ELEC9712 High Voltage Systems 1.2 Heat transfer from electrical equipment The basic equation governing heat transfer in an item of electrical equipment is the following incremental balance equation, with dθ and dt being incremental changes in temperature and time respectively: where: P = m = c p = h = A = T = p ( ) Pdt = m c dθ + Ah T T dt power dissipation (generation) level in the equipment equipment mass specific heat of the equipment material overall heat dissipation coefficient surface area available for heat dissipation from equipment surface temperature T o = ambient temperature Thus Pdt is the energy dissipation in time dt dθ is the temperature rise of the equipment in dt mc p is the thermal capacity of the equipment Ah( T T 0 ) is the heat dissipated to the ambient in dt. 0 ELEC9712: Heat Transfer p. 1/32
Under steady state heating conditions, we have constant temperature θ, and so ( ) Pdt = Ah T T dt This equation is the basis for determination of continuous thermal rating of equipment. Under short circuit conditions, the heating is adiabatic ht ( T o ) = 0 and thus we have Pdt = mc dθ p This equation is the basis for determining transient rating of electrical equipment. We will look at these two cases separately. 0 Steady state heating and thermal transfer Cooling of electrical equipment is usually achieved by a combination of the three standard modes of heat transfer mechanisms, viz. conduction, convection and radiation. In order to quantify the actual heat transfer rates by any of the above mechanisms or combination of mechanisms, a number of material and configuration (geometrical) factors must be taken into account. These include the following important parameters: ELEC9712: Heat Transfer p. 2/32
A surface area of equipment c p specific heat of the major material involved k thermal conductivity of the material ρ material density ε radiative emissivity of the surface β expansion coefficient of the material being heated v fluid flow velocity past hot surfaces µ fluid viscosity (dynamic) L (or X) a characteristic length of the surface dissipating heat For determining the overall heat transfer rate for an item of equipment under steady state conditions, we must determine an overall heat transfer coefficient h (W/m 2 / o C), such that the total heat dissipation H t from the equipment surface is: ( ) H = Ah T T watts t t 0 where H t (and thus h t ) are the total dissipation including all mechanisms. The determination of h t is very important for any electrical equipment and its calculation must take account of the environment and the equipment considered together. The calculation of h t is achieved by a simple addition of the three individual coefficients for conduction, convection and radiation, or whichever ones are applicable. ht = hcond + hconv + hr W/m 2 / o C ELEC9712: Heat Transfer p. 3/32
The determination of the heat transfer coefficient for conduction is relatively simple. For the radiation coefficient, it is also relatively simple. However, it is very complex for the thermal convection dissipation coefficient. We must develop methods for calculating them according to the following methods: (i) thermal conduction, using Newton s law of cooling (ii) thermal radiation, using Stefan s law emission (iii) thermal convection: a. natural convection empirical equations b. forced convection empirical equations 1.2.1 Thermal Conduction This is generally the most inefficient method of the three, as most materials that have to dissipate the heat in electrical equipment are electrical insulators and these are also generally good thermal insulators and poor thermal conductors. However, in some cases thermal conduction may be the only dissipation mechanism available (for example in power cables). Thermal conduction is only significant for solid materials: in liquids and gases, convection loss will be much more efficient and dominant than conduction loss. dh T 2 l da A dx A T 1 dh ELEC9712: Heat Transfer p. 4/32
We are mainly interested in steady state heat conduction, where we can write: dt dh = k da dx ka H = T2 T1 watts l A H = T2 T1 gl i.e. ( ) or: ( ) H A = k dt dx where: i.e. where: 1 g = = thermal resistivity k H = ( T T ) 2 1 G gl G = = thermal resistance A This is the Thermal Ohm s Law and the analogy with the electrical Ohm s law is obvious, with the following correspondence: H heat flow flux quantity (cf current I) ( T T ) 2 1 temperature difference potential quantity (cf voltage difference V) G thermal resistance medium characeristic (cf electrical resistance R) We can devise thermal equivalent circuits that can be handled by exact analogy with electrical circuits once the thermal ELEC9712: Heat Transfer p. 5/32
resistances are determined for the materials and the configurations involved in the equipment. Example: H T 1 T 2 T 3 T A 4 G 1 G 2 G H 3 L 1 L 2 L 3 We have: and: G G 1 G 2 G 3 H H T 1 T 2 T 3 T 4 gl A 1 1 2 2 1 = ; G2 gl gl 3 3 = ; G3 = A A [ ] T T = H G + G + G 1 4 1 2 3 Example: H 1 2 3 4 5 6 7 H G 2 H G 1 G 3 G 4 G 6 G 5 H G 7 ELEC9712: Heat Transfer p. 6/32
Calculation of the thermal resistance. G can be calculated using exactly the same methods as are used to determine the electrical resistance for the same geometry. The only change needed is to use the thermal g = 1 k instead of the electrical resistivity ρ. resistivity ( ) For example, for a coaxial geometry, the two equations for electrical and thermal resistance for radial electrical current and heat flow through the cable insulation are: r r = b = a R G ρ b = ln 2π a g b = ln 2π a (electrical) ohms thermal ohms For any thermal insulation geometry we can determine the thermal resistance by using the electrical-thermal analogy so that we can simply use the known electrical resistance expression for any geometry to calculate the thermal resistance. This is done simply by replacing ρ by g. All that is required is the value of thermal resistivity, g. ELEC9712: Heat Transfer p. 7/32
The most common area of use of thermal resistance is in the rating calculation of power cables, where the above coaxial geometry is used. It is also possible to use the standard line to plane electrical resistance formula for determining the effective thermal resistance between a buried cable to the ground surface above (representing the ambient infinite heat sink). Some typical thermal resistivities (g) of various materials are: Material PVC insulation Mica Porcelain & Glass Moist soil Asphalt Sand (dry) (moist) Air Concrete Rubber Water Oil-paper Wood Polyethylene Copper g ( o Cm/W) 7.0 2.0 1.0 1.0 1.4 3.0 0.9 41 1.0 5.0 1.7 6.0 4 8 4.0 0.002 Conduction heat transfer coefficient [ h cond ] The general equation heat transfer equation is: ELEC9712: Heat Transfer p. 8/32
cond ( ) H= h AT T 2 1 For thermal conduction along a constant cross-section tube of length L, Thus: A = [Th. resistance G = gl A] ( ) H T2 T1 gl h cond 1 = W/m 2 / o C [g is thermal resistivity] gl For most thermal conduction calculations, h cond is not much used as H cond can be calculated easily from the thermal equivalent circuit. Also, thermal conduction usually acts by itself, so that h conv and h r are not relevant. 1.2.2 Thermal Convection Heat transfer from a surface by a fluid (liquid or gas) by either natural or forced convection flow dissipation depends on the following factors: Surface temperature Length of surface exposed to the fluid flow Surface configuration (shape) Surface texture (rough or smooth) Properties of the fluid, particularly the following o Thermal conductivity k o Density ρ ELEC9712: Heat Transfer p. 9/32
o Specific heat c p o Viscosity µ o Thermal expansion coefficient β The first three of the above fluid properties define the thermal diffusivity ( k ρ c p ) of the fluid which is the main thermal characteristic of importance for convection heat removal from surfaces. As before, we need to find a heat transfer coefficient h conv for dissipation by convection. This is a very complex calculation (estimation!) for convection because the processes involved are not able to be quantified analytically. Empirical equations must be used to determine H conv and h conv. Empirical equations are available for most commonly encountered surfaces, but for more complex situations we must use a more basic empirical fluid flow approach in the estimation, using the various defined numbers to calculate heat transfer. Empirical equations for h cond from typical surfaces are given below. A Natural Convection The method of natural convection flow establishment is the expansion of the fluid as it heats up, thus leading to a reduction in density and hence rising buoyancy forces that cause the hot fluid to rise and be replaced by cooler fluid at the bottom. The following give empirical equations for the ELEC9712: Heat Transfer p. 10/32
heat transfer coefficients by such a process in simple geometries. Dissipation from flat surfaces and cylinders by natural convection (a) Vertical plates and cylinders in air 0.25 T h = 1.37 W/m 2 / o C L [L = plate length] (b) Horizontal plate in air (top surface) 0.25 T h = 1.32 W/m 2 / o C L (c) Horizontal plate in air (bottom surface) 0.25 T h = 0.59 W/m 2 / o C L (d) Horizontal cylinder (in air) 0.25 T h = 1.32 W/m 2 / o C D [D = diameter] (e) Vertical plate & cylinder in water at 21 o C 0.25 T h = 127 W/m 2 / o C L (f) Vertical plate & cylinder in transformer oil at 21 o C ELEC9712: Heat Transfer p. 11/32
h T = 59 L 0.25 W/m 2 / o C Note that in all of the above expressions the heat transfer coefficient calculated will be temperature dependent. This is a significant complication in determining the total heat transfer using the equation, H = ha T watts = constant ( ) 54 A T watts compared to thermal conduction, for example, where the heat transfer coefficient h was constant and independent of temperature. A typical magnitude of the natural convection heat transfer coefficient in air at normal operating temperatures is about h conv 5-6 W/m 2 / o C Examples of natural convection in electrical equipment include: Natural circulation of oil in transformers Natural air cooling of pipes on transformers Heat sinks on power semiconductor components Natural convection from exposed overhead lines and busbars ELEC9712: Heat Transfer p. 12/32
B Forced Convection Dissipation from a flat surface by forced convection flow hconv hconv = 5.7 + 4.5v [for v < 5 m/s - laminar flow] 0.8 = 7.8v [for 5 v > m/s - turbulent flow] or, we can use the following more general formulation: Nu = C m 13 Re Pr where Re is the Reynolds number, Nu is the Nusselt Number, Pr is the Prandtl Number, m is a general exponent dependent on the situation, C is a constant. and we then use the equation: Nu = hl k where k is thermal conductivity, L is a general length. Thus: hconv = 13 ( klc ) Re m Pr The calculation becomes even more complex for cylindrical geometries. Techniques to enhance heat dissipation rates from electrical equipment ELEC9712: Heat Transfer p. 13/32
From the above equations and the parameter dependencies, some simple techniques of enhancing heat dissipation rates from electrical equipment can be determined as follows: (a) Natural convection Optimisation of surface orientation Increasing the surface area for dissipation Use of a fluid with high thermal diffusivity Keeping ambient temperature low (b) Forced convection Use a high flow velocity Optimise surface orientation Use a large surface area Use a fluid with high thermal diffusivity Using low ambient temperature Thermal convection dissipation requires the removal of heat from a hot surface by using fluid flow past the surface to remove heat and dissipate it by the fluid flow. This requires of course that the fluid temperature is kept lower than that of the surface. Such fluid flow past a surface results in a boundary layer of the fluid adhering to the surface with the inner part of the boundary layer at the surface being attached to the surface and essentially static. The fluid velocity then increases in the layer as away from the surface. ELEC9712: Heat Transfer p. 14/32
The result of this layer attachment is that the thermal transfer then has to take place via the boundary layer, so that even thermal conduction in the fluid material is important in convection dissipation, with the initial transfer being by thermal conduction through the attached static inner fluid layer. Thus convection dissipation is determined by fluid parameters such as the viscous drag of the fluid at the surface, the specific heat of the fluid, the thermal conductivity of the fluid, the viscous drag of the fluid, the fluid flow velocity, fluid density, etc, as well as the thickness of the layer. All of these factors are incorporated in a series of characteristic numbers derived by dimensional analysis of fluid flow and these are used in fluid flow problems to determine general convective dissipation. The following gives a general discussion of the use of such numbers. Use of non-dimensional characteristic numbers to describe and calculate convective heat dissipation from surfaces In less well-defined geometries than those for the equations given previously, we have to use the non-dimensional fluid flow numbers to quantify the heat dissipation coefficient. ELEC9712: Heat Transfer p. 15/32
The fluid flow numbers of importance for convection dissipation are: The Reynolds Number [ Re ] The Grashof Number [ Gr ] The Prandtl Number [ Pr ] The Nusselt Number [ Nu ] Reynolds Number (Re) Re ρvx vx = = µ υ ρ = fluid density, v = fluid flow velocity, µ = fluid (dynamic) viscosity, υ ( = µ ρ) = kinematic viscosity, and X is some characteristic length. The Reynolds number represents the ratio of the fluid dynamic forces ( ρv 2 ) of the flow to the viscous drag forces µ v X retarding the fluid flow motion. ( ) The magnitude of Re indicates whether the flow is laminar or turbulent, giving a numerical criterion for laminar flow. Laminar or turbulent flow is an important factor in determining the convective dissipation of heat by a fluid. The Reynolds number relates primarily to forced flow situations. It is not easy to define a specific Re value for the transition from laminar to turbulent flow, however a value of Re = 10 5 ELEC9712: Heat Transfer p. 16/32
is often used for general flow conditions, but the geometry will affect this transition value. Thus Laminar flow Re < 10 5 Turbulent flow Re 10 5 A low Re means that viscous forces dominate the flow. A high Re means that inertial forces dominate the flow. Typical values of Re: For cylindrical pipes, the flow is turbulent when Re 2000. For a pipe of diameter 10 mm, and a flow velocity of 10 mm/s: Re = 6 for air Re = 100 for water Re = 1 for oil Grashof Number ( Gr ) Gr 2 3 gβρ X = T 2 µ g = the gravitational acceleration constant, β = coefficient of volume thermal expansion, T = the temperature difference between the surface and the fluid. Other symbols are as previously defined. ELEC9712: Heat Transfer p. 17/32
The Grashof number represents the ratio of the fluid buoyancy force of natural convection heat dissipation to the viscous drag of the fluid flow. It represents the effects of the hydrostatic natural buoyancy lift force of the heated fluid and the viscous force opposing the flow during natural convection. Gr is used instead of Re when the convection is naturally generated in the fluid. Laminar flow Gr < 10 5 Turbulent flow Gr 10 5 Typical values of Gr: For a surface with X = 20 mm, and for a temperature difference T = 10 o K Gr = 7500 for air Gr = 240,000 for water Gr = 20 for oil. Prandtl Number ( Pr ) Pr = µ k c p c p = fluid specific heat, k = fluid thermal conductivity, µ = viscosity. The Prandtl number represents the ratio of momentum diffusivity( µ ρ ) of the flow to the thermal diffusivity ELEC9712: Heat Transfer p. 18/32
( k ρ c p ). Note that it does not involve any geometry factors: it is a purely physical property of the fluid. It represents the efficiency relationship of the heat transfer process of the fluid to the fluid motion during convective loss processes. Typical values: Pr = 0.7 in air Pr = 6.5 in water Pr = 500 in oil Nusselt Number ( Nu ) Nu = hx k h = convective heat dissipation coefficient of the fluid flow (ultimately this is what we need to calculate to determine heat loss) and k = thermal conductivity of the fluid. The Nusselt number is the basic parameter which quantifies convective heat transfer by a fluid. It represents the ratio of ha T to conductive heat transfer convective heat transfer ( ) ( ka T X ) by the fluid. In order to find h, the Nusselt number must be evaluated from the Re, Gr and Pr numbers (as appropriate to the flow conditions). The Re, Gr and Pr numbers are themselves determined from the fluid properties ELEC9712: Heat Transfer p. 19/32
and the geometry using experimental data for various fluids and geometries. We use the following form of relationship: Nu = f (Gr, Pr ) Nu = f 1 ( Re, Pr ) for natural convection for forced convection The functional nature of f and f 1 are determined by the situation and fluid properties. For natural convection, Nu Gr 1/4 Nu Gr 1/3 for laminar flow for turbulent flow For forced convection, Nu Re 1/2 Nu Re 4/5 for laminar flow for turbulent flow For natural convection the following equation is often used: Nu = 0.53 ( Gr. Pr) 0.25 = 0.275 X 0.75 ( T ) 0.25 for air ELEC9712: Heat Transfer p. 20/32
= 1.2 X 0.75 ( T ) 0.25 for water = 0.4 X 0.75 ( T ) 0.25 for oil From such relations we can determine H: for example, for natural air convection h conv k Nu 0.275 k T = = 0.25 X X 0.25 ELEC9712: Heat Transfer p. 21/32
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1.2.3 Thermal radiation All bodies radiate electromagnetic radiation. At normal temperatures under consideration here, [T 100 o C or 373 o K] this radiation is primarily in the infra-red region of the spectrum. For a body at an absolute temperature of T kelvin, the heat radiated is given by Stefan s Law: H 4 = σεt W/m 2 where: σ is Stefan s constant (5.7 x 10-8 W/m 2 K 4 ) ε is the surface radiative emissivity (0 < ε 1) T is the absolute temperature of the surface [K] However at the same time as radiating heat, the body will also absorb heat from the ambient, which is at temperature T 0 o K, so that the net quantity of heat lost by the body at T o K will be given by: H = σε T T A 4 4 0 watts where T is the surface temperature ( o K) T 0 is the ambient temperature ( o K) Here we assume that the radiation absorption coefficient of the surface is the same as the emissivity coefficient. This is generally a good assumption, though not always strictly correct. ELEC9712: Heat Transfer p. 24/32
The value of emissivity ε is very dependent on the surface condition: ε 0.8 0.9 for a matt (eg weathered) surface ε 0.2 0.3 for a shiny (eg new) surface We can calculate from the above an expression for the radiative heat transfer coefficient h r, using the usual general expression We get: r [ ] H= hat T watts [ ] 2 2 hr = σε T + T o T + T o W/m2 / o C It can be seen that, as with the convective heat transfer coefficient, the radiative heat transfer coefficient is also temperature dependent. Typically, h r is about 4-6 W/m 2 / o K at normal operating temperatures. At normal operating temperatures we can approximate h r by the expression: For example, ( ) 0.17 h = 2.9ε T T W/m 2 / o K r for T = 333 o K, T o = 293 o K and ε = 0.8 the heat transfer coefficient is: h = 2.9 0.8 40 4.3 W/m 2 / o K r o ( ) 0.17 o ELEC9712: Heat Transfer p. 25/32
1.2.4 Examples of calculations (a) Temperature distribution in a transformer winding: Heat flow transformer core The winding has three layers of insulated copper as shown. The space factor = 0.75 = (area of copper) / (total area) The 2 I R loss = 17.8 W/kg in the winding. The thermal resistivity of the insulation over the copper is g = 7.5 o Km/W. The total thickness of the three layers = 30 mm. It is required to find the temperature distribution across the winding. Assume no core loss. y x ELEC9712: Heat Transfer p. 26/32 1
The space factor = 0.75, i.e. if the insulation thickness is y and the copper area is x 2, with total side taken as unity: 2 x 0.75 2 = x = 0.866 units, and 2y = 0.134 units 1 The insulation covers 1 x = 0.134 of the overall thickness. To calculate thermal transfer and temperature rise (fall) we need an effective thermal resistivity of the copper and insulation. ( 1 ) ( ) geff = gins x [assume g cu = 0] = 7.5 1 0.866 = 1.01 o Km/W If power loss = p watts/m 3 then in the volume ( 1 1 x) : loss = px watts This flows radially from left to right dx dt x= 0 x= a x eff Hence: dt = ( px) 1 1 g dx o K ELEC9712: Heat Transfer p. 27/32
and: where: = px g dx a T = dt 0 a 0 eff 1 = pgeff xdx = geff pa 2 2 o K above ambient p = space factor x Cu density x 17.8 W/m 3 ( ) 3 5 = 0.75 8.9 10 17.8 = 1.19 10 W/m 3 Hence: g eff = 1.01 a = 0.03 m 1 1.01 1.19 10 5 0.03 2 54 ( ) T = = o K 2 and the variation is parabolic. T 54 o K T o 0 x = a x (b) Loss calculation for a transformer: Transformer is oil-filled with circulation of oil through pipes mounted on walls. ELEC9712: Heat Transfer p. 28/32
oil flow There are 2 heat loss mechanisms: radiation and convection. The transformer rating is 7.5 MVA and has a rated core loss = 14 kw and copper loss = 46 kw. Total = 60 kw. The tank is 1.4m x 2.5m x 3.0m high. It has 925m of external tubes to circulate oil. Tube diameter = 50mm. Determine the average surface temperature rise above 25 o C ambient for rated current. Assume ε = 0.9 for all surfaces and: For flat top surface For vertical surface For vertical pipes hconv 0.25 = 2.6 T W/m 2 /K 0.25 hconv = 2.0 T W/m 2 /K 0.25 0.25 h = 1.3d T W/m 2 /K conv (i) Radiation loss and: p P r r 4 4 ( T 298 surf ) ( 5.7 10 8 ) 0.9 ( T 4 298 4 surf ) = σε W/m 2 = = pa r ELEC9712: Heat Transfer p. 29/32
In this case, A is the effective area presented to the ambient. Thus: A = top surface + 4 sides P (ii) Convection loss r = 1.4 2.5 + 2 1.4 3 + 2 2.5 3 = 26.9 m 2 6 4 4 = 1.38 10 Tsurf 298 watts Top surface: h = 2.6 θ 25 W/m 2 /K ( ) 14 ( θ ) 54 p = 2.6 25 W/m 2 A = 3.5 m 2 ( ) 54 Ptop = 9.1 θ 25 watts Flat sides: h = 2.0 θ 25 W/m 2 /K Tubes: h ( ) 14 ( θ ) 54 p = 2.0 25 W/m 2 A = 23.4 m 2 ( ) 54 Psides = 46.8 θ 25 watts ( ) ( θ ) ( ) ( θ ) 14 14 = 1.3 0.05 25 W/m 2 /K 14 54 p = 1.3 0.05 25 W/m 2 A = 925 π 0.05 = 145.2 m 2 = 399 θ 25 watts Ptubes ( ) 54 ELEC9712: Heat Transfer p. 30/32
Thus, total loss is: Ptot ([ ] 4 ) 6 4 θ ( 9.1 46.8 399)( θ 25) 54 6 [ ] 4 4 = 1.38 10 + 273 298 + + + ( ) ( ) 54 P = 1.38 10 θ + 273 298 + 454.9 θ 25 tot = 60 kw Have to solve by iteration. Try θ = 70 o C as first trial. Then heat loss: ( ) 1.25 = 455 45 + 1.38 10 6 ( 343 4 298 4 ) = 53.03 + 8.22 = 61.25 kw Try θ = 69 o C, we get = 51.56 + 8.00 = 59.56 kw Try θ = 69.2 o C, we get = 51.71+ 8.02 = 59.73 kw Try θ = 69.3 o C, we get = 52.00 + 8.06 = 60.06 kw θ = 69.3 o C is close enough. Thus the surface temperature is: θ = 69.3 25 = 44.3 o C (or o K) above ambient To consider the effect of overloading, we try the calculation assuming a load of 1.25 pu, losses will be: 14 + 46( 1.25) 2 = 85.9 kw Apply same iteration procedure. Try θ = 85 o C as first trial. Then heat loss: ELEC9712: Heat Transfer p. 31/32
= 75.98 + 11.79 = 87.77 kw Try θ = 84 o C, we get = 74.40 + 11.53 = 85.93 kw θ = 84 o C is close enough. Thus the temperature rise above ambient is: θ = 84 25 = 59 o C (or o K) To consider the effectiveness of the cooling pipes, we try the calculation assuming only the tank sides, i.e. Try θ = 100 Try θ = 140 Try θ = 150 Try θ = 155 conv ( ) 54 = 55.9 θ 25 with same radiation o C: 12.34 conv P =, P = 15.83, P = 28.17 kw o C: 21.05 conv r P =, P = 29.27, P = 50.32 kw o C: 23.36 conv r P =, P = 33.30, P = 56.66 kw o C: 24.54 conv r P =, P = 35.43, P = 59.97 kw Thus the temperature is θ = 155 o C and the temperature rise above ambient is: θ = 155 25 = 130 o C (or o K) as compared to 44.3 o C with pipes. It can be seen that the pipes provide a significant improvement in cooling. r tot tot tot tot ELEC9712: Heat Transfer p. 32/32