Energy Conversion Examples
The following examples are solved using SI units, as per the existing international standards. The SI units make the energy conversion calculations much simpler. If the application data is specified in units other than SI units, then it is preferable to convert the data values into SI units before starting the calculations. The final results can be converted to the specified units, if required. This process may be required if working with American clients!!
Example 1 – Mechanical work
A mass of 20 kg is lifted to a height of 10 m. Calculate the work done or energy spent.
We have the following equations:
Force (F) = mass (m) x acceleration (a) newtons (kg-m/s2)
Work done (W) = Force (F) x distance (d) joules (Nm)
In this case, the ‘Force’ is the gravitational force. Hence, the acceleration is the acceleration due to gravity (g = 9.81 m/s2). The mass is being lifted against the gravitational force.
W = F x d = (m x g) x d
= (20 x 9.81) x 10
= 196.2 x 10
= 1,962 J (1.962 kJ)
The above calculations are simple, but real life is not! There are some interesting factors to consider while calculating the force in practice.
- Force is a vector. It has a magnitude and direction. So, it is necessary to use the component of force acting along the distance (motion) while calculating the work done.
- When the mass is pulled along a horizontal surface, the gravitational force is not directly relevant as it is acting at 90 degrees to the direction of motion. But it influences the value of frictional force.
Consideration of the above factors complicates the calculation of the work done. We will leave such complications to the mechanical engineers!
Example 2 – Electric Lift
An electric motor lifts a mass of 500 kg to a height of 30 meters in 12 seconds. Calculate the power developed in ‘kW’ and ‘hp’. (Neglect all losses).
Energy spent (E) = F x d = (500 x 9.81) x 30
= 147.15 kJ
Power = Rate of Energy = Energy / time duration
= E / t
= 147.15 / 12
= 12.262 kW
Power in ‘hp’ = kW / 0.746 = 16.437 hp (1 hp = 0.746 kW)
Example 3 – Water Pump
It is required to specify a pump to lift 135 cubic metres of water per hour (m3/h) to a height of 30 metres. Calculate the required rating of the motor in kW and hp. Assume the efficiency of the pumping system to be 60%.
We have,
ρ = 1000 kg/m3 (density of water)
In the CGS system, water was used as the reference, that is, a unit volume of water corresponds to one unit of mass. By this definition, the mass of 1 cc (cubic centimetre or milli-litre) of water is 1 gm. Hence, the mass of 1 litre of water is 1 kg and the mass of 1 m3 of water is 1000 kg.
Flow rate of water (q) = 135 m3/h (Given data)
Volume of water (v) = 135 m3
Mass of water (m) = v x ρ = 135 m3 x 1000 kg/m3
= 135,000 kg
Force required to lift the water (F) = m x g
= 135,000 kg x 9.81 m/s2
= 1,324,350 N
Work done or Energy required = F x h
= 1,324,350 N x 30 m
= 39,730.5 kJ
This work needs to be done in 1 hour, since the required flow rate is 135 m3/h.
Pumping power (Ph) required in kW = Energy in kJ / Time in seconds
= 39,730.5 kJ / 3,600 s
= 11.03625 kW
Motor rating = Pumping Power / Efficiency
= 11.03625 / 0.6
= 18.39375 kW
Power in ‘hp’ = kW / 0.746
= 18.39375 / 0.746
= 24.66 hp
The pumping power (Ph) in kW can be written in a simplified form as below:
Ph = ( q ρ g h / 3600 ) x 10-3 kW
Example 4 – Water Heater
It is required to heat 70 litres of water from 20ºC to 75ºC using an electric heater. Assume the heating efficiency to be 100%.
(a) Calculate the energy required.
(b) Calculate the power required if the final temperature is to be reached in 1 hour.
The equation for the quantity of heat (Q) required is as given below:
Q = m x c x (t2 -t1)
where,
Q – quantity of heat (J)
m – mass of the liquid (kg)
c – specific heat of the liquid (J/kg/ ºC)
t1 – Initial temperature (ºC)
t2 – Final temperatures (ºC)
The specific heat of water is 4184 J/kg/ ºC.
How did we get this magic value? The popular unit for quantity of heat was a ‘calorie’. The term ‘calorie’ is derived from the Latin word ‘Calor’ for ‘heat’. Water was again used as the reference. One ‘calorie’ of heat is defined as the heat required to raise 1 gram of water by 1 degree centigrade. Hence, the ‘specific heat’ of water is 1 cal/gm/ ºC in CGS units.
The Americans still use ‘calorie’ as the unit of heat. To make things more confusing, they also use the term ‘Calorie’ with capital ‘C’ – especially for food packaging. 1 Calorie is equal to 1000 calories (1 kilocalorie)!
In SI units, 1 calorie of heat is equivalent to 4.184 Joules. This equivalence is established by experiments. The value ‘calorie’ itself is dependent on atmospheric pressure and initial temperature. So, it can result in slightly different values. To avoid confusion, the SI standards committee has defined 1 calorie as equal to 4.184 joules.
We know that the density of water (ρ) is 1000 kg/m3.
Hence the mass of 1 litre of water is 1 kg, since 1 m3 = 1000 litre.
(a) Amount of heat energy required = (70 L x 1 kg/L) x 4184 x (75 – 20)
= 16,108.4 kJ
(b) Power required = Energy / Time = 16,108.4 kJ / 3600 s
= 4.475 kW
Example 5 – Cost of Energy
The electric heater in Example 4 is supplied from an electric power outlet at 230 V. (For the sake of simplicity, let us assume that the voltage is equivalent DC voltage)
(a) Calculate the current flow
(b) Calculate the monthly energy bill if the heater is ‘ON’ for 3 hours in a day. The energy tariff is given as $0.30 for each kilo-watt-hour (kwh) of energy.
(a) We can calculate the current in Amperes (A) using Equation 2.
Current (I) = P / V = 4.475×103 W / 230 V
= 19.46 A
(b) Assuming a 30-day month,
Total monthly energy consumption = 4.475 kW x (30 x 3) hours
= 402.75 kwh
Monthly energy bill = 402.75 kwh x $0.30 / kwh
= $120.83
(The amount is in Australian dollars!)
Important Note
The objective of the above examples is to illustrate the concepts and equations for energy conversion calculations using SI units. It is not intended for use as the basis for specification of equipment in practice. For example, motors are often rated based on the starting torque requirements, which can be much higher than the normal load conditions. One should refer to the appropriate literature for specifications of equipment for a given application.
