Your browser does not support script   Updated : 16/10/2016 # Energy Calculations

## Power Costs

All electrical appliances by law must be marked with a voltage and power rating. The voltage rating of an appliance must be suitable for the supply voltage ie the voltage supplied to your home by the power supplier. The power rating, generally given as Watts or KiloWatts (1000 watts) states how much power the appliance will used when switched on.
The power supplier charges for the power used in terms of a cost per kWhr. So how much does it cost to run an electrical item?

Example
An electric heater rated at 2kW is switched on for 4 hours. If the power supplier charges R1.24 per kWh how much will it cost?

Cost = Power rating (kW) X time (hrs) X Charge Rate (R/kWh)

Cost = 2 X 4 X 1.24 = R 9.92

## Hot Water Heating

Many homes have immersion heaters for heating the water and the example given here shows how the heating effects of immersion heaters can be calculated.

## Gather Data

Before we can analyse the electrical cost of heating water we require information regarding the type, size and power rating of the installed hot water geyser. In many cases all the information required is presented on the geyser serial pale as shown in the figure below. For my geyser the data on the serial plate is as follows:
• Model no R4200U3G1
• Power 3kw
• Volume 200l
• Pressure 400kPa
• Code 02-A-5 = Feb 1997
• Serial no 482897
• Standing loss per 24 hrs 2.6kWh
• Plate colour Red = 400kPa

## Heating Calculation

Assuming that the all the electrical energy is converted to heat energy we can write: Electrical energy input = Heat energy absorbed by the object

But:

Heat energy = mass(m) x specific heat capacity (Cp) x temperature change(dT)

and

Electrical energy = Volts(V) x Amps(A) x time(t)

= Electrical Power (w) x time(t)

Therefore:

Heat energy = mass(m) x specific heat capacity(C) x temperature rise(dT)

= VAt = Power x time

Example - Geyser

A 3kW immersion heater is used to heat 200kg of water. The water starts at 20°C and is heated to 60°C - a rise of 40°CThe cost of electrical power is R1.23 per kWh. How long will this take and how much will it cost? Mass of water needed = 200kg

Rise in temp needed = 40°C

Power of heater = 3 kW = 3000W = 3000(J/sec)

Specific heat capacity of water = 4200 J/(kg°C)

Heat energy required = 200 x 4200 x 40 = 33 600 000 J

Time for heater to produce this amount of energy

= 33 600 000 /3000 = 11 200 seconds

= 186.7 minutes

= 3 hours 7 minutes

The energy cost to heat the water = 3 (kw) x 3.13 (h) x 1.24 (R/kWh) = R11.58 Example - Kettle

A 2kW electric kettle is used to heat 1 litre (1kg) of water. The water starts at 20°C and is heated to 100°C - a rise of 80°C.;The cost of electrical power is R1.23 per kWh. How long will this take and how much will it cost?

Mass of water needed = 1kg

Rise in temp needed = 80°C

Power of heater = 2kW = 2000W = 2000 (J/s)

Specific heat capacity of water = 4200 J/(kg°C)

Heat energy required = 1 x 4200 x 80 = 336 000 J

Time for heater to produce this amount of energy

= 336 000 / 2000 = 168 seconds

= 2.8 minutes

= 0.0466 hours

The energy cost to heat the water = 2 (kw) x 0.04666 (h) x 1.24 (R/kWh) = R0.12

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