The economic effect of introducing an automatic condenser installation consists of the following components:
- Saving on the payment of jet energy. Payment for reactive energy ranges from 12% to 50% of the payment for active energy in various regions of Russia. As practice shows, the cost of a capacitor unit for reactive power compensation pays off six months to a year after implementation
- For existing facilities, reduction of electric energy losses in cable lines due to reduction of phase current values;
- For the designed facilities, the introduction of a capacitor unit at the design stage allows you to save on the cost of cable lines by reducing their cross-section.
On average, 10 … 15% of the active energy consumed is lost in operating facilities in the supply cables. Losses are proportional to the square of the current flowing through the cable. For calculations, we take the loss coefficient Kp = 12%.
Consider the economic component of the operation of a compensating installation using the example of an existing facility.
Before the introduction of the automatic capacitor installation cos φ = 0.60
After the introduction of the automatic capacitor installation cos φ = 0.97
The relative active component of the current (coinciding in phase with the voltage) is assumed to be equal to unity.
The relative total current is up to implementation I _{ 1 } = 1 / 0.6 = 1.667
The relative total current after implementation is I _{ 2 } = 1 / 0.97 = 1.03
The reduction in active power consumption will be: ΔWc = [(I _{ 1 } ^{ 2 } -I _{ 2 } ^{ 2 } ) / I _{ 1 } ^{ 2 }] · CP · 100% = 7.42%
i.e. in this example, the cost of active energy decreased by 7.42%.
In the general case, for an active facility, a decrease in active energy consumption due to an increase in cos φ ΔWc = {[1 / cos ^{ 2 } φ _{ 1 } – 1 / cos ^{ 2 } φ _{ 2 }] / [1 / cos ^{ 2 } φ _{ 1 }]} · CP · 100%
where:
- cos φ1 – cosine phi before compensation (for example 0.6)
- cos φ2 – cosine phi after compensation (for example, 0.97)
- Kp – loss coefficient Kp = 0.12
Then, for our example, ΔWc = 7.40%
Annual savings C in electricity bills C = (ΔWc / 100%) · T = 0.074 · T
where:
- T – the cost of electricity consumed per year
Payback period, years: Tr = Stu / S
where:
- Stu – the cost of the capacitor unit (CRM or UKM 58 04);
- C – annual savings for electricity.