FORCED COOLING OF UNDERGROUND ELECTRIC POWER TRANSMISSION LINES PART II OF IV HEAT CONDUCTION IN THE CABLE INSULATION OF FORCE-COOLED UNDERGROUND ELECTRICAL POWER TRANSMISSION SYSTEMS by Joel V. Sanders Leon R. Glicksman Warren M. Rohsenow Energy Laboratory in association with Heat Transfer Laboratory, Department of Mechanical Engineering MASSACHUSETTS INSTITUTE OF TECHNOLOGY Sponsored by Consolidated Edison Co. of New York, Inc. New York, New York Energy Laboratory Report No. MIT-EL 74-004 Heat Transfer Laboratory Report No. 80619-88 May 1974 I ACKNOWLEDGEMENTS The authors express their appreciation to Dr. H. Feibus and to Mr. M. D. Buckweitz of Consolidated Edison for their kind assistance. The generous support of Consolidated Edison Co. of New York, Inc. who sponsored this work is gratefully acknowledged. C ABSTRACT Forced-cooled systems for oil-filled pipe-type cable circuits have recently been considered. In such systems the conduction resistance through the paper insulation of the cables is the limiting thermal resistance. Assuming bilateral symetry, steady-state conditions, and two-dimensional heat transfer, a FORTRAN IV computer program was written to solve the heat conduction problem in the cable insulation for arbitrary configurations of a three-cable system. For a steel pipe, a cable system is most susceptible to overheating in the equilateral configuration with the three cables touching. Proximity effects are very significant in forced cooling, especially when cable a nt provided with a copper tape under the insulation moisture seal assembly, accounting for as much as 21% of the total oil temperature rise between refrigeration stations. This figure, however, is reduced to 8% when 0.005 inch thick copper tape is present. de 4 TABLE OF CONTENTS . ..... * .. . * Page TITLE PAGE . . . . . . . ABSTRACT ..... ACKNOWLEDGMENTS TABLE . . LIST OF SYMBOLS LIST OF FIGURES . . . . . . . OF CONTENTS . . . . ... .. LIST OF TABLES . . .. CHAPTER 1. INTRODUCTION CHAPTER 2. FORMULATION The Cable Ins .ulation . . Path The Inter-Cable Conduction The Solution Domain ........ ihe Conduerosition Domain . . . PaMeth . . Boundary Cond[itions... itions ........ Variations onL the Statement Problem OF. .SOLUTIO. . . . . . . . . .. .. . 12 . . 3 . . . 4 . . . 7 . . . 9 . . . 11 * . . 12 . . . 19 . . . 19 . . . 19 .od . . . ... 1 . . 20 . . . 22 . . . 30 roblem . . . . . . .. ... ,ulatiProblems Nondimensiona1l Formulation . . .. . . .. * * 31 I OFSOLUTIONS . .. . . . 33 . . . 33 . . 34 . . . . . . . . . . 37 . . . 41 .* CHAPTER 3. SUPERPOSITION General . . * . . . The Overall Problem . . . . . . . . The Component tProblems Validity . . . . . . of t.he Supoerposition Method Determination of Conductor Temperatures CHAPTER 4. . . . . . . . . . 45 THE FINITE-DIFFERENCE METHOD . . . . . . . 49 Discretization . . . . of Domains .. . . * 49 Difference Form of Governing Equations and Boundary Conditions , . . * 55 5 Page CHAPTER 5. THE COMPUTER PROGRAM . . . . . . . . . . . . . The Coefficient Matrix . . . . . . Forcing General . . . . . . . Vectors Verification CHAPTER 6. . . . . 61 . . . . 61 . . . . . . . . . . 64 . . . . . . . . . . . . . 68 . . . RESULTS AND CONCLUSIONS 75 Evaluation . . . . . . 75 . . . . . . . . . . . . 76 Criteria Results Conclusions . . . . . . . .. Recommendations for Further Work . . . . . 92 . . . . . 92 . 95 CHAPTER 7. REFERENCES APPENDIX A. THE RELATIVE MAGNITUDE OF CONDUCTION AND APPENDIX B. . C. . . . . . . . . . . CONVECTION RESISTANCES . . . . . . . 96 INVESTIGATION OF THE CABLE-CONDUIT BOUNDARY APPENDIX 61 . CONDITION . . . . . . . . . . . 99 THE SOLUTION FOR MAXIMUM CURRENT . . . . . 108 . . . . 108 The Superposition Method for Solution 2 . . . .. . . . .. Maximizing Current in Solution 2 . APPENDIX D. E. . 117 THE DIFFERENCE FORM OF THE CONDUCTOR BOUNDARY APPENDIX . USER CONDITION INSTRUCTIONS . . . . . . . . . Geometry and Mesh Size ... Input Variables Output Variables Array Data Dimensions Card Example 126 . 126 .. . . 1. 132 . . . . . . . . 138 . . . . . . . . . . . . . 142 . . . . . . . . 143 . . . . . . 149 . 155 . . . . . . . 157 . . . . . . . . Assembly Problem 121 . . . . . . . . . . . . Capabilities and Limitations of the Computer Program Program Modifications . . . . . . . . . . . 6 Page APPENDIX F. LISTING OF THE SOURCE PROGRAM APPENDIX G. ONE-DIMENSIONAL SOLUTIONS FOR TEMPERATUREAND CURRENT . The H. Solution . . . . . . 220 M . 2220 . Genera . The APPENDIX Temperature 162 Current Solution 224 . . . . . . . CONSERVATIVE APPROXIMATE SOLUTIONS FOR MAXIMUM TEMPERATURE AND CURRENT . 227 * .* . 227 * . . 228 * . . 229 o . . . . . . . . .. The Temperature Solution The Current The Effective Solution . Perimeter .. . . ... . . . . . · . . 230 7 LIST OF SYMBOLS 2D = height of D3, measured along y-axis. D1 = solution domain consisting of the cable insulation of Cable 1. D 2 = solution domain consisting of the cable insulation of Cable 2. D 3 = solution domain consisting of the inter-cable conduction path. h = local convective film coefficient. h r = dimensionless radial spacing between mesh points in D 1. h = dimensionless azimuthal spacing between mesh points in D 2 . h = dimensionless azimuthal spacing between mesh points in D1 . h = dimensionless radial spacing between mesh points p in D2 I = current. jl = radial index in D. j2 = radial index in D2' j3 = normal index in D 3. k = thermal conductivity. kl = azimuthal index in D 1. k2 = azimuthal index in D 2. k3 = tangential index in D 3 . q = heat flow per unit length. q = heat generation per unit volume. 8 r = radial r coordinate = inner radius in D 1. of Cable 1 insulation. r2 = outer radius of Cable 1 insulation, r = dimensionless radial coordinate in D 1 . T = temperature, Toi = mixed-mean oil temperature outside the convective Oil boundary layer. T = maximum allowable system temperature. max Wd = dielectric loss per unit volume. W = arbitrary loss per unit length. W c = conductor loss per unit length. W d = dielectric loss per unit length. W s = sheath loss per unit length. x = normal coordinate x = dimensionless in D 3 . normal coordinate in D 3 y = tangential coordinate in D3 y = dimensionless tangential coordinate in D 3. a = azimuthal coordinate in D2. p = radial coordinate in D 2 P1 = inner radius of Cable 2 insulation. P2 = outer radius of Cable 2 insulation. p = dimensionless radial coordinate in D 2 . T- Toi 9 - W/k - dimensionless temperature. = azimuthal coordinate in D1 . 9 LIST OF FIGURES Page Figure 1.1 Cross-Section of Underground Pipe-Type Cable Figure 1.2 System . . . . . . . . . . . . . . . . . . . . . . . System . . . . . . . . . . . . . Figure 2.1 The Inter-Cable Conduction Path Figure Coordinate Systems Figure 2.4 The Cable-Cable and Cable-Conduit Boundary - 17 Conditions . . Surfaces of D . . . . . . . . . . . Figure 3.1 Curves Ci(x) Comprising the Boundaries of the Solution 21 Domain . . 24 . . . Convective . . . * * * - 27 * - 29 * - 35 A Regular, One-Dimensional Finite Difference Mesh . . . . . . . . . Figure 4.2 Discretization of the Domain D Figure 4.3 A Typical Mesh for the Equilateral Configuration . . . . . . . . . . . . . . 0 . . * * . . * 50 52 - 54 . 58 . 63 . 67 Nomenclature in the Neighborhood of Pj2,k2 in D 2 ........ . Points Affected by the Application of a Governing Equation at Various Locations in Figure 5.2 * 23 Figure 2.5 Figure 5.1 16 in D 1 , D 2 , and D 3 Regional Divisions in D1 and D 2 Figure 4.4 - . . . Figure 2.3 Figure 4.1 * Bilateral Symmetry of the Underground Cable 2.2 15 - * Cross-Section of Underground Power Cable Figure 1.3 . . a Discrete Network . . . . . . . . . A Radial Mesh Illustrating the Area Associated With Each Mesh Point . . . Page Figure 5.3 Percent-Error in Temperature vs. N2, the Number of Radial Subdivisions in D 2 . . . . . . . . . 71 1. . System System 1. Figure 6.1 Nomenclature for Cable Configurations Figure 6.2 Temperature Distribution for the . . 77 Equilateral-Pipe Configuration . . . . . . . . Figure 6.3 Figure 6.4 85 . Isothermal and Adiabatic Lines for the Open Configuration - System 1 . . * .* 88 . Isothermal and Adiabatic Lines for the 89 Cradled Configuration - System 1 . Figure 6.5 Isothermal and Adiabatic Lines for the 90 Equilateral Configuration - System the1 . . Figure 6.6 . Isothermal and Adiabatic Lines for the Equilateral-Pipe Configuration . . . Figure B.1 . . . . . . . . * . . 91 Fin Geometry for the Cable-Conduit Boundary Condition . . . . . . . . Figure B.2 Thermal Model of the Conduit Wall Figure D.1 Nomenclature for the Discretized . Boundary Condition Figure E.1 Effective Surfaces of Figure E.2 A Discrete Model of the Equilateral-Pipe Configuration D3 . . . . . Conductor Figure E.3 . . . . . . . . . . . . . . . . . . . . . . - 100 102 123 129 151 Computer Solution Printout for Example Problem . . . . . . . . . . . . 156 11 LIST OF TABLES Page Table 1 - Boundary Conditions in D1 and D 2 Table 2 - Values for the Physical Parameters of Table Systems 1 and 2 . . . . . . . . 80 4 - Solution 3 for Four Cable 81 Configurations - System 1 Table 5 - Solution 2 for Four Cable Configurations - System 2 Table 82 6 - Solution 3 for Four Cable Configurations - System 2 Table 78 3 - Solution 2 for Four Cable Configurations - System 1 Table 25 . . . . . . . . . 83 7 - Specification of Subdivisions Throughout Progra . Progra . D 1 , D 2 , and D 3 in Terms of the Computer Variables Table N(J) and M(J) . . . . . . . . 8 - Input Variables for the Computer . . . . . . . . . . . . . . Table 130 133 9 - Output Variables from the Computer . . . . . . . . . . . . . . Table 10 - Array Table 11 - Data Dimensions Card Assembly . . . . . . . . . . . . . . . . . . . . . . Table 12 - Input Data for Example Problem . . . . . 139 144 147 152 j_ CHAPTER 1 INTRODUCTION High-pressure oil-filled pipe-type cable circuits have been used for underground electrical power transmission for a number of years. Such circuits employ a steel conduit inside which are several cables, each consisting of a copper conductor wrapped with porous, oil-soaked paper insulation and a protective outer covering. The space between the cables and the pipe is filled with a dielectric oil which is under high pressure. The oil, which impregnates the paper wrapping on the cables, provides electrical insulation for the cables and also transfers the heat generated by losses in the cables to the conduit and the surrounding soil. Pressurization of the oil prevents vapor formation in the paper insulation and ensures proper electrical insulation of the cables. In this non-circulating type of system, heat which is generated in the cables is transferred from the insulation to the pipe wall by natural convection through the oil, and then from the pipe to the atmosphere by conduction through the soil. The power-carrying capacity of underground cables is limited by the maximum allowable cable temperature, which depends on the rate of heat removal from the system. Force-cooled systems for oil-filled pipe-type cable circuits, which appear to have power capacities significantly larger than those of non-circulating systems, 12 13 have recently been considered. In force-cooled systems, chilled oil is circulated through the pipe, and heat is transferred from the oil to the atmosphere at refrigeration stations. Most of the heat is transferred from the cables to the flowing oil, heat transfer secondary importance [1]. to the soil being of The cable-to-oil temperature difference for a given current and voltage is determined by the overall cable-to-oil heat transfer resistance, which is due to two effects: the resistance to conduction heat transfer through the cable insulation, and the resistance to convection heat transfer from the surface of the insulation to the bulk of the oil. Based on results of the natural convection experiments performed by Orchard and Slutz [2], it is demonstrated in Appendix A that the conduction resistance for the type of system which was considered is an order of magnitude larger than the convection resistance. Therefore the rate of heat removal from the system depends primarily on conduction, and an accurate conduction model of the cable insulation is required in order to confidently predict the cable temperature. Conduction within the insulation is complicated by the proximity of one cable to another. When two cables come into direct contact, their mutual presence causes a large increase in the resistance to heat transfer near the point of contact. Consequently, the cable insulation near the contact point experiences a sharp increase in temperature, 14 which in turn elevates the conductor temperature, and thermal failure of the system will ensue unless the oil temperature is appropriately adjusted. Given a system with a maximum allowable cable temperature, it is therefore desirable to know the maximum oil temperature which should be allowed in order to avoid thermal failure of the system. This involves determining the two-dimensional (i.e., radial and circumferential) steady-state temperature distribution within the cable insulation for various cable configurations, especially those which produce the most severe operating conditions. This heat conduction problem is too complicated to be solved analytically. However, the solution for arbitrary cable configurations is readily obtained by means of numerical methods. The particular system which was studied consists of three circular conductors inside a circular conduit. The dimensions of this system are shown in Figures 1.1 and 1.2. In addition to the outer moisture seal, the cables are wrapped with skid wires, which protect the cable coverings and reduce friction when the cables are pulled into the conduit. In order to simplify the geometrical problems which arise in handling configurations of three cables, it was assumed that the system possesses bilateral symmetry, as shown in Figure 1.3. This assumption reduces the system to one and one-half cables inside half a conduit, while permitting arbitrary configurations of the one and one-half 15 w L1J L3 -I L') (9 w J a0 co -i -Jw Wl W a F- O - E E Q) r- E1 2 U a) 0 0 * H C) '~ U .,H C: t,) a)4 04 U) O O FO 4-,U 9 .H a) I En O U To sq LULI LJ LU I _J -J U 0Uf) 16 U C) r-3 L~) "~ 3 w 0*.H 0 z X r0 0a "4 0 P~ ro ir 14 UU 4R w 0 o~~~~~~~ U> 0 UH 4i u 17 )le 2 Cable 1 ;ymmetry FIGURE 1.3 Bilateral Symmetry of the Underground Cable System 18 cables. As Figure 1.3 indicates, the half- and whole cables are referred to as Cable 1 and Cable 2, respectively. In Chapter 2 a complete formulation of the conduction problem is presented, followed in Chapter 3 by a discussion of the superposition methods which were employed in obtaining final solutions. Chapters 4 and 5 are concerned with discretizing the conduction model and with translating the discretized model into a computer program. In Chapter 6 the results of several problems are discussed, and conclusions are stated. I~ ~ ~ ~ ~ ~ ~~~~~~~~~~~~~~~~~~ CHAPTER 2 FORMULATION The Cable Insulation In developing a conduction model for the cable insulation, the following assumptions were made: any axial conduction along the length of the cable is negligible, thus reducing the problem to two dimensions; steady-state conditions prevail in the system; the thermal conductivity throughout the insulation is taken to be uniform. Using these assumptions, an energy balance on an infinitesimal element in a cylindrical coordinate system yields the following expression, which is a special form of Poisson's equation [3]: 1T r r 2T + 2 1 2 ar 2T 2 - 2 a2 2 r (2.1) (2.1) ~$ This equation governs the temperature distribution in the cable insulation, together with appropriate boundary conditions which operate around the various portions of the cable surface. The heat generation term q in Equation 2.1 is due to a dielectric loss which occurs throughout the insulation. The Inter-Cable Conduction Path In order to model the situation which exists when Cables and 2 are lying together in direct contact (skid wires overlapping), a special conduction path was placed 19 20 between the cable and half-cable. A conduction path was used because there is a small region between the cables in which the oil is essentially stagnant. The thermal conduc- tivity of the path was taken to be the same as that of the insulation. The width of the path is usually taken to be the thickness of a skid wire, since this is as close as the cables come to actually touching. As an estimate of how large an angle the path should subtend along the cable surfaces, it was decided to use the angle subtended by the overlapping skid wires. For the system which was studied, this angle is approximately 25 ° . The inter-cable conduction path is thus an extension of the cable insulation, joining Cable 1 to Cable 2, as depicted in Figure 2.1. Since no heat sources are present within the conduction path, the governing equation for its temperature distribution is Laplace's equation [4: a 2T a2T ° -__ +T a7 2- = 0, ax (2.2) ay where x and y are the normal and tangential coordinates, respectively, and where "normal" denotes an axis which joins the cable centers. The Solution Domain Two additional assumptions underlie the conduction model. The first is that the oil is assumed to be well- mixed, so that the oil temperature outside the convective 21 -i2 FIGURE 2.1 The Inter-Cable Conduction Path (Shaded) 22 boundary layer is uniform at a given cross-section in the The second system. conductivity is that, because of the very high of copper, each conductor is assumed to be at a single, uniform temperature (though the two conductor These two temperatures are not, in general, equal). temperatures are obtained from a knowledge of the losses at each conductor, and this is discussed in Chapter 3 under the subject of superposition. The point to be made here is that the two conductor temperatures are not unknown quantities in Therefore the conduction problem has the temperature field. as its solution domain only the paper insulation surrounding the conductors and the inter-cable conduction path. purposes of nomenclature, referred to as D 1 as D2 the insulation of Cable 1 is (Domain 1), that of Cable 2 is referred (Domain 2), and the region conduction path is called D domains D For comprising (Domain 3). to the inter-cable The solution D2 , and D3 , together with their associated coordinate systems, are shown in Figure 22. Boundary Conditions The solution regions of varying condition which domains D 1 and D size according are divided into to the type of boundary is acting at the cable surface. A set of regional divisions for both cables is illustrated in Figure 2.3, and the boundary conditions associated with the various regions are listed in Table 1. Regions II of D 1 and D2 are not included in the table, because they join the inter-cable conduction path and therefore have no surface 23 FIGURE 2.2 Coordinate Systems in D 1 , D 2, and D 3 24 optional FIGURE 2.3 Regional Divisions in D 1 and D 2 25 TABLE 1 BOUNDARY CONDITIONS IN D 1 AND D2 D1 Region Convection I Region III Convection Region IV Cable-Conduit I Convection D2 Region Region III Region IV Convection (if used) Region V Region VI Region VII Cable-Cable Convection (if used) Cable-Conduit Convection 26 The cable-cable and cable-conduit boundary boundaries. are depicted conditions in Figure 2.4. There are several different surface boundary conditions, but it can be shown that they are all convective in form. The convective boundary condition itself is obtained from an energy balance at the cable surface: =-k q h[T(r2,) - Toil] (2.3) r2 P21 where r2 is the outer radius of the Cable 1 insulation, P is the perimeter of the cable, and h is the local film coeffi- cient, which may vary around the periphery of the cable. The cable-cable boundary condition occurs when the three cables are in an equilateral configuration. A small arc along the surface of Cable 2 then lies immediately adjacent to the line of symmetry. The arc length is taken to be the same as that for the inter-cable conduction path. Since no heat flow crosses a line of symmetry, this boundary is taken to be an insulated one, which is just a convective boundary with a local film coefficient of zero. While this boundary condition differs considerably in form from the conduction mechanism operating in the inter-cable conduction path, both mechanisms have the same effect on the temperature distribution. For in the equilateral configuration, symmetrical conditions on either side of the conduction path act to prevent any flow of heat across the tangential axis 27 U-, Q ) z 0 F- -i D 0 3.) _L (9 z LL 0 0 U 3 _) 0 0 04 0 H rH U ro z Q -Q :D Qj UlU z 0-- cr L a w ci) U0 O 28 (y-axis) of D3 . The trilateral symmetry of an equilateral configuration is thus preserved by modelling the cable-cable effect as a convective boundary. An alternative method would be to employ an optional conduction path for the cable-cable effect, but this would introduce unnecessary complication. The cable-conduit boundary condition, which exists when either cable is lying directly against the conduit, is influenced by the following factors: convection cooling near the point of contact; the thermal conductivity of the conduit, which for steel is large; potentially large AC losses in the conduit itself; and heat conduction from the conduit to the adjacent soil. This situation is examined in Appendix B, where a portion of the conduit wall is thermally through modelled as a fin, and the thermal the fin is compared to the thermal resistance resistance across the cable insulation for a given set of AC losses. In the most conservative case, the resistance from the fin base to the oil is an order of magnitude smaller than the resistance across the insulation. Thus the cable-conduit boundary condition for a steel conduit is essentially a convective one with a slightly modified film coefficient. The arc on the cable surface affected by this boundary condition is again taken to be the same as that for the inter-cable conduction path. There are two surface boundary conditions in D3, each a convective one. These are depicted in Figure 2.5. 29 CONVE( FIGURE Convective 2.5 Surfaces of D3 30 An energy at the surface balance of D 3 gives -k a Ti] ay sufc surface = +h[T(surface)ITol (2.4) where (+) and (-) apply for positive and negative y, respectively, and h is a variable film coefficient. In addition to the surface boundary conditions, there are two internal boundary conditions. the line of symmetry boundary condition symmetry bisects The symmetry occurs in Cable 1, where the line of the cable and forms a portion insulation boundary. insulated. and at the conductors. These are at of the This boundary, of course, is perfectly The boundary condition at each conductor, as was stated previously, is one of uniform temperature. The aforementioned governing equations and boundary conditions, along with the requirement that the temperature between D1 and D3 and between D2 and D 3 be single-valued, constitute a complete formulation of the conduction problem. Variations on the Problem Statement Although the heat conduction problem was originally posed with the oil temperature as the unknown quantity, it is possible to specify solve for other quantities. the oil temperature and Assuming that the voltage is constant for a given system, there are three variables: maximum allowable oil temperature, maximum cable temperature, 31 and maximum allowable current. third may be found. Given any two of these, the In subsequent discussions it will be necessary to specify which of the three variables is unknown, and so the following solutions are defined for reference: Solution 1 finds the maximum cable temperature, given the current in the circuit and the oil temperature; Solution 2 finds the maximum allowable current, given the oil temperature and the maximum allowable cable temperature; Solution 3 finds the maximum allowable oil temperature, given the current in the circuit and the maximum allowable cable temperature. Nondimensional Formulation In preparation for numerical solution, the heat conduction problem is cast into nondimensional form. Such a formulation can be obtained by introducing the following dimensionless variables: -r --p -x - oil - r r2 r - P = ~~~~~~~~~~T T .l X x=2A Y W/k 2D' (2.5) where the minimum width of D 3 2A denotes 2D is the height of D 3 (at the x-axis), (along the y-axis), and W is an arbitrary loss per unit length (Btu/hr-ft). The form of the governing equation in D 1 and D 2 is then 1 2~~~~2 - -2 32e 2 r 2 - r Dr 520 3~2 - (r2 r) 2 q (2.6) 3 whereas the governing equation in D 3 becomes 4A 2 D2a a2e ;:X 2 ~-x 2 D2 22 ay (2.7) The form of the standard convective boundary condition becomes k 2 6 r = he(1,4) (2.8) CHAPTER 3 SUPERPOSITION OF SOLUTIONS General In the solution of linear problems, such as this problem of conduction with uniform thermal conductivity, it is often convenient to employ the principle of superposition. This reduces the overall problem to a number of simpler problems, each having the same geometry as the overall problem, whose individual solutions may be linearly combined to form the overall solution. The required number of separate solutions is equal to the number of nonhomogeneities, or potentials, in the overall problem. In the conduction problem which has been posed, there are three potentials: the two conductor temperatures and the volumetric heating effect, The overall problem may thus be decomposed into three component problems. Solutions to these component problems need to be generated only once for a particular cable geometry and voltage (dielectric loss); the total solution for any arrangement of current-produced losses can then be achieved by suitably combining the three component solutions. In the following sections the superposition technique for obtaining Solution 1 (which finds the cable temperature) is presented. It is then rigorously demon- strated that the overall governing equation and boundary conditions are obtained from a linear combination of the 33 34 governing equations and boundary conditions of the three component problems. introduced: For brevity the following notation is x is a generalized position vector for the overall solution domain (comprised of D1 , D2 , and D3); x C3-1,_ (x) denotes all points in the solution domain which lie on the curve C (x); n is an outward normal to the curve Ci(x); v2 is the Laplacian operator. The nine curves Ci(x) which comprise the boundaries of the solution domain are shown in Figure 3.1. less: The nine normals n 1 are all dimension- normals to curves in D1 are nondimensionalized with r2, normals to curves in D2 with P2 , and normals the two curves in D3 with the length 2D. The Overall Problem The governing equation for the overall problem is the following: V20(x) = f(x) , (3.1) where f(x) describes forcing effects throughout the domain. 0(x) also satisfies boundary conditions on the nine curves C i (x). On the curve C (x) the condition 0(x)IXEC (x) = 1 where is 01 , (3.2) 01 is some uniform (as yet unknown), dimensionless temperature. On the remaining curves the boundary 35 C2 C. FIGURE 3.1 Curves Ci(x) Comprising the Boundaries of the Solution Domain 36 conditions are ae x) hr 2 (3.3) e(x) an2 ECC2 XEC2 (x) hr 2 2x) 3 k an3 (3.4) OC xEC (x) ~ 3~ xEC (x) ~ 3 ~ ae(x) an 4 =0 (3.5) = (3.6) XEC4 (x) a (x) n5 0 XEC (x) 0(x) where x 5 x) nc 6 W I 02 is a uniform 02 (3.7) 02 ' (as yet unknown), dimensionless temperature. ae (x) hP2 an7 k O(x) xCc 7 ae (x) an 8 2Dh - xeC 8~(x) ~ k (3.8) C7(X) 6(x) (3.9) xEC8 (x) 37 ao (x) _ an9 2Dh (x) k (3.10) ~xC 9 (x) xEC9 (x) The Component Problems The overall problem is decomposed into three component problems, each of which has only one potential and is individually solvable. OA(x), OB(x), and The component solutions are C(x). eA(x) is the solution in which the Cable 1 conductor for the physical situation is hot, the Cable 2 conductor is cold (at the oil temperature) and in which there is no dielectric loss. A(x) satisfies the homogeneous governing equation V2OA(x) = 0 , (3.11) and it satisfies the following boundary conditions: OA(x) where A 0 (X) =A (3.12) ' is some arbitrary dimensionless temperature. a A(x) - n2 xEC 2 (x) hr2 k OA(x) (3.13) xEC 22 (x) - 38 3CA(x) 3 hr 2 - kOA (3.14) (x) xcC (x) 3 xEC3 (x) a 3n3 A(x) 4 = 0 (3.15) ) = 0 (3.16) xEC4 4 (x) aOA(x) an5 xeCc (x) = 0 0A(x) [x£C XE 66 ( x) OA(x) (3.17) hP2 - an) an 7 k GA(x) xcC7(x) xEC7 (x) - (3.18) 7 SeA(x) 2Dh OA(x) an8 xeC (3.19) xEC (x) a OA(x) 8 x) 2Dh - n9 x£C (x) 9 k GA(x) (3.20) x C 9 (x) OB(x) is the solution for the physical situation in which the Cable 1 conductor is cold (at the oil temperature), the Cable 1 conductor is hot, and in which there is no dielectric loss. The component solution OB(x) satisfies the homogeneous governing equation 39 =0 V OB(x) (3.21) as well as the following boundary conditions: OB(x) = xc1 ( (3.22) 0 a0B (x) I hr an 2 k ~ 2 (3.23) OB (x) 'xEC (x) 2 X hr2 k xC 3B(x) (3.24) xEC3(x) 3 (x) aOB(x) 39Bn4 (x) 1 an4 I1Ix =0 (3.25) = 0 (3.26) C 4 (x) DOB (x) an xc C(x) 5 5B(x) OB(x) where B 0 CC6( is an arbitrary aOB (x) (3.27) 0 dimensionless temperature. hP 2 an7 1 XE = B k 7 W OB(x) k (3.28) xCC 7 (x) 40 3eB(x) 2Dh aeB (x) an 9 Finally, eB(x)[ xEC k 8xeC (x) (3.29) 8 (x) 2Dh (3.30) 2kh 6B(x) x£C 9 (x) ~c9_ C(x) is the solution for the physical situation in which both conductors are cold (at the oil temperature), but in which there is a prescribed dielectric loss. The component solution C(x) satisfies the nonhomo- geneous governing equation V2 C(x) = f(x) (3.31) and the following boundary conditions: ec(x)IXEC(X)- (3.32) 0 aec(x) hr 2 (3.33) an2 xcC XE 2 2 (x) W aec (x) an 3 x£C3 (x) kr 2 e(x) ~ - xEC3 (x) (3.34) 41 a6c (x) 0 (3.35) = 0 (3.36) = 4xEC n 4 Wx - OC (x) 5 xEC x ~5xc -- ec(x) 5 (x) 6 hP2 6c (x) k ~C aOC (x) an 7 (3.37) 0 EC () - - (3.38) xeC (x) 7- aec (x) 8 -an8 xeC Cx k ~ x£C8 xC 8( I 2Dh (3.39) 8(x) a OC (x) _ 2Dh 0C (x)I an 9 xc 9 (3.40) - xC 9 (x) (x9) Validity of the Superposition Method Having described the overall problem and the three component problems, it remains to demonstrate the validity of the superposition method. The three component solutions are linearly combined to form the total solution according to [5]: 6(x) = a 1 OA(x) + a 2 OB(x) + C(x) , (3.41) 42 where a 1 and a2 are two arbitrary constants, to be determined from two additional boundary conditions in the That Equation 3.41 is indeed valid is overall problem. proven by substituting it directly into the overall governing equation and boundary conditions. The following results are then obtained: V 0(x) = V [al0A(x) +a 2 0B(x) +C(x)] 1 2 -_ = f(x) Check (3.42) e (x)I~ xC (x~~) = [alA = alA provided alA = 01 (x) +a 2 eB( ~ x) + C (x) ]jI I x -~~ Check, (3.43) This presents no problem, since 01 is unknown, and a1 and A ° are both arbitrary. XEC ()x) 2 a [a1 OA(x) +a 2 6B(x) +C(x)] n - ~~~~~~~~xC2 - hr2 - -- k [a 1 OA(x) - + a2 OB(x) - + OC(x)] hr = - k e (x) Check xeC2 (x) E (x) xEC2 (3.44) 43 ae (x) an 3 a an 3 xEC [a1 OA(x) + a2 B(x) + 0C(x)] I xeC _ J (x) 3-~ .. , (x) \,= ~I hr 2 [a1 6A(x) + a2OB(x) + OC(x)] (x) xcC3 hr2 0 (x) k xC 3(x) Check (3.45) ae (x) [a1 A(x) + a2OB(x) +C(x)] (x) xcC 4-~ xnC4 () = 0 0e (x) a -an5 3n5 xEC5 (x) = 0 6(x) I EC 6 (x) a2Bo = 002' e02 is unknown, [a1 0A (x) + a 2 B(x) + ec (x)] xC 5 x) (3.47) Check = [a1 OA(x) + a2 eB(x) + a2B o provided (3.46) Check C(x) ] xC 6( Check, This also causes no difficulty, and a 2 and B ° are arbitrary, (3.48) since 44 a (x) [a1 OA(x)+ a 2 B(x) + ec (x)] an7 xEC - 7-~ (x) hp 2 [alOA(x) + a 2 OB(x) + C(x) ] k xEcC7 (x) hp 2 a 1 GA(x) + a 2 B(x) +OC(x)] -n xcc8 (x) (3.49) xcC 7 (x) ae(x) an 8 Check 6 (x) k a8 xEC 2Dh 8 (x) [a1 eA(x) + a2 OB(x) + OC(x)] k xC C 8 (x) -- 2Dh k a (x) 8(x) an 9 xC 9 (x) Check =-2Dh 2kh 2Dh k (3.50) CxC 8 (x) [a l OA(x)a2x) - +Bx) 2 an9 Dn 9 8 - +C(x) ] xEC9 (x) [a1 OA(x) + a2 OB(x) + C(x) 6 (x) W xcC (x) XEC 9 Check. I XC 9 (X) (3.51) 45 It is thus established that the superposition technique just described is indeed valid. temperatures However, the two conductor 01 = alA ° and 02 = a2B have yet to be found. Determination of Conductor Temperatures All that has been said of the conductor temperatures up to this point is that each one is uniform. These two temperatures, though, are uniquely determined by two additional conditions in Solution 1: the specified loss per unit axial length at each conductor. Returning momentarily to dimensional variables, let WCl and WC 2 be the specified conductor losses per unit axial length in Cables 1 and 2, respectively. Equating W to the total heat flow per unit length transferred from the conductor of Cable 1, the following result is obtained: ,- ar r 1dd=w Cl (3.52) 0 Likewise, for Cable 2 [ 2Tr = -k 2~ Pd ap Plia = WC 2 C 0 These two expressions are rendered dimensionless and rearranged to give (3.53) 46 r2Wc 38 r2 Clw = (3.54) 0 2 7T da = - P2Wc2 (3.55) 1 l 0 Finally, in terms of the present vector notation, Equations 3,54 and 3.55 become the following: a (x) an 1 xl cI(x) WCl r 2 dC (x) = +Wr xC - 1 (x) dC 6 (x) n (x) xC 6 (3.56) 1- a0(x) C I = + WC 2 P2 P (3.57) ~~~~~ (x) 0(x) may now be eliminated from these two equations in favor of OA(x), B(x), and C(x) by substituting into Equations 3.56 and 3.57: Equation 3.41 47 dC1 (x) an- [al OA(x) + a2 B(x) + C(x)] 1(x) c (x) WCl r2 =+ Wr- (3.58) ' 1 a an 6 C 1 C 6 C(x) 6 dC6 (x) OB(x) + 6C(x)] [a OA(x). +a 2... _ WC2 P2 + . 0Wp 1 (3.59) Since the component solutions OA(x), B(x), and 6C(x) are each known, Equations 3.58 and 3.59 are two simultaneous equations from which the arbitrary constants a1 and a2 are determined. Furthermore, since A0 and B0 are known quantities (the arbitrary dimensionless temperatures which were used in solutions A(x) and B(x), respectively), the dimensionless conductor temperatures follow directly from Equations q 3.43 and 3.48: 801 = alA ~~~~01 1 and 602 = a2B. 2 o* and 602 This then completes a description of the superposition technique for Solution 1. The technique for obtaining Solution 2 (which finds the maximum current) is somewhat more complicated, owing to the fact that current is then a variable. This makes necessary a separation of current-produced losses from 48 voltage-produced losses, as well as a subsequent procedure for maximizing current with respect to the allowable cable temperature and the oil temperature. A full description of this solution is presented in Appendix C. Solution 3 (which finds the oil temperature) is nearly identical to Solution 1, the former requiring only a minor extension of the latter. In particular, Solution 3 is obtained by using an arbitrary oil temperature in Solution 1 and then by equally incrementing all temperatures (including the arbitrary oil temperature) until the maximum temperature in the field has reached the prescribed allowable value. The two temperature distributions therefore have the same shape, differing only by a constant. CHAPTER 4 THE FINITE-DIFFERENCE METHOD Discretization of Domains The numerical method used to generate solutions for the various component problems described in Chapter 3 is the finite-difference method. It has as its first basic step discretizing the solution domain. Discretization is the reduction of a continuous system into a system which has a finite number of degrees of freedom. The basic approxi- mation involves the replacement of a continuous domain by a network of discrete points within the domain. A one- dimensional example of this is shown in Figure 4.1. Instead of obtaining a continuous solution defined throughout the domain, approximations to the true solution are found only at these isolated points Discretization of D 1 and D 2 is accomplished by defining a network of radial and circumferential mesh points. Since it is desirable in terms of computational labor for the mesh to be as regular as possible, conventions were adopted: uniformly the following points along a radius are spaced, though the spacing in D 1 nmay be different from that in D2 ; circumferential spacing of points within a particular region is uniform; the number of circumferential subdivisions in both Regions II of D 1 and D2 and the number of tangential subdivisions in D 3 are constrained to be equal, thereby avoiding mismatches at the D 1 -D3 and D2-D3 49 50 CONTINUOUS ONE-DIMENSIONAL DOMAIN j-1 0 Pj Pj+ i.. 0 DISCRETE 0 * * REPLACEMENT OF ONE- DIMENSIONAL DOMAIN FIGURE A Regular, 4.1 One-Dimensional Finite-Difference Mesh 51 interfaces. It should be noted that, just as the conductors were not included in the continuous domains D and D2 , so are they not included in the corresponding discrete domains. The dimensionless radial spacings h and h are p from obtained obtained from h r = 1- r - h N1 = 1.- Pl 1- (4 f N2 1) where N1 and N2 are the number of radial subdivisions in D1 and D2 , respectively. .~~ spacings h and h The dimensionless circumferential vary in magnitude depending on which region is involved, since the regions generally are of varying size. Thus there are four values seven values of h of h in D 1 and in D2 , or one for each region. A typical spacing is calculated according to (h ) n _ n- n-l (4.2) M n where 4n and 4n-1 are the values of 4 at the bounding lines of Region n, and M subdivisions n is the number radial of circumferential in Region n. Discretization of the domain D3 is difficult because of its irregular siape. For this reason it was approximated by the more regular shape shown in Figure 4.2. The nondimensional normal and tangential spacings hx and h y 52 I CONTINUOUS MODEL 2A: skid wir-ethickness c: ftE-c os(2)] B-2 (c2 ct) B: V i(£ 7) D: sin() DISCRETE FIGURE Discretization 4.2 of the Domain D !DEL DE 53 are obtained by dividing the nondimensional width and height, respectively, by the appropriate number of subdivisions. It is seen in Figure 4.2 that these spacings are not uniform: h h x yx on y and h depends are small compared value for h y y is a function to changes in h, was assumed. Since changes of x. a uniform, The linear dependence in mean of h x on y was retained in the model. A subtlety regarding the two convective surfaces of domain D3 is also mentioned here briefly. At each corner of D3 there exists a discontinuity in the area available for conduction heat transfer. This discontinuity is most conveniently accounted for in the following manner: the regular form of the governing equation, which itself assumes no discontinuity in area, is applied at each corner point. Four effective corner locations, which lie outside the corner mesh points, are thereby established, and these corner locations define the two effective surfaces for convection in D3. The mesh points along either convective surface thus lie inside the conduction path, rather than on the boundary itself. are discussed Details of this modelling procedure in the user's instructions in Appendix E. The discretized domains are shown with a typical mesh in Figure 4.3. The number of points to be used in a given problem is dictated by the level of accuracy required in the solution; as the mesh becomes finer, it more nearly approaches the original continuous domain. Also it is economical to use a coarse network in regions where the 54 FIGURE 4.3 A Typical Mesh for the Equilateral Configuration 55 temperature gradient is small, switching to a finer mesh where there are rapid variations in temperature. Note, for example, how the mesh points in Figure 4.3 are arranged: since the temperature distribution away from points of contact should be nearly one-dimensional, most of the points are concentrated between the cables, where large gradients are expected. Difference Form of Governing Equations and Boundary Conditions The reduction of a governing equation and boundary conditions for a continuous domain to those of its discrete replacement may be accomplished physically or mathematically. In the mathematical approach, which was used by the author, the continuous formulation is reduced to a discrete formulation by simply replacing derivatives with finiteWhen this is done, the original difference approximations. system of governing partial differential equations is reduced to a set of n simultaneous algebraic equations, where n is the number of discrete points in the mesh. Since the original continuous system is linear, the algebraic system will also be linear. In preparation for replacing differential equations with finite difference relations, two basic onedimensional finite-difference expressions are listed. extension Figure to two dimensions 4.1 let the points is straightforward. ... , Pj-l P The In Pj+ 1 ,..' be separated by a dimensionless spacing h, and let the value of 56 '(z) at P. be denoted by derivatives Then the first and second .. '1 J at P. may be ap,,roximated by the following J finite-difference expressions [6]: (d,) + (h2 ) ; (4.3) 1-22 + O(h (4. 4) 2j+lJJ = '+ 2 ~~~h2 where O( ) denotes the order of the error. With the availability of these computational formulas, the process of ec1.,,tionand boundary conditions of replacing the governin the heat conduction roblem with approximate algebraic equations is simple and irect: at each internal point the to finite-difference aproxii,:.i.rion :. d ential equation p,(v)Msi the values sex.' l_-n.graic equatlon connecting J points. .. r1iqborn a. a typical equation example, Region of u at ti. he governing dffer- the poLnt For (j2,k2) in D2, IV is: 2 - ,2 __ j22-liik L+r 2hJ 2 + e- 4 : [i2 2 -22tj, 40+rL -2 +h-- ej2,k2_ Ot4 + h2 0~~4 j2,k2+i 2' Wf' 2 j2+lk2 4 57 where 2 is the outer radius of Cable 2, ()j2,k 2 is the local volumetric loss, W is an arbitrary loss per unit length, and the remaining symbols are explained in Figure 4.4. This result was obtained by substituting the two-dimensional forms of Equations 4.3 and 4.4 into a typical governing equation, such as Equation 2.6. Two types of exceptional situations can arise in applying this equation. The first is that on the boundaries, not all the neighboring points of a governing equation will lie within the domain. It is then necessary to introduce finite-difference approximations to the given boundary conditions and thereby to eliminate the need for any point that lies outside the domain. For example, the standard convective boundary condition in D 1 reduces to the following equation after discretization: [2h ejl-l,kl+ where jlkl 6jl+lkl hr2.6 k 1jl,kl + ajl+l,kl = is a temperature (4.6) on the surface of D1 . is then a fictitious temperature outside D1 . However, when the governing equation is applied at the point Pjlkl (whose temperature is jlkl), simultaneous equations in the unknown there will then be two jl+l,kl' and this fictitious temperature may be eliminated in favor of real temperatures within D1. The same problem occurs at the conductors, where a finite-difference expression for the 58 J2,k2 +1 1' I FIGURE 4.4 Nomenclature in the Neighborhood of P j2k2 in D2 59 first derivative is required. In this case, however, the situation is less easily resolved. Since no governing equation is applied at the conductor, there is no way of eliminating a fictitious point within the conductor. This problem is addressed in Appendix D, where a suitable approximation to the conductor boundary condition is derived. The basic method involves satisfying the boundary condition at a slight distance from the conductor, and then relying on the fact that the temperature distribution is nearly onedimensional in the immediate vicinity of the conductor. The remaining exceptional case occurs at mesh points whose neighboring points on either side have different dimensionless spacings. This happens for radial spacings at the D1 -D 3 and D2 -D 3 interfaces, and it happens for circumferential adjacent regions spacings at the interfaces in D 1 and D 2. of all In such situations it is necessary to have finite-difference approximations which have been modified to fit an irregular mesh. The expressions for the first and second derivatives in a nonuniform mesh which are used in this study are readily derived, either from a Taylor's series expansion about the central point or by deduction from the mean value theorem of differential calculus. dzj They are the following [7]: [h2 (hl+h2 )] j+l+[12 2 ] j - hlh+h2 j-1 )+ ;h (4.7) 60 - dz = F 2 [2] 2 1 2 *j+ [lh + jp +0(h); 2 ]-i 2ih(hlh2 2h (4.8) where h 1 is the dimensionless spacing between and h2 is the spacing dimension between ij and j-1 and j+l j, Also it is noted that these expressions reduce to the standard form of Equations 43 and 4.4 when the dimensionless spacings are uniform (h1 = h 2 ). CHAPTER 5 THE COMPUTER PROGRAM General The result of transforming the continuous formulation of the conduction problem into the corresponding finite-difference formulation taneous algebraic equations. is a linear set of simul- A FORTRAN IV computer program written by the author generates this system if equations, performs the matrix inversion and multiplication to obtain various component solutions, and then combines component solutions to produce a final solution according to the superposition principle. User instructions for the program are discussed in Appendix E, and a complete listing of the source program is given in Appendix F. The Coefficient Matrix The set of simultaneous equations for the conduction problem may be written in the form [A]{e} = {B} , where [A] is an nxn matrix of coefficients, (5.1) {} is a vector of n unknown dimensionless temperatures, and the right-handside vector {B} is a vector of n forcing elements. Referring back to Figure 4.4, a typical algebraic equation was shown to be of the form 61 62 (ajlk) + (ajk) jl,k jk +(aj+l,k) j+l, k +(ajk) jk 1 +(a.j,k+l k+l 52 jk+ where jk bj,k '(5.2) is the central point at which the governing equation was applied, and the an are the coefficients m,n which were given in Equation 4.5. A typical equation thus involves five points - a central point and its four neighbors - and a typical row of the coefficient matrix accordingly has five non-zero elements. However, the application of a governing equation at certain mesh points produces rows with fewer than five non-zero elements. These situations are depicted in Figure 5.1, together with a typical mesh point. In this figure, points P3 and P4 initially had the full complement of four neighboring points, but fictitious points outside the domain were eliminated by incorporating the boundary conditions at those locations into the governing equations. It is a simple matter to assemble the various coefficients a into a matrix. The only requirement is m,n that the rows be arranged so as to place the coefficients of central points (Pi, P2 ' P3 , or P4 in Figure 5.1) on the main diagonal of the matrix. Since a governing equation will necessarily involve the central point at which it is applied, it is therefore ensured that only non-zero elements will appear on the main diagonal, a necessary condition prior to matrix inversion. 63 -MESH O POINT CONDUCTOR POI P. 4 P,TYPICAL POINT WITH FOUR NEIGHBORING MESH POINTS P 2 -POINT WITH THREE NEIGHBORING MESH POINTS ONE NEIGHBORING CONDUCTOR POINT P3 -POINT WITH THREE NEIGHBORING MESH POINTS P-POINT WITH 4 TWO NEIGHBORING MESH AND POINTS FIGURE 5.1 Points Affected by the Application of a Governing Equation at Various Locations in a Discrete Network 64 The coefficient matrix is inverted by the packaged subroutine RMINV, which uses the standard Gauss-Jordan An automatic feature of this subroutine is the algorithm. calculation of the determinant of the matrix. In order to keep the order of magnitude of this determinant within a range acceptable to FORTRAN IV, each row of the matrix and the corresponding elements of the forcing vectors are scaled so that the largest element of every row is unity. matrix has been inverted, the temperatures {} Once the are available from {8} = [A] {B} ,(5.3) where [A]- 1 denotes the inverse of [A]. Forcing Vectors In the conduction problem the total heat flow is comprised of the conductor losses, the dielectric loss, and the sheath loss. potentials: heating. This heat flow is driven by two types of the conductor temperatures and the dielectric These potentials are accounted for in the right- hand-side vectors {B} of Equation 5.1. reference the following vectors For purposes of are defined: {B}1 is the forcing vector for the component problem in which a conductor temperature is driving the heat flow; {B} 2 is the forcing vector for the component problem in which the heat flow is driven by dielectric heating. 65 The procedure for generating {B}1 is suggested by point P2 of Figure all zero. points 5.1. The elements of {B}1 are initially However, when a governing equation is applied at adjacent the neighboring to the conductor, points such as point P 2 , one of is the conductor itself. The conductor temperature is not an unknown, though, and when the governing equation is written in the form of Equation 5.1, the term involving the conductor temperature is carried over to the right-hand side and becomes a forcing term. The non-zero elements of {B}1 are therefore comprised of conductor temperature-terms which have been referred to the forcing vector. The elements of {B}2 appear as volumetric heating terms in the difference form of Poisson's equation, Equation 4.5. It can be shown [8,9] that the dielectric loss per unit volume is of the form: WdW= -C 2' r (5.4) _ where C is a constant for a given system, and r is the radius at a point in the insulation where the local dielectric loss per unit volume is wd. the dielectric loss is known, Since the distribution of it is possible to integrate Equation 5.4 over any particular area to obtain the total loss per unit axial length within that area. A typical radial mesh and the areas associated with each mesh point 66 The dielectric loss per unit are shown in Figure 5.2. length in a typical (Wd)2 (d)2 = {dd= d = say A 2 , is given by area, ~~~~Cr (A)(2rr)dr d W d 3 = 2C A ln r= (5.5) Jr 2 r2 The total dielectric loss per unit length is: r6 wddA = 2C Wd = (5.6) ln(r)6 1 The loss per unit length in any particular area may be expressed as a fraction of the total loss per unit length just from information about the radial mesh. For example, ln (__3) (Wd)2 ( Wd (5.7) rl The computer program distributes the dielectric loss in this manner. The fraction of the total loss per unit length which occurs in each discrete area is calculated from the shape of the radial mesh. The dielectric loss per unit length for each area is then obtained by multiplying the 67 & FIGURE MESH POINT 5.2 A Radial Mesh Illustrating the Area Associated With Each esh Point 68 various fractions by the total loss per unit length, Wd, a number which is supplied as input data. Each mesh point therefore has an associated dielectric loss and the volumetric loss terms in Equation 4.5 can be generated accordingly. A question then arises, however, regarding the disposition of the loss in the innermost discrete area, A1 in Figure 5.2. It is noticed in this figure that there is no mesh point associated with A1. For this reason the loss which occurs in A is added to the current-produced heating of the conduc- tor. This suitably accounts for the loss, providing a con- servative approximation to the true dielectric distribution. The sheath loss is a current-produced loss which occurs on the surface of the cable. In the finite- difference model, this loss is placed in the outermost discrete area of insulation, which is associated with the mesh point on the cable surface. In Figure 5.2, for example, the sheath loss would be placed in A5 . treated as a volumetric type of heating dielectric loss for that area. introduced, which is accounted It is then in addition to the A third potential is thereby for in a third vector: {B}3 is the forcing vector for the component problem in which sheath heating drives the heat flow. Verification During the course of developing the computer program, periodic tests were performed to ensure its correctness. One general technique used to verify a computer program is to solve a problem with it whose 69 solution is already known and then to compare the two results. Several such problems, as well as a simpler checking procedure, were employed in this study. The first major verification was a check on the coefficient matrix. This test was accomplished by printing out the matrix for a 111-point mesh and then by verifying each element by hand computation. The 111-point mesh was of sufficient size for the matrix-generation portion of the program to pass through all its decision branches. This test was repeated for meshes of decreasing size, until the matrix for the smallest possible mesh had been verified. In partic- ular, the coefficient matrices for the following mesh sizes were verified: 16-point. 111-point, 57-point, 24-point, 17-point, and Forcing vectors were checked in a similar manner. The second major verification of the computer pro- gram was accomplished by solving the following problem: the conductor of Cable 2 was maintained at a specified hot temperature, while that of Cable 1 was maintained at the oil temperature. Sheath and dielectric losses were not included. The height of the inter-cable conduction path was chosen so as to include an angle of 6 in either cable. This angle was judged to be sufficiently small so as to minimize the thermal effect of Cable 1 on Cable 2. Temperatures in Cable 2 diametrically opposite the intercable conduction path (and hence far-removed from the limited two-dimensional effect) were then compared to corresponding analytical temperatures from the one-dimensional solution. This comparison 70 was repeated for successively finer meshes in order to examine the convergence of the solution. In performing the test, temperatures at similar radial points were compared, and a root-mean-square error was defined: .1/2 N -1 IN~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ i [(Tj)computer- (Tj)analytical] RMS-error (F) = j-l . . N r N (5.8) where N is the number of mesh points along a radius in Cable 2. This error was then expressed as a percentage of the total temperature drop through the insulation. The re- sult for five different mesh sizes is shown in Figure 5.3. The errors demonstrate the typical (1/h2)-dependence on mesh size, which is expected, since nearly all the finite difference expressions used in this problem are tions. 2 (h ) approxima- It is further seen that the computer solution clearly converges to the analytical solution, having less than one percent error in temperature with as few as three radial subdivisions. A final run was then made with this problem to check whether a symmetrical cable configuration would produce a symmetrical temperature distribution. A coarse mesh of 18 points was used with a cable geometry such that the line joining the cable centers was a line of symmetry. In the resulting temperature distribution, the seven temperatures 71 10 6 0 0o Cr 4 LLJ o° 2 v 3 2 5 4 z(2) FIGURE Percent-Error 5.3 in Temperature vs. 2, the Number of Radial Subdivisions in D22Z 72 above the line of symmetry agreed with their mirror images to six significant figures. The third major verification of the program was a cross-check, using the output from Solution 1 (which finds temperature) as the input for Solution 2 (which finds current). For this test a 79-point mesh was employed, and all losses in both cables were included. Using an oil temperature of 140°F and a current of 942 amperes in each cable, Solution 1 predicted a maximum temperature of 189.451°F for the system. This maximum temperature, together with the oil temperature of 140°F, was then used as input for Solution 2, which predicted a maximum allowable current of 942 amperes in each cable. The solutions mutually agreed to six significant figures, thereby demonstrating the reciprocal validity of the program. The final major verification procedure was to approximate the one-dimensional solution for a single cable with all losses, by shrinking to a minimum the height of the inter-cable conduction path. In the computer program the conduction path is presumed to have some non-zero height, so the included angle in either cable was taken to be 2°. Again temperatures in Cable 2 opposite the inter-cable conduction path were compared to corresponding analytical temperatures, and the same error criteria (RMS-error as a percentage of the total drop through the insulation) were employed. The test was made for three cases: loss only, conductor losses dielectric only, and then all losses. 73 Using a mesh comprised of four radial subdivisions in the insulation, the following results were obtained: dielectric loss only - 2.6% error in temperature solution; conductor losses only - 0.4% error in temperature solution, 1.0% error in current solution; all losses - 0.8% error in temperature solution, 2.0% error in current solution. Temperatures in the dielectric solution are elevated above their analytical counterparts because of the referral of loss near the conductor into the conductor itself. However, when the dielectric loss is considered proportionately with all other losses, the temperature small error is observed (less than one percent predicted to be reasonably for this mesh). The currents in this test are seen to be less accurate corresponding predicted temperatures. than the This circumstance, though, reflects a limitation of the test itself rather than one of the model. is expected That is, the temperature distribution to smooth out into a one-dimensional region far-removed from disturbing effects. form in a However, the current solution depends on the entire temperature distribution. Since an inter-cable conduction path of any non-zero size will inevitability produce local distortions in the temperature distribution, it is ultimately futile to expect very close agreement with a one-dimensional current solution. Agreement could be demanded if the size of the conduction path could be made identically zero, but the model does not possess this capability, having been designed for the solution of real two-dimensional problems. Given 74 this limitation, the current solutions demonstrate excellent agreement with the one-dimensional results. The credibility generated by this test, together with all the evidence previously stated, suffices to establish the validity of the computer program. 74.1 CHAPTER 5A COPPER TAPE EFFECTS Effective Conductivity Many cable systems used in the underground power transmission have a thin copper tape wrapped around the cable insulation directly under the moisture seal assembly. The tape is included to circumferentially smooth out the electric potential and to provide electric ground. Having a very high thermal conductivity the copper tape provides a mechanism for transIt therefore causes a ferring heat away from high temperature regions. redistribution of the temperature, tending toward the one-dimensional form. Order of magnitude calculations indicate that the presence of a five mil (t = 0.005") copper tape (thermal conductivity k CU = 220 BTU/hr ft°F) under the moisture seal assembly have a significant effect on the temperature distribution of the cable: Consider a typical system of Fig. 5A.1, r = 0.9125 in r 2 = 2.0675 in k = 0.1153 BTU/hr ft°F The thermal resistance per unit length in the circumferential direction without the tape R is approximately c R = c 1 k(r 2 -rl) = 90 hr°F/BTU The circumferential thermal resistance including the copper tape R c is 74.2 1 k(r2 -r 1-t) + kt C c . R c Since R c = 1 1 9.7 hr°F/BTU k(r2 -r1-t) + ktht- and R' are significantly different, the presence of a highly con- c ductive medium under the cable moisture seal assembly must be taken into account. An order of magnitude calculation also indicates that the copper tape effect in the radial direction is negligible: whereas in the circumferential direction the resistance of copper and insulation are parallel,in the radial direction the two resistances are connected in series. The radial thermal resistance per unit length without the tape R is r R = in r/r 2 = 1.129002568 hr F/BTU k The radial thermal resistance per unit length including the tape, R ' r -t i~n r R ~~~~~~~~~~~~~r2_ r in is r + r 1 -t = 1.12900432 hr°F/BTU 2 2 rk 2 k cu Since the tape is very thin only the mesh points located around the outside circumference of the cable insulation are affected. In Fig. 5A.1 the thermal conduction path between point 0 and 1 consists of the copper tape (thickness t) and a layer of insulation (thickness L -t). The path length is L r thermal resistance of the two conducting media L - = k(Lr-t) ____ + k R rCU eff r t = k L R is . The 74.3 A Oil Circle A angle 4 angle _- I th Lo l Fig. 5A.1. Development of the expression for the combined effective conductivity of a layer of the cable insulation and the thin copper tape. 74.4 where k ff is the effective conductivity of the entire thermal conduction path betwee 0 and 1. In the computer program L = (r -rl)/N; r where 21 N is the number of radial divisions. Hence, t+k( rk - t) 2 r k eff - cu N (5A.1) r 2 - r1 N Equation (5A.1) applies only for points on the insulation outer circumference and in the circumference direction only, al other mesh poinst are unaffected. Finite Difference Equations For the mesh points located on the outside circumference some of h -J z4t _O 4 . r i of the governing equation (2.6) and the boundary condition must be adjusted to account for the higher conductivity in the circumferential direction. Thus for the point 0 in Figure 5A.1 the finite difference approximations in non-dimensional form (L r a20 40- 200 + 03 r2 2 r 04 4 - 03 3 2h r = hrr2, r 2 = h r2) L a a 2 74.5 a20 a _= 0 2 - 20% + 2 0 ~-ct-2 h 0 2 1 (5A.2) 2 0 2 h 0 (hcl+h 2 ) halh o9 + 2 0 ho (ha+h a2) Twhere 0 is given by (2.5) and O Toi W/k W/keff = Then using (5A.2), the equation (2.6) about a regular boundary point 0 (h = h -2 ) becomes, 0 -200 [4 r +r[ 4]+I['1 _-0 + 2 03 2h r 02 - 20 + 4 - 200 + 03 r h . 4 -03 +h r [ 2h ] II h2 L L J h (5A.3) 2 For a boundary point at a regional interface (ho -2 le (r2r) 2 r # ho ): 2 2 (hDi+h02) ocio I 0 r -2 (r 2 r) 2 h (h +h 1 ) W (5A.4) The boundary condition (2.8) is unaffected for all boundary points; k r 04 - 3 ~= 2h k2~ 2 Eliminating r h2r 0 (5A.5) 04 from (5A.3) and (5A.4) using (5A.5) and noticing that from 0= k AT e AT 0 = keff W obtain 0'-= keff 0 k (5A.6) 74.6 Hence for h = h 2hhr k = h 2 (-2 rk2( k - (h2 -2 eff ) 2h + k r k k 2(r I eff k 2r) 2 h h cx and for h r a + [1 eff k + 2r 2 q (5A.7) (5A.7) r] h hhrcx r h2 2h hr -2 r 2( + +-[ ff k k khh k +eff_ _ ~ rr k ) 2h r 2 r + + 2r h2+ ft k k + )1® 2 + h ah lc, 3 r k_ eff ( 22r +h h )®1+ h, 1 (h eff h k _ _ 2_ ]9 ,o - +[2 ] - 0 2]2 2 ] r -2 . _(2 ) q (5A.8) W Computer Program Modification From (5A.7) and (5A.8) it is apparent that to account for the presence of a thin high-conductivity tape under moisture seal of the cable insulation it is only necessary to multiply the original coefficients (corresponding to k~~~ ! the homogeneous insulation material) of ®1's and eff 2's by k and to add k 2 ftihft (1 follows thatt followsth can be eased eff) k to both 2 h f (5A.7) and (5A.8) for if = hh e =h c2 , it 2 2. -hc a The simple complementation of this procedure by the following manipulation. 74.7 k From of 0 (5A.8) the coefficients 's and O's for 1 2 h h= 0l'S and 02 s in (5A.7). = 1 (original coefficient k which are the original coefficients of 2 reduces to ci h~~~~~~~~h2 hc for h eff It is possible therefore to use (5A.8) for all circumferential boundary points. Further, let the original coefficients of 's and 0 's be 1 2 h Ul(hc+h =0 ) (5A.9) 1 2 =0 h o2(h 2 +h o) Also, let the new coefficients of 0l's and 1 2 f =P (h D+h a ) h 2 2's be 2 =o. P1 01 (5A. 10) 2 h2 P f (h l+h where f- ) 0 2 2 f keff k k 2 Let Q h = h (1 - f = factor to be added to the original coefficient alo02 02's. 2 Then from (5A.9) and(5A.10) 0 or h (01 - P1) 4 (- + 1) = P + hCa (h o2+h 1 h 2(l -t) h Q ) 1 (5A. 11) h and similarly (2 - P2) ( ) (12 = Q (5A.12) of 74.8 Eliminating h and h Q = 01 + 02 -P1 between (5A.ll) and (5A.12) obtain, (5A.13) - P2 Thus to modify the original matrix of coefficients of O's it is necessary I!I keff to multiply the original coefficients of Ol's and 02 s by f = k add the difference (1 0 1 and 0 2 of 0 1'sand 1 - P 1 ) and (2 - P 2 ) between the original coefficient 0 's and the new coefficients P 2 to the original coefficients of 0 s. and to 1 0 f and P 1 = 0 2f 2 2 An important property of the matrix 0 Awas used to find the locations of 's and 0 s, s: the numbering system is such that all coefficients of 0 's lie on the main diagonal and 0 the coefficients of 0 's immediately precede 0's in each row and the ~~~~~~0 1 coefficients of 02 's immediately follow O 's in each row with the exception 2 0 or tne row corresponding to the governing equation written about the point in cable 2 at an angle = 0. Also all 's are circumferential boundary 0 points. Verifications The first major verification was a check on the coefficient matrix. This test was, as in Chapter 5, accomplished by printing out the matrix for a 42 and 17-point mesh and then by verifying each element by hand computations. Tha second major verification was an energy balance performed on selected mesh elements, again by hand computation. The final major verification was accomplished by solving the following problem: the intercable conduction path was reduced to a very small size so that approximately uniform convective boundary conditions existed at every circumferential point. The problem was run with and without the tape and both solutions were converging to the 1-D solution as the intercable conduction path was being reduced. CHAPTER 6 RESULTS AND CONCLUSIONS Evaluation Criteria As a first criterion for evaluating the severity of a given set of system operating con.ditions, the numerical solutions produced by the computer program are compared to the corresponding analytical solutions for a single, undisturbed cable. The one-dimensional temperature and current solutions for a single cable are presented in Appendix G. Since the undisturbed cable represents an optimum operating condition, the one-dimensional solution provides an upper bnini fnr system performance. A second criterion for evaluation may be obtained from a modification of the one-dimensional solutions, in which a conservative allowance for two-dimensional effects is made. manner: This modification is effected in the following it is reasoned that an effect on any portion of the cable surface which disturbs the one-dimensional temperature distribution is less severe than the effect of insulating that portion of the surface. The conservative approximation is then made that such disturbances do indeed effectively insulate appropriate portions of the surface, and that the entire sector defined by such an insulated arc is likewise insulated. All losses which would have occurred in the insulated sector are then placed into the undisturbed fraction of the cable, and the one-dimensional solutions are r7 P. 76 employed with appropriately scaled-up losses. This proce- dure is explained fully in Appendix H, where the conservative approximations for maximum temperature and current are derived. Since by physical argument the disturbed portions of the cable surface cannot have a more stringent condition imposed than that of being insulated, the approximate formulas of Appendix H provide a lower bound for system performance. Results While the primary product of this study is the computer program itself, a total of 16 cable problems were solved by the author in order to provide preliminary inforThese mation about some typical operating conditions. 16 problems break down into the following: Solution 2 (for maximum current) and Solution 3 (for maximum oil temperature) were employed for four configurations of cables, and two cable systems were considered. The four cable configura- tions were equilateral, cradled, open, and equilateral-pipe. These are depicted in Figure 6.1. The systems considered were a 2500 MCM system (System 1) and a 2000MCM system (System 2). Values for the physical parameters associated with these two systems are listed in Table 2. The thermal parameters were taken to have the following values for all 16 problems: 0.1153 Btu/hr-ft-°F for the thermal conduc- tivity of the insulation, and 5.0 Btu/hr-ft2-°F for the thermal film coefficient in convective regions. Also the conservative assumption of a thermally nonconducting conduit 77 EQUIL ATERAL CRADLED OPEN EQUILAT ERAL PIPE FIGURE 6.1 Nomenclature for Cable Configurations 78 TABLE 2 Values for The Physical Parameters of Systems 1 and 2 System 1 System Inner Radius of Cable Insulation (in) 0.9125 0.8155 Outer Radius of Cable Insulation (in) 2.0675 1.9675 Skid Wire Thickness (in) 0.10 0.10 5.36 6.63 at Conductor 1.19 1.13 at Sheath 1.24 1.18 Typical Dielectric Loss (watts/conductor-ft) 3.18 3.18 942 888 Thermal Conductivity of Insulation (BTU/hr-ft°F) 0.1153 0.1153 Thermal Conductivity of Copper Tape (BTU/hr-ft°F) 220.0 220.0 Copper Tape Thickness (in) 0.005 0.005 DC Resistance of Conductor (/ft) AC/DC Ratio Typical Current (amps) 2 79 was made. This latter assumption is significant only for the cradled and equilateral-pipe configurations, where the . t cables actually touch the conduit wall. A nonconducting wall substantially increases the thermal resistance in the vicinity of the contact point, thereby producing a local region of high temperature and hindering the removal of heat from the cable. Results for the 16 problems are given in Tables 3 -6. In these tables the first column of percent- ages is a comparison of computer solutions to corresponding one-dimensional solutions. These negative percentages are a measure of how much worse the given operating condition is than the best pssihl nondi;on. The second column of percentages is a comparison of computer solutions to the conservative approximate solutions from Appendix H. These percentages, which are positive, provide a measure of how much better the given operating condition is than the estimated worst condition. Upon examining the four tables, the equilateralpipe configuration is immediately identified as the most severe operating configuration. This is expected, since the greatest obstruction of cable surface area occurs in this configuration. sequence: The other configurations follow in logical equilateral, cradled, and open. It is shown in Appendix B and also in the heat transfer report [2] that a steel pipe is very effective in conducting heat away from the cables to the bulk of the oil. So had the problems 80 TABLE 3 Solution 2 For Four Cable Configurations System 1 1-D I-I I (amps) 1-D 1075.4 -5.0% +0.2% Open -- tape 1105.0 -2.4% +0.3% Cradled (Nonconducting Pipe) -- no tape 1030.2 -9.0% +2.0% Cradled (Nonconducting Pipe) -- tape 1086.0 -4.1% +7.5% 978.2 -13.6% +3.7% 1071.3 -5.4% +13.5% 946.4 -16.4% 1057.0 -6.6% (1132.1) - Open -- no tape Equilateral -- no tape Equilateral -- tape Equilateral-Pipe (Nonconducting Pipe) -- no ape Equilateral-Pipe (Nonconducting-Pipe) -- tape One-Dimensional T T oil max = 140 ° F = 185°F I - current from computer solution I - current from one-dimensional solution I* - current from conservative approximate solution I,- current from conservative approximate solution * To . 6 +21.3% 81 TABLE 4 Solution 3 For Four Cable ConfigurationsSystem 1 T .- (T il)l Uj-j- Toil (F) oil -DD To (Toil)l-D Toil- (Toil)* T- o (Toil)* 0.0% Open -- no tape 148.3 -8.9% Open -- tape 149.9 -4.2% Cradled (Nonconducting Pipe) -- no tape 145.7 -16.6% Cradled (Nonconducting Pipe) -- tape 148.9 -7.1% Equilateral -- no tape 142.4 -26.4% Eqi-I1teral -- tape 168. Equilateral-Pipe (Nonconducting Pipe) -- no tape 140.3 -32.6% +11.5% Equilateral-Pipe (Nonconducting Pipe) -- tape 147.3 -11.9% +25.4% One-Dimensional I = 942 T max Toi oil (151.3)' amps - oil temperature from computer solutions (Toil)* O r = 185°F (Toil)l T -9 D - oil temperature from one-dimensional solution oil temperature from conservative approximate solution - conductor temperature +4.4% +2.8% +10.7% +4.8% +17.8% 82 TABLE 5 Solution 2 For Four Cable Configurations System 2 I-ID 1 -D Ii_ D I (amps) I-I* 1I* Open -- no tape 950. 2 -5.0% +0.3% Open -- tape 975.3 -2.5% +3.0% Cradled (Nonconducting Pipe) -- no tape 911.9 -8.8% +2.4% Cradled (Nonconducting Pipe) -- tape 959.2 -4.1% +7.7% Equilateral -- no tape 864. 6 -13.6% +4.0% Equilateral -- tape 946.6 -5.4% +13.9% 837.2 -16.3% 938.8 -6.7% Equilateral-Pipe (Nnn-onductin_ Pine) -- no taDe Equilateral-Pipe (Nonconducting Pipe) -- tape One-Dimensional T T oil max (1000.4) = 140 F = 185°F I - current from computer solution I1-D - current from one-dimensional solution I* - current from conservative approximate solution (0.0%) +9.3% +21.9% 83 TABLE 6 Solution 3 For Four Cable Configurations System 2 _ T (Toi) Toioil ilD-D(Toi TOil-(T ° Toil ( F)oil D il)ll-D To-(T oil~~T -( l)* -( oil T (Toil), Open -- no tape 144.4 -8.8% +0.3% Open -- tape 146.1 -4.3% +4.5% Cradled (Nonconducting Pipe) -- no tape 141.8 -15.8% +3.4% Cradled (Nonconducting Pipe) -- tape 145.1 -7.0% +10.8% Equilateral -- no tape 138.2 -25.5% +5.8% Equilateral -- tape 144.2 -9.4% +17.9% Equilateral-Pipe (Nonconducting Pipe) -- no tape 136.0 -31.4% +12.4% Equilateral-Pipe (Nonconducting Pipe) -- tape 143.3 -11.8% +25.4% One-Dimensional (147.7) - I = 888 amps T max = 185°F T il = oil temperature from computer solution oil (T il)l D - oil temperature from one-dimensional solution (T il)* -oil T temperature from conservative approximate solution - conductor temperature O 84 been solved using a thermally conducting conduit material such as steel, then the results for the equilateral-pipe configuration would have been essentially the same as those for the equilateral case, and the latter would have been the most severe configuration. A complete temperature distribution for the equilateral-pipe configuration without the high tape under the cable moisture seal assembly of System 1 conductivity is displayed in Figure 6.2. Two observations are made regarding the conservatke solutions. approximate The first is that in cases without tape they are reasonably accurate, being conservative by 9.3% in the least accurate case (Solution 2, System 2, Equilateral-Pipe). They are therefore useful whenever a highly refined solution is not required. The second observation is that the conservative solutions become less accurate as the configurations become more severe. This tendency is readily exnlained. for as the sur- face area of a cable is increasingly obstructed, two-dimensional effects grow stronger. Since the conservative approximations are based on the one-dimensional solutions, they become increasingly deviant with the severity of the configuration. So despite the conservative nature of the approximate solutions, they are not recommended for design purposes. Also it is noticed that there is a large discrepancy between corresponding percentages in Solution 2 and Solution 3: temperature deviations are somewhat larger than current deviations. crepancy, however, should not be a surprising one. solutions are applied to a given configuration. Such a dis- Consider that both In Solution 3 the heat flow is constant, and the temperature distribution must be linearly adjusted so as to account for the two-dimensional constraints imposed 85 ~A: A 47.8 18C 18C 18 18. 'ES IN I,= = 942 AMPS the n - 1473 14 7 3 F. 86 by the configuration. In Solution 2, though, the heat flow is the adjustable quantity of the total heat flow, only the current-produced fraction (usually about 2/3) is variable, and thus the current-produced losses must be disproportionately adjusted so as to align the overall heat flow according to the two-dimensional constraints of the configuration. Furthermore, the current itself varies as the second root of the variable heat flow. Therefore the relationship between the two solutions is a complicated one, and there is no reason to expect any similarity between their respective deviations. Finally, it is evident from the tables that cable proximity effects are very significant in forced cooling especially for cables without the high thermal conductivity tapes. In System 2 with no tape, for example, the maximum allowable oil temperature is 11.7°F lower for the equilateralpipe configuaLiolL haff ot the one-dimensional Cat. Since force-cooled systems are typically designed for an axial oil temperature rise of about 45°F between refrigeration stations, the 11.7°F difference itself would accout for 26% of the axial oil temperature rise. On the other hand, in the same system with the tape present the maximum allowable oil temperature is only 4.4°F lower than in ID case, or about 10% of the axial oil temperature rise. For the more realistic equilateral configuration the same figures are 9.5°F or 8% with the tape. This means r 21.6% without the tape and 3.5°F that depending on whether the tape is or is not present under the cable moisture seal assembly System 2 in the equilateral configuration would require either an 8% or 21% higher flow rate, or either an 8% or 21% shorter axial distance between refrigeration stations then the same system in a completely free (one-dimensional) configuration. 87 Approximate allowance for cable proximity effects must therefore be made in the overall design of force-cooled systems. Since there is a sig- nificant improvement in the results when the copper tape is wrapped around the cable insulation, such cables are from thermal considerations, the more suitable for force-cooled power transmission work. Isothermal lines for the four oil temperature solutions of System 1 are shown in Figures 6.3 - 6.7. One-dimensional portions of the various solutions may readily be identified in these figures by isotherms which are circular arcs. As expected, all the distributions smooth out into one-dimensional form away from points of cable contact and conduit contact. Regions of high temperature within the insulation are identi- fied by isotherms which depart significantly from the circular shape, protruding outward from the cable centers. This effect is observed to be most prevalent in the equilateral-pipe configuration, decreasing in strength in the equilateral, cradled, and open configurations, respectively. Thus the isothermal lines in themselves provide a vivid illustration of the severity of the various configurations. adiabatic lines. Shown with the isotherms are These lines are everywhere normal to the isotherms, and they represent curves along which the heat flow travels. that not all adiabatic lines originate at the conductors. It is noted This circum- stance is attributable to the dielectric heating, which occurs within the insulation itself. 161 183.5 I = 942 amps TOIL = 148.3 --- ISOTHERM ADIABATIC LINE 17 N (NO COPPER TAPE) 16 16 FIGURE 6.3 Isothermal and Adiabatic Lines for the Open Configuration - System I 89 i7t 178 3 183.4 163 i3 169 I = 942 amps 185.0 TOIL 146.5 - ISOTHERM --- ADIABATIC LINE ALL TEMPERATURES 178 IN F 173 HOMOGENEOUS INSULATION (NO COPPER TAPE) 161 FIGURE 6.4 Is(thermal and Adiabatic Lines for the Cradled Configuration - System 1 90 4 I INE ATURES SULATION E) FIGURE Isothermal 6.5 and Adiabatic Lines fr the Equilateral Configuration - System 1 9i URES II LATIC N FIGURE 6.6 Isothermal and Adiabatic Lines for the Equlilateral-Pipe System Configuration 1 - 92 156 160 15 6 160 160 180 174 168 162 156 Figure 6.7 Isothermal and Adiabiatic Lines for the Equilateral-Pipe Configuration with copper tape in the moisture seal assembly - System 1. 93 Conclusions On the basis of the results discussed, the following conclusions are drawn: 1. For a thermally nonconductiong conduit, a cable system is most susceptible to thermal failure in the equilateral-pipe configuration. For a thermally conducting conduit, the equilateral- pipe and equilateral configurations are equally severe, and the latter represents the worst operating configuration. 2. The conservative approximate solutions developed in Appendix H are useful for obtaining good estimates of maximum temperature and current. However, recourse should be made to the computer solutions whenever design information is required. .able ..iguration, ... _r . , ... o21 configu 3. Cable proximity effects are important in forced cooling. The BV with no copper tape) to 26% (equilateral-pipe configuration with no tape) of the total oil temperature rise between refrigeration stations. 4. The presence of a thin copper tape in the cable insulation moisture seal assembly significantly smooths out the temperature distribution in the cable insulation and thus higher maximum allowable oil temperature and higher currents are permitted than if a homogeneous cable insulation is used. Numerically the improvement is from about 4.3% for the oil temperature and approximately 9.5% for the current. If this figure is unacceptable, thicker copper tapes would smooth out temperatures even more. 94 Recommendations for Further Work In order to fully exploit the convenience of having a separate region associated with each boundary condition, a provision should be written which would allow Regions IV and VI of Domain 2 to be used simultaneously. Such a provision does not presently exist, because there is only one configuration for which it would be desirable. However, the exceptional configuration is the equilateral-pipe case, in which both cable-cable and cable-conduit boundary conditions act simultaneously. Since this is the most severe condition for a thermally nonconducting conduit, it is expected that this configuration will be frequently used, and the change is probably warranted. It is noted, however, that in the present program all boundary conditions are independently specified by means of a variable film coefficient. The equilateral-pipe configuration can therefore be modelled as accurately as the user desires, and the proposed modification represents only a convenience in laying out the system geometry and mesh size. Instructions for implementing the change are given in Appendix E. Finally, some improvement could be effected in the input-output formats of the computer program. Throughout the development of the program, attention was continually given to simplicity of I/O procedures and to ease of user operation. Yet certain aspects of the final I/O formats are less convenient than is desirable. are offered, again in Appendix E. Suggestions for their improvement CHAPTER 7 REFERENCES 1. Long, H. M., Notaro, J., and Webster, D. J., "Economic Analysis of a Generalized Design for a Forced Cooled Cable," IEEE Trans., Vol. PAS-90, No. 3, May-June, 1971, pp. 1225 - 1231. 2. Glicksman, L. R., Orchard, W. P., Rohsenow, W. M., and Slutz, R. A., "Cooling of Underground Transmission Lines: Heat Transfer Measurements," Unpublished Report for Consolidated Edison, Energy Laboratory and Heat Transfer Laboratory, Department of Mechanical Engineering, MIT, January, 1974. 3. Kreith, F., Principles of Heat Transfer, International Textbook Company, Scranton, Pennsylvania, 1965, p. 80. 4. Ibid., 5. Arpaci, V. S., Conduction Heat Transfer, Addison-Wesley Publishing Company, Inc., Reading, Massachusetts, 1966, pp. 126 -129. 6. Crandall, S. H., Engineering Analysis, McGraw-Hill Book Company, New York, 1956, p. 246. 7. Forsythe, G. E., and Wasow, W. R., Finite-Difference Methods for Partial Differential Equations, John Wiley & Sons, Inc., New York, 1960, pp. 186, 188. 8. Von Hippel, A. R., Editor, Dielectric Materials and Applications, The Technology Press of MIT and John Wiley & Sons, Inc., New York, 1954, pp. 1- 10. 9. Pugh, E. M., and Pugh, E. W., Principles of Electricity and Magnetism, Addison-Wesley Publishing Company, Inc., p. Reading, 10. 79. Massachusetts, 1970, p. 39. Holman, J. P., Heat Transfer, McGraw-Hill Book Company, New York, 1968, p. 28. 11. Ibid., p. 29. 12. Feibus, H., Consolidated Edison Company of New York, Inc., New York, Personal Source, January, 1974. 95 APPENDIX A THE RELATIVE MAGNITUDE OF CONDUCTION AND CONVECTION RESISTANCES The equation which defines the resistance per unit length to heat transfer is the following: AT AT q (A.1) where q is the total heat flow per unit axial length, AT is the temperature difference driving the heat flow, and R is the resistance per unit length to heat transfer. The heat flow per unit length due to convection from the outer surface of the insulation to the oil is [10] q = 2fr h(AT) 0 (A.2) , where h is the convective film coefficient, and r0 is the outer radius of the insulation. The convective resistance per unit length is therefore -h 1 2ffr0 h' h 2rr~h (A.3) The heat flow per unit length due to conduction in the cable insulation, which is assumed to be one-dimensional in this type of calculation, depends not only on AT, but 96 97 also on the distributed dielectric loss. Because of this additional dependence on the dielectric loss, it is not analytically possible to model the conduction path as a simple resistance. However, a conduction resistance may be ob- tained numerically, by substituting appropriate values into Equation A.1. The (AT) for a given set of losses (q) is available from the one-dimensional solution presented in Appendix G. A typical set of values for (q) and (AT) yields R AT hr-ft-°F (q) = 0.93 Btu ' (A.4) where AT is the temperature drop across the insulation, and (q) is the heat flow per unit length emanating from within the insulation (the conductor and dielectric losses). The corresponding value for Rh from Equation A.3 is 1 Rh 2r hr-ft-°F 0 h 0.18 Btu (A.3) where the most conservative value for the natural convection film coefficient (5.0 Btu/hr-ft -°F) was used [2]. relative magnitude of the two resistances is Rh _ Rk 0.18 0.93 (hr-ft-°F/Btu) (hr-ft-°F/Btu) The therefore _ 19 A.5 (hr-ft-01F/Btu)-*(.5) 0.93 98 Thus the natural convection resistance, which is always larger than the combined forced and natural convection resistance, is small when compared to the conduction resistance, and the latter is the limiting resistance to heat transfer. APPENDIX B INVESTIGATION OF THE CABLE-CONDUIT BOUNDARY CONDITION In order to examine the cable-conduit boundary condition described in Chapter 2, a portion of the conduit wall is thermally modelled as a fin. which to determine The geometry from is shown in Figure B.1. a fin length It is based on a cradled configuration of cables, since that configuration produces the largest conduit loss. possible lengths L and L 2 , the latter is chosen maximize the temperature drop through the fin. Of the two so as to L2 is found from the following relations: sin r2 + R 1 2 -r p 0.660 , (B.l) 3 from which 1 = sin -l (0.660) = 41.3 ° . (B.2) Then 2 = 180- 2Bl 99 = 97.4 (B.3) 100 FIGURE B.1 Fin Geometry for the Cable-Conduit Boundary Condition 101 and L2 = (Rp + = 8.92" (B.4) One end of this fin is insulated by symmetry, and as a conservative approximation, the side of the fin which is adjacent to the earth is likewise taken to be insulated. Furthermore, the pipe loss is taken to be concentrated at the end of the fin which touches the cable (again to maximize the temperature drop through the fin). model is illustrated in Figure B.2. This thermal Considering the heat flow to be one-dimensional, the fin temperature is governed according to [11] d2T dz h P -2kA dz(T-T p oil ) =0 (B.5) where P is the convective perimeter, and A is the crosssectional area. Using the dimensionless variables z n = ' 2' e = T-T T oil W/k (B.6) p p Equation B.5 becomes the following: d2 _2 -9-d2 d-- P eh 0= 0 2 kp A( -- . B 7 (B.7) 102 W, " Wp kp PIPE LOSS/I UNIT LENGTH CONDUCTIVITY OF PIPE - THERMAL FIGURE B. 2 Thermal Model of the Conduit Wall 103 Making the substitution m 2 2 2 ~h hpP = L2 k' (B.8) P p the governing equation is just 2 d20 --- 2 - m 0 = 0 dn The boundary conditions . (B.9) are -k t dT P dz = (B. 10) W 1 z=O0 p and dT dz = 0 . z=L (B.il) 2 These are rendered dimensionless to give L2 da a -TIIr (B.i2) and dO! 7,1 -= 0 . n=1 (B. 13) 104 The general solution of Equation B.9 is a () = Cle-m +m C2 e (B.14) The two arbitrary constants C 1 and C2 are found from the boundary conditions to be L2 L2 tm(l-e-2m) ' C 2 =- (B.15) tm(e2m-1) When these are substituted into the general solution, Equation B.14 can be manipulated into the following form: ( =L2 cosh[m(1-n) tm ] (B.16) sinh (m) The following values are then taken: = 3.3 Btu/hr-ft 2-F P = 1 ft h P = 1 ft = 25 Btu/hr-ft-°F (conservative, based on [2]) (B.17) k (1% C steel) A = 0.0208 ftP 2 A = 0.0208 ft Equation B.16 finally becomes 6(n) = 5.73 cosh[l.92(1-n)] . (B.18) 105 It is desired to know the temperature drop from the fin base to the oil, so it is now convenient to return to the form W T(rl)-Ti oil = 5.73 - (B.19) cosh[1.92(l-ri) k p ~~~~~~~(B.19) p The desired quantity is obtained by evaluating Equation B.19 at = 0 and by substituting W (8.85 watts/system-ft). The quantity in parentheses is a typical loss value, which is divided by 6 to account splitting of the heat flow to either cable contact. oil 5.73 = and for the side at the poin of The temperature drop is then 1 T(0)-T for the 3 cables 1 (8.85 watts/ft) 25.413 (25 Btu/hr-ft- = 3.9°F F) (3.413 tu) Btu) cosh(.92) (watt-hr) . (B.20) It is now desired to compare this temperature drop to the drop across the cable insulation. Using once again the one-dimensional solution for a caole with all losses (Appendix G), the temperature distribution in the insulation is given by W (W d O~)d d(~ 4fW(ln (ln E)221 +W (+2 d c 2irWc (W in ~ 4± +W +W Wdc++W c d s 2-7Whr2 ) k (G.16) 106 where Wd, Wc, and W s are the dielectric, conductor, and sheath losses per unit length, respectively, and W is an arbitrary loss per unit length. The dimensionless temperature drop across the insulation is given by (~1) -86(1)= 61 ln - Wd\ 1 2W 2TrW (2 +W c)(.1. (B.21) Returning to the dimensional temperature by means of Equation G.6, this result becomes: T 1) ( -T(1) = -- ln ~ 1 /W d W )) (B.22) The dielectric and conductor losses corresponding to the W P used previously are 3.18 and 5.66 (watts/conductor-ft), respectively. These values, together with k = 0.1153 (Btu/hr-ft-°F) give T(Ci) - T(1) = 31.1°F . (B.23) The relative magnitude of the two temperature drops is therefore (AT) fin390 (Afin = 3.9 0 F = 0F (AT) insulation 31.-1 12.5% . (B.24) 107 It may be inferred from this result, which represents the most conservative comparison of the two effects, that contact between the cables and the steel conduit does not significantly alter the overall temperature distribution. The cable conduit boundary should thus be modelled as a convective one, with at most a slightly modified film coefficient. _ _ _ _ APPENDIX C THE SOLUTION FOR MAXIMUM CURRENT The Superposition Method for Solution 2 In Solution 2 it is necessary that current- produced losses be treated separately from voltage-produced losses. This makes it possible to distinguish the variable component of the temperature distribution from the stationary component, and subsequently to adjust the variable component so as to maximize the current. When the losses are so separated, there are two complete problems, each one having three components and resembling the problem presented for in Chapter 3. In fact, the solution in Chapter 3 (x) may be taken as one component of Solution 2, provided that the forcing term f(x) is clearly identified, either with stationary losses or with variable losses. Accordingly, let f(x) describe all forcing effects in the domain which are attributable to current. The solution (x) then denotes that part of the total temperature distribution which is current-dependent. It is now necessary to determine the stationary portion of the temperature, that portion which depends on voltage. called Let this part of the total temperature solution be D(x). The solution D(x) satisfies the governing equation V2 OD(x) = g(x) , 108 (C.1) 109 where g(x) describes all forcing effects in the domain which are attributable to voltage. D(x) also satisfies the following boundary conditions (using the notation introduced in Chapter 3): D(X)l xEC (x) 1 - where = (C.2) D1 D01 is some unknown dimensionless temperature. hr2 OD (x) k - OD(x) xcC 2 (x) a ED(x) an2 hr2 = xEC - a an4 D(x) 43n (C.3) x eC 2 (x) -- (x) 3- (C.4) OD(x) k xcC - (x) 3 I xC -CC (x) 4 = 0 (C.5) = 0 (C.6) = 8D 0 2 (C.7) W DOD(x) an 5 xcc5 (x) 8D(x)xeC6(x) - 6 - _ _ 110 where OD02 is some unknown dimensionless temperature. hp2 3aOD(x) = - -- an7 (C.8) OD(x) x C7(x) 7 a OD(x) an8 (C.9) 2Dh OD(x)i D xkCc8(x) xC8 C8(x) aOD(x) -mx Dn - 9 xEC 9 (x) 2Dh (X [D C( 2kh OD(x) I xC (x ) 9 - (C.10) As before, OD(x) may be decomposed into three component problems. However, it is not necessary to introduce three new components, for the solutions A(x) and eB(x) of Chapter 3 already describe the homogeneous components required. So only one additional component is needed, and let it be referred to as E(x). This component satisfies the nonhomogeneous governing equation V2OE(x) = g(x) (C.11) The boundary conditions satisfied by OE(x) are identical to those satisfied by OC(x), Equations 3.32- 3.40. 111 Once again the three component solutions are linearly combined according to OD(x) = b OA(x) + b 2 OB(x) + OE(x) , 1 where b 2 (C.12) _ and b2 are two new arbitrary constants. The validity of Equation C.12 is readily established by direct substitution into the appropriate governing equation and boundary conditions, Equations C.1- C.10. Since this procedure is identical to that followed in Chapter 3 (see Equations 3.42 -3.51), constants b it is not repeated The two and b2 are again determined from a knowledge of the losses at the conductors. Equations here. By analogy with 3.56 and 3.57, a3OD(x) dC(x) 3n 1 C xEC - 1 (x 1 = 0 , (C.13) (x) _ and OD(x) I an6 C6 X () XE dC 6 (x) = (C.14) 0 . 6 W The zero right-hand sides of these equations reflect that there are no voltage-produced conductor losses. When 112 Equation C.12 is substituted into Equations C.13 and C.14, two simultaneous algebraic equations result which uniquely determine b and b2. The stationary part of the total temperature solution, OD(x), is therefore available. Attention is now turned to the total solution, 0I(x), which includes both stationary and variable losses. 0I(x) satisfies the nonhomogeneous governing equation V2 I(x) = f(x) +g(x) , (C.15) as well as the following boundary conditions: I(x)IxeC1 where 01 = (x) 01 (C.16) ' is some unknown dimensionless temperature. aoI (X I an2(x) hr = IxEC I 2 k ( x) - aeI (x) (C.17) OI (x)I 0~x IxfC (x) 2 hr 2 = an3 (C.18) ei(x) xC 3(x) xC3() =0 XEC4( aeix (x) cC k (C. 19) 113 3aI(x) n5 C(x) = 0 (C.20) xc55 I(x) xcC 6 (x) - 6 02 = ,01~~~ ~(C.21) where eI02 is some unknown dimensionless temperature. TOI (x) hp 2 an 7 - k 61(x) xEC77 (x) M6I x) 2Dh an 8 k xEC -, 8 2Dh an 9 k xcC (C.23) eI(x)[ XEC8 (x) (x) aeI (x) (C.22) xcC 7 (x) (C.24) eO (x) x (x) C 9 (x) The total solution eI(x) is obtained as a simple sum of the particular solutions 6(x) and 6D(x): eI(x) = (x) + D(x) (C.25) The validity of C.25 is established by direct substitution into Equations C.15 -C.24: V 2 6I(x) = V 2 [0(x) + 6D(x)] = f W + W Check (C.26) 114 I(x) Ix(c (X) = [(x)+(x)x) ]lX C(x) = 1 + 0Dl Check, (C.27) 01 + OD0 provided 01 directly =(a from the linearity 1 +b1 )A0. This result follows of the problem: al and b were determined by the variable and stationary components, respectively, of the loss at the conductors. Since the variable and stationary losses may be added to give the total loss at the conductors, the temperatures alA and l~~~~~~~~o blA° may likewise be added to give the true conductor temperature. 36I(x) 3 an2I=) x¢ 2 ( 2 2E [0(x) + 2 -n D(x)] x'C2 (x) - hr 2 [6(x) + 2x D(x)] 'x~C2 (X) hr 2 6I(x)i k Check x (EC () x£ 22 (x (C.28) 115 30I (x) [ (x) + OD(x)] an ~C xcC3 (x) hr2 2 k 3 x) [0(x) + ED(x) I xC 3(x) hr2 OI (x) k aOI (x) a 3n4 xC 4 Check [0(x) + D(x)] xEC4 (x) (x) = 0 M6I (x) Check _a 9n 5 (C.30) [8(x) + D(x)] xEC 5 (x) x XEC5 (x) = 01(x)j xE C = (C.29) xeC 3 (x) [0(x) + 0 (C.31) Check D(x)] XC 6 (x) 6 = 02 + 0D02 Check, (C.32) This result is provided e102 = (a2 +b 2 )Bo. = 002 + D02 analogous to Equation C.27, again following from linearity. 116 aeI (x) a dn 7 xEC 7 (x) - [0(x) + OD(x)] an7 xCC(x) - hp. [6 (x) + OD(x)] C7 (x) h p2 I I - I (x)i I Check C (C. 33) (x) 7- aOI (x) an 8 xC 8 (x) an 8 I 0(x) + eD(x) ] Ix,C (x) -.. ,,,., 2Dh [6 (x) + 6D(x)] XEC 8(X) 2Dh k 6I (x) Check ix C 88 (x) (C. 34) aeI (x) an 9 XEC 9 (x) n_9 3n9 [6(X) + D(x) ] xcc 2Dh k 2Dh k 9 (X) [6(x) + D(x)]i xC 6I (x) 9 (x) Check XtC 9 9 (X) - (C. 35) 117 It is thus established that the overall solution OI(x) is available from its stationary and variable components according to Equation C.25. variable component It now remains to adjust the (x) so as to maximize current with respect to the allowable cable temperature and the oil temperature. Maximizing Current in Solution 2 The current I is introduced into the temperature solution through the relation O(x) = 1 (x)I 2 (C.36) where yl(x) is a constant of proportionality whose magnitude depend on position x. Equation C.36 follows directly from two elementary facts: 1. Because of linearity, the cable temperature is directly proportional to the cable loss. 2 2. Current-produced losses are directly proportional to I. (x) is available, It is now recalled that the solution provided that the current-produced losses (and hence I) have been specified. be an arbitrary current Accordingly, let I 0 from which a temperature distribution e0o(x)is determined. Then from Equation C.36, y (W)I e0 0 0oX == ¥1(x I2 , (C.37) 118 and hence e (x) l = (C.38) I 20 Equations C.38, C.36, and C.25 may be combined to give the result 6 I( 2 2 + (x) = (x) , (C.39) 0 where (x), I, and D(x) are all known. It is desired allowable value, say C.39, I(x) take on some maximum to have ma x . Inserting this value into maxEquation Equation C.39, mx= o (x) 2 + D(x) . (C.40) 0 Upon rearrangement this relation yields m max- D) (x) I o where the dimensionless scalar brevity. (C. 41) 0- (x) has been introduced for 6(x), which is known throughout the domain, 119 determines the ratio I /I2 which will produce a temperature 0 of It is now necessary to choose max at the location x. max~ the particular ratio I2/I2 (and hence the particular value ~ of 6(x)) which will yield a maximum temperature This is accomplished simply by taking distribution C.39. 2 2 the smallest possible ratio I/I 0 Let denote the minimum The desired temperature distribution is over all x of 6(x). ~ in the max .~ then 0I(x) = 60 (x) + D(x) Proof is as follows: let x o (C.42) be the location at which the minimum value of 6(x) occurs: ~ = 6(x o ) (C.43) It then follows from Equation C.41 that eI(xo ) = max (C.44) . Now consider any other location in the domain, say x1 . It is known from the definition C.41 of 6(x) that 6(x 1 )0o (x1 ) +D(x 1) = max (C.45) 120 It is also known that < 6(x 1 ), (C.46) m since 6 is the minimum over the entire domain of 6(x). It then follows directly from the relations C.45 and C.46 that 6o6 (x 1) + D(x 1) = I(x1) < 6max (C.47) The distribution C.42 is therefore proven to be the correct one, and the maximum allowable current is determined from Equation C.41 with 6(x) = 6: I = i / 0 . (C.48) I- APPENDIX D THE DIFFERENCE FORM OF THE CONDUCTOR BOUNDARY CONDITION In Chapter 3 the heat flow emanating from the conductor of Cable 1 was given as Tar =~~~ -kf q= rirlrd = wCl .(3.52) 0 It is now convenient to express this in the dimensionless form 7r I_~8 - 0 e ld~ rldP WC - w (D.1) rl1 where r1 denotes the dimensionless inner radius of the As the discussion insulation. of Chapter 4 indicated, a problem is incurred in the discretization of this boundary condition. mation For if the standard central difference approxi- 4.3 is substituted for the derivative at the conductor, a fictitious temperature within the conductor is introduced. Since no governing equation is applied at the conductor, there is no way to eliminate such a fictitious temperature. In approximating the boundary condition D.1, 121 122 it is therefore necessary to have a difference expression which involves only real temperatures. There are a number of methods for approximating this boundary condition. As a reasonable compromise between accuracy and simplicity, the following method is chosen, where reference is made to Figure D.l: location r in between r and r2 some central is sought, at which location a good approximation to the derivative can be achieved. The boundary condition D.1 is then satisfied at the location r, [l| WCl (D rather than at the conductor: n r~d% 8 ~ WC1 _ 0 r, The difference form of the derivative is constructed according to a| ar where ~ l,kl r,,~ r 2 01 -r (D.3) 1 01 denotes the conductor temperature. to ascertain the location r, attention corresponding one-dimensional problem. In order is turned to the The analytical temperature distribution for the half-cable with prescribed conductor loss W is readily found to be I_ 123 1 )k1 FIGURE D.1 Nomenclature for the Discretized Conductor Boundary Condition 124 = - 0(r) is available: r n kn r - ) D.4, a criterion With the distribution location W (k Cl r* is chosen (D.4) (. 4 . for determining the such that, as the true temperature distribution approaches the one-dimensional solution (which it does in the vicinity of the conductor), the difference form of Equation D.2 becomes exact. This is accomplished by replacing the discrete temperatures of Equation D.3 with their analytical expressions and by then The difference substituting the result into Equation D.2. form of Equation D.2 is (kl kl \ (D.5) r*(Al)kl = -W 2-r1 Replacing the discrete temperatures with analytical ones gives ... ()r*)| ( kl A W (D.6) r2-rl Upon expanding through Equation D.4, this becomes CWE 1 kl kl L\ ln - rW-r1 KrA)ll1-l 1 r= - W. (D.7) r2_ r 1 - 125 does not depend on kl, the Since the term in brackets summation can be carried out. D.7 then yields, Equation upon rearrangement, - r2 i r= in r r1 (D.8) - in r1 When this result is substituted back into Equation D.5, the difference form of the conductor boundary condition becomes 6 lkl- 01 wmm~N m kl The procedure in - 2 1 1 Akl] 0 alk2 ln - -- for Cable 2: analogous is of course E[(1'k k2 r1 r2-n WCl ~~~ (Aa)k2 J02 =- WC2 W (D.10) J -ln P The results D.9 and D.10 need not be weakened by the assumption that the temperature distribution becomes one-dimensional near the conductor. The distribution is always one-dimensional right at the conductor, since it has a uniform temperature. So the only requirement is to choose the radial mesh size so as to place r* out of the range of strong two-dimensional effects. This choice is a matter of judgment, and it depends on the given problem. _ _· I_ I APPENDIX E USER INSTRUCTIONS Geometry and Mesh Size In setting up a problem for computer solution, it is necessary to provide information about the region size within the cables and about the number and distribution of mesh points. This section discusses specification of region size, subdivision of regions, special considerations for D3 , and weighting of the mesh so as to have a good expression for the gradient at each conductor. Reference is now made back to Figure 2.3, where a set of regional divisions is depicted, and to Figure 2.2, which shows the origin of cylindrical coordinates for both cables. The domain D1 always has four regional divisions. The domain D2 employs only six regional divisions, since Regions IV and VI are never used simultaneously. The angles included by the various regions are determined from the azimuthal coordinates of their bounding radial lines. For example, the angle ~1 specifies the location of the boundary between Regions I and II in D1 , and it thus determines the size of Region I. The angle ~2 likewise specifies the location of the boundary between Regions II and III of D. The angle included by Region II is then (c2 -~1 ). The sizes of all the regional divisions are therefore specified by three angles in D and by five angles a in D2. However, since D1 and D 2 share a single inter-cable conduction path, 126 127 only six of the above eight angles are independent. The orientation of the two cables is determined by al and by ~1 and ~2) (1/2)(a1 +a2). . 2 (or Specifically, the orientation angle is The convention of establishing regional divisions within the cables actually has two purposes. It first of all provides a way to clearly identify a given portion of the cable surface with a given boundary condition. The second purpose of the divisional convention is to provide a mechanism for varying the azimuthal distribution of mesh points, so that they may be concentrated where the largest gradients are expected. Mesh points inherently exist at all regional boundaries. They are placed inside a given region by specifying the number of subdivisions within that region, both in the radial and in the azimuthal direction. Here the term "subdivision" denotes the smallest element of the region, rather than the act of subdividing. Thus if a region has three azimuthal subdivisions, it is uniformly divided into three sectors by two radial boundaries, and two azimuthal mesh locations within the region are thereby introduced. The number of radial subdivisions does not vary from region to region; it is uniform within a particular domain. Thus a choice of four radial subdivisions in D2 places four uniformly spaced radially points at every azimuthal location in D2 . Placement of mesh points inside D3 proceeds in similar fashion, by specifying the number of normal and tangential subdivisions within the domain. In -~~~~__11 128 the computer program the various numbers of subdivisions throughout the solution domain are denoted by the variables These are described in Table 7, together N(J) and M(J). with a column containing the minimum allowable value of each variable. It is noted that the variables M(2) and M(11) cannot be chosen to be less than two. This is necessary in order to preserve the basic trapezoidal structure of D 3. Attention is now returned to the discretized model of D 3. The width of D 3 is taken to be equal to the skid wire thickness, a number supplied directly as input data. The height of D 3 is determined from the outer cable radius and the angle (a2- 1), as was shown in Figure as the discussion of Chapter 4 indicated, 4.2. However, the height so designated is only an apparent height and not the effective height. For when the regular form of governing equation is applied at the four corner points of D3, four effective corner locations are produced which lie outside the corner mesh points. These effective corner locations then define two effective surfaces, as shown in Figure El. The effective upper and lower surfaces of D 3 extend halfway to the neighboring mesh points above and below the domain, as suggested by the figure. A conduction resistance based on this extended length is implicitly added in series with the convection resistance for boundary points in the numerical model. Of primary concern to the user is that an appro- priate allowance must be made for this extension in specifying the height of D3 Say, for example, that each 129 rr-rrr-- I\ /r- UPPER SURFACE FF. LOWER SURFACE H = APPARENT HEFF = EFFECTIVE HEIGHT HEIGHT FIGURE E. 1 Effective Surfaces of D3 130 TABLE 7 SPECIFICATION OF SUBDIVISIONS THROUGHOUT D D2 AND D 3 IN TERMS OF THE COMPUTER VARIABLES N(J) AND M(J) Type and Location of Subdivision Number of Subdivisions Minimum Allowable Value Radial N(1) 1 N(2) 1 M (1) 1 Region II M(2) 2 Region M(3) 1 Region IV M (4) 1 Region I M(5) 1 Region II M(2) 2 Region III M(6) 1 Region IV M(7) 1 Region V M(8) 1 Region VI Region VII M(9) 1 M(10) 1 Normal - D3 M(11) 2 Tangential - D3 M(2) 2 D1 Azimuthal D1: Region I III D2: 131 azimuthal subdivision in Figure E.1 happens to be 10 size. in The effective height of D 3 is then based on an included angle of 50° , whereas the apparent included angle is only 40° . So in order to achieve this true included angle of 50 ° , the apparent angle (a2 -al) = 40° would have been specified, and the mesh points above and below D3 would have been chosen so as to place neighboring points at an azimuthal distance of 10° . An additional consideration is that the four surface mesh points in D 1 and in D2 which are adjacent to the corner mesh points of D3 should be reasonably symmetrical about the y-axis of D3. This is necessary so that the effective surfaces of D 3 remain parallel or nearly parallel to the normal axis. Some degree of foresight is therefore required in laying out the regional divisions and in choosing appropriate numbers of subdivisions. The final topic of this section concerns the conductor boundary condition. It is recalled that the discrete form of this boundary condition involves a summation of temperatures around the innermost discrete ring of mesh points. In the summation each temperature is weighted according to the azimuthal sector associated with the given mesh point. Attention is now called to the physical circumstance that Regions II of D1 and D 2 are regions of elevated temperature, owing to the presence of the inter-cable conduction path. The temperature trails off rapidly on either side of these regions, tending toward the - : 1.32 one-dimensional distribution. Since the gradient at the conductor is constructed numerically by means of summing discrete temperature differences around the cable, it is essential that a good sampling of temperatures near Regions II of D1 and D2 be taken. This ensures that the elevated temperatures in those locations will not be unduly weighted. Based on comparisons with one-dimensional solutions, the following convention for weighting mesh points has been found to produce a sufficiently accurate numerical expression for the gradient at the conductor: number of azimuthal subdivisions in Regions the I and III is chosen so as to place a minimum of two radial mesh locations adjacent to Regions II, each at an azimuthal spacing equal (or nearly equal) to the azimuthal spacing of points within Regions II. In Figure E.1, for example, this means that there should be a minimum of two 10°-sectors on both sides of both Regions II. This convention should also be followed for all regions whose surfaces are insulated, for the same argument then applies. Input Variables The input variables used by the computer program are listed in Table 8, together with a brief description of each variable. The five angles ALPHA(J) of D2 are specified sequentially, skipping over any region not present. So if Region IV is used, ALPHA(3) denotes the III- IV boundary 133 TABLE 8 INPUT VARIABLES FOR THE COMPUTER PROGRAM Variable Name Description and Units ( ) ALPHA (J) = the five angles a which specify the boundaries of the six regions of D 2. (degrees) FILMP (J) = the variable film coefficients for the surface (Btu/hr-ft 2-F) mesh points of D2. FILMR(J) = the variable film coefficients for the surface (Btu/hr-ft2-F) mesh points of D 1. FILMX3 (J) = the variable film coefficients for the mesh points on the upper surface (+y) of D 3. (Btu/hr-ft2-OF) FILMX4(J) = the variable film coefficients for the mesh points on the lower surface (-y) of D 3. (Btu/hr-ft2-°F) IPARAM = 1 or 2: 1 denotes that Region VI of D 2 is present; 2 denotes that Region IV of D 2 is present. (unitless) IPROB = 1, 2, or 3, corresponding to Solution 1 (for maximum cable temperature), Solution 2 (for maximum current), or Solution 3 (for maximum oil temperature). (unitless) M(J) = various numbers of subdivisions, as per Table 7. (unitless) N(J) = various numbers of subdivisions, as per Table 7. (unitless) PHI(3) = the angle in D 1 which specifies the boundary between Regions III and IV. (degrees) RAD1 = the inner radius of the insulation of Cable 1. (inches) 111111111111--1 ------ 134 TABLE 8 (Continued) Description and Units Variable Name ( ) RAD2 = the outer radius of the insulation of Cable 1. (inches) REST1 = the DC resistance of the conductor of Cable 1. (Q/ft) REST2 = the DC resistance of the conductor of Cable 2. (pa/ft) RHO1 = the inner radius of the cable insulation of Cable 2. (inches) RHO2 = the outer radius of the cable insulation of Cable 2. (inches) SKID = the skid wire thickness. (inches) TMAX = the maximum allowable temperature in the cable system. (F) TOIL = the oil temperature outside the convective boundary layer. (°F) WD1 = the total dielectric loss per unit length in Cable 1. (watts/ft) WD2 = the total dielectric loss per unit length in Cable 2. (watts/ft) XHFILM = the thermal film coefficient for the one-dimensional solution. (Btu/hr-ft2-oF) XI = the current in Cable 1. XI2 = the current in Cable 2. XI20I1 = the ratio of the current current in Cable 1. (k-amps) (k-amps) in Cable 2 to the (unitless) XK = the thermal conductivity of the insulation. (Btu/hr-ft-°F) - 135 TABLE 8 (Continued) Variable Name YC1 Description and Units ( ) = the AC/DC ratio at the conductor of Cable 1. (unitless) YC2 = the AC/DC ratio at the conductor of Cable 2. (unitless) YS1 = the AC/DC ratio at the sheath of Cable 1. (unitless) YS2 = the AC/DC ratio (unitless) ICUTAP = 0 denotes that no tape is present; any other integer at the sheath of Cable 2. indicates that tape is wrapped around either Cable (unitless) THICK1 = thickness of tape wrapped around Cable 1 (in) THICK2 = thickness of tape wrapped around Cable 2 (in) XKCU1 = conductivity of tape wrapped around Cable 1 (BTU/hr-ft°F) XKCU2 = conductivity of tape wrapped around Cable 2 (BTU/hr-ft°F) ______l_·___rl ________ 136 line. All angles ALPHA(J) are measured from the vertical, as was shown in Figure 2.2. For all the variable film coefficients FILMP(J), FILMR(J), FILMX3(J), and FILMX4(J), a separate value is specified for each convective boundary point. The total number of values specified in each case is thus determined by the total number of mesh points on the respective surfaces. The sequence for specifying the various coef- ficients is as follows: FILMR(J) starts at = 0 and proceeds clockwise around D1; FILMP(J) starts at = 0 and proceeds clockwise around D2; FILMX3(J) starts at (-A,D) and ends at (+A,D); FILMX4(J) starts at (-A,-D) and ends at (+A,-D) The inter-cable conduction path is merely skipped over in specifying FILMR(J) and FILMP(J). The variable IPARAM specifies whether Region IV or It is convenient to use Region VI of D2 is present. IPARAM = 1 for the cradled configuration and IPARAM = 2 for the equilateral configuration, even though those boundary conditions have not been implicitly programmed. For the open and equilateral-pipe configurations the choice is arbitrary, because the strict correspondence between regions and boundary conditions then no longer applies. Of the 11 variables M(J), only ten are specified as input. The omitted region of D2 is skipped over, and the program subsequently assigns a value of zero subdivisions for the region not present. - 137 Of the three regional necessary to specify PHI(3). angles in D 1 , it is only PHI(l) and PHI(2) are deter- mined automatically from ALPHA(l), ALPHA(2), and the outer radii of the two cables. It is noted that the variables REST1 and WD1 describe only half a cable. So if Cable 1 and Cable 2 had identical properties and losses, REST1 and WD1 would be exactly half of REST2 and WD2, respectively, The variables XI1, YC1, and YSl are not affected by this distinction. Should it be desired to compute the total dielectric loss per unit length W d rather than to specify it directly, the following integrated-out form is available [12]: Wd = 2 -12 (SIC)(df) watts conductor-ft' log1 0 (D) t(7-354)(101) 3 (E 1) where VZk = line-to-line voltage (volts) w = 2f = frequency (Hz) (SIC) = specific inductive capacitance, or relative dielectric constant (df) = dissipation factor d = inner radius of insulation D = outer radius of insulation. _ _I___~ 138 The conductor and sheath losses are computed according to W I2RY c (E.2) c and Ws =I2R(YY ), c $ (E.3) respectively, where R is the DC resistance per unit length of the conductor, and Y Y is the AC/DC c is the AC/DC ratio at the conductor, ratio at the sheath. Output Variables A listing and brief description of the output variables from the computer program are presented in Table 9. In the computer printout six values of ALPHA(J) are written rather than five. However, two of the six are always equal, reflecting that one of the regions in D2 (either Region IV or Region VI) has an ncluded angle of zero degrees. Because of the matrix scaling method employed in the program, that the determinant it is expected o the coefficient matrix will never attain an unwieldy order of magnitude. The variable DETEP4 is nevertheless so that its magnitude may be rinted out onitored for each problem. The user need not be concerned with this variable so long as it lies in the general range 10 -50 50 +r50 to 10 5 . However, f if it takes on values significantly outside this range, another matrix scaling procedure may be required in order to avoid an underflow or overflow. If the value of DETERM is ever 139 TABLE 9 OUTPUT VARIABLES FROM THE COMPUTER PROGRAM Variable Name ALPHA(J) Description and Units ( ) = the regional angles of D2 , as per Table 8. (degrees) DETERM = the determinant of the coefficient matrix. (unitless) IERR = 0, 1, or 2 This is a condition code from the matrix inversion subroutine. IERR = 0 denotes no difficulties encountered in the inversion. IERR = 1 denotes that the matrix is not dimensioned correctly or that the subroutine is not called correctly. singular matrix. IERR = 2 denotes a (unitless) M(J) = various numbers of subdivisions, as per Table 7. (unitless) N(J) = various numbers of subdivisions, as per Table PHI(J) 7. (unitless) = the regional angles of D1 , as per Table 8. (degrees) TAMAX1 = the maximum (conductor) temperature from the one-dimensional solution, based on the properties of Cable 1 - Solutions 1 and 3 only. (F) TAMAX2 = the maximum (conductor) temperature from the one-dimensional solution, based on the properties of Cable 2 - Solutions 1 and 3 only. TANALl(Jl) (F) = the analytical temperature distribution from the one-dimensional solution, based on the properties of Cable 1. (F) -- 140 TABLE 9 (Continued) Variable Name Description and Units ( ) - TANAL2(J2) = the analytical temperature distribution from the one-dimensional solution, based on the properties of Cable 2. (F) TCON1I = the conductor temperature of Cable 1 Solution 2 only. (F) TCON2I = the conductor temperature of Cable 2 Solution 2 only. (F) TCOND1 = the conductor temperature of Cable 1 - Solutions 1 and 3 only. () TCOND2 = the conductor temperature of Cable 2 Solutions 1 and 3 only. (F) THETA(J) = the temperature distribution for the entire solution domain - Solutions 1 and 3 only. (°F) THETAI(J) = the temperature distribution for the entire solution domain - Solution 2 only. (F) TOIL = the maximum allowable oil temperature Solution 3 only. (°F) XANAV1 = the maximum allowable current from the one-dimensional solution, based on the properties of Cable 1 - Solution 2 only. (k-amps) XANAV2 = the maximum allowable current from the one-dimensional solution, based on the properties of Cable 2 - Solution 2 only. (k-amps) 141 TABLE 9 (Continued) Variable Name Description and Units ( ) XIlMAX = the maximum allowable current in Cable 1 Solution 2 only. (k-amps) XI2MAX = the maximum allowable current in Cable 2 Solution 2 only. (k-amps) . ~~ ~ ~ "~---I-- 142 identically zero, the matrix is then singular. Provided no underflow has occurred, the probable cause is either that the matrix has been dimensioned incorrectly, or that the calling statement for RMINV is not corrects The printout of the M(J) includes all 11 values, with a null value inserted for the region not present. All three regional angles PHI(J) of D are printed out. The analytical temperatures TANALl(Jl) and TANAL2(J2) are printed out for each radial mesh point in D and D2 , respectively, They are written sequentially, moving radially outward; the first temperature in each sequence is the conductor temperature. The complete temperature distributions THETA(J) and THETAI(J) are printed out in the following sequence: starting with the mesh point nearest to the origin of coordinates, all temperatures in D1 are written, the azimuthal index moving through its entire range for each increment of the radial index; the identical procedure is then followed for all temperatures in D2 ; finally, all temperatures in D3 are printed, beginning in the lower lefthand corner of the domain (-x,-y) and moving through the entire range of the normal index for each increment of the tangential index. Array Dimensions A number of subscripted variables, or arrays, are used in the computer program. _ __ These arrays and the 143 variables which determine the size are listed in Table 10, together with their dimensions in the present program. For brevity the following computer variables have been used in the table: M14 = M(1) + M(2) +M(3) +M(4) ; (E.4) M5210 = M(5) +M(2) +M(6) +M(7) +M(8) +M(9) +M(10) ; NM3 = N(l)x[M14l4+ 1] +N(2)xM5210+ (E.5) [M(2) + l]x[M(11) - 1 o (E.6) Arrays or portions of arrays which have no variable dimension listed in the table have been permanently dimensioned at their present size. Data Card Assembly Instructions for assembling data cards for the computer program are listed in Table llo While most of this table is self-explanatory, a few additional remarks are offered here. Attention is called to the integer variables N(J) and M(J), which employ the I-format for their input. It is necessary that all these entries be right-justified to their respective columns. Values of the variable film coefficients FILMR(J), FILMP(J), FILMX3(J), and FILMX4(J) begin on card 9, as the table indicates. The total number of cards required for these variables depends on the mesh size chosen for the I 144 TABLE 10 ARRAY DIMENSIONS Name of Array Variable Dimension (s) Present Dimension(s) ALFNT (J) J = 7 ALFSQ (J) J = 7 ALFTN (J) J = 7 ALPHA (J) J= 6 ClFRAC (J) J = M14 +2 J = 20 C2FRAC(J) J = M5210 J = 36 COEFF (J,J) J = NM3 J = 168 FACTOR(J) J = NM3 J = 168 FILMP (J) FILMR (J) J = M5210-M(2) J = M14-M(2) - 1 J = 33 J = 16 FILMX3(J) = M(11) - 1 J =1 FILMX4(J) = M(il) - 1 J =1 IWORK(J,K) J = NM3 J = 168 K= 2 M(J) J = 11 N(J) J = 2 P(J) J = N(2) +1 J = 5 P1(J) J = N(2) +1 J = 5 P2(J) J = N(2) +1 J = 5 P1HX (J) J = M(2) +1 J = 3 P2HX (J) J = M(2) +1 J = 3 P3HX(J) J = M(2) +1 J = 3 P4HX (J) J = M(2) +1 J = 3 PALF (J,K) J = N(2) +1 J = 5 K = 7 PALFNT(J,K) J = N(2) + 1 J = 5 K = 7 PHI (J) J = 3 PHINT (J) J = 3 PHISQ (J) J = 4 PHITN (J) J = 3 145 TABLE 10 (Continued) Name of Array Variable Dimension(s) PMID (J) J = N(2) R(J) J = N(1) + 1 J= 4 J= 5 R1(J) J = N(1) + 1 J= R2 (J) J = N(1) + 1 RlFRAC (J) J = N(1) + 1 J= 5 J= 5 R2FRAC(J) J = N(2) + 1 R1HX (J) J = M(2) + 1 R2HX (J) J = M(2) + 1 R3HX (J) J = M(2) + 1 R4HX (J) J = M(2) + 1 J= 3 J= 3 J= 3 RATIO1 (J) J = N(1) + 1 J = 5 RATIO2 (J) J = N(2) + 1 J= 5 RMID (J) J = N(1) J = 4 RPHI (J,K) J = N(1) + 1 J= 5 RPHINT(J,K) J = N(1) + 1 Present Dimension(s) 5 J= 5 J= 3 K= 4 J= 5 K= 3 (J) J = N(1) + 1 J= 5 TANAL2 (J) J = N(2) + 1 J= 5 THET1 (J) J = NM3 J = 168 THET2 (J) J = NM3 J = 168 THET3 (J) J = NM3 J = 168 THET4 (J) J = NM3 J = 168 THETA (J) J = NM3 J = 168 THETAD (J) J = NM3 J = 168 THETAI (J) J = NM3 J = 168 VECTR1 (J) J = NM3 J = 168 VECTR2(J) VECTR3(J) VECTR4(J) J = NM3 J = 168 J = NM3 J = 168 J = NM3 J = 168 TANALl XHALF (J) J= 7 XHPHI (J) J= 4 ------ 146 TABLE 10 (Continued) Name of Array Variable Dimension(s) XHX (J) J = M(2) + 1 J = 3 XHXHY (J) J = M(2) + 1 J XHXSQ (J) J = M(2) + 1 XIVAR (J) _ .- - - - J = N 3 Present Dirrmension (s) = 3 tj = 1 , XM (J) = 11 XN (J) = 2 147 TABLE 11 DATA CARD ASSEMBLY Column Card (s) 1 2 (s) 1 IPROB 2 IPARAM 1- 10 SKID 11- 20 RAD1 21- 30 RAD2 31 - 40 RHO1 41- 50 RH02 51- 60 XK 61 - 70 REST1 80 REST2 71- Format Variable I I FF F F F F F F F 3 1- 10 YC1 11- 20 YC2 21- 30 YS1 31- 40 41- 50 YS2 WD1 51 -60 61 - 70 4, 4, Solution Solution 1 2 Solution 3 F F F F F F WD2 XHFILM 1 - 10 XI1 F 11- 20 XI2 F 21- 30 TOIL F 1- 10 F F F 21- 30 TMAX XI20I1 TOIL 1- 10 XI1 11- 20 XI2 F TMAX F 11 - 20 4, F 21 - 30 _ F _ _ ^1·----^---1------· 11(1118 148 TABLE 11 (Continued) Card(s) 5 Column(s) 1 - 10 ALPHA (1) F 11 - 20 ALPHA (2) F 21- 30 ALPHA (3) F 31 - 40 ALPHA (4) F 41 - 50 ALPHA (5) F PHI (3) F 1 - 7 1 -5 N (1) I 6 - 10 N (2) I 1-5 M (1) I 6 - 10 M (2) I 15 M (3) I 20 isi(4) I 21 - 25 M (5) I 26 - 30 M (6) I 31 - 35 M (7)or(8) I 36 - 40 M (8)or (9) I 41 - 45 M (10) I 46- M (11) I 11 6lb- 10 50 9- (10) FILR (J) F FLIMP (J) F (14) FLIMX3(J) F (15) FLIMX4(J) F ICUT AP I 11 - 20 XKCU1 F 21 - XKCU2 F 31 - 40 THICK1 F 41 - 50 THICK2 F (11-13) (16) I I __ _ _ Format 6 8 __C Variable __ 1- 10 30 149 particular problem. The individual coefficients are entered sequentially across each data card, with each value occupying ten columns. When all the values of a particular variable have been specified, the next variable begins on a new data card. The numbers in parentheses in the Card(s)- column are typical for common mesh sizes. If it is desired to run more than one problem time, data decks may be assembled in series. at a Each separate deck should be arranged according to Table 11. A final data card having a zero in columns one and two must always be placed at the end of the overall data deck. This double-zero card tells the program that there is no more data to be transmitted. Example Problem This section illustrates the solution of a particular cable problem using the computer program. For the example problem it is desired to know the maximum allowable oil temperature for an equilateral-pipe configuration of System 1 cables, given their current and the maximum allowable system temperature. In particular let the current in each cable be 942 amperes, and say that the maximum allowable system temperature is 185°F. The thermal conductivity of the insulation is taken to be 0.1153 Btu/hr-ft-°F, and the film coefficient on convective surfaces is taken as 5.0 Btu/hr-ft2 - ° F. Also the con- servative assumption of a thermally nonconducting conduit is made. A complete set of input data for this problem is 1_1 150 listed in Table 12, and the resulting discrete model is depicted in Figure E.2. A number of observations are made about the discrete model used in this problem. The effective included angle in either cable associated with the inter-cable conduction path is seen to be 30° , a slightly conservative angle. An insulated arc of this size is centered about the point of cable-conduit contact which, from elementary geometrical calculations, is found to occur at a = 220 °. Also it is seen that there is no particular association between regions and boundary conditions: Region III of D2 is primarily concerned with the cable-cable effect, while Region V is involved with the cable-conduit interaction. This breakdown of convention is necessary in modelling the equilateral-pipe configuration, because in that configuration there are too many separate effects operating around D 2 for the available number of regions. (It is noted, however, that the thermal model is equally effective.) The angle PHI(3) = 170° is likewise arbitrary in this problem, since there is no cable-conduit contact on the surface of D1. The radial mesh size of four subdivisions has been found from experience to be sufficiently fine to produce an accurate solution; use of a finer radial mesh does not significantly alter the temperature distribution. It is finally noted that the azimuthal distribution of mesh points adjacent to insulated regions follows the convention outlined in the first section of this appendix; such a 151 ONAL DIVISION _ATED FIGURE E2 A Discrete Model of the Equilateral-Pipe Configuration 152 TABLE 12 INPUT DATA FOR EXAMPLE PROBLEM Card 1 2 3 4 5 6 Column(s) Data 1 3 2 2 1 - 10 0.10 11- 20 0.9125 21- 30 2.0675 31- 40 0.9125 41- 50 2.0675 51- 60 0.1153 61 - 70 2.68 71 - 80 5.36 1 - 10 119 11 - 20 1.19 21 - 30 1.24 31 - 40 1.24 41 - 50 1.59 51 - 60 3.18 61 - 70 5.0 1 - 10 0.942 11 - 20 0.942 21 -30 185.0 1 - 10 20.0 11 - 20 40.0 21 - 30 120.0 31 - 40 190.0 41- 50 250.0 1 - 10 170.0 153 TABLE 12 (Continued) Card Column(s) Data 5 4 10 4 7 8 5 2 10 2 14,15 9 10 13 20 1 25 2 30 8 35 1 40 6 45 3 50 2 1-10 0.0 11- 20 0.0 21 - 30 5.0 31 - 40 5.0 41 - 50 5.0 51 - 60 5.0 61- 70 5.0 71 - 80 5.0 1- 10 5.0 11 -20 5.0 21- 30 5 0 31- 40 5.0 41 - 50 5.0 51- 60 5.0 61- 70 5 0 71 - 80 5.0 154 TABLE 12 (Continued) Card Data Column(s) 5.0 1- 10 11 11 -20 5.0 21- 30 0.0 40 31- 12 41- 50 0.0 51 -60 0.0 61-70 0.0 71- 80 0.0 1- 10 5.0 11 - 20 5.0 21- 30 5.0 31- 40 5.0 41- 13 - 50 US 51- 60 0.0 61-70 0.0 71- 80 5.0 1- 10 5.0 11-20 5.0 21- 30 5.0 14 1- 10 0.0 15 1- 10 0O0 16 5 17 1,2 0 00 155 distribution should produce an accurate numerical expression for the gradient at each conductor. The computer solution for this problem is shown in Figure E.3, where the desired oil temperature 140.3°F is printed. The present version of the program requires approximately 220 K of core memory for execution. It requires 310 K of core memory for compilation on the FORTRAN IV Gl-compiler; it is too large to permit optimization on the FORTRAN IV H-compiler. Capabilities and Limitations of the Computer Program The present computer program has two notable capabilities which have not yet been specifically mentioned. The first is that Cables 1 and 2 need not be the same size. In the case of unequal cable radii, the included angle associated with the inter-cable conduction path is still designated by ( 2 -X 1 ); the angle ( 2- ) is then auto- matically adjusted so as to equalize the lengths of the D 1 - D3 and D2 - D3 interfaces. The second capability not yet mentioned is that there is no restriction to alternating current; either one or both of the two cables may carry direct current. This is handled by merely setting to zero the appropriate AC/DC ratios and dielectric losses. In addition to these two features, it is noted that, while only certain orientations of the two cables are physically realizable, the computer program permits arbitrary configurations. Finally, attention is called to two automatic tests which will facilitate the location of certain input 156 FIGURE E.3 Computer Solution Printout for Example Problem An NW Wo0 o 7 0 UN - " Om .-O (01 U f Q 0?7-MO m 10 IA W N 7 - I *0 0 4w 'I m N 7-LOM lo g*e 0 * 1 In4'N 7t- 0* * U'IAO 4 2 '0 U Q IW IQ W 7- UW -4 - -_4-4 ol Li - -* 0 '_ Q WU'% m %d 4 - 0 0 U)4t U) W -'0 4r u1n I - U U) W f. ff 0 - Sf U) 4-4-_44- WIRI 4 4) I N NU_ § .2 Mf W Q 7r- - QV - 0 U-, rW 4 M a N1N 7@ a a, W _ M M a, - 11I . W la ' 0 N - a, 0 N N 4 r- r-rQ N 0of L N 0 IQ MN MoIAC) co IA IAU) N OM Uwi N N WM -_ U o o - r- N N CU) 0'?7- 0 _ _ W i_ _ _0 -0 _)I _-7-7 o0 V N '4N f0 ICU f -ffO O) O' 0 N 0vW 7--47- LA I0 tD . Li M'N U 4'W N - U U * IA7F 40U IA NN INO 0 N tf) - 4f 10 tD U) 00 U' N 40% tD U) U)40 NoN RI N u' I 'C "4 4 4 1-1-' 1-, r0 0 00 . 0 0 N- M 0N _- -4_ N 0 * UN U)U ) O Q U)r W r? UIr M w W Q - W 'o W 0 V U RI r- N 4 0 f IA - N 0 D ) -' 1 - en 0W 0 0 00 * SO * U' N _-' 0 P-'- CDO 0n (' -4 0 M V _ _ 1U RI) o N 0 *O. '4 ) 7- 7- q' .) 0v 7- U' 0 Nx7--_) D '0 IA7- - - T N n 0 7 ' 70 '05le W O4 4 0 _ 0 o ' 0' 4''U X) 1- 7- 0 '' N a, * S 0 ff1 0 0 0 * * 00 00 0 W '0 N a, '0 Q V I IA. 0' 0 U)7- A in IA4 4 000. N Oi 4n U' N1 U ff N 40NI- A f' 'CN N- ~4 0* 5 * O 0 0 I_ N IA- - -4 p'0 00 0 O0 . 0 c- 0 0 N t- U)ff1 4 * I 0 a, I W M It - -40 LUi U N' Q - 0 ' 7- U A) N 0 0s o -4 N N' 0 f 7- IT N N 0' 7a 0f 0 1 * *00 * WI U) N N eo N W g 0 O 7- 7-4" 4 4- * N N i r * U' N N IA 7 f] * A ' ' * 7IA IA * S 4' '0 N U) M 00 000O 0 O 0 '3.4 1. 8 Li Li RI1 O O 0 0 0 ~~~ - or r.0W v WI VI P3 IA W '10 - t N 0 I cr ff1o U) N 7o 0 N .0ff f r1- Li U'' 0 -U 'U A7 7 U .0Lg7-'00. V O O tS 4) M M S _A t- U .) Li 0 W I- U- 04 It : " a. L3 _ 0 L W o -4 -4 U. u. .. LD ,) 0- _ -4 _ 71 Li) I M _X n "-4---"_ _ 0 U II 0 H:If -- M 7 W 0N U. M 7- le N N U..4 -.._4 * vtD 0 _ 0; - 0O 7- 7- '07-IA U) r N LA N IA IA 7-U 0 v) 0 .3 rN 7CD PU o N U'U 7-A U) U' N ff '1 - IA U4 U' Nb ff'* 'C V~~ o~~~~~~~~~~~ o"NU f )5'4 )7 NU' U) -' '7-IAU' 3 ?U 7 _0 '44 C) - l- U) L3 LD Li, U Li (a - '- ,U) N Li Lic Z -7 4 -X -IJ zr z I- I-- -4,- t.,,- - ^it N -4 X X 4 _ A. -,.. '.- 157 errors. Should the input specify that the maximum allowable system temperature is less than or equal to the oil temperature, an appropriate error statement is then printed, and the program passes to the next problem. temperature A similar procedure is followed if the maximum allowable system is too small for the given dielectric loss. Two limitations of the computer program are also called to the attention of the user. The first is that there are constraints on the admissible size of D3 . In particular, the thickness of D 3 must be non- zero, and its included angle in D1 , ( 2- )' must be less than two radians. Secondly, the included angles of the various regions must all be non-zero. have been used successfully by the author, but Included angles of 2 regions smaller than this are not recommended. Program Modifications briez suggestions for effecting modifications in the present computer program are offered in this section. are the following: The modifications to be considered provision for simultaneous use of Regions IV and VI or D 2, and simplification of I/O procedures. A modification which would permit simultaneous use of both Regions IV and VI of D2 could be effected without much difficulty. However, before making such a change, consideration should be given to the handling of boundary conditions. Li the present program all boundary conditions are specified by means of a variable film coefficient. This was done in order to retain the greatest amount of flexibility in treating boundaries. If it is desired to preserve this feature, then there is no substantial advantage in performing the above modification. For when boundary conditions are specified separately by means of the variable coefficient, regional divisions 158 are important only in varying the azimuthal distribution of mesh points; there need be no correlation between the various regions and specific If, on the boundary conditions (as the example problem illustrated). other hand, it is desired to treat some or all of the boundary conditions implicitly, then the modification under consideration may be necessary. Boundary conditions may be implicitly written into the program by inserting the appropriate values of the various film coefficients directly into the matrix-generation portion of the program. This would be convenient for boundary conditions which never change from problem to problem. For example, an insulated surface might be identified with a particular region (such as D2, Region IV). Then upon specifying the size and location of that region, The number of such the appropriate boundary would be inherently insulated. permanent kinds of boundaries depends on the particular problems the user elects to solve with the program. This method for handling boundaries, though, would make it necessary to have seven available regions in D2 for the equilateral-pipe configuration. The modification required for this involves the input format for ALPHA(J) and the matrix-generation statements for Region III through VII of D2 . Provisions for generating the variables associated with all seven regions of D2 presently exist; certain statements are merely bypassed at IPARAM-type decision branches. It would probably be convenient to introduce a third category, IPARAM = 3, for a new branching criterion. This criterion could then be used in the matrix generator to choose such branches from the (IMARAM = )-type and (IPARAM = 2)-type tests so as to move sequentially through all the regions of D2 . The (IPARAM =3)-test could likewise signal a special input format for ALPHA(J), indicating that six rather than five angles are to be read. The program so modified would 159 be most convenient, provided that boundary conditons were treated implicitly. Finally, some brief suggestions are given for simplifying the I/O procedures of the computer program. could be improved: Concerning input, two primary areas specification of the variable film coefficients and specification of the effective size of D3 . The former is bothersome because so many values must be entered, and because of the need to keep track of all the individual surface mesh points. Implicit treatment of boundary conditions would completely eliminate this inconvenience. specification is retained, a provision might at least the input. If explicit be written to simplify For example, it might be desirable to merely specify a single coefficient and direct that it apply for all the mesh points of a given surface (when appropriate). Or since most of the surface points are con- vective, it might be more convenient to read in just the non-standard coefficients. angle (- Specification of the height of D 3 by means of the apparent ) is akward; much foresight is required in order to end up with the desired effective height. with ( - It would be much more convenient to work ) as the effective included angle, referring the associated aximuthal adjustments in D1 and D2 to the computer. three suggested improvements are mentioned. Concerning output, First of all, in its present ver- sion the program prints out very little of the input data. Such procedures as solution identification, error location, and output analysis would be facilitated if more of the original data were written. Secondly, it would be a simple matter to include the conservative approximate solutions in the program. These could be written alongside the one-dimensional solutions, thereby making the upper and lower bounds for system performance immediately available. - Lastly, the overall temperature distribution is not very des- 160 criptive in its present format. Separating the temperatures in the print- 23 out according to the three domains D1 , D2 , and D would present no problem, and this would help considerably in identifying features of the distribution. Also effective use could be made of plotter routines for illustrating the temperature distribution graphically. I___·__ APPENDIX F LISTING OF THE SOURCE PROGRAM 162 163 U -0 O ° V O O tDO O ° U. O C O - O U ' Q 0 - I' O (- U----O -. - O,C) C.;ON N U' F U U O 0" ' (, j UDO O- N )O( J On U n Ln O mm m . . Ux -JZ -- 4 C CL N9. I- C 4.. LV) * U-' U. . -- i,_ 0.. ,m[ C F L o. - 11. 4') )¢ L ',. x ) 0.. I" 1-. O-4 .4 9. OC _*?' 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W x U Z I-I) X wb N Z - ,-, Z auJ Z C W D I O C t-- 4 U] *\j (V mk r Y) (Y)~ APPENDIX G ONE-DIMENSIONAL SOLUTIONS FOR TEMPERATURE AND CURRENT The Temperature Solution In Chapter 2 the equation which governs the temperature distribution of the cable insulation was given as 1 T + r r 2T r22 ? 2 k (2.1) Considering now only the radial dependence of the temperature, Equation 2.1 reduces to 1 dT + rr 2T dr2 k (G.1) and this may be consolidated to rdr r TdrJ 1 d k (G.2) Expressing the volumetric heating term q in terms of the total dielectric loss per unit length Wd, the governing equation G.2 becomes finally 220 221 d r- dT dr dr Wd = (G.3) 2rrk in ) where r1 and r2 denote the inner and outer radii of the The boundary condition at the insulation, respectively. conductor is dkr qc = -2Trlk dr 1 (G.4) = W c 1 where W is the conductor loss per unit length. As for the boundary condition at the surface, consider for the present that the surface is at some arbitrary uniform temperature: T(r 2 ) = T 2 (G.5) . A dimensionless formulation may be obtained by introducing the variables r T - Toi 6 (G.6) W/k . r2 where W is some arbitrary loss per unit length. governing equation is then The 222 d dE ( where Wd d) dE 27rW (G.7) n 1 = r/r 2 , and the boundary conditions are W dO c (G.8) I 2-W~ 1 and T e (1) 0 = -T . w/k - o (G.9) W/k' The general solution to Equation G.7 is Wd 6() 4rW in (ln ) 2 + Cln (G.10) + C2 Substituting this solution into the boundary conditions G.9 and G.10, the arbitrary constants 1 Cl= -2-W (Wd+W) C 1 and C 2 are found to be C2 = (G.11) 00 Putting this result back into the governing equation gives Wd 0(U~~ e1 = 47W ) i )2 n (ln (Wd+W ) d - ln 21rW + o . (G.12) 223 Equation G.12 is then the temperature distribution for the case in which the cable surface is at some arbitrary temperature T . 0 However, this arbitrary temperature may now be eliminated by applying an energy balance at the cable surface: Wc +Wd +Ws = 2r where W 2 h(T (G.13) -Toil) is the sheath loss per unit length. The nondimensional form of this is W +W +Wd = W/k 2r (G.14) 2 hO0 from which (Wc+Wd+Ws) / k (G.15) hr 2 W 6o= The one-dimensional temperature distribution for the cable is therefore Wd 4W=n 4W n 2 1 (ln ) (Wd+Wc) - 2W Wn + (Wc+Wd+Ws) k 2TW hr2 (G. 16) 224 In terms of dimensional temperatures, T(~) - Toil Wd 4rk ln (ln E) 2 - (Wc+Wd+Ws) (Wd+Wc) 27rkin + 1 2wr2 h (G.17) The Current Solution In order to obtain the one-dimensional current solution, the stationary and variable components of the temperature distribution are again separated: 0(E) = D(i) + C(E) , (G.18) where Wd OD(E) 4 W in 2 1 (in ) W n Wd Wd in l2W + k (G.19) and (w +w ) eC(E) = - 2wW in E + 2rW (G.20) The current I is introduced into OC(E) following the reasoning of Appendix C. By analogy with Equation C.39, 2 ec(E) = ec ( ) I0 (G.21) 225 Co(E) is the distribution obtained by using the where arbitrary current I0 in Equation G.20. 0() The total solution is then 2 e(E) + OC0 () = OD(E) (G.22) 2 I0 Again it is desired to have O(E) take on some maximum value emax . However, in the one-dimensional case, since there are no heat sinks within the cable, 0max must occur at the max location = 1 (at the conductor). Substituting this information into Equation G.22 gives 2 emax = D(g 1 ) + C( 1) . (G.23) 0 Upon rearrangement, Equation G.22 yields the current /I \ = 0 max e-D(1) (G.24) eco ( kIo It is then only necessary to insert the appropriate values for D(C1 ) and C (C1) 2 from Equations G.19 and G.20: emax Wd -W 2 (k hr2 In -_ 27TW hr2 in E1 1) 2 I 20W 2) + (G.25) so 2 kwWhr2 226 where Wco and Wso are the conductor and sheath losses, respectively, produced by the arbitrary current Io . This expression may be simplified by using the relations W where Y C = I2Ry and Y S W =2 IR(Y Y) (G.26) denote the AC/DC ratios at the conductor and at the sheath, respectively. Making these substitutions, Equation G.25 becomes ( md max 2TrW k ln hr2 2 )w 276 (--2)n(G.2 7) 2 2) in terms c hof kFinally, temperatures, 2~rW r dimensional 12 W F- (~h which may be further simplified to 2W - Wd h nr l _ (max /Ysk 2= R RFr2- n l) 2 )(G.28) i= 2 (G.29) Yi Finally, in terms of dimensional temperatures, 2,Fk(T -T )_ max oil i2= /Ysk Wd~ ~ i d -2 1 2 ~~~~~~~~~~~~(G. Rhr2 Y nE 29) APPENDIX H CONSERVATIVE APPROXIMATE SOLUTIONS FOR MAXIMUM TEMPERATURE AND CURRENT General Conservative approximations for maximum temperature and current in the two-dimensional conduction problem may be achieved from a suitable modification of the one-dimensional solutions presented in Appendix G The conservative assumption to be employed is that cable-cable and cable-conduit interactions effectively insulate appropriate portions of the cable surface, thereby reducing the perimeter available for heat transfer. Furthermore, it is assumed that the entire cable sector subtended by an insulated arc on the perimeter is also effectively insulated. Any losses which occur in the insulated sector are then referred to the remaining undisturbed portion of the cable. Consider, for example, that a 60°-arc of the cable perimeter is taken to be insulated. for heat transfer is then reduced and all losses in the ()-cable The perimeter available to (5) its original size, must be scaled up by () in order to have the same heat flow or temperature as in the original problem. The maximum current or temperature is then computed from the one-dimensional solution, using (6) of the original one-dimensional losses. 227 228 The Temperature Solution In Appendix G the one-dimensional temperature distribution was given as 2 Wd 4Trkin oil 1 (n (W +Wd+W ) 2in 2r2 r+h (Wd+W) - 2k ) (G.17) The temperature drop from the conductor to the oil is then T() Toil 1~~~~ - - Toil = T - in 1 4Tk W+Wd+W ) (W (Wd+2Wc) + 2 hr 2' 2~ihr (H.1) where T 0 denotes the conductor temperature. Equation H1 that the temperature drop (To -T oil ) varies Now let the cable perimeter linearly with the cable losses. available It is noted in for heat transfer take on the value P' = fP, where P is the total perimeter, and f is some fraction. The losses in the undisturbed portion of the cable are then scaled up according drop to q' = ()q. (T -T oil) varies o oil up by (/f), linearly Since the temperature with loss, it too is scaled and the conservative expression is given by = T0-Toil)* = (To-Toi (o i) (T0 -Toil where the temperature drop on the right side is that (H.2) 229 produced by the one-dimensional solution. Equation H.2 may then be used to conservatively estimate either the maximum allowable oil temperature or the maximum cable temperature in the two-dimensional conduction problem. The Current Solution From Appendix G the one-dimensional current solution is 2Tk(T 2 i2= -T 2 Trk(Tmax ) W _ in k 1) hr oil l~sk = Rth - \ nY n Y hr 2 c1 2 (G.29) 2 . n 1 Again consider that the effective perimeter takes on the value P' = fP, and that the losses are scaled up according to q' = ()q. Since current-produced losses vary as 12, the latter quantity must itself be linearly scaled, along with the dielectric loss, Inserting this into Equation G.29 gives the result 2 I* f 27Tk (T max -T oil ) * f /Y k R 1______1_111________----------.--_111 -d (hr k in \~~~~h2 _ t(H.3) JI s- c1 Y in ---------- _ 230 which may be rearranged to give -T ) 27kf(T max oil 2 I.~~Y ( W k d)h-r 2 _ 2 i n 41 (H.4) /Ysk hr2 -Y in i Equation H4 is then the conservative approximation for current. The Effective Perimeter The size of the inter-cable conduction path is a reasonable guide in selecting the amount by which to reduce the cable perimeter for cable-cable and cable-conduit interactions. This convention was followed in generating the conservative comparisons tabulated in Chapter 6. For those 16 problems the inter-cable conduction path was chosen so as to subtend an angle of 30° on either surface. cable The following cable perimeters were therefore used for the various configurations: open - 330° effective; cradled - 300° effective; equilateral - 270° effective; and equilateral-pipe - 240 ° effective. effective perimeter of 360 ° It is noted that for an (f = 1), the one-dimensional solutions are recovered in Equations H.2 and H.4.