ELEC4611-14-Lec 5 - Equipment Thermal Rating

ELEC4611: Equipment Thermal Rating p.1

ELEC4611 Power System Equipment

EQUIPMENT THERMAL RATING

The thermal rating of electrical power equipment is determined, in general, by the maximum permissible operating temperatures of its components (usually the insulation material which is the item most at risk of failure) under the conditions at which the rating is to be determined. As all electrical equipment generates heat in its operation, the thermal rating is an important characteristic of equipment. Thermal rating is determined from thermal balance considerations of heat generation and heat dissipation from equipment. In the following we consider only steady state rating, which is determined from a thermal balance equation at constant temperature. There are many variations of transient ratings that can be determined for loads that are not constant or not sustained in their operation. These transient ratings will not be considered here. 1. Heat Dissipation Mechanisms The possible heat dissipation mechanisms are: Conduction. Convection. Radiation.


We usually try to determine a heat transfer coefficient, for each loss mechanism, in watts per unit area of exposed surface per degree of temperature rise above ambient: i.e. hi [W/m2/oC] We then use the sum of all of the contributing hi coefficients to determine the total heat loss at the maximum permissible temperature using the appropriate specified maximum temperature, the heat dissipation surface area and the ambient temperature. Thus, in the case with all three loss mechanisms in operation, the total heat loss coefficient htotal will be: htotal = hcond + hconv + hrad 1.1 Enhancing dissipation The dissipation mechanisms can be either natural or can be intentionally enhanced.

For example in electrical equipment, convection dissipation is frequently enhanced by the use of pumps or fans to increase flow rates (of air, water or any other appropriate cooling fluid) past surfaces. This has the effect of increasing hconv. Power transformers use this method. Additional enhancement can be achieved by increasing the effective surface area presented to a cooling fluid flow.

Radiation loss can be enhanced by increasing radiative emissivity of surfaces by painting them with matt black paint for example or by simply increasing the surface area.


Conduction loss can be increased by using materials with high thermal conductivity, although there are normally considerable constraints on this as the materials may be there for particular insulation purposes. Note that air and most gases are very poor thermal conductors. Conduction is generally the least efficient of the heat dissipation processes, although in some cases such as cables, it is the only available means of heat loss. 2. Heat Generation Mechanisms The possible mechanisms by which heat is generated in electrical equipment are: Ohmic heating in conductors. Eddy current heating. Electrical contacts Semiconductor junctions Iron losses - in steel cores, etc (including hysteresis

and eddy currents). Dielectric losses - in dielectric insulation.

Here, the first three heating mechanisms scale as I2, the fourth scales as I, while the last two heating mechanisms depend on the magnetic core flux density and the electric field in the insulation respectively and as these two field quantities are constant and independent of load, the iron and dielectric losses are constant whenever the equipment is energised.


Thus, the balance of heat generation and heat dissipation reduces to evaluating the current level I which generates losses such that the equipment dissipates all of the heat generated at the specified constant maximum temperature. This value of I is then the steady state thermal rating of the equipment (Ithermal). As with developing available techniques for enhancing heat dissipation, good equipment design requires that steps must be taken to minimise the heat generation in equipment as far as possible. This may require, for example: Minimising resistance by increasing conductor cross-

section area Using non-magnetic materials for structural purposes

to limit eddy current generation Using low loss steel laminations to reduce iron core

losses in motors and transformers Using low dielectric loss insulation (such as

polyethylene) to limit DDF loss 3. Thermal Transients Suppose P = rate of total power (loss) generation in the equipment. Then, (a) As the equipment temperature rises above the ambient level, it causes dissipation (heat loss) of some of the heat generated (P watts) to the ambient. We assume a total heat


dissipation coefficient, h watts/m2/oC, so that, knowing the surface area and ambient temperature, the total heat dissipation can be calculated from the product of h, the surface area, and the temperature rise above ambient. (b) In addition to the heat loss to the ambient, the equipment also heats up as its own thermal capacity absorbs some of the heat (P) being generated. Knowing the thermal capacity: mc (c= specific heat) and the temperature rise, this contribution can be calculated also: it is the product of mass, specific heat, and temperature rise. This will be zero at constant operating temperature. Thus, during a time dt within the transient (temperature varying) period, the total heat generated, Pdt, is partly lost to the ambient (Ahodtand is partly retained by the thermal capacity (mcd Thus we can write: Pdt = mcd + Ah [(t) – o] dt where (t) is equipment temperature at time t, A is surface area available for heat dissipation ois the ambient temperature s is the final steady state temperature For constant P (and this is not always a valid assumption), the solution of the above equation is (see Appendix): (t) – os (1 – e –t/T)


The diagram below shows the temperature variation corresponding to this.

where SPAh

steady-state temperature rise

S o

and mcTAh

thermal time constant

We are interested in the particular case when the final steady-state temperature has been reached: (i.e. mcd = 0). This then allows calculation of the thermal or steady-state rating Ith, which is the condition when all heat generated is dissipated to the ambient.


i.e. Pdt = Ah [(t) – 0] dt or P = I2R + k = Ah [(t) – o] or Ith

2 R + k = Ah [max – o] where Ith is the thermal rating and max is the maximum

permissible operating temperature. k is a constant term for any constant heat generation

which does not scale with current (such as iron loss or dielectric loss). Such constant losses may not always be present in some cases.

In many cases the equipment operation will be cyclic and the temperature variation will also be cyclic. In this case the balance equation must be modified to take the duty cycle of heat loss generation into account. Note that the maximum permissible operating temperature max is not the same as the maximum permissible short circuit temperature. The latter is allowed to be higher because it is only of very short duration whereas max is of very long duration. 4. Steady State Temperature Rise Note that the steady state temperature rise magnitude is given by the ratio of the power generation (P or P + k) to


the rate of heat loss (Ah). In the following, we want to look at the considerations needed for ratings calculations for some different common items of equipment. 4.1 Overhead Line

For a bare conductor overhead line, the heat generation in the conductor is caused by simple Ohmic heating and also by a constant level of solar heat absorption (during daylight). Thus solar heating is the constant “k” term in this particular case. The temperature rise above ambient is thus proportional to a current dependent term (I2R) and a fixed term determined by the intensity of solar radiation.

i.e. 2 I SolarR H


It is inversely proportional to the rate of heat loss, which is comprised of a radiative (hr) and a convective (hc) component in this case:

i.e. r

1 ch A h A

Note that the convection heat may be due to either natural thermal convection or to wind forced convection (or a combination of the two). 4.2 Cables XLPE insulated cable (132 kV)


Paper-oil insulated (230kV)

For an insulated cable, the temperature rise is determined by a current dependent ohmic heating term (I2R) and a fixed level heating term which is determined by the dielectric heating. For voltages below 66 kV, such dielectric heating is negligible. Thus temperature rise (I2R + Hd) In the cable the only dissipation mechanism within the cable structure is by thermal conduction from the inner conductor to the outer surface.


The temperature rise is inversely proportional to the conducted thermal transfer within the cable to the outer surface. Other dissipation mechanisms may occur at the surface depending on the ambient medium and whether the cable is buried or exposed in air, in which case radiation and convection play a part in addition to thermal conduction inside the cable. 3-Phase 66kV oil-paper cable

For three phase cables, as shown, the conduction paths for heat are more complex and mutual heating of the various phases can occur. This makes the calculation more complex.


4.3 Transformers For a transformer, the temperature rise is determined by a current dependent term (I2R - the copper loss) and a fixed term determined by the hysteresis and eddy current losses in the core (the iron losses). The temperature rise is thus determined by:

(I2R + Hm)

Suppose a particular transformer has a core loss which is 25% of the copper loss at full load current (a relatively common ratio for transformers). Even when such a transformer is not loaded, it will still operate at an elevated temperature whenever it is energised and the core is thus magnetised. Thus with copper losses at full load 4 times the iron losses, if the temperature rise at full load is 60oC, then the temperature rise at no load but with the core magnetised is 60/5 = 12oC rise above ambient. The temperature rise is also inversely proportional to the convective (internal and external), radiative (external) and conducted losses (between the winding conductor and insulation). Transformers have very complex structures and many different internal and external components, including the magnetic core, the copper windings the solid (paper) insulation, the oil (or epoxy resin in dry-type transformers), the steel tank, the insulating winding spacers, the external heat exchangers etc. These can all be involved in the thermal operation of the transformer and as a result


transformers have quite complicated methods of heat dissipation. These can include forced oil cooling of the windings, forced fan cooling and radiative cooling of the oil in large external radiator heat exchanger banks for oil-cooled transformers. Dry-type transformers have no oil in them and their cooling is thus only possible by thermal conduction in the inner structure, with radiation and convection playing a part at the external surface. There is also an increase in the use of SF6 gas-insulated transformers which have to rely on the SF6 gas to provide the internal heat dissipation. Fortunately, SF6 has much better thermal dissipation characteristics than air. Oil-paper insulated transformer


Dry-type epoxy resin insulated transformer

4.4 Other equipment Similar procedures as outlined above are required for any item of electrical equipment to determine its thermal rating. Other such items may include, for example, high current power electronic components such as

thyristors, IGBTs (insulated gate bipolar transistors) all types of rotating machines


all circuit breakers and fuses low voltage indoor busbar systems switchboards electrical contacts (both static and dynamic) integrated circuits

5. Equipment Rating The thermal rating of electrical equipment is determined by the maximum permissible operating temperatures of the equipment components. The actual temperature attained during operation is determined by the balance of the heat loss and the heat generation as discussed before. The heat generation within the equipment will be determined (in total or in part at least) by the current level in the equipment and the specification of a maximum permissible operating temperature of some part of the equipment will, in turn impose, via the heat balance equation, a maximum permissible operating current. This is then the specified thermal rating of the equipment item. In almost all cases the maximum permissible temperature of electrical equipment is determined by the properties of the dielectric insulation material used, although in some cases (for example for overhead lines) other factors may play a part (for example, thermal expansion, sag, annealing etc. determine upper temperature limits for normal overhead lines, while for busbar systems the limiting temperature might be determined by the melting point of


solder or by tin coating used to improve contact resistance). 5.1 Electrical Insulation and Temperature The great majority of electrical insulation materials are organic in their chemical nature and are thus subject to relatively low permissible temperatures because of the chemical reactions which occur continually in such materials. The rate of these chemical reactions is exponentially dependent on temperature, following Arrhenius' law. Chemical change will result from these reactions and the effect will be a gradual deterioration of insulating properties of the material. For example paper in transformers will have its cellulose chain structure broken down by heat and the material then loses tensile strength and is longer able to act as an insulant. Thus the operating temperature of the insulation is of primary importance in determining the effective lifetime of the insulation and hence of the equipment. The higher the temperature the greater is the reaction rate and the faster the deterioration and the shorter the lifetime. The operating temperature must thus be chosen carefully to attain a satisfactory equipment lifetime. The basis of the temperature limit determination for insulation materials is the dependence of chemical reaction rate on temperature:

Reaction rate = (a constant) x exp ()


[where is the absolute temperature (K) = + 273 where is the Celsius temperature] The insulation lifetime (T) variation with the Celsius temperature (oC) is often written in either of two forms for electrical equipment application: (a) T = X exp (-a)

where X and a are constants which depend on the insulation material thermal class. Here is the Celsius temperature.

For example, for Class A insulation (with the lowest temperature withstand level): X = 7.15 x 104 and a = 0.088

(b) ln273

BT A

where A and B are constants, and is the Celsius temperature. With equation (b), the constants for PVC and XLPE insulation are:

PVC: A = –31.4 ; B = 15,000 20,000T hours at 90oC

XLPE: A = –27 ; B = 14,500 20,000T hours at 120oC 2.3 years 48T years at 90oC


The specification of a nominal required lifetime then means a specified maximum temperature and this is used to calculate the maximum current carrying capacity - the thermal rating. It should be noted that inorganic materials such as mica, ferrites, porcelain and glass and also some organic materials such as silicone rubber and teflon have much higher temperature limits than normal organic dielectrics. Gases also have generally high temperature limits but other factors such as decrease of voltage breakdown strength with decreased density at high temperatures may need to be considered for gases. 5.2 Thermal (Steady State) Rating This is the steady state rating when the temperature has reached a constant value. Then, the thermal balance equation becomes: Ahdt Pdt

and thus s = s o = PAh

wheres = final steady state temperature.


To calculate the steady state thermal rating, we specify a maximum s and calculate Ah, and the equation then gives I, the thermal rating. 5.3 Typical maximum allowable temperatures There are a number of general temperature classes specified for electrical insulation and theses are listed below with some examples of materials in each class. Each class is simply designated by its temperature limit (the letter designations included are the old classification method). Class Typical materials (not all!) in the class 90 (Y) Un-impregnated cellulose: paper cotton etc. 105 (A) Oil-impregnated cellulose etc. 120 (E) Epoxy resins varnishes etc. 130 (B) Combinations of mica chips, glass fibre etc

with bonding (eg resin).155 (F) Combinations of mica chips, glass fibre etc

with epoxy bonding. 180 (H) Silicones and combinations of mica etc with

silicone. 180 plus (C) Mica, porcelain, glass etc with or without

an inorganic binder: PTFE.


A Cable insulation limits

Type Normal use

Maximum permissible

Polyvinyl chloride (PVC) (oC) (oC) V-75 75 75 V-90 75 90 V-105 75 105 Elastomeric (rubber) R-75 75 75 R-EP-90 90 90 R-CSP-90 90 90 Cross linked polyethylene (XLPE)

90 90

Oil impregnated paper 80 80 Mineral insulated metal-sheathed (MIMS)

90 250

High temperature type Silicone rubber 150 150 PTFE PolyTetraFluoroEthylene

200 200

B Overhead Lines Copper 90 110 Aluminium Galvanised Steel

90 75

105 110


C Transformers (oil-paper) Hot spot 100 140Top oil 80 105 Average oil 70 95 Top winding 90 110 Average winding 60 85 Note: In A, B and C the limits are for normal continuous operation only. In emergencies and for short-term overload and short circuits the allowable limits will be higher than above. Example of rating calculation: A (bare) rectangular copper busbar: 50mm x 5mm in cross-section. Maximum allowable temperature is o110 CS Ambient temperature: o

0 30 C Copper resistivity: 8 o1.68 10 -m at 20 C Temperature coefficient of resistivity is o 10.0043 C Skin effect and proximity effect give an increase in effective AC resistance of 5%:

i.e. RAC / RDC = 1.05 The heat dissipation rates from each side of the busbar, obtained from empirical data, are:


Convection: 2 o7.66 W/m / Cch Radiation: 2 o7.93 W/m / Crh Use the above to find the thermal rating for the above maximum temperature limit. Calculation The total heat dissipation coefficient is

h = hc + hr = 15.59 W/m2/oC. Consider one-metre length of the busbar. Surface area for heat dissipation (two sides and neglecting edges):

A = (2) x 1.0 x 0.05m2 = 0.1 m2 Thus, total heat loss (neglecting the 5mm edges):

hA = 15.59 x 0.1 = 1.56 W/ oC. But:

sPAh

P = sAh = 124.8 W This is the allowable level of power loss generation. In this case, P = I2R only and the resistance is able to be calculated from the data.


DCc

lRA

[ Ac is cross-section area ]

8 11.68 10 1 0.0043 110 200.05 0.005

= 9.32 x 10-5 and with a 5% increase due to skin effect, RAC = 1.05 RDC = 9.79 x 10-5 Im

2 x 9.79 x 10-5 = 124.8 and Im = 1130 A = thermal rating. 6 Calculation of Steady State Thermal Ratings of:

Overhead Lines, Underground Cables and Power Electronic Devices

6.1 Overhead Lines The overhead line rating is determined by the specified maximum permissible temperature which is determined by considerations of sag and statutory clearances and by annealing effects. Heat is generated by:


(i) I2R heating - I is the unknown to be determined. (ii) HS : Solar Absorption For (i) we need to calculate resistance R at the maximum operating temperature m: We use the relation R = Ro (1 + (m20)) [Use R per unit length to get I2R in watts/metre] For (ii) we calculate HS = s S.d W/m where s = absorptivity of the line conductor surface S = Solar heat flux in W/m2 d = projected area of conductor (= diameter x 1 (m2) per metre length). Heat is dissipated by:

(i) Radiation: (we use Stefan’s equation)

8 4 45.67 10 W/mR m oH T T d where: .d = surface area of a 1m length of overhead line. = emissivity of the line conductor surface

Note that and m oT T must be in kelvins in Stefan’s equation: i.e. 273m mT where m is in oC.

(ii) Convection: (We have to use empirical equations

based on experimental data).


(a) Natural: 1.230.672.60 W/mc m oH d [due to buoyancy flow of air] (b) Forced: 0.458.54 W/mc m oH vd [wind generated flow] where: d = conductor diameter (m) m o = maximum temperature rise permissible v = transverse wind velocity (m/s) Equate heat generation to heat dissipation to obtain a heat balance for unit length of line: i.e. I2RAC + Hsolar = Hr + Hc We then solve this to find the rating I for the maximum permissible conductor temperature. 6.2 Cables The rating of insulated cables is determined by the maximum permissible temperature of the insulation around the conductor: this temperature is determined by the potential damage to insulation resulting from thermal ageing due to increased temperature. We use this temperature to specify a maximum permissible conductor temperature for use in the calculations.


Heat is generated by the following mechanisms:

(i) Ohmic heat in conductors: I2R. (ii) Dielectric loss in insulation (constant – not

affected by load current); (CV2tan) (iii) Eddy current loss in metallic sheath and armour

(constant 2I R ). To quantify these heat losses, we need:

(i) I2RAC: use skin effect and proximity effect multipliers to get RAC from RDC (typically RAC = 1.04 RDC).

(ii) 2 tan W/mdH CV Dielectric loss: this is significant only for V>66kV. We need tan values (dielectric loss factor) and cable capacitance per metre, C, from the manufacturer.

(iii) Need to get sheath loss information from the manufacturer. Typically sheath losses are ~ 5% of the main conductor I2R losses.

Heat is dissipated by the following:

(i) Thermal conduction. [Thermal Ohm’s Law] through the various coaxial layers of the cable to the external surface where the heat is dissipated.


(ii) Radiation and convection from the outer surface (possibly depending on the cable installation method and location).

Usually only (i) is the only possible dissipation mechanism if the cable is buried in-ground or is in a small duct. However, if it is in open air or in a large air duct, radiation and convection dissipation from the surface are also possible. The commonly used techniques of solution for the thermal rating of cables is to use a thermal equivalent circuit as shown below, to determine the thermal rating for a maximum conductor temperature m.

o

G

m

The equivalent source is I2R + Hd watts

Thermal Equivalent Circuit Here: m o temperature difference (the potential function) 2

dI R H the heat flow in watts (the flux function)

I2R + Hd


G the thermal resistance of the circuit (the medium characteristic).

From the above circuit we have:

m – o = (I2R + Hd)G This is the Thermal Ohm’s Law. To find I, we need to find Gi for each component of the cable and also for any other components which will limit heat flow, for example the earth in which a cable is buried, as shown below.

In fact, a cable is composed of a number of separate layers of different material and each of these layers will have a


thermal resistance which must be calculated before the rating determination can be carried out. The diagram below shows a typical structure of a single-phase cable with metallic sheath and armour.

Example: for the single-core cable configuration shown above and for a unit length of cable:

Insulation layer: 2

1

ln2

ii

g rGr

thermal ohms

Bedding layer: 3

2

ln2

bb

g rGr

thermal ohms

Serving layer: 4

3

ln2

ss

g rGr

thermal ohms


where gi, gb and gs are the thermal resistivities (c.f. electrical resistivities) of the materials which make up the various layers: eg the insulation may be paper or cross-linked polyethylene (XLPE), the bedding may be textile with pitch and the serving may be PVC. The thermal resistances of any metal sheaths or armouring are negligible. The full thermal circuit is thus:

where: m = maximum permissible temperature of conductor s = metal sheath temperature. A = armour temperature. surface = cable outer surface temperature. o = ambient temperature.


oG is the thermal resistance of the substance between the cable surface and ambient conditions [This substance may be air, other cables, water, earth, sand-cement, etc]. The heat sources are thermal analogues of current sources and 1 and 2 represent eddy current heating factors in the sheath and armour respectively. We can use the Thermal Ohm’s Law applied to the various parts of the thermal circuit to get the following equations:

2

21

21 2

20 0 1 2

12

[ (1 ) ]

[ (1 ) ]

[ (1 ) ]

m s i i d

s a b d

a surface s d

surface d

G I R G H

G I R H

G I R H

G I R H

whence

12

0 0

1 1 2 0

[1/ 2 ( )](1 ) (1 )( )

m d i s b

i b s

H G G G GIRG R G R G G

We know 0 1 2, , , , ,m dH R and we can calculate

, , .i b sG G G Thus the only unknown is Go. This thermal resistance depends on the environment of the cable surface (i.e. whether it is in air, water, earth, etc.).


Go can be determined in various configurations as follows: (a) If the cable is buried in earth In this case Go is then the thermal resistance of the earth mass between the cable surface and the ground surface.

It can be shown that the thermal resistance for a unit length of cable:

2ln2

eo

g hGa

thermal ohms

where ge= earth (thermal) resistivity h = depth of burial of the cable. a = cable overall outer radius. (b) If the cable is in open air Then the thermal resistance Go represents effective radiation and convection losses and we thus need to know


heat radiation and heat convection coefficients at the cable surface to calculate the total heat transfer coefficient h.

h = hr + hc W/m2 /oC We have: surface o oHG

where H is the total heat flow to ambient. But we also have:

surfacer c oH h h A

where A = surface area for dissipation.

Combining these two equations, we get:

1 r c oh h AG

and thus:

1o

r c

Gh h A

thermal ohms

or: 1oG

hA thermal ohms

where h = hr + hc is the total surface heat dissipation coefficient of the outer surface of the cable, whatever the loss mechanisms involved. This is an extremely useful equation for use in rating calculations. 6.3 Semiconductor Devices For semiconductors such as thyristors or IGBTs, the limiting rating factor is the junction temperature of the


device. As with insulation materials, there is deterioration of the junction caused by thermally induced chemical change and the device lifetime can be related to junction temperature by an equation similar to that for insulation lifetime:

lnj

BT A

where T is lifetime, j is junction temperature and A and B are constants of the device. In terms of its thermal dissipation of the power loss, the semiconductor device is similar to a power cable in that the heat generated at the junction from power (VjI) loss at the junction contact resistance or voltage drop has to be thermally conducted out to the device surface where it can then be dissipated in a variety of ways: e.g. by use of an attached ribbed heat sink to enhance convection and radiation loss. The method of rating determination for semiconductors also uses an equivalent thermal circuit as shown below. The various designations used in the diagram are: j = junction temperature, c = case temperature, s = heatsink temperature, o = ambient temp.

Gjc = thermal resistance between junction and case Gcs = thermal resistance between case and heat sink


Gca = thermal resistance between case and ambient Gsa = thermal resistance between heat-sink and ambient

heat-sinkcasevirtual junction

ambient air

j cG c sG

s aG c aG

dP

j c s

a

This is the full thermal circuit. In most cases, Gca is very large compared to the other resistances and so the total thermal resistance simplifies to:

Gja = Gjc + Gcs + Gsa If there is no heat-sink, then the total thermal resistance is: Gja = Gjc + Gca Typically, maximum junction temperatures are about 120 – 140oC. For switching semiconductors, steady state operation at constant power is not used and the requirement is thus a pulse rating. In this case if switching is at high enough frequency, there is no significant decay of junction temperature between on pulses. Then the average temperature is effectively the peak temperature and the


same calculation procedure as for cable thermal circuits is used. The power loss is the total average energy loss per second. Thus energy loss per pulse and the switching frequency must be known. Typical switching pulse “on” time might be 1 millisec. with power dissipation of 100 W [200 amps x 0.5 volts] during that pulse, The duty cycle (“on” time/”off” time) might typically be = 0.4. Thus the average power dissipation at the junction would be 100 x 0.4 = 40 watts.


APPENDIX A

Calculation of Coaxial Insulation Resistance and its application to calculation of thermal resistance in coaxial geometries

The electrical resistance of the insulation layer between the cylindrical surface at radius r and at radius r + dr and of length l is:

2drdR

l r

Hence the total resistance is:

2

1

2

1

ln2 2

r

r

dr rRlr l r

ohms

We consider a unit length of cable, 1 ml and thus:

2

1

ln2

rRr

ohms


By the electrical - thermal analogy, where the heat flow and current flow are taken as analogous field quantities, the thermal resistance is:

2

1

ln2g rG

r

thermal ohms

where: g = thermal resistivity. Note that the expression quoted for Go for a buried cable (at depth h) is exactly analogous to the electrical resistance of a conductor - plane arrangement in an electrically resistive medium, obtained from the general equation

RC = and the equation:

2

ln 2C

h a

farads for a rod-plane gap

which gives:

2ln2og hG

a

thermal ohms


APPENDIX B

Problem: find solution to

x td x t Kdt

Eq.1

This is a first-order differential equation. The parameter is called the time constant. We solve this equation by separating the variables and then integrating. Rearrange:

dx K xdt

dx dtx K

1dx dt Dx K

where D is a constant of integration. Integrate:

ln tx K D

exp tx K D

D tx t K e e

At t=0: 0 Dx K e 0De x K

Hence: 0 tx t K x K e Eq.2


Now consider:

oPdt mcd Ah dt Rearrange:

td t Kdt

where: mcAh

and 1o

PKAh

Eq.3

Thus from Eq.2 and initial condition: 0 o :

tot K K e Eq.4

As t : o ssPKAh

ss o ssPAh

Thus, ss oK Eq.5 Hence Eq.4 becomes:

tss o sst e

i.e.

1 t

o sst e 1 to sst e


If the initial condition is 10 then Eq.2 becomes:

1tt K K e

1t

ss o ss o e

1t

ss o o ss o o e

1t

ss ss e Hence: 1 1 1

tss sst e

i.e.

1 1 1 t

sst e 1 1 1 tsst e

Documents

ELEC4611-14-Lec 5 - Equipment Thermal Rating