# Chapter 2: Fundamental Thoughts¶

## Example 2.1¶

In [2]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.1
'''What is the weight of a 1 kg mass at an altitude where the local acceleration of gravity is
9.75 m/s2?'''

#Variable Declaration:
m = 1 		#Mass in kg
g = 9.75 		#Acc.due to gravity in m/s**2

#Calculation:
F = m*g 		#Weight of 1 kg mass in N

#Result
print "Weight of person is:",round(F,2),"N"

Weight of person is: 9.75 N


## Example 2.2¶

In [3]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.2
'''A1m3 container, shown in Fig. 2.9, is filled with 0.12 m3 of granite, 0.15 m3 of sand, and
0.2 m3 of liquid 25◦C water; the rest of the volume, 0.53 m3, is air with a density of 1.15
kg/m3. Find the overall (average) specific volume and density.'''

#Variable Declaration:
Vliq = 0.2 		      #volume of liquid in m**3
dliq = 997 		      #density of liquid in kg/m**3
Vstone = 0.12 		#volume of stone in m**3
Vsand = 0.15 		#volume of sand in m**3
Vair = 0.53 		#voume of air in m**3
dstone = 2750 		#density of stone in kg/m**3
dsand = 1500 		#density of sand in kg/m**3
Vtot = 1 		      #total volume in m**3
dair = 1.1 		      #density of air in kg/m**3

#Calculation:
mliq = Vliq*dliq 		#mass of liquid in kg
mstone = Vstone*dstone #volume of stone in m**3
msand = Vsand*dsand 	#volume of sand in m**3
mair = Vair*dair 		#mass of air
mtot = mair+msand+mliq+mstone 		#total mass in kg
v = Vtot/mtot 		#specific volume in  m**3/kg
d = 1/v 		      #overall density in kg/m**3

#Results:
print "Average specific volume is: ",round(v ,6),"m**3/kg"
print "Overall density is:",round(d),"kg/m**3"

Average specific volume is:  0.001325 m**3/kg
Overall density is: 755.0 kg/m**3


## Example 2.3¶

In [4]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.3
'''The hydraulic piston/cylinder system shown in Fig. 2.11 has a cylinder diameter of D =
0.1 m with a piston and rod mass of 25 kg. The rod has a diameter of 0.01 m with an
outside atmospheric pressure of 101 kPa. The inside hydraulic fluid pressure is 250 kPa.
How large a force can the rod push within the upward direction?'''
from math import pi

#Variable Declaration:
Dcyl = 0.1 		#Cylinder diameter in m
Drod = 0.01 	#Rod diameter in m
Pcyl = 250000 	#Inside hydaulic pressure in Pa
Po = 101000 	#Outside atmospheric pressure in kPa
g = 9.81 		#Acc. due to gravity in m/s**2
mp = 25 		#Mass of (rod+piston) in kg

#Calculation:
Acyl = pi*Dcyl**2/4 		#Cross sectional area of cylinder in m**2
Arod = pi*Drod**2/4 		#Cross sectional area of rod in m**2
F = Pcyl*Acyl-Po*(Acyl-Arod)-mp*g 	#Force that rod can push within the upward direction in N

#Result:
print "Force that rod can push within the upward direction is:",round(F,1),"N"

Force that rod can push within the upward direction is: 932.9 N


## Example 2.4¶

In [5]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.4
'''A mercury barometer located in a room at 25◦C has a height of 750 mm. What is the
atmospheric pressure in kPa?'''

#Variable Declaration:
dm = 13534 		       #Density of mercury in kg/m**3
H = 0.750 		       #Height difference between two columns in metres
g = 9.80665 		 #Acc. due to gravity  in m/s**2

#Calculation:
Patm = dm*H*g/1000 	 #Atmospheric pressure in kPa

#Result:
print "Atmospheric pressure is:",round(Patm,2),"KPa"

Atmospheric pressure is: 99.54 KPa


## Example 2.5¶

In [6]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.5
'''A mercury (Hg) manometer is used to measure the pressure in a vessel as shown in
Fig. 2.13. The mercury has a density of 13 590 kg/m3, and the height difference between the
two columns is measured to be 24 cm.We want to determine the pressure inside the vessel.'''

#Variable Declaration:
dm = 13590 		      #Density of mercury in kg/m**3
H = 0.24 		      #Height difference between two columns in metres
g = 9.80665 		#Acc. due to gravity  in m/s**2
Patm = 13590*0.750*9.80665 #Atmospheric Pressure in Pa

#Calculation:
dP = dm*H*g 		#Pressure difference in Pa
Pvessel = dP+Patm 	#Absolute Pressure inside vessel in Pa

#Result:
print "Absolute pressure inside vessel is:",round(Pvessel/101325,2),"atm"

Absolute pressure inside vessel is: 1.3 atm


## Example 2.6¶

In [7]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.6
'''What is the pressure at the bottom of the 7.5-m-tall storage tank of fluid at 25◦C shown
in Fig. 2.15? Assume that the fluid is gasoline with atmospheric pressure 101 kPa on the
top surface. Repeat the question for the liquid refrigerant R-134a when the top surface
pressure is 1 MPa.'''

#Variable Declaration:
dg = 750 		#Density of gaasoline in kg/m**3
dR = 1206 		#Density of R-134a in kg/m**3
H = 7.5 		#Height of storage tank in metres
g = 9.807 		#Acc. due to gravity in m/s**2
Ptop1 = 101 	#Atmospheric pressure in kPa
Ptop2 = 1000 		#top surface pressure in kPa

#Calculation:
dP1 = dg*g*H/1000	#Pressure difference in kPa
P1 = dP1+Ptop1
dP2 = dR*g*H/1000#Pressure difference in kPa
P2 = dP2+Ptop2

#Result:
print "Pressure at the bottom of storage tank if liquid is Gasoline:",round(P1,1),"Kpa"
print "Pressure at the bottom of storage tank if liquid is R-134a:",round(P2),"Kpa"

Pressure at the bottom of storage tank if liquid is Gasoline: 156.2 Kpa
Pressure at the bottom of storage tank if liquid is R-134a: 1089.0 Kpa


## Example 2.7¶

In [8]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 2.7
'''A piston/cylinder with a cross-sectional area of 0.01 m2 is connected with a hydraulic
line to another piston/cylinder with a cross-sectional area of 0.05 m2. Assume that both
chambers and the line are filled with hydraulic fluid of density 900 kg/m3 and the larger
second piston/cylinder is 6mhigher up in elevation. The telescope armand the buckets have
hydraulic piston/cylinders moving them, as seen in Fig. 2.16.With an outside atmospheric
pressure of 100 kPa and a net force of 25 kN on the smallest piston, what is the balancing
force on the second larger piston?'''

#Variable Declaration:
Po = 100		#Outside atmospheric pressure in kPa
F1 = 25 		#Net force on the smallest piston in kN
A1 = 0.01 		#Cross sectional area of lower piston in m**2
d = 900 		#Density of fluid in kg/m**3
g = 9.81 		#Acc. due to gravity in m/s**2
H = 6 		#Height of second piston in comparison to first one in m
A2 = 0.05 		#Cross sectional area of higher piston in m**3

#Calculation:
P1 = Po+F1/A1 	#Fluid pressure in kPa
P2 = P1-d*g*H/1000 #Pressure at higher elevation on piston 2 in kPa
F2 = (P2-Po)*A2 	#Balancing force on second piston in kN

#Result:
print "Balancing force on second larger piston is:",round(F2,4),"N"

Balancing force on second larger piston is: 122.3513 N