Using the Neher McGrath method for the calculation of the rating of buried cables under cyclic loading conditions allows for a daily, weekly, and yearly load factor in the same calculation.

Posted 2016-08-16
Categories: New feature, Theory

In a previous post, we explained the method by IEC and Neher McGrath for the calculation of the rating of buried cables under cyclic loading conditions. Cableizer uses the Neher McGrath method and now extended it with a daily, weekly, and yearly load factor in the same calculation.

### Introduction

Burying cables at great depths is more and more frequent using directional drilling and laying cables in deep tunnels becomes attractive. Several new tunnels with EHV cables were recently built (e.g. in London: 4 tunnels with a total length of 40 km built between 2009 and 2016).

The thermal environment for deeply installed cables is different than for installation at conventional depths and the IEC 60287-3-1 gives no guidelines for cable installed at shallow (<0.5 m), or larger depths (>3 m). The soil tends to have a higher moisture content at large depth due to the penetration of ground waters. Daily, weekly, and yearly load factors have a profound effect on the ratings of deeply buried cables, an effect which is not so pronounced for cables buried at the usual depths. In addition, the large amount of soil above the cables results in a large time constant of the thermal circuit and a slow heating of the cable conductor.

The new edition of the IEC 60287-2-1 from 2015 has not incorporated any new method related to deep burial of cables. We, however, implemented a method published by Mr. Eric Dorison in 2010 as a new feature in Cableizer.

### Theory

On top of the already existing load factor $LF$ we introduced the possibility to add a weekly load factor $LF_w$ and yearly load factor $LF_y$. If the user selects this option, the existing load factor will be set to a fixed transient period $τ$ of 24 hours, overriding a previous selection of time period from the drop down list.

#### External thermal resistances

The formula for $T_4$ with daily, weekly, and yearly load factor becomes:

 $- T_{4d} \left(- \mu + 1\right) + T_{4ss} \left(- \mu \left(\mu_{w} \mu_{y} - \mu_{w} - \mu_{y} + 1\right) + 1\right) - T_{4w} \mu \mu_{y} \left(- \mu_{w} + 1\right) - T_{4y} \mu \mu_{w} \left(- \mu_{y} + 1\right)$

This formula contains the three loss factors $μ$.

 $μ = LF^{2} \left(- k_{LF} + 1\right) + LF k_{LF}$ $μ_{w} = LF_{w}^{2} \left(- k_{LF} + 1\right) + LF_{w} k_{LF}$ $μ_{y} = LF_{y}^{2} \left(- k_{LF} + 1\right) + LF_{y} k_{LF}$

There is a transient thermal resistance for each load factor or time period respectively.

 $T_{4d} = \frac{\rho_{4}}{2 \pi} \log{\left (\frac{D_{x}}{Do_{d}} \right )}$ $T_{4w} = \frac{\rho_{4}}{2 \pi} \log{\left (\frac{D_{x w}}{Do_{d}} \right )}$ $T_{4y} = \frac{\rho_{4}}{2 \pi} \log{\left (\frac{D_{x y}}{Do_{d}} \right )}$

With characteristic diameters for daily, weekly, and yearly load variation.

 $D_x = D_{e} e^{\frac{2000}{D_{e} q_{x}} \operatorname{abs}{\left (\frac{\operatorname{k0}{\left (\frac{D_{e} q_{x}}{2000} \right )}}{\operatorname{k1}{\left (\frac{D_{e} q_{x}}{2000} \right )}} \right )}}$ $D_{xw} = D_{e} e^{\frac{2000}{D_{e} q_{xw}} \operatorname{abs}{\left (\frac{\operatorname{k0}{\left (\frac{D_{e} q_{xw}}{2000} \right )}}{\operatorname{k1}{\left (\frac{D_{e} q_{xw}}{2000} \right )}} \right )}}$ $D_{xy} = D_{e} e^{\frac{2000}{D_{e} q_{xy}} \operatorname{abs}{\left (\frac{\operatorname{k0}{\left (\frac{D_{e} q_{xy}}{2000} \right )}}{\operatorname{k1}{\left (\frac{D_{e} q_{xy}}{2000} \right )}} \right )}}$

With the factors for the characteristic diameters $q_{x}$:

 $q_{xd} = \frac{\sqrt{3} \sqrt{\pi}}{360} \sqrt{\frac{1}{\delta_{soil}}}$ $q_{xw} = \frac{\sqrt{21} \sqrt{\pi}}{2520} \sqrt{\frac{1}{\delta_{soil}}}$ $q_{xy} = \frac{\sqrt{487} \sqrt{\pi}}{87660} \sqrt{\frac{1}{\delta_{soil}}}$
These three formulae contain the corresponding time period $τ$ multiplied by 3'600.
 $τ =$ 24 hours (= 1 day) $τ =$ 168 hours (= 7 days = 1 week) $τ =$ 8766 hours (= 365.25 days = 1 year)

Actually, the formula for $T_4$ is calculated with four different formulae depending if loading is steady, with a general load factor, with a daily and weekly load factor or with a daily, weekly and yearly load factor. Please note that the fourth formula is also valid if you have a daily and yearly load factor and set the weekly load factor to 1.0 (continuous load):

CaseFormula
without load variation (steady state)$T_{4ss}$
with transient load variation$T_{4d} \left(- \mu + 1\right) + T_{4ss} \mu,$
with daily and weekly variation$- T_{4d} \left(- \mu + 1\right) + T_{4ss} - T_{4w} \mu \left(- \mu_{w} + 1\right)$
with daily, weekly, yearly variation$- T_{4d} \left(- \mu + 1\right) + T_{4ss} \left(- \mu \left(\mu_{w} \mu_{y} - \mu_{w} - \mu_{y} + 1\right) + 1\right) - T_{4w} \mu \mu_{y} \left(- \mu_{w} + 1\right) - T_{4y} \mu \mu_{w} \left(- \mu_{y} + 1\right)$

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