WORK AND POWER: BASIC CONCEPTS ABOUT PHYSICAL QUANTITIES

The Work - Velocity and power - Types of energy - Efficiency - Measure units - I S - Dimensional analysis

These notes are intended to enable all persons whith poor physics knowledge - and who are going to read the notes on the laser LLLB (Low Level Laser Beam) applied to medical treatments LLLT (Low Level Laser Treatment) - to acquire a minimum notion of the physical quantities involved in the process. It is therefore good to speak advisely of thermal power, light output, duration of LLLT treatment and dose.

NOTION OF 'WORK'

 With the help of this figure, I explain what is the work (otherwise known as 'energy') by an example of a calculation made neglecting frictions, efficiency, and acceleration. We have here a hoist with which we want to lift two meters a weight of 1000 Kg. To lift up the weight of 2 meters as indicated in figure a force of 1000 kg must be employed to win the gravity. Using such measurement units we say that the minimum ideal required mechanical energy is 2000 kgm. So we define the 'WORK', otherwise known as 'energy'. Figure 1

WORK: IN THIS CASE LET'S DEFINE 'WORK' [L], THAT IS THE NECESSARY ENERGY, AS PRODUCT OF THE USED FORCE [F] AND DISPLACEMENT [S] AGAINST GRAVITY

L = F x S

In summary, so far have defined the following variables:

FORCE F = 1000 kg in our case it is to lift 1000 kg
SHIFT   S = 2 m in our case, it is  2 meters
WORK L = 2000 [kgm]. In our case the work is 2m x1000kg

We will see the end units that are used in various systems and their conventional equivalents. About the time to lift the weight above, it is clear that we can employ different times. If for example we want to spend 60 seconds or 30 seconds or 10 seconds, what does it change? It seems obvious that it should change the speed, and therefore the required power of the winch motor-reducer complex. The lifting speed and the power required are inversely proportional to the time that we want to use for the lifting operation. Ie. a lower lifting time corresponds to greater lifting speed and then to increased required power.

VELOCITY (Speed):
The average speed is the ratio of the space S and the time T required to tread it. We have the experience of when we drive a car or a motorcycle.

V = S/T

If for example we went by car from Milan to Rome in six hours along 700 Km, our average speed was:

Vm = 700/6  that is about 117 km / h

POWER
Taking an elementary calculation of the power required to lift our weight of 1000 Kg , if wanting to perform the work above in 10 seconds, in our case  the needed power is approximately

W = 200 Kgm/s    (2000/10)

with an average lifting speed V = 2/10 = 0.2 meters per second.

THEN POWER [W] IS DEFINED AS RELATIONSHIP BETWEEN WORK (in this case mechanical energy required to lift the weight) AND TIME REQUIRED TO PERFORM IT. Obviously, the power is inversely proportional to the lifting speed, or the time taken in lifting the same.

W = L/T

The power required is the ratio between the needed energy  L and the time T spent, thus can also be written as a product of the force F for the lifting speed:

W = (F x S) / T in in fact W = F x (S / T) and therefore W = F x V

THEREFORE WE HAVE DEFINED THE FOLLOWING QUANTITIES

1. DISPLACEMENT S = 2 m
2. REQUIRED TIME T = 10 s
3. LIFTING FORCE   F = 1000 kg
4. ENERGY (WORK) L = 2000 kgm in our case the work is 1000kg x 2m
5. LIFTING SPEED   V = S / T (0.2 m / s or twenty centimeters per second)
6. POWER ( F V )     W= 200 kgm / s

KINDS OF ENERGY

The device  of the upper shown example transforms the  ELECTRIC energy to MECHANICAL, that is to lift the weight.

The energy thus presents itself in different major forms. Depending on how it is manifested as MECHANICAL, ELECTRICAL, THERMAL, CHEMICAL, ENERGY, and the twentieth century also atomic energy.
ALL FORMS OF ENERGY CAN BE TRANFORMED TO ANOTHER, in MANY WAYS. Let me offer some simple examples:

1. Eolic energy (wind energy) is a form of mechanical energy in turn generated by combination of other energy types. It can be transformed into electricity through wind turbines. Hydropower is an analogous energy's type, but much more conspicuous.
2. Solar energy (sunlight energy), roughly speaking, is a form of thermal and luminous energy together, because composed of electromagnetic radiation of different wavelength. IT GENERATES ALL CHEMICAL AND ENERGETIC PHAENOMENA REQUIRED TO LIFE ON EARTH. It can be transformed to heat by use of SOLAR HEAT EXCHANGER, or to electric power WITH VOLTAIC SOLAR PANELS.
3. Energy from fossil fuels. is obtained by the chemical reaction between the fuel and the oxygen content in the air (oxidizing). Remember that in turn the fuels were generated from vegetable and animal matters and along the geological ages, thus by a combination of chemical, mechanical and thermal energy,and stored in the earth's crust. Fossil fuels are extracted and also used to the direct heat generation  or of other derived types of energy.
4. Chemical Energy that STEMS FROM CHEMICAL REACTIONS, AND THAT FOR EXAMPLE in fuel cells or other types of batteries used to generate electricity. Strictly speaking also the combustion is a type of chemical energy, resulting from the oxidation of fuel elements.
5. Nuclear energy (atomic energy) is a form of energy resulting from profound modifications of the matter's structure, because it's derivative by fission or fusion of the atoms' nuclei  of particular chemical elements.
6. Mechanical energy it is energy of various kinds that the matter has due to the body speed (kynetic) or position (position energy) and all the energy type due to elastic deformation and so on. Please, make yourself an example that comes to your mind, or define another type of mechanical energy.

POTENTIAL ENERGY
Going back with reference to the example in Figure 1, after lifting the weight of 2 m, we have transformed the electrical energy into potential mechanical energy. In physics, the potential energy stored in an object is the energy which it possesses due to its position or its orientation with respect to a field of forces (in the case of the example it is the conservative force acting on the weight due to gravity).

More generally - regardless of the physical definitions of orthodoxy - one can think  the potential energy as the ability or capacity of an object (or system) to transform its own energy into another form of energy.

PERFORMANCE OR ENERGY EFFICIENCY

ENERGY in every processing is not all converted to useful energy to the purpose for which the transformation occurs, but a part of it is lost forever and is not useful. In each transformation we can measure therefore the TRANSFORMATION PERFORMANCE or the EFFICIENCY. Taking the example of the winch in the picture, let's put a few numbers:

Input Energy = 2250kgm
 Trasformation
Useful energy =2000 kgm

Lost Energy
250= kgm

In this case we have - very optmistically indeed - assumed that the energy lost due to friction in the linkage, the efficiency of the motor, and other factors is 11% of the input and is lost as heat. In such case, the YIELD (or efficiency) is the ratio of useful energy and total supplied energy. In this case is 2000/2250 = 0.89 (in percentage 89%)

EXAMPLE: a solid state laser module receives incoming electrical energy and emits a beam of electromagnetic polarized energy (polarized light). Even In this case, the polarized light's power of the ray is less than the input electrical power, and then a part of the energy goes to heat the module itself and is lost in the environment .

MEASUREMENT UNITS OF QUANTITIES

The units of measurement of quantities must be such as to make consistent and measurable all types of energy transformation. The basic of the system are the measurement units of the fundamental variables, including those used to measure space, time and strength, from which all the others derive.

A system of units of physical quantities must be consistent, that is, its derived quantities must be obtained by processing of fundamental physical quantities. There are various measuring systems.

For example, in continebtal Europe we are accustomed to use the metric system and is commonly know that the force is measured in Kg, the space in meters, and the time in seconds, minutes and hours;  in the anglosaxon countries for the time the measure is the same, while to the rest it's still in use instead the imperial system, ie lbs, feet, inches.

Is aroused  then the need to have a unified reference units system,  and with time it has come to the INTERNATIONAL SYSTEM, which is used mainly for scientific and technical purposes.
Of course one still can use his own units, thus conversion factors are used to express measurements units of  other systems.
To get a complete idea of all the existing units we can see on wikipedia or other sites. We give here a conversion table of the only units of magnitudes LENGTH FORCE ENERGY and their derivatives.

 FORCE Newton Kg Force Pound force Newton 1,00 1 / 9,81 1/ 4,48 Kg Force (Kgf) 9,807 1,00 2,24 Pound force 4,48 0,454 1,00 LENGHT meter Centimeter Millimeter Foot Inch Meter 1 100 1000 3,28 39,37 Centimeter 0,010 1 10 0,0328 0,3937 Foot 0,3048 30,48 304,8 1 12 Inch 0,0254 2,54 25,4 0,0833 1 ENERGY Kgfxm Kcal British Thermal Unit Kwh JOULE (J) N x m Kgf x m 1 0,0023 0,0093 1/367092 9,807 9,807 Kcal 426,93 1 3,97 0,0012 4,1868KJ 4186,8 BTU 107,58 0,252 1 0,00029 1,0551KJ 1055,1 Kwh 367092 859,8 3412,14 1 3600 KJ 3600(KN) JOULE (J) 0,102 0,00024 0,00095 1/3600KJ 1 1 Newton x m 0,102 0,00024 0,00095 1 1

INTERNATIONAL SYSTEM

Today the SI is based on seven fundamental physical quantities and the corresponding units with which are defined the derived quantities and the corresponding measurement units. Also it defines the prefixes to be added to the measurement units to identify multiples and submultiples.
The international system is consistent as its derived quantities are obtained as a product and relationship of physical fundamentals.
Each physical quantity and the unit of measure is a combination of two or more physical units of measurement and the corresponding basic or the reciprocal one. All units are defined by measuring natural phenomena except kilogram. In addition, the kilogram is the only unit of basic measurement containing the prefix K because the gram is a unit of measure too small for most applications.
The physical fundamentals of the international system involved in the calculations for the example of Figure 1 are to date the following:

 MEASURE SYMBOL UNIT NAME UNIT SYMBOL Lenght l meter m Mass m Kilogram Kg Time t Second s

We write, for example l = 10 m to indicate a length of 10 meters
The derived quantities that interest us especially are:

 QUANTITY SYMBOL UNIT NAME UNIT SYMBOL Derivation from the basic units Frequency f, ν Hertz Hz s−1 Force F Newton N kg · m · s−2 Velocity v Meter per second m/s m · s−1 Energy E,Q Joule J N · m = kg · m2 · s−2 Power W,P Watt W J · s−1 = kg · m2 · s−3

For complete information on all the units of measure you can click and open this link http://it.wikipedia.org/wiki/Sistema_internazionale_di_unit%C3%A0_di_misura

APPLICATION EXAMPLE OF dimensional analysis:

when we write a formula or any relationship with certain symbols, to check if it is correct we have to make the dimensional analysis.

For example, let's calculate the mean velocity V of the flow of an incompressible liquid in a tube with diameter D, knowing the mass flow Q and the specific weight of the fluid itself.

We have to verify immediately that our relationship for calculating the velocity V give the result  in meters per second

 4 Q V = ------------------ [ m/ s] 2 π γ D

To check whether the formula is correct let's put the following dimensions for the involved variables and then use the dimensional analysis

Flow in weight       Q = [kg/s] (Kilograms per second)
Specific Gravity      γ = [Kg m−3] (Kilograms per cubic meter)
Pipe's diameter  D = [m] (diameter in meters)

Substituting in the previous formulas the symbols of measurement's units to those of magnitudes, and neglecting of course the numbers 4 and π which do not have size, we will then get

 3 m Kg ------------------ = [ m/ s] 2 kg s m

The symbols kg in the numerator and the denominator cancel out, while the ratio of m cubed and  m squared gives m,  then the formula, if the dimensionless constants is attributed to the correct value, is exact.

It seems clear that if the result was not [m / s] the formula would be wrong or if the unit of measurement of the variables were not homogeneous the result of the calculation would be wrong too.
We cannot, for example, express the diameter in centimeters, or the specific weight of the pounds per cubic inch, or the flow rate in tons per hour. The units of measurement must all be consistent to the unit you want for the result of the formula, in this case [m / s]

Lino Bertuzzi - novembre 2014