I am utilizing one of Schneider electric's energy meters for measuring energy consumption. This meter shows both active and reactive components. I decided a small test of measuring Watts / Hour. At 3:30 pm i got reading (Active = 1.8 kWh ) (Reactive= 614 kVArh) while at 4:30 pm i noted (Active= 148.6 kWh ) (Reactive=600 kVARh). In terms of apparent power we know its a component of active+reactive however the consumed one is the active power.
What could be the role of this reactive power in the load consumption or power loss if only active power is charged by the utility?
Dear Mahmmood,
welcome,
by definition the active power Pa= IV cos phi for single phase source, and the corresponding reactive power= Pr= IV sin phi,
So, knowing both, one can get the phase angle phi and the IV , the volt ampere supplied by the source. Accordingly, here in your example at 3:30, one has: tan phi = Pr/Pa= 614/ 1,8 one can get phi= 89.832 and the IV= Pr/ sin phi= 614/.999= 614.6 kVA.
at 4:30,
phi= 76.09 degrees, and the IV= 616 VA,
we see that the IV decreases as the active power increases relative to the reactive power. This fact is well known. So, as the power factor cos phi increases the the IV required will decrease for the same active power.
Unfortunately the power station must generate the required volt ampere. Which means more generated volt ampere for smaller power factor which rises the cost of the generators.
Therefore one undertakes the improvement of the load power factor to decrease the voltampere required by the power stations.
Best wishes
As Abdelhalim states, the generators don't like generating reactive 'power' because nobody (usually) pays for it. So the higher Cos phi, for them, the better. In addition the higher current levels of a given real power ( due to cos phi
Role of reactive power in the load consumption - Reactive power directly affects the voltage at a node. Any mismatch in Generation & demand of reactive power change the voltage.
Role in power loss - More reactive power means poor power factor/more inductive current for same real power demand, that imply more current which increases the loss. And, less reactive power means good power factor or less inductive current for the same real power demand, that means less loss.
There is indirect pricing scheme for reactive power known as power factor tariff. Also, separate mechanism for reactive power pricing comes in to picture after implementation of the concept of deregulated power system.
Hi
there are many type of calculation regarding power consumption .. depending on many variable and condition, maybe you should run some testing trough experimentation and get more result.
I don't see that deregulating the electricity power supply system can have anything to do with costing reactive power. The only thing deregulation of a public utility does is vastly increase the cost of power to the end user. Why? Because before you had a utility supplying power at a notionally not-for-profit basis; now you have four or more entities all 'publically' (privately) owned making as much money as they can while incidentally supplying power to a captive market. It's a bonanza for them; costing reactive power in the midst of this veritable torrent of cash is the least of their concerns.
hi, everybody,
in the 3-phase system meter, firstly how to calculate the PF? and then this PF to be use in Pa equation.
Hi everyone I have one day data for active power,reactive power,power factor values for every second for 24 hours. How can I calculate connected load with that data
- Badges
- Science topic
- Similar topics
- Social Science
- Sociology