# Commercial Unit of Electrical Energy

The ability and requirement for the body to conduct labour are referred to as **Energy**. Energy may be found in a variety of places and in many forms. We’ve noticed that on food packages and ready-to-cook foods, the amount of energy provided is always stated; as energy is defined as the ability to perform labour, it must be quantified in units.

Energy exists in different forms such as thermal, Kinetic, nuclear, potential, chemical, electrical, etc.

Its SI unit is Joule which is represented by J. When a force of 1 N acts on a body with a displacement of 1 m, 1 joule is defined as the quantity of work done. When large amounts of energy need to be expressed, commercial units of energy are required, and this is utilised in industries and factories.

### Commercial Units of Energy

Here, we must define the commercial unit of electric energy and explain:

- How it differs from the home unit of electric energy.
- A kilowatt-hour is equal to one joule.
- In a commercial setting, such as factories or other business sectors, how electrical equipment utilizes one kilowatt of energy for one hour.

Kilowatt-hour(kWh) is a commercial unit of energy. A kilowatt-hour is an amount of energy consumed by a machine working at a steady rate of one kilowatt for one hour. As a result, Joule is a SI unit of energy, but Kilowatt-hour is aCommercial unit.

**Electrical Energy**

The energy generated from the electric potential or kinetic energy of charged particles is known as electrical energy. It’s the energy that’s been transformed from electric potential energy, in general. The energy created by the passage of electrons from one location to another is known as electrical energy. Current or electricity is the flow of charged particles along/through a media (such as a wire).

**Relation between the commercial unit and SI unit of energy**

Kilowatt-hour gives a direct relationship between energy and power. Electric power in terms of energy is defined as the amount of electrical energy consumed per unit of time, i.e., electrical energy divided by time is equal to electrical power. SI unit of power is Watt and represented by W.

The formula of power in terms of energy is given as:

P = E ⁄ tHere,

- P is the power generated
- E is the energy consumed
- t is the time
Rearrange the formula in terms of energy.

E = P tRepresenting the Kilowatt-hour in terms of above formula.

Consider the power equal to 1 kW and time equal to 1 h.

E = P t1 kW h = 1kW × 1 h

1 kW h = 1000 W × 1 h

1 kW h = 1000 (J / s) × 3600 s

1 kW h = 3.6 × 10

^{6 }J.

Therefore, the kilowatt-hour (kW h) is a unit of energy equal to 1 kilowatt of power expanded for 1 hour. It does not refer to the number of kilowatts you consume every hour. It’s just a unit of measurement that represents the amount of energy used if a 1000 Watt appliance was left on for one hour. It would take 10 hours to accumulate 1 kW h of energy if you switched a 100 Watt light bulb.

Although the kilowatt-hour is not a recognized system unit, it is widely used in electrical applications. The unit of electrical energy is the kilowatt-hour (kWh), which is the same for commercial and household use. Commercial apartments have distinct rates of charge, which are higher than residential unit charges. It is mostly used to reduce power usage in factories and other industries that use a lot of electricity.

Note:The kilowatt-hour is the commercial unit of electric energy here, and 1 kW h = 3.6 × 10^{6}J. So, be cautious while converting kilowatt-hours to joules.

### Sample Problems

**Problem 1: Calculate the number of Joules contained in 1 unit of energy.**

**Solution:**

1 unit of energy is equal to 1 kW h.

1 kW h = 1000 W × 1 h

=1000 (J / s) × (60 × 60) s

= 3600000 J

= 3.6 × 10

^{6}J.Hence, the number of Joules in 1 unit of energy is equal to

3.6 × 10.^{6}J

**Problem 2: Why does the Kilowatt-hour is used as the commercial unit of energy?**

**Solution:**

In practice, the unit kWh is used for the measurement of electrical energy, rather than joule. This is because Joule is a very small unit and the energy consumption in day to day life is very large, i.e., it comes in figures of 10

^{6}to 10^{8}. Thus, to reduce the complexity of handling such large figures, a bigger unit was required. This bigger unit used for the measurement of electrical energy is kWh and is related to Joule as: 1kWh = 3.6 ×10^{6 }J. Hence, the energy reading commercially became simpler by using this bigger unit instead of joule

**Problem 3: Name the unit used in selling electric energy to consumers. An electric motor takes 10 A from a 110 V line. Determine the power of the motor and energy consumed in 3 hrs. Calculate the cost of electric energy for the month of June at a rate of Rs. 9 / unit.**

**Solution:**

Kilowatt-hour (kWh) unit is used in selling electric energy to consumers.

Given:

Current drawn to motor, I = 10 A

Voltage across the motor, V = 110 V

Time for energy is consumed, t = 3 h

Power generated by motor, P = V I

= 110 V × 10 A

= 1100 W

Energy consumed by motor, E = P × t

= 1100 W × 3 h

= 3.3 kW h

Now,

Energy consumed, E = 3.3 kW h

Cost of energy per unit, c = Rs. 9 ⁄ unit

Number of days, n = 30

Total Cost, C = E c n

= Rs. (3.3 × 9 × 30)

= Rs. 891.00

Hence, selling unit of energy is

Kilowatt-hour, power generated by motor is1100 W, energy consumed by motor is3.3 kW h, and cost of energy isRs. 891.

**Problem 4: Convert 72000 J of energy into kilowatt-hours.**

**Solution:**

Since, 3.6 × 10

^{6}J = 1 kW hSo, 1 J = (1 ⁄ 3.6) × 10

^{-6}kW hTherefore, 72000 J = (72000 ⁄ 3.6) × 10

^{-6}kW h= 2 × 10

^{-3}kW hHence, 72000 J of energy is equal to

2 × 10.^{-3}kW h

**Problem 5: Differentiate between watt and watt-hour.**

**Solution:**

Watt is the unit of power while watt hour is the unit of work/energy, since energy = power × time.

**Problem 6: An electrical appliance consumes 7.2 MJ of energy to generate power equal to 2 kW. Find the time in minutes for which the given energy is consumed.**

**Solution:**

Since, 3.6 × 10

^{6}J = 1 kW hSo, 1 J = (1 ⁄ 3.6) × 10

^{-6}kW hTherefore, 7.2 × 10

^{6}J = (7.2 ⁄ 3.6) × 10^{6}× 10^{-6}kW h= 2 kW h

Now,

Energy consumed, E = 2 kW h

Power generated, P = 2 kW

Time for which appliance runs, t = E ⁄ P

= 2 kW h ⁄ 2 kW

= 1 h

= 60 min

Hence, the time for which appliance runs is

60 minutes.

**Problem 7: An electric motor of 1.5 HP runs continuously for 3 hours. Find the energy consumed in kilowatt-hour. (1 HP = 0.746 kW)**

**Solution:**

Given,

Power generated by motor, P = 1.5 HP

= 1.5 × 0.746 kW

= 1.119 kW

Running time of motor, t = 3 h

Energy consumed, E = P × t

= (1.119 × 3) kW h

= 3.357 kW h

Hence, the energy consumed in kilowatt-hour is

3.357 kW h.

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