simple energy conversion from hydroelectric power plant to the house
Water-->Turbine-->Generator-->Transformer-->House Circuit Breaker--->House Power line-->Motor-->Electric Fan
Water
Hydropower, hydraulic power or water power is power that is derived from the force or energy of moving water
Turbine
A turbine is a kind of engine that extracts energy from a fluid and used to turn generators
From Latin turbo + binis (whirling object)
Generator
A generator converts mechanical energy into electrical energy
Transformer
An electrical device for changing the voltage of an alternating current
Circuit breaker
A circuit breaker is an automatically-operated electrical switch designed to protect an electrical circuit from damage caused by overload or short circuit
Power line
Power is transmitted through power lines then to wall sockets
Motor
A motor is a device for converting electrical energy into mechanical energy
From Latin motor = mover, "movere" = move
Electric Fan
A fan run by an electric motor used for cooling
Saturday, November 17, 2012
SCIENCE: Physics - Charles' law, Gay-Lussac's law, gas law
Charles' law
Also known as the law of volumes, Gay-Lussac's law
V T curve is a straight line
At constant pressure, the volume of a fixed mass of gas varies directly with the absolute temperature
At constant volume, the gas pressure varies directly with the absolute temperature
V1/T1 = V2/T2 ---> if p = constant
p1/T1 = p2/T2 ---> if v = constant
where:
p1 = absolute pressure at point1
p2 = absolute pressure at point2
V1 = gas volume at point1
V2 = gas volume at point2
T1 = absolute temperature at point1
T2 = absolute temperature at point2
1. Problem:
Air occupies a volume of 70 L at 300 K. If the pressure remains constant, find the temperature of the air when the volume increased to 140 L.
find:
T2 = temperature of the air at V2 = 140 L
given:
p = c
V1 = 70 L
T1 = 300 K
V2 = 140 L
solution:
V1/T1 = V2/T2
70/300 = 140/T2
T2 = 300 * 140/70
T2 = 600 K
thus, direct proportion between volume and absolute temperature. The initial volume of 70 L was doubled to 140 L and therefore 300 K is also doubled giving 600 K.
2. Problem:
In a constant volume process, oxygen has a temperature of 80 C under a pressure of 4 atm. If the pressure is halved, calculate the final temperature.
find:
T2 = final temperature
given:
v = c
T1 = 80 + 273
T1 = 353 K
p1 = 4 atm
p2 = 2 atm
solution:
p1/T1 = p2/T2
4/353 = 2/T2
T2 = 353 * 2/4
T2 = 176.5 K ---> T2 also halved when p2 is halved
Thursday, November 15, 2012
SCIENCE: Physics - Mechanical work (variable force)
Work
- amount of energy transferred by a force acting through a distance
- Work can be zero even when there is a force
W = S(a b) F * dx
where:
W = mechanical work done
F = variable force applied
dx = differential distance (delta x, change in x)
S(a b) = integral from a to b
a = lower limit, initial point
b = upper limit, final point
1. Springs - as springs get stretched, the harder it is to pull
The car suspension system steel spring has a spring constant of 12 lb/in. Find the work needed to stretch from an initial amount of 2 in to a final amount of stretch of 4 in.
find:
W = work done by stretching the spring 2 in to 4 in amount of stretch
given:
k = 12 lb/in
a = 2 in
b = 4 in
solution:
The force needed to stretch the spring increases as the spring gets extended
F = kx ---> equation1
W = S(a b) F * dx
substituting,
W = S(a b) kx * dx
W = k * S(a b) x * dx
W = 12 * S(a b) x dx
W = 12/2 [x^2, (2 4)]
W = 6 * (4^2 - 2^2)
W = 6 * (16 - 4)
W = 6 * 12
W = 72 in.lb
thus,
W = 1/2 k (b^2 - a^2) ---> for springs
SCIENCE: Physics - Mechanical work (constant force)
Work
- amount of energy transferred by a force acting through a distance
- Work can be zero even when there is a force
W = (F cos A) * d
where:
W = mechanical work done
F = constant force applied
A = angle between F and d
d = distance travelled as a result of the application of force
1. Force and distance on the same line
A toy cart is placed in a rail that is flat on a table. A pulling force of 10 N is applied resulting in a distance of 2 m travelled by the toy cart. Calculate the work done.
find:
W = work done
given:
F = 10 N
A = 0 degrees
d = 2 m
solution:
W = (F cos A) * d
W = (10 * cos 0) * 2
W = (10 * 1) * 2
W = 20 N m
2. Force is at an angle with the direction of motion
If the force is applied with an angle of 30 degrees, calculate the work done.
find:
W = work done
given:
F = 10 N
A = 30 degrees
d = 2 m
solution:
W = (F cos A) * d
W = (10 * cos 30) * 2
W = 17.3 N m
3. The rail is at an angle of 15 degreees with the table
If (in problem #2) the toy cart is moved down, calculate the work done.
find:
Wd = work done if the toy cart is moved down
given:
F = 10 N
a1 = 30 degrees, angle of the force
a2 = 15 degrees, angle of the rail
d = 2 m
solution:
reference is the rail --> direction of the displacement
A = a1 + a2
A = 30 + 15
A = 45 degrees
Wd = (F cos A) * d
Wd = (10 * cos 45) * 2
Wd = 14.1 N m
If (in problem #2) the toy cart is moved up, calculate the work done.
find:
Wu = work done if the toy cart is moved up
given:
F = 10 N
a1 = 30 degrees, angle of the force
a2 = 15 degrees, angle of the rail
d = 2 m
solution:
reference is the rail --> direction of the displacement
A = a1 - a2
A = 30 - 15
A = 15 degrees
Wu = (F cos A) * d
Wu = (10 * cos 15) * 2
Wu = 19.3 N m
Monday, November 5, 2012
C++ TUTORIAL FOR BEGINNERS: Program to calculate Circumference, Area of a Circle, and the Volume of a Sphere
The Basketball ball of the NBA has a standard dimension of 4.7 inch in radius, 9.4 inch in diameter, 29.5 inches in circumference, 22 ounces (size 7) in weight, and 7.5 - 8.5 pounds per square inch of air pressure.
Objective: Calculation of the Circumference, Area of a Circle and the Volume of a Sphere
Directive: #include <iostream>, #include <cmath>
Namespace: std
Function: main
Variables:
CircleRadius
double CircleCircumference
double CircleArea
double SphereVolume
Data types of variables: double
Statements: cout, cin, return
Input: Radius of the circle
Output:
Circumference of the Circle
Area of the Circle
Volume of the Sphere
C++ source code:
/*
PROGRAM : Circumference, Area of a Circle; Volume of a Sphere
AUTHOR: eternaltreasures
DATE: 2010 September 1
*/
#include <iostream>
#include <cmath>
using namespace std;
#define PI 3.14159
#define NEWLINE '\n'
int main ()
{
double CircleRadius;
double CircleCircumference;
double CircleArea;
double SphereVolume;
cout << "Enter the Radius of the Circle = ";
cin >> CircleRadius;
// Circumference of a Circle
CircleCircumference = 2 * PI * CircleRadius;
cout << "Circumference of the Circle = " << CircleCircumference;
cout << NEWLINE;
// Area of a Circle
CircleArea = PI * pow(CircleRadius, 2);
cout << "Area of the Circle = " << CircleArea;
cout << NEWLINE;
// Volume of a Sphere
SphereVolume = 4 * PI/3 * pow(CircleRadius, 3);
cout << "Volume of the Sphere = " << SphereVolume;
cout << NEWLINE;
return 0;
}
Objective: Calculation of the Circumference, Area of a Circle and the Volume of a Sphere
Directive: #include <iostream>, #include <cmath>
Namespace: std
Function: main
Variables:
CircleRadius
double CircleCircumference
double CircleArea
double SphereVolume
Data types of variables: double
Statements: cout, cin, return
Input: Radius of the circle
Output:
Circumference of the Circle
Area of the Circle
Volume of the Sphere
C++ source code:
/*
PROGRAM : Circumference, Area of a Circle; Volume of a Sphere
AUTHOR: eternaltreasures
DATE: 2010 September 1
*/
#include <iostream>
#include <cmath>
using namespace std;
#define PI 3.14159
#define NEWLINE '\n'
int main ()
{
double CircleRadius;
double CircleCircumference;
double CircleArea;
double SphereVolume;
cout << "Enter the Radius of the Circle = ";
cin >> CircleRadius;
// Circumference of a Circle
CircleCircumference = 2 * PI * CircleRadius;
cout << "Circumference of the Circle = " << CircleCircumference;
cout << NEWLINE;
// Area of a Circle
CircleArea = PI * pow(CircleRadius, 2);
cout << "Area of the Circle = " << CircleArea;
cout << NEWLINE;
// Volume of a Sphere
SphereVolume = 4 * PI/3 * pow(CircleRadius, 3);
cout << "Volume of the Sphere = " << SphereVolume;
cout << NEWLINE;
return 0;
}
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