With the advent of systems at frequencies above 60 Hz, we've seen increased interest in the rating of cables in these applications. Mainframe computer systems, running at 400 Hz, certainly come to mind here. There are several factors that significantly influence conductor resistance as the frequency increases. Let's talk about each in detail and provide a suggested method for determining conductor

With the advent of systems at frequencies above 60 Hz, we've seen increased interest in the rating of cables in these applications. Mainframe computer systems, running at 400 Hz, certainly come to mind here.

There are several factors that significantly influence conductor resistance as the frequency increases. Let's talk about each in detail and provide a suggested method for determining conductor ampacity at 400 Hz as related to that at 60 Hz.

### Key factors

In general, you would design 3-phase, 400-Hz power systems the same way as 60-Hz systems, but you must realize that the higher frequency will increase the skin and proximity effects in the conductor. This, in turn, will increase the effective conductor resistance. The increased frequency will also increase circuit reactance, which when combined with resistance, will increase voltage drop.

The higher frequency will also increase the effect of magnetic materials on cable reactance and heating. It's for this very reason you should not run 400-Hz conductors in magnetic conduit or too close to magnetic structures in a building.

The losses due to frequency are proportional to the square of the line current. So for very small currents, these losses may be negligible. For large currents, however, they can be significant.

### Cables in air and nonmagnetic conduit

We'll limit our discussion here to single conductor 600V cables. For frequencies up to about 1,000 Hz, you can assume that reactance is directly proportional to frequency. You can neglect the reduction in inductance due to frequencies up to this magnitude, because it's minimal — and the purposely introduced error is negligible. For frequencies higher than 1,000 Hz, however, you should include the inductance correction. In these cases, the inductance is defined by the following equation:

L = ( [0.1404 × log_{10} 2S/D_{C}] + [0.0153 × L/L_{O}] ) × 10^{-3} Henries per 1,000 ft

Where,

L = inductance to neutral

L/L_{O} = correction factor from **Table 1** (click here to see **Table 1**)

S = axial spacing between conductors (in.)

D_{C} = Conductor diameter (in.)

### Calculating resistance ratios

From the original Neher McGrath paper titled “The Calculation of Temperature Rise and Load Capability of Cable Systems,” we know that for any cable system, the AC/DC resistance ratio of the conductor (R_{AC}/R_{DC}) is expressed by the following equation:

R_{AC}/R_{DC} = 1 + Y_{C} + Y_{S} + Y_{P} (Eq. 1)

Where Y_{C}, Y_{S}, and Y_{P} are the effects due to the conductor, sheath, and conduit, respectively. Because we're considering only nonmetallic sheaths in air or nonmetallic conduit, we have to be concerned only with Y_{C} in our discussion.

The Neher McGrath paper also shows that you can express Y_{C} as follows:

Y_{C} = Y_{CS} + Y_{CP}

Where Y_{CS} is the conductor component due to *skin effect* and Y_{CP} is the conductor component due to *proximity effect*.

So taking into consideration the above comments, Equation 1 becomes:

R_{AC}/R_{DC} = 1 + Y_{CS} + Y_{CP} (Eq. 2)

Both the skin and proximity effects are a function of F_{(x)} which for solid and concentric round conductors is given in Table 1. Thus, for Y_{CS} and Y_{CP}, we have the following equations:

Y_{CS} = F_{(x)} (Eq. 3)

Y_{CP} = F_{(x)} K^{2} × [ (1.18/F_{(X)} + 0.27) + 0.312K^{2} ] (Eq. 4)

Where,

K = D_{C}/S,

x = 0.0276 × √f/R_{DC}

f = frequency in Hz

R_{DC} = conductor direct current resistance at operating temperature (ohms per 1,000 ft)

D_{C} = conductor diameter (in.)

S = axial spacing between conductors (in.)

F_{(x)} = function of x from Table 1

**Table 2** (click here to see **Table 2**) lists the AC/DC resistance ratios as calculated from Equations 3 and 4. *Note*: B = f/√R_{DC }and K = D_{C}/S

You can determine the ampacity of a given conductor size at 400 Hz by multiplying the 60-Hz rating by the corresponding derating factor. For the larger sizes, the deratings are on the conservative side in that they are actually based on DC ratings of the conductors. You can consider the small error, on the safe side, as negligible in view of the possible variations in insulation thicknesses, etc.

In addition, the resistance ratios provided here do not include losses in any metal sheath, armor, or conduit. Losses in a thin aluminum armor tape may be small, but those in a metal conduit could be large enough to cause trouble if currents are large.