ELEC971 High Voltage Systems Electrical Power Cables Part Cable Rating Calculations The calculation of cable ratings is a very complex determination because of the large number of interacting characteristics and parameters involved in the establishment of a heat balance (for steady state ratings) or general heat dissipation/heat generation equation (for transient ratings). The cable heat dissipation is achieved by means of thermal conduction to the cable surface (an inefficient heat transfer process compared to convection and radiation transfer). From the cable surface heat is then dissipated by a variety of means, depending on the cable installation, the ambient conditions and the general configuration. The cable thermal properties and also the thermal characteristics of the environment of the cable installation are thus very important in the rating calculations. 1. Cable installation methods Cables are installed in a very large variety of environments, including the following examples: ELEC971: Power Cables Part p. 1/3
Direct burial in the ground with no special backfill material Burial in a trench packed with thermally enhanced dissipation material Installation in ducts or pipes buried in the ground Installation in a duct or pipe situated in open air Installation in large diameter tunnels Installation directly in open air situations Undersea: open or buried In troughs with circulating water In addition to the usual case of internal heat generation in a single cable (self-heating), it will generally be the case that cables will be installed in closely-coupled three phase groups or even double circuit arrangements of six cables. The result of this close interaction will be that mutual heat generation between cables must also be taken into consideration when determining the heat generation level. In all of these cases the thermal properties, including the thermal resistances, of the material surrounding the cable must be known in detail so that the heat dissipation rates from the cable surface to the infinite heat sink at ambient temperature can be determined. It is also necessary to determine such details, particularly the internal thermal resistances, for the cable and its component parts and materials. ELEC971: Power Cables Part p. /3
In particular, the following are the important material characteristics required: thermal resistivity of insulation (g: o C/W or thermal Ω.m) specific heat of insulation and of the metal (c: J/kg/C) mass density of the insulation and the metal (δ : kg/m 3 ) The thermal resistivity (g) of the various materials is of major importance in determining the thermal resistance of various parts of the thermal circuit used for steady state rating calculations. The thermal diffusivity of the insulation, defined as: 1 α = [m /s] gc.. δ is also an important parameter for transient rating calculations of buried cables. In this case the heat storage characteristics are important. Typical values of the above quantities for some relevant materials are shown in the accompanying table. The thermal diffusivity of different soil materials does not vary very much, being generally in the range of about 1.0 ± 0.8 m /sec. An average value of about 0.6 is often used. A value of soil thermal resistivity of 1. thermal Ω.m is often used. Typically insulation materials will have thermal resistivities of about 5-6 thermal Ω.m. ELEC971: Power Cables Part p. 3/3
. Heat generation in cables In addition to the heat dissipation details, the heat generation characteristics of all of the relevant cable materials are required. These details have been outlined previously in these notes. Typical values of thermal properties of materials used in the design and installation of power cable systems. ELEC971: Power Cables Part p. 4/3
3. Cable component thermal characteristics The general thermal characteristics of the component parts of cable systems that are of importance in the rating calculation process are: (1) Main conductor Ohmic heating losses No thermal resistance () Main insulant Dielectric losses at high voltages (electrical) Calculable thermal resistance (3) Metallic sheath Eddy current losses No thermal resistance (4) Insulating sheath No losses Calculable thermal resistance (5) Bedding No losses Calculable thermal resistance (6) Armour Eddy current losses No thermal resistance (7) Serving (insulating No losses (outer sheath) Calculable thermal resistance (8) Metal pipe (duct) Eddy current losses No thermal resistance (9) Insulating duct No losses (pipe) Calculable thermal resistance ELEC971: Power Cables Part p. 5/3
In addition to the thermal characteristics of the cable components, the thermal characteristics of the cable installation environment are also required: these include, for example, the ground (soil) thermal resistance, the air thermal resistance and the associated heat capacities and diffusivities etc. If the cable is in open air, the radiative and convective dissipation properties are also needed. 3.1 Equivalent Steady State Thermal Circuit The general thermal equivalent circuit used for steady state rating calculation is shown below. It is an electrical-thermal analogy using the heat generation as the equivalent current sources, the thermal resistances for electrical resistances and the temperature as the voltage potential analogue. From the equivalent circuit it can be seen that it is necessary to calculate accurately the thermal resistances and the heat generation losses and to be able to specify the fixed temperatures at the conductor and at the ambient condition (the ground surface, for example). The unknown is the current level (the thermal rating) which will be determined by the heat balance at steady state operation. ELEC971: Power Cables Part p. 6/3
conductor insulation G i surface bedding serving dissipation sheath armour surface ambient T s G T b G s G o c T a T o T A I R W d λ 1 I R λ I R Thermal equivalent circuit for a single core cable with metal sheath and armour. T c = maximum allowable conductor temperature T A = ambient temperature T s = sheath temperature T a = armour temperature T o = cable outer surface temperature W d = main insulation dielectric heat loss per unit length I R = conductor ohmic heat generation at rated current (I is the unknown) R = AC resistance of the conductor per unit length λ 1 = (conducting) eddy current sheath loss coefficient λ = armour eddy current heat loss coefficient G i = main insulation thermal resistance G b = bedding thermal resistance ELEC971: Power Cables Part p. 7/3
G s = serving or insulating sheath thermal resistance G o = effective thermal resistance between the cable surface and the ambient G o will depend on the nature of the cable installation and may ultimately be the most difficult quantity to evaluate accurately, particularly when the cable surface is open to air or fluid environments. Note that the (distributed) dielectric heat generation W d is normally included at the mid-point of the insulation resistance, so that W d passes through only half of the thermal resistance. It can be shown mathematically that the temperature rise of insulation due to dielectric heating is: 1 Δ T = W G d i Thus it is necessary to place the dielectric heat input at the midpoint of G i in the equivalent circuit. 3.1.1 Specified Conductor Temperature The maximum conductor temperature to be used in the calculation will be that specified by the manufacturer as the maximum permissible steady state temperature of the insulation material that is in contact with the main insulation. Some typical upper limits of temperature are shown over. ELEC971: Power Cables Part p. 8/3
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3.1. Ambient Temperature The ambient ground temperature (or the general main heatsink temperature in other environments) to be used for rating determinations is normally specified as a weighted average. For example, in Australia, an ambient ground temperature of 5 o C is normally used whenever rating tables such as those given in Standards are utilized. If the rated values of a Standards table of ratings are to be varied for different ambient temperatures, the following multiplying factor is used to adjust from an ambient of 5 o C to another ambient of T o C. I T A = I 5 T T T cmax A cmax 5 where T cmax = maximum allowable conductor temperature T A = the new ambient temperature applicable 3. Equivalent Thermal Circuit for Transient Effects When considering transient thermal ratings, which are perhaps a more common requirement than steady ratings, it is necessary to take thermal storage elements in the cable structure into account in the cable s equivalent circuit. In the most general case this means determining an equivalent thermal capacitance Q (i.e. ability to store the heat) for all of the cable components, as shown in the circuit ELEC971: Power Cables Part p. 11/3
diagram below. This will require use of the material thermal capacities and mass densities. I R W sh Q c G i dielectric pq d G b ( ) 1 pq d P T A (ambient) conductor sheath Equivalent circuit for transient heating of cable. In the particular case of the main dielectric, it is necessary to split its thermal capacitance Q d and divide it between the conductor and the sheath, as below. The sharing factor p (Van Wormer coefficient) is determined assuming a logarithmic temperature distribution in the dielectric material. The factor p can be determined from the following equation: 1 1 p = ln R R 1 r o ro R is the outer dielectric radius and r o is the inner dielectric radius. Note that p = 0.39 when R/r o =.0. ELEC971: Power Cables Part p. 1/3
4. Rating Calculations Steady state rating calculations for cables are performed according to the International Electrotechnical Committee Standard IEC 6087. This specifies uniform methods of calculation. The preceding details and equivalent circuits are based on this Standard. For coaxial cables, the various thermal resistances used in the equivalent circuit are determined from the general formula: G i g i d i = ln thermal ohms/metre π dc where d i is the outer diameter of the insulation layer, and d c is the inner diameter of the insulation layer. This comes from the analogy with the shunt electrical resistance of a coaxial insulation system. From the thermal equivalent circuit on the previous page, the determination of the current rating I for a maximum conductor temperature proceeds by logical steps: T = T + W + 1+ λ + λ I R G e.g. we use ( ) a o d 1 s Note that: T T = ( T T ) + ( T T ) + ( T T ) c A c s s a a A Proceeding in this manner, we find the thermal rating is given by: ELEC971: Power Cables Part p. 13/3
I 1 Tc TA Wd Gi + [ Gb + Gs + Go] = RGi + R( 1+ λ1) Gb + R( 1+ λ1+ λ)( Gs + Go) 1/ The thermal resistance G o, representing the heat dissipation from the cable surface to the ambient, must be determined from the installation configuration and the environment details. The various possibilities that may occur are discussed below. 4.1 Cables in air In this case thermal dissipation from the cable surface will be by a combination of convection and radiation and G o will be determined from the total heat dissipation coefficient of those two mechanisms. ht = hc + hr H = hat T = haδ T we have ( ) T A T but also GH o =Δ T [thermal Ohm s law] thus G o = 1 ha T For example, for a cylindrical cable of diameter D in open air, ELEC971: Power Cables Part p. 14/3
hence: and: ( π )[ ] 5/4 H = k D T T s [ ] 1/4 h = k T T G T s A o 1 1 = = ha k D T T A ( π )[ ] 1/4 T s A Thus, G o is a function of T s and this relationship must be found and used in order to get G o, unless we can specify h T exactly. [Sometimes a general value of h T 10 W/m /K is used for normal operating temperatures and ambient conditions, but more accurate estimations should be used whenever available.] A more accurate rating calculation technique is to use iteration to find the surface temperature T o. The method used is as follows: From the equivalent circuit: [ ] T T = I R G + G + G + G c A i b s o + W Gi + G + G + G + λ + + d b s o [ b s o] [ G ] 1I R G G G I R Gs o + λ + This can be rewritten as follows: ELEC971: Power Cables Part p. 15/3
[ ] T T = I R G + G + G c A i b s Gi + W + G + G d b s λ1i R[ Gb Gs] λi R[ Gs ] [ T T ] + + + + o A Hence: Gi ( T T ) ( T T ) W + G + G c A o A d b s ( 1 λ ) ( 1 λ λ ) = + + + + + I R Gi Gb 1 Gs 1 [eq.1] But the total heat dissipation from the surface is: ( + λ + λ ) + = ( π )[ ] 5/4 I R 1 1 Wd k D To TA Thus: I R = ( π )[ ] 5/4 k D T T W o A d 1+ λ + λ 1 Substitute into eq.1 above and then solve by iterative means (e.g. method of bisection) for T o. ELEC971: Power Cables Part p. 16/3
4. Cables in ducts, pipes etc. In this case G o is the sum of the various thermal resistances in series: i.e. G o = G cable-duct + G duct-ground surface + etc 4.3 Direct buried cables Here, G o is the thermal resistance between the cable surface and the ground surface: the ground thermal resistance as shown below. We use the method of images, the electrical thermal analogy and the two-wire line capacitance expression to determine the thermal resistance as follows: Capacitance of a line to ground: ELEC971: Power Cables Part p. 17/3
C = πε ln / ( h a) Using the general equality RC insulation material, we have ( h a) ρε ln / ρ R = = ρε = ln / C πε π = ρε for an electrical ( h a) Thus, using the electrical-thermal analogy, we have, for the ground thermal resistance: G o g h = ln thermal ohms/m π a where: g = soil thermal resistivity h = depth of burial a = cable outer radius ELEC971: Power Cables Part p. 18/3
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4.4 Multi-core cables For three phase cables, there will be mutual heating between the cores and the thermal resistances G i will be affected by this mutual heating. This is particularly true of belted cables. The calculation will be more complex as a result, in this case. t 1 r 0 t A 3-core belted-type cable For the case shown above, an approximate formula (Simons formula) is given by: G i where: g t t t = + + π 1 1 0.85 0. ln 8.3. 1 t t ro t 1 = insulation belt thickness t = conductor insulation thickness r o = conductor radius The G i above are for each core s thermal resistance to the outer surface of the belt. ELEC971: Power Cables Part p. 0/3
G i G i G b G s G i sheath 4.5 Cables in ducts For cables in ducts, an empirical expression is used, followed by iteration to get the rating. A G = 1 + ( B + CTm) De where: values of A, B and C are given in IEC 6087. T m is the mean temperature of the duct filling material D e is the equivalent diameter of the cores. [For a single core cable: D e = OD of cable two cores cable: D e = 1.65 x OD three core cable: D e =.15 x OD ] The thermal resistance is the thermal resistance between the pipe and the cable surfaces. The calculation procedure requires an initial educated guess of T m and then iteration. Following is an example of a calculation by this method. ELEC971: Power Cables Part p. 1/3
Example: Determine the rating of a 500kV, 60Hz, three-phase oil pipe type cable buried in ground. The details are: Pipe loss factor = 0.8 (eddy current loss) Conductor OD = 4.14 cm Insulation thickness = 3.40 cm Pipe diameter = 30 cm Burial depth = 100 cm Soil resistivity = 0.9 o C.m/W Ambient temperature = 5 o C AC resistance = 0.044 Ω/ph/km Insulation: g = 5.0 o C.m/W tanδ = 0.00 T max = 80 o C Thermal resistance from pipe to cable is obtained from: G cp A = 1 + B + CT D ( ) max where (from previous table) for an oil pressure pipe type cable: A = 0.6 ; B = 0 ; C = 0.006 D e =.15 x O.D. of core =.15 x (4.14 + x 3.4) cm = 3.5 cm e ELEC971: Power Cables Part p. /3
We take T m as 55 o C for first guess. Then: 0.6 G cp = 1 + 0.006 55 3.5 ( ) = 0.06 thermal Ω/m For the cable insulation: 5 3.5 G i = ln = 0.77 thermal Ω/m π 4.14 For the ground: 0.9 100 G o = ln = 0.37 thermal Ω/m π 15 Thus the equivalent circuit is: G i I R AC 0.77 G cp G o I R AC I R AC 0.77 0.77 0.06 0.37 T A = 5 o C 80 o C ( ) I R = I 0.044 10 = 4.4 10 I W/m/phase 3 5 AC 5 Pipe loss 3 0.8 ( 4.4 10 I ) = W/m Dielectric loss: Po = ωcv tanδ ELEC971: Power Cables Part p. 3/3
but: πε C = = 00 4.14 + 3.4 + 3.4 ln 4.14 hence: P o 3 pf/m/ph 1 500 10 = 314 ( 00 10 ) 0.00 3 = 10.47 W/m/core W/m T C T = 80 5 = 55 A o C Hence: ( 5 I )( ) ( 5 I )( ) 55 = 4.4 10 0.77 + 3 4.4 10 0.06 + 0.37 0.77 + 10.47 + 3 10.47 ( 0.06 + 0.37) + 3 0.8 4.4 10 5 I 0.37 i.e. 37.46 10.44 10 I ( ) I = A 5 = 599 then use this to check T m guess. T m 5 ( ) ( ) = 5 + 0.37 3 10.47 + 3 1+ 0.8 4.4 10 599 = 5 + 34 = 6 o C [c.f. 55 o C] This is too high. Thus try T m =58 o C next and repeat until agreement is reached. ELEC971: Power Cables Part p. 4/3
5. Transient Heating of Cables When operated under non-steady state conditions, we can define three different transient ratings for cables. Because of the high thermal capacity of power cables, particularly when buried in the ground, these transient ratings are more relevant and more often used than the steady state rating determinations because of the long time constants associated with the cables. The three transient ratings used are: (i) Short circuit rating: Determined from adiabatic heating of the cable under short circuit. (ii) Cyclic rating: Determined using specified daily load cycles (iii) Emergency Rating: Usually calculated for 1- hours of overload conditions 5.1 Short circuit rating This is calculated for a specified maximum (short circuit) temperature T sc and a specified protection operating time (t sc ) to clear the fault current. The equation previously quoted for adiabatic heating is used: ELEC971: Power Cables Part p. 5/3
Itsc ( T( t ) T ) ( ) δ c 1+ α sc o = ln A ρα 1 + α ( T 0 To ) where: δ = conductor mass density c = insulation specific heat ρ = conductor electrical resistivity α = temperature coefficient of resistivity for ρ I = short circuit current rating t sc = short circuit duration (protection operating time) T o = cable surface temperature T(0) = initial conductor temperature T(t sc ) = T sc = maximum permissible temperature For example, typical maximum permissible (short circuit) temperatures are: Main conductor: 10 o C Aluminium sheath: 00 o C Lead sheath: 00 o C In some cases the sheath temperature limit may be the determining factor rather than the conductor maximum temperature. The following example illustrates the case. Example: A 500 mm three-core Al conductor, screened cable with lead sheath and steel armour has a continuous rating of 480A for a ELEC971: Power Cables Part p. 6/3
T c of 70 o C. It is an 11 kv cable. The sheath O.D. is 6.96 cm and the sheath thickness is 0.9 cm. At rated current, T c = 70 o C, T s = 66. o C, and the ambient T A = 0 o C. Find the maximum short circuit currents for T ( max) = 10 o C and T c ( max) = 00 o C if the protection operating s time is 1 second. (a) Al conductor: I sc A δ c ( ) ( ) 3 3.7 10 0.19 4.18 10 1 + α 10 0 = 8 ln.83 10 0.00403 1 + α 70 0 ρ α 1 4 8 1.403 = 5 10 1.47 10 ln 1.0 i.e. I sc ( ) = 8.9 ka (b) Lead sheath: Cross-section area Initial T s = 66. o C. Thus: = π d thickness = π 6.818 0.9 = 6.1 cm 1 ELEC971: Power Cables Part p. 7/3
I sc A δ c ( ) ( ) 3 3 11.34 10 0.0305 4.18 10 1 + α 00 0 = 8 ln 0.65 10 0.004 1 + α 66. 0 ρ α 1 i.e. I sc ( ) 4 7 1.7 = 6.1 10 4.18 10 ln 1.18 = 15.8 ka Thus, the lead sheath is the limiting factor. (c) Note that if an Al sheath is used instead: I sc A ( ) ( ) 8 1+ α 00 0 = 1.47 10 ln 1 + α 66. 0 i.e. I sc ( ) = 4 8 6.1 10 1.47 10 0.61 = 56 ka 1 1 5. Cyclic rating of cables The main problem in determining the rating in this case is the part played by the thermal capacity of both the cable and ground materials. Thermal diffusion into the ground must be considered. Thus the soil thermal diffusivity (α D ) is important: this quantity incorporates thermal conductivity ELEC971: Power Cables Part p. 8/3
(resistivity), density and specific heat of the soil: α D = k ( cδ ). To do this we must use a thermal equivalent circuit which includes energy storage elements, with the thermal capacitance included as well as thermal resistance. The following diagram shows the equivalent circuit used: T c T s T A G i G o Q 1 Q Q 3 where: Q 1 is the thermal capacitance of the conductor plus a fraction (p) of the thermal capacitance of the electrical insulation. Q is the thermal capacitance of the sheath plus a fraction( 1 p) of the thermal capacitance of the electrical insulation. Q 3 is the effective thermal capacitance of the earth (this is the difficult estimate). The fraction p is calculated from: ELEC971: Power Cables Part p. 9/3
where: 1 1 p = ln ( R/ r) ( R/ r) 1 R = the outer radius of the electrical insulation r = the conductor radius The problem is the determination of Q 3, the ground thermal capacity. This is dependent on the soil thermal diffusivity which can vary with soil condition. The simplest way is to divide the ground around the cable into a distributed set of coaxial cylinders, as shown below: T A G 1 G G 3 G 4 Q 1 Q Q 3 Q 4 For more accurate analysis the problem is modeled using a full distributed system which then requires use of the exponential integral Ei(x), defined as: ( ) Ei x e = t x t dt This function arises when a linear heat source is considered to be a continuous distribution of point sources of heat along a line. ELEC971: Power Cables Part p. 30/3
Using this function and the model of the cable and its mirror image, the cable surface temperature rise as a function of time is given by the equation: gw T d h To () t = Ei + Ei 4π 16αDt αdt plus any contributions from mutual heating by other cables where: d = cable outer diameter h = burial depth W T = total losses per unit length g = soil thermal resistivity [ = 1 k] = 1 gcδ α D = soil thermal diffusivity ( ) c = soil specific heat δ = soil mass density t = time 5.3 Emergency ratings Emergency ratings are determined for 1 or hours of overload with the temperature of the insulation not to exceed the maximum limit. It is also complicated by the thermal diffusivity factor as with the cyclic rating. Emergency rating are an important consideration and are often used. The overload rating can be quite high because of ELEC971: Power Cables Part p. 31/3
the high thermal capacity and thermal diffusivity of the cable and surrounding ground. ELEC971: Power Cables Part p. 3/3