Does Voltage Stabilizer increases Electricity Bill ?
The answer is No.
How much energy we are consuming decides our electricity bill i.e. billing is done on kilowatt-hour basis (after some fixed charge) used by our electrical equipment to perform mechanical work or lighting. A kilowatt-hour is the unit of electricity, known as Board of Trade Unit. If your machine is at full load for 1 hour consumes more power to do useful work as compared to that machine runs at half load for the same amount of time. Reactive demands are also charged in industry. Tariffs are fixed by regulatory commissions. The more is the kilowatt intake i.e. active or useful power consumption, the more we have to loosen our pocket.
Electrical Input = Output + Losses.
If Losses rise for the same output, we require more input power to our equipment, so that we end up spending more on losses. Here, my scope is to give an explanation of power consumption trends owing voltage fluctuations.
Case 1– Normally the static load used are resistive load – They are supposed to work at 220V.
Suppose an equipment is running is designed for 1000w and 220v supply.
It will consume 1000W power at 220 V, then the corresponding current will be 1000/220=4.54A. These values are the rated values.
If the voltage drops, the load current will be increasing for the same amount of work. The increase in current shall compensate the voltage drop across the equipment, performance of the equipment shall go down. (If voltage=150V, current= 1000W / 150V = 6.66A).
4.54A to 6.66A is the rise in current. Flow of current occurs due to the flow of electrons inside a conductor in a closed path. When electrons collide, they generate heat. As this heat is dissipated to the atmosphere it is a loss termed as copper loss of the conductor.
Copper loss = I2 *R ;
I = current flowing in the circuit; R = Resistance offered by the conductor to current flow
This means that with a rise in current flow, heat dissipation I2 *R loss increases. For the same reason incandescent bulbs get heated up at low voltage range. This will add to the power consumption levels. That is why it is not encouraged to turn on high current drawing devices during peak hours of the day.
Similarly on higher voltage side, the current drawn shall be reduced that incurs less loss in the line on account of heat loss above rated kilowatt. However, there will be a higher amount of stress on insulation beyond voltage limit. Sometimes, short circuit condition prevails on such condition.
Case 2– Electronic devices like TV, UPS come with Switch Mode Power Supply. So, voltage stabilization against high or low voltage levels is absolutely unnecessary. SMPS itself does that job. The reason behind is- as they have voltage control elements like transistor or zenor diode which dissipates power equal to the voltage difference between an (unregulated input voltage and a fixed supply voltage) multiplied by the current flowing through it. The switching regulator acts as a continuously variable power converter and hence its efficiency is negligibly affected by the voltage difference.
Case 3 – Inductive load of induction motor- It is called work horse of the industry. It has operating voltage range as well as mechanical load condition on it, which are to be considered. There is a catch here. We need to choose a motor according to your expected load.
For lower voltage case, we definitely see the voltage parameters going down to make power input higher, however, increasing the power consumption. The difficulty is, heat due to the copper loss in the rotor. Rotor copper loss of the induction motor is slip times (0.02 to 0.04) rotor input.
Rotor copper loss α Slip (for constant for constant load torque)
And Slip α 1/(voltage)2
∴ Rotor copper loss α 1/(voltage)2
So, using the rotor copper loss and voltage relationship, it is found out that, for a constant load on the motor, 10% reduction in voltage can cause 23.46% increase in rotor copper losses as compared to specified ratings. Hence power consumption is on the higher side.
We have to pay for, Load power + 123.46% of full load copper loss –> Rate of consumption increasing
As a thumb rule every 10-degree rise in temperature (heat dissipated due to copper loss) can reduce the life of insulation to 50 percent. Motor torque should meet the load torque for useful work. Again, Torque is the important mechanical parameter of a motor. The force due to rotation of the shaft is the torque. The motor has to drive or rotate the load. It has to meet the demand. However, we see significant changes in motor torque, and other associated torque specification as in starting torque, stalling torque etc. Even it is not possible to start the motor due to a deficit in starting torque at lower voltage operation.
As Torque is proportional to the square of the voltage, even a small change in negative side below rated value in voltage can make a substantial reduction in torque values.
Torque α voltage2
Form this relationship, we can conclude, 10% reduction in voltage, can cause 19% reduction in torque. The motor is not run under this situation. Overload protection relays always trip to kick the motor offline way below alarming condition and the motor won’t generate horse power.
At full load and high voltage cases, exceeding the voltage by a small range is not very harmful, the motor magnetizing current and loss increases a bit. (It is noted that magnetizing current is that current component, which energizes the motor electromagnet.) Copper losses will be less due to high voltage. It is normally true that motors tend to run cooler at specified horse power at voltages exceeding rated voltage by up to 10%. It will reduce power input to some extent. However, if the Over-voltage is beyond 110% (approx..) of rated, magnetic characteristic of core goes to saturation and saturation is the condition in which, after a magnetic field strength becomes sufficiently large, further increase in magnetic field strength produces no additional magnetization in the magnetic material. It can raise the current and temperature. The core loss is 20 to 30% greater than normal and could cause the machine to overheat.
Thus, High voltage even across the lightly loaded motors may cause to operate in the saturation region of B-H curve. Saturation region of B-H curve is that part where magnetic materials get saturated. Obviously it is not good for its health due to more copper losses in the machine. Power factor also drops sharply with it (undesirable). Thus there will be an increase in Electricity Bill
It is always wise not to play with the specified ratings of the induction motor. Rather play safe. Electricity bill will be calculated on power consumption level according to your use. It up to owner’s wish to co-relate the functionality and consequences involved and make a wise choice.
When estimation of the load is done, then rated value is calculated for every practical reason expecting that the machine would work at the specific design value most of its lifespan. Creep life of your equipment will be following normal creep curve as expected. If we have to do more work implies we have to increase specification and pay the corresponding electricity fixed and variable cost. The better way is to reduce reactive power demand. Power factor should be maintained near unity. One can have the reduced Electricity Bill by installing Servo Voltage Stabilizer.
Save Electricity, Save Penny.