:
# Gage Pressure:
Pg=-34000; #Pa

# Solution:
# Atmospheric Pressure,
Po=101000; #Pa
# Absolute Pressure(Pa) =Gage Pressure + Atmospheric Pressure
Pa=Pg+Po; #Pa

# Results:
print "  Results:  "
print " The Absolute Pressure is  Pa.",Pa

  Results:
The Absolute Pressure is  Pa. 67000


## Example 2.17 pgno:67¶

In :
# Aim:To find % change in volume of the oil
# Given:
# Volume of original oil:V=164 #cm^3
# Initial Pressure:
P1=687.0; #kPa
# Final pressure:
P2=13740.0; #kPa
# Bulk Modullus:
betaa=1718.0; #MPa

# Solution:
# Change in pressure,
delP=P2-P1; #kPa
betaa=betaa*1000; #kPA
# % Change in volume,
delV=-(delP/betaa)*100; #% ,- sign indicates oil is being compressed

# Results:
print "  Results:  "
print " The Percentage change in volume of oil is ",round(delV,2)

  Results:
The Percentage change in volume of oil is  -0.76


## Example 2.18pgno:68¶

In :
# Aim:To find absolute viscosity of oil in Ns/m**2 and cP
# Given:
# Area of moving plate surface in contact with oil:
A=1; #m**2
# Force applied to the moving plate:
F=10; #N
# velocity of the moving plate:
v=1; #m/s
# oil film thickness:
y=5; #mm
y=5*0.001; #m

# Solution:
# absolute viscosity of oil,
mu=(F/A)/(v/y); #Ns/m**2
# absolute viscosity of oil in cP,
mu_P=(F*100000*y*100)/(v*100*A*10000); #poise
mu_cP=mu_P*100; #centipoise

# Results:
print " Results:  "
print " The viscosity of oil is  Ns/m**2.",mu
print " The viscosity of oil is  cP.",mu_cP

 Results:
The viscosity of oil is  Ns/m**2. 0.05
The viscosity of oil is  cP. 50.0

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