Type a line (power factor correction 250 kW cos phi 0.78 to 0.95 440 V 60 Hz) or fill in the fields: the line and the fields stay in step.
Enter the load, in kW or kVA, with its present and target cos phi, and the calculator gives the capacitor kvar to raise the power factor: Qc = P x (tan phi1 - tan phi2). With the line voltage it gives the line current before and after and how much it falls, the capacitance per phase of the bank in delta or star and its rated current. A capacitor rating in kvar gives its capacitance and current, or its output at another voltage or frequency. It shows the working with your numbers, draws the power triangle, and gives a line you can copy or add to a request.
For a ship switchboard or a single motor, on 50 Hz and 60 Hz networks. The calculator gives no target cos phi, sizes no detuning reactor and does no harmonic study: it works at the fundamental frequency, and where the network carries harmonics a capacitor bank needs a separate check.
The reactive power of a load is Q = P x tan phi. To raise cos phi from 0.78 to 0.95 at 250 kW: tan phi1 = 0.8023 and tan phi2 = 0.3287, so the load takes 200.6 kvar now and 82.2 kvar after, and the bank gives the difference: Qc = 250 x (0.8023 - 0.3287) = 118.4 kvar.
The apparent power S = P / cos phi falls from 320.5 kVA to 263.2 kVA. On a 440 V three-phase network the line current I = S / (root 3 x U) falls from 420.6 A to 345.3 A, 75.3 A or 17.9 % less; the bank itself takes In = Qc / (root 3 x U) = 155.4 A.
In delta each capacitor takes the line voltage: C = Qc / (3 x 2 pi f x U^2) = 540.7 uF per phase for 118.4 kvar at 440 V 60 Hz. In star it takes the line voltage divided by root 3, so C = Qc / (2 pi f x U^2), three times as much: 1622.2 uF. The kvar of a capacitor rises with the square of the voltage and in step with the frequency.
| Quantity | Relation | Note |
|---|---|---|
| Capacitor kvar | Qc = P x (tan phi1 - tan phi2) | kvar; P in kW, phi1 and phi2 the angles of the present and the target cos phi |
| Active power from apparent | P = S x cos phi1 | kW; S in kVA |
| Reactive power | Q = P x tan phi | kvar, before and after |
| Apparent power | S = P / cos phi | kVA, before and after |
| Line current | I = S / (root 3 x U); single-phase I = S / U | A; U the line voltage |
| Capacitance per phase, delta | C = Qc / (3 x 2 pi f x U^2) | each capacitor on the line voltage |
| Capacitance per phase, star | C = Qc / (2 pi f x U^2) | three times the delta value; single-phase the same |
| Rated current of the bank | In = Qc / (root 3 x U); single-phase Qc / U | A |
| Another voltage or frequency | Q2 = Q1 x (U2 / U1)^2 x (f2 / f1) | 50 Hz to 60 Hz: x 1.2 |
U is the line voltage. The capacitance is per phase of a three-phase bank: delta unless star is given. Frequencies 50 and 60 Hz; with no frequency given, the capacitance is shown for both.
The relations are the standard ones for alternating current power: Qc = P (tan phi1 - tan phi2), Q = P tan phi, S = P / cos phi and S = root 3 U I are given in independent engineering references and study guides on power factor correction, and the calculator reproduces their worked examples (5 kW from 0.75 to 0.90 needs 1.99 kvar; 100 kW from 0.80 to 0.95 needs 42.1 kvar; 2000 kW from 0.72 to 0.95 needs 1270.5 kvar; 75 kW at 400 V from 0.72 to 0.97 falls from 150.35 A to 111.60 A). The relations for the capacitance per phase in delta and in star, for the rated current of a bank (200 kvar at 11 kV: 10.5 A) and for the change of capacitor kvar with the square of the voltage are each given in two independent references. The rise with the frequency, 20 % from 50 Hz to 60 Hz, follows from Q = 2 pi f C U^2, which two of them also give. That harmonics in the network call for a separate check of the capacitors is stated in the same references.
From the active power and the two power factors: Qc = P x (tan phi1 - tan phi2), where phi1 is the angle whose cosine is the present cos phi and phi2 the angle of the target. A load of 250 kW at cos phi 0.78 has tan phi1 = 0.8023; at 0.95, tan phi2 = 0.3287; so Qc = 250 x (0.8023 - 0.3287) = 118.4 kvar. From an apparent power the active power comes first, P = S x cos phi: 500 kVA at 0.8 is 400 kW, and raising it to 0.96 takes 183.3 kvar.
The current follows the apparent power, S = P / cos phi: on a three-phase network I = S / (root 3 x U), on single-phase I = S / U. The 250 kW load on 440 V takes 320.5 kVA and 420.6 A at cos phi 0.78, and 263.2 kVA and 345.3 A at 0.95: 75.3 A less, or 17.9 %. The fall is 1 - cos phi1 / cos phi2 of the present current, the same at any voltage.
It depends on the connection. In delta each capacitor takes the line voltage: C = Qc / (3 x 2 pi f x U^2) per phase. In star each takes the line voltage divided by root 3: C = Qc / (2 pi f x U^2) per phase, three times the delta value. A 50 kvar bank on 400 V 50 Hz has 331.6 uF per phase in delta and 994.7 uF per phase in star. The calculator assumes delta unless star is given, and says so.
Its rated current is In = Qc / (root 3 x U) on a three-phase network and Qc / U on single-phase. A 50 kvar bank on 400 V takes 72.2 A; 118.4 kvar on 440 V takes 155.4 A.
The reactive power of a capacitor is 2 pi f C U^2, so it changes with the square of the voltage and in step with the frequency: Q2 = Q1 x (U2 / U1)^2 x (f2 / f1). A 50 kvar 400 V 50 Hz capacitor gives 60 kvar at 400 V 60 Hz, 20 % more, and 50 x 1.21 x 1.2 = 72.6 kvar on 440 V 60 Hz, where the bus voltage is 10 % above its rated voltage.
The calculator gives no target: it works out the kvar for the target entered. It works at the fundamental frequency, and where the network carries harmonics, from drives and other non-linear loads, a capacitor bank needs a separate check that this page does not do.
The calculator gives the kvar, the capacitance and the current. When capacitors, capacitor contactors or other parts of a bank are needed, send the line from the calculator with the switchboard details and a photograph of the capacitor rating plate, and Vessel Core checks the details before an offer goes back.
