# Chapter 1 : Fundamental Concepts¶

## Example 1.1 Page No : 8¶

In [1]:
#C = (5/9)*(F-32);
#F = 32+(9*C/5);
#Letting C = F in equation;
#F = (5/9)*(F-32);
#Therefore
F = -160/4;           			#fahrenheit

# Results
print "Both fahrenheit and celsius temperature scales indicate same temperature at %.2f"%(F);

Both fahrenheit and celsius temperature scales indicate same temperature at -40.00


## Example 1.2 Page No : 18¶

In [3]:
#Given
Mm = 0.0123			#Unit:lb 			#Mass of the moon;
Me = 1 	    		#Unit:lb 			#Mass of the earth;
Dm = 0.273 			#Unit:feet 			#Diameter of the moon;
De = 1. 	    		#Unit:feet 			#Diameter of the earth;
Rm = Dm/2; 			#Radius of the moon; 			#Unit:feet
Re = De/2; 			#Radius of the earth; 			#Unit:feet

#F = (K*M1*M2)/d**2 			#Law of universal gravitation;
#Fe = (K*Me*m)/Re**2; 			#Fe = Force exerted on the mass;
#Fm = (K*Mm*m)/Rm**2; 			#Fm = Force exerted on the moon;
F = (Me/Mm)*(Rm/Re)**2; 			#F = Fe/Fm;
print "Relation of force exerted on earth to mass is"
print "Fe/Fm  = %.2f"%F

Relation of force exerted on earth to mass is
Fe/Fm  = 6.06


## Example 1.3 Page No : 20¶

In [4]:
#Given
M = 5;            			#Unit:kg 			#mass of body;
g = 9.81;         			#Unit:m/s**2 			#the local acceleration of gravity

# Calculations
W = M*g;          			#W = the weight of the body 			#Unit:Newton 			# 1 N =  1 kg*m/s**2

# Results
print "The weight of the body is %.2f N"%(W);

The weight of the body is 49.05 N


## Example 1.4 Page No : 21¶

In [5]:
print "Solution for a";
#given
M  =  10.            			#Unit:kg 			#mass of body;
g  =  9.5           			#Unit:m/s**2 			#the local acceleration of gravity
W  =  M*g;          			#W  =  the weight of the body; 			#Unit:Newton 			# 1 N  =   1 kg*m/s**2
print "The weight of the body is %.2f N"%(W);

print "Solution for b";
#Given
F  =  10;           			#Unit:Newton 			#Horizontal Force
a  =  F/M;          						#			#newton's second law of motion
print "The horizontal acceleration of the body is %.2f m/s**2"%(a);

Solution for a
The weight of the body is 95.00 N
Solution for b
The horizontal acceleration of the body is 1.00 m/s**2


## Example 1.5 Page No : 25¶

In [6]:
#Conversion Problem
# 1 inch = 0.0254 meter so, 1 = 0.0254 meter/inch    			#Eq.1
# 1 ft = 12 inch so, 1 = 12 inch/ft.........			#Eq.2
#Multiplying Eq.1 & Eq.2  			# We get 1 = 0.0254*12 meter/ft
#Taking Square both side
#1**2 = (0.0254*12)**2  meter**2/ft**2
print "1 ft**2 = %.2f meter**2"%((0.0254*12)**2);

1 ft**2 = 0.09 meter**2


## Example 1.7 Page No : 33¶

In [9]:
#The Specific gravity of mercury is 13.6 			#Given
#Converting the unit of weight of grams per cubic centimeter to pounds per cubic foot
# 1 lbf = 454 gram 			#1 inch =  2.54 cm
#So 1 gram = 1/454 lbf and 1 ft = 12*2.54 cm
#Gamma = (gram/cm**3)*(lb/gram)*(cm**3/ft**3) = lb/ft**3
#Gamma = (1 gram/cm**3)*(1 lbf/454 gram)*(2.54*12)**3 *cm**3/ft**3
Gamma = (1./454)*(2.54*12)**3; 			#lbf/ft**3 			#conversion factor
print "Conversion Factor = %.2f"%Gamma
p = (1./12)*(Gamma*13.6); 			#lbf/ft**2 			#gage pressure
p = (1./12)*Gamma*13.6*(1./144) 			#ft**2/inch**2 			#gage pressure
print "Guage Pressure is %.2f psi"%(p);
print "Local atmospheric pressure is 14.7 psia";
P = p+14.7; 			#Pressure on the base of the column 			#Unit:psia
print " So Pressure on the base of the column is %.2f psia"%(P);

Conversion Factor = 62.37
Guage Pressure is 0.49 psi
Local atmospheric pressure is 14.7 psia
So Pressure on the base of the column is 15.19 psia


## Example 1.8 Page No : 34¶

In [10]:
#Given
Rho = 13.595; 			#Unit: kg/m**3 			#The density of mercury
h = 25.4; 			#Unit: mm 			#Height of column of mercury
g = 9.806; 			#Unit:m/s**2 			#the local acceleration of gravity

#Solution
p = Rho*g*h; 			#P = Pressure at the base of a column of mercury 			#Unit:Pa

# Results
print "Pressure at the base of a column of mercury is %.2f Pa"%(p);

Pressure at the base of a column of mercury is 3386.14 Pa


## Example 1.9 Page No : 34¶

In [11]:
#Given
Patm = 30.0; 			#in. 			#pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature
Vacuum = 26.5;  			#in.  			#vaccum pressure
Pabs = Patm-Vacuum; 			#Absolute pressure of mercury 			#in.
# 1 inch mercury exerts a pressure of 0.491 psi
p = Pabs*0.491; 			#Absolute pressure in psia

# Results
print "Absolute pressure of mercury in is %.2f psia"%(p);

Absolute pressure of mercury in is 1.72 psia


## Example 1.10 Page No : 35¶

In [12]:
#Given
Rho = 2000; 			#Unit: kg/m**3 			#The density of fluid
h = -10; 			#Unit: mm 			#Height of column of fluid 			#the height is negative because it is measured up from the base
g = 9.6 			#Unit:m/s**2 			#the local acceleration of gravity

#Solution
p = -Rho*g*h; 			#P = Pressure at the base of a column of fluid 			#Unit:Pa

# Results
print "Pressure at the base of a column of fluid is %.2f Pa"%(p);

Pressure at the base of a column of fluid is 192000.00 Pa


## Example 1.11 Page No : 35¶

In [13]:
#Given
Patm = 30.0 			#in. 			#pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature
Vacuum = 26.5  			#in.  			#vaccum pressure
Pabs = Patm-Vacuum; 			#Absolute pressure of mercury 			#in.

# (3.5 inch* (ft/12 inch) * (13.6*62.4)LBf/ft**3 * kg/2.2 LBf * 9.806 N/kg)/((12 inch**2/ft**2) * (0.0254 m/inch)**2)
p = (3.5*(1./12)*13.6*62.4*(1/2.2)*9.806)/(12**2*0.0254**2*1000); 			#kPa 			#Absolute pressure in psia

# Results
print "Absolute pressure of mercury is %.2f kPa"%(p)

Absolute pressure of mercury is 11.88 kPa