These were some of the modern examples of energy conversion. Now, let’s take a look at how the energy conversion technology has evolved over the years and how the early humans used energy conversion by harnessing the natural forms of energy.

#### The Evolution of Energy Conversion

**The Use of Fire:** The first evidence of energy conversion from the early ages is when the ancient humans discovered the use of fire. They burned dried plant, wood, and animal waste and used the resultant energy for cooking and heating. This was followed by the use of mechanical energy that was used to harness the energy of flowing water and wind. However, this generation of mechanical energy with the invention of simple devices came much later around 2000 years after humans discovered the use of fire.
**Waterwheels: **The earliest evidence of a machine being used to grind grain is that of a waterwheel, which was later also used to drive sawmills and pumps, driving tilt hammers for forging iron, etc. Waterwheels were the primary means of mechanical power production, rivaled only occasionally by windmills. Therefore, many industrial towns, especially in early America, were set up at locations that assured water flow all year.
**Windmills: **Apart from waterwheels, windmills were also used as a source of power to replace the animal muscle. Windmills were used in various parts of the world to convert the wind energy into mechanical energy for grinding grain, pumping water, and draining the lowland areas.

#### Energy Conversion during the Industrial Revolution

**Steam Engine:** With the rapid growth of the industry around the mid-18th century (and somewhat later in various other countries), there was a pressing need for new sources of motive power, particularly those independent of geographic location and weather conditions. To meet the demands of the industry, the steam engine was developed as the first device for converting thermal energy into mechanical energy.
**Newcomen Engine: **A few years later, a better and more efficient version of the steam pump which consisted of a cylinder fitted with a piston was developed. Although engines of this kind converted only about one percent of the thermal energy in the steam to mechanical energy, they remained unrivaled for more than 50 years.
**Watt’s Engine: **Subsequently in the later years, the Newcomen engine was modified by adding a separate condenser to make it unnecessary to heat and cool the cylinder with each stroke. Since the cylinder and piston remained at steam temperature while the engine was operating, the fuel costs dropped by about 75 percent.

#### Conservation of Energy

Lastly, let’s understand the law of conservation of energy to further strengthen your fundamentals.

The Conservation of energy is the principle that energy can neither be created nor destroyed. It can only be converted from one form to another or transferred from one object to another. The energy transformation is applicable to all types of energy such as kinetic energy, potential energy, light energy, heat energy, sound energy, nuclear energy, gravitational potential energy etc. These transformations would occur again and again till the time that resource is present there. The friction effect and the air-pressure are the other factors which can affect the transformation to some extent.

Thus, we can safely say that in the whole system, the total energy remains constant but only its transformation takes place.

We hope this page proved useful to you in understanding energy conversion and the various aspects related to it.

#### Electron conversion

Decay by internal conversion (IC) results in the emission of an atomic electron. This electron, called the internal conversion electron, is emitted from an atom after absorbing the excited energy of a nucleus.

#### DC amps to kilowatts calculation

The power *P* in kilowatts (kW) is equal to the current *I* in amps (A), times the voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = *I*_{(A)}* *×* V*_{(V)} / 1000

#### AC single phase amps to kilowatts calculation

The power *P* in kilowatts (kW) is equal to the power factor *PF* times the phase current *I* in amps (A), times the RMS voltage *V* in volts (V) divided by 1000:

*P*_{(kW)} = * PF *×* I*_{(A)}* *×* V*_{(V)} / 1000

#### AC three phase amps to kilowatts calculation

#### Calculation with line to line voltage

The power *P* in kilowatts (kW) is equal to square root of 3 times the power factor *PF* times the phase current *I* in amps (A), times the line to line RMS voltage *V*_{L-L} in volts (V) divided by 1000:

*P*_{(kW)} = *√*3
× * PF *×* I*_{(A)}* *×* V*_{L-L (V)} / 1000

#### Calculation with line to neutral voltage

The power *P* in kilowatts (kW) is equal to 3 times the
power factor *PF* times the phase current *I* in amps (A), times the line to neutral RMS voltage *V*_{L-N} in volts (V) divided by 1000:

*P*_{(kW)} = 3 × * PF *×* I*_{(A)}* *×* V*_{L-N (V)} / 1000

#### Single phase amps to kVA calculation formula

The apparent power S in kilovolt-amps is equal to current I
in amps, times the voltage V in volts, divided by 1000:

*S*_{(kVA)} = * I*_{(A)}* *×* V*_{(V)}* */ 1000

#### 3 phase amps to kVA calculation formula

#### Calculation with line to line voltage

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to line RMS voltage V_{L-L} in volts, divided by 1000:

*S*_{(kVA)} = *√*3
× * I*_{(A)}* *×* V*_{L-L(V)}* */ 1000

#### Calculation with line to neutral voltage

The apparent power S in kilovolt-amps is equal to phase current I
in amps, times the line to neutral RMS voltage V_{L-N} in volts, divided by 1000:

*S*_{(kVA)} = 3 × * I*_{(A)}* *×* V*_{L-N(V)}* */ 1000

#### Single phase amps to VA calculation formula

The apparent power S in volt-amps is equal to current I in amps, times the voltage V in volts:

*S*_{(VA)} = * I*_{(A)}* *×* V*_{(V)}

#### 3 phase amps to VA calculation formula

The apparent power S in kilovolt-amps is equal to square root if 3 current I in amps, times the line to line voltage V_{L-L} in volts:

*S*_{(VA)} = *√*3
× * I*_{(A)}* *×* V*_{L-L(V)} = 3
× * I*_{(A)}* *×* V*_{L-N(V)}

#### eV to volts calculation with elementary charge

The voltage *V* in volts (V) is equal to the energy E in electron-volts (eV), divided by
the electric charge
*Q* in elementary charge or proton/electron charge (e):

*V*_{(V)} = *E*_{(eV)}* */* Q*_{(e)}

#### eV to volts to calculation with coulombs

The voltage V in volts (V) is equal to 1.602176565×10^{-19} times the energy E in electron-volts (eV), divided by
the electrical charge Q in coulombs (C):

*V*_{(V)} = 1.602176565×10^{-19}
× *E*_{(eV)}* */* Q*_{(C)}

#### Joules to watts calculation

The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):

*P*_{(W)} = *E*_{(J)}* */* t*_{(s)}

#### Joules to volts calculation

The voltage V in volts (V) is equal to the energy E in joules (J), divided by the charge Q in coulombs (C):

*V*_{(V)} = *E*_{(J)}* */* Q*_{(C)}

#### Single phase kVA to amps calculation formula

The current I in amps is equal to 1000 times the apparent power S in kilovolt-amps, divided by the voltage V in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ *V*_{(V)}

#### 3 phase kVA to amps calculation formula

#### Calculation with line to line voltage

The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power S in kilovolt-amps, divided by
the square root of 3 times the line to line RMS voltage V_{L-L} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (*√*3
× *V*_{L-L(V)}* *)

#### Calculation with line to neutral voltage

The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power
S in kilovolt-amps, divided by 3 times the line to neutral RMS voltage V_{L-N} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (3 × *V*_{L-N(V)}* *)

#### kVA to watts calculation

The real power P in watts (W) is equal to 1000 times the apparent power S in kilovolt-amps (kVA), times the power factor PF:

*P*_{(W)} = 1000 × *S*_{(kVA)} × *PF*

#### kVA to kW calculation

The real power P in kilowatts (kW) is equal to the apparent power S in kilovolt-amps (kVA), times the power factor PF:

*P*_{(kW)} = * S*_{(kVA)} × *PF*

#### kVA to volt-amps calculation

The apparent power S in volt-amps (VA) is equal to 1000 times the apparent power S in kilovolt-amps (kVA):

*S*_{(VA)} = 1000 × *S*_{(kVA)}

#### DC kilowatts to amps calculation

The current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by the voltage *V* in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / * V*_{(V)}

#### AC single phase kilowatts to amps calculation

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by the
power factor *PF* times the RMS voltage *V* in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (*PF* × * V*_{(V)}* *
)

#### AC three phase kilowatts to amps calculation

#### Calculation with line to line voltage

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by square root of 3 times the
power factor *PF* times the line to line RMS voltage *V*_{L-L} in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (*√*3
× * PF* × * V*_{L-L(V)}* *)

#### Calculation with line to neutral voltage

The phase current *I* in amps (A) is equal to 1000 times the power *P* in
kilowatts (kW), divided by 3 times the
power factor *PF* times the line to neutral RMS voltage *V*_{L-N} in volts (V):

* I*_{(A)} = 1000* *×* P*_{(kW)} / (3 × * PF* × * V*_{L-N(V)}* *)

#### DC kilowatts to volts calculation

The voltage *V* in volts (V) is equal to 1000 times the power
*P* in
kilowatts (kW), divided by
the current *I* in amps (A):

*V*_{(V)} = 1000 × *P*_{(kW)}* */*
I*_{(A)}

#### AC single phase kilowatts to volts calculation

The voltage *V* in volts (V) is equal to 1000 times the power
*P* in
kilowatts (kW), divided by the
power factor *PF* times the current I in amps (A):

*V*_{(V)} = 1000 × * P*_{(kW)} / (*PF* × * I*_{(A)}* *
)

#### AC three phase kilowatts to volts calculation

The line to line RMS voltage *V*_{L-L} in volts (V) is equal to 1000 times the power
*P* in kilowatts (kW), divided by square root of 3 times the power factor
*PF* times the current *I* in amps (A):

*V*_{L-L(V)} = 1000 × * P*_{(kW)} / (*√*3
× * PF* × * I*_{(A)}* *)

≈ 1000 × * P*_{(kW)} / (1.732
× * PF* × * I*_{(A)}* *)

#### KW to kilowatt-hours calculation

The energy E in kilowatt-hours (kWh) is equal to the power P in kilowatts (kW), times the time period t in hours (h):

*E*_{(kWh)} = *P*_{(kW)}* *×* t*_{(h)}

#### kW to VA calculation

The apparent power S in volt-amps (VA) is equal to 1000 times the real power P in kilowatts (kW), divided by the power factor PF:

*S*_{(VA)} = * *1000 × *P*_{(kW)} / *PF*