1.5 Thermal considerations 1.5.1 Heat flow The energy transfer by transformers and electromechanical energy conversion by the rotating electrical machines absorbs currents in the conductors and fluxes in the ferromagnetic parts.
The current flow in conductors produces I2 R losses in the windings and core losses in the ferromagnetic cores. Further, the additional losses occur in the tank walls, end plates and covers on account of leakage flux. The above mentioned losses appear as heat which increases temperature above the ambient medium in every affected part of the machine. The dissipation of heat by the parts of an electrical machine has been carried out by conduction, and convection assisted by radiation. Amongst the two methods of heat dissipation, the convection through air, liquid or steam is the most significant method. Forced convection through the coolant system is used for heat dissipation.
However, the design for forced convective cooling is a straightforward and the designer should ensure for a large enough amount of coolant flows through the machine. This means that the cooling channels have to be large enough. If a machine with open-circuit cooling is of IP class higher than IP 20, using heat exchangers to cool the coolant may close the coolant flow. On the other hand, the proportion of heat transfer by radiation is usually not appreciable, yet not completely insignificant. Heat transfer by radiation can be done by a providing black surface to the machine.
Thermal Resistance: The thermal resistance is defined as the thermal resistance which causes a drop of 1oC per watt of heat flow. The thermal resistance, like electrical resistance, can be written as: R_e= ?t/S —- (1.8) = t/?S —- (1.9) ? = thermal resistivity of material, ? (thermal) m or oC-m/W ? = 1/? = thermal conductivity, W/oC – m t = length of medium, m S = area of surface separated by the medium, m2 The heat flow equation between two conducting surfaces is expressed as, Q_con= (?_1-?_2)/R_c W . The above equation can be written as, Q_con= (S(?_1-?_2))/?t W —- (1.10) Heat dissipated per unit surface area by conduction is q_con= (?_1-?_2)/?t W/m^2 —- (1.11) The temperature difference across the conducting medium, ?= ?_1-?_2 ?= Q?_con R_c —- (1.12) ?= Q?_con (?t/S) —- (1.13) From the equation (1.13), it is understood that a material with a high value of thermal resistivity will dissipate less amount of heat or alternatively for the dissipation of same heat the temperature rise will be higher. 1.5.2 Newton’s law of Cooling It is described previously that the machine parts temperature rises by the increase in losses.
After sometime, the parts of a machine attain a steady temperature rise. The produced heat is dissipated by radiation and convection. At this temperature, the heat produced in the machine parts equal the heat dissipation. The equation relating the heat dissipation by radiation is given by, Q_rad=?_rad ?S watt The above expression can be modified if temperature rise varies between moderate limits.
The heat dissipated by convection is given by, Q_rad=?_con ?S watt Therefore, the sum of heat dissipation by radiation and convection results the total heat dissipation. It is expressed as, Q=Q_rad+Q_rad= ?_rad ?S watt+?_con ?S ?=(??_rad+?_con ) ?S= ??S watt —- (1.14) where, ?=?_rad+?_con —- (1.15) = specific heat dissipation or emissivity due to radiation plus convection Equation (1.14) represents the Newton’s law of cooling. Newton’s law of cooling varies where the body is acted upon by a uniform current of air. Therefore, this law applicable to the natural process of cooling for restricted range of temperature. 1.5.3 INTERNAL TEMPERATURES (HOT SPOT TEMPERATURES): The loss in electrical machines occurs inside the iron cores and windings. The produced loss is dissipated to the surface by the cooling medium.
The internal flow of heat, from the parts in which it is actually generated to the cooling surface from which it is transferred to the coolant, is important in determining the hot spot temperatures and the temperatures to which the insulating materials would we subjected. If the cross-section of a coil, which produces electrical losses, is very large or if the insulation around the coil or the core is very thick, then there is always danger of exceptionally high internal temperatures developing, even when the temperature of external surface is below the maximum specified limit. In order that there should not be any injury to the insulating materials we must determine the temperature of the hottest spot, a place where the local temperature is the highest. Hence, the problem is to find out the difference in temperature between the outside surface from which heat is carried away and the hottest spot inside the winding, from which the heat must travel through the conductors and insulation before it can be dissipated away. 188.8.131.52 Calculation of internal temperature The Figure 1.6 shown below has a very large plate of thickness ‘t’, consisting of a homogeneous material. It is assumed that the length and width of the plate are very large as compared with its thickness.
The direction of heat flow across its thickness is shown through the Figure 1.6. The two surfaces are assumed to be at the same temperature. Fig. 1.6 Temperature gradient Let us consider, l = length of the plate, m w = width of the plate, m ? = thermal resistivity of material along the direction of heat flow, m q = heat produced per unit volume, W/m3 ? = temperature rise, oC d? = the temperature difference across the walls of this strip dx = elementary strip thickness x = distance from the centre We get, Qx = per unit volume x volume = q x lwx The temperature difference between the walls of this strip is, d? = heat conducted x thermal resistance of the strip = q lwx x ?dx/lw=q?x dx ? The difference in temperature between the centre and any point at a distance x from the centre along the path of heat flow, ?= ?_0^x???x dx=(q?x^2)/2? —- (1.16) Thus the temperature difference curve is a parabola shown in Figure1.6.
From equation (1.16), the difference in temperature between the centre of the plate and the outer surface (i.e. x=t/2) is, ?_?=(q?t^2)/8? —- (1.17) Hence the temperature of the hottest spot (centre of plate in this case) ?_m=?_?+?_s=(q?t^2)/8+?_s —- (1.18) where, ?S = temperature the surface. Temperature gradients in cores The built up area of laminated core in electrical machine is shown in Figure1.7. It consists of steel laminations insulated from each other by varnish. The heat is dissipated duo to the iron loss at the core when it is subjected to alternating magnetization.
The point O shown in the Figure 1.7 is the hottest part of the laminated core and the generated heat is to be conducted to surfaces A and B. The path of heat flow along X axis is across the laminations while the heat flow along Y axis is along the laminations. Figure 1.7 Built up laminated core Consider that all the heat flows alongside the direction OX. Therefore, from equation (1.16), temperature difference between O and A is given by ?_OA=(q?_x x^2)/2? Where, ?x is the thermal resistivity across the laminations. Considering the total heat to flow alongside the laminations (along OY), temperature difference between O and B is given by ?_OB=(q?_y y^2)/2? Where, ?y is the thermal resistivity along the laminations.
When comparing the two laminations, the value of thermal resistivity along the laminations is low as compared with that across the laminations. Therefore, at first sight it would indicate that all the heat should be taken along the laminations in order to keep down the internal temperature or indicates that axial ventilation wherein air is blown across the laminations would be most effective. 184.108.40.206 Heat flow in two dimensions In practical, the heat does not flow along parallel paths and the dissipating surface as discussed in the previous sections. Actually, it flows in different directions and the windings and the cores have insulation in addition to copper and iron respectively.
The thermal resistivity of built up windings and cores depends upon relative thickness of insulation to copper or iron. Let us consider a coil or a core which has a large axial length comparing its width or thickness. Figure 1.8 shows the cross section through the coil. It is obvious that the entire outer surface must be maintained at constant temperature.
Figure 1.8 Heat flow in two dimensions Let, l = length of the coil, m w = width of the coil, m t = thickness of coil, m ?_y = thermal resistivity along aa, ?m ?_x = thermal resistivity along aa, ?m Q = heat produced per unit volume, W/m3 It is assumed that the heat flows outward through walls of successive imaginary spaces of rectangular. The boundary of the rectangular space is specified as ABCD, the thickness of the wall is dx in the direction bb and dx w/t in the direction aa. Thermal resistance of horizontal elementary strips, =(?_y dx w/t)/l(AB+CD) =(?_y wdx)/4lxt Thermal resistance of vertical elementary strips, =(?_x dx)/l(AD+BC) =(?_x t dx)/(4 lwx) Therefore, the total thermal resistance of walls is expressed as, (as the above mentioned two paths are in parallel), dR_?=((?_y wdx)/(4 lxt) . (?_x tdx)/(4 lxw))/((?_y wdx)/(4 lxt)+(?_x tdx)/(4 lxw))=(?_x ?_y tw dx)/4lx(?_y w^2+?_x t^2 ) q_x be the heat produced in ABCD, q_x=ql(2x.2x w/t)=4qlx^2 w/t Temperature difference between inner and outer walls of ABCD d?=heat loss in space ABCD x thermal resistance of walls =4qlx^2 w/t.(?_x ?_y tw dx)/4lx(?_y w^2+?_x t^2 ) =(qw^2 ?_x ?_y x dx)/((?_y w^2+?_x t^2 ) ) Difference of temperature between O and outer surface ?=(qw^2 ?_x ?_y)/((?_y w^2+?_x t^2 ) ) ?_0^(t?2)??x dx=(qw^2 t^2 ?_x ?_y)/8(?_y w^2+?_x t^2 ) ? —- (1.19) Let Q be the total heat produced in the coil, Q = ql wt Substituting this value of Q in equation (1.19) we get, ?=Q/8l wt (?_x ?_y)/((?_y w^2+?_x t^2 ) )=Q/(8l ( w/(t?_x )+t/(w?_y ) ) ) —- (1.20) 220.127.116.11 Thermal gradients in conductors placed in slots The conductors placed in slots carry current which produces heat in them on account of copper loss. The temperature in the centre of the core is assessed by considering the following two cases. Case i: Considering the slot insulation to be very thick as compared with that on the end connections, the heat produced in the embedded portions of the conductor is conducted along its length to the end windings.
Figure 1.9 shows a conductor placed in a slot. It is preferred to find the temperature difference between its centre O and the overhang. Fig. 1.9 Temperature gradient across the length of embedded conductor Let Iz = current carried by the conductor, A L = length of the embedded conductor, m az = area of conductor in m2 ? = current density in the conductor, A/m2 ?_c = thermal resistivity of the conductor, ?m ? = electrical resistivity of the conductor, ?m Consider a strip of width dx at a distance x from O Heat conducted through the strip, Q_x=I^2 R loss between O and strip=I_z^2 ?x/a_z Thermal resistance of the strip dR_?=(?_c dx)/a_z Temperature across the strip, d ?=I_z^2 ?x/a_z .(?_c dx)/a_z = ? (I_z^2)/(a_z^2 ) ?_c xdx Temperature difference between O and overhang ?=?_0^(L?2)??? x ? (I_z^2)/(a_z^2 ) ? ??_c x dx =? (I_z^2)/(a_z^2 ) ? ??_c L^2/8? —- (1.21) =? ?^2 ?_c L^2/8? —- (1.22) Case ii: Consider that the overhang is considerably hot and therefore the heat produced in the embedded portion of the conductor is to be conducted through the slot insulation to the iron core. we have now to compute the difference between copper and surrounding iron.
Let, Ws = width of the slot, m ds = depth of the slot iron, m t = thickness of insulation, m ?i = thermal resistivity of insulation, ?m Fig. 1.10 Temperature gradient across the insulation of conductor embedded in a slot. Heat produced in the conductor Q=I_z^2 ?L/a_z Area to the path of heat flow = L (2 ds + Ws ) Resistance of thermal insulation R_?=(?_i t)/L(2d_s+W_s ) Temperature gradient across the insulation ?=I_z^2 ?L/a_z . (?_i t)/L(2d_s+W_s ) From equation (1.12), temperature across the strip is expressed as, ??=I?_z^2 ?L/a_z x (?_i t)/L(2d_s+W_s ) ??=I?_z^2 ?/a_z x (?_i t)/((2d_s+W_s ) )? —- (1.23) putting I_z ?a_z , we have ?=?^2 a_z ? (?_i t)/L(2d_s+W_s ) ? —- (1.24) 18.104.22.168 Heating of turbo-alternator rotors: The temperature difference between the conductors in a slot and its iron walls is ??=I?_z^2 ?/a_z . (?_i t)/L(2d_s+W_s ) ? (see from eqn.1.23) We can neglect the term Ws for the case of strip copper laid flat in the rotor slots of turbo-alternators because the heat will not travel down the layers of insulation.
? For turbo alternator rotors ??=I?_z^2 ?/a_z . (?_i t)/((2d_s ) ) —- (1.25) 1.5.4 THERMAL STATE IN ELECTRICAL MACHINES: 22.214.171.124 THEORY OF SOLID BODY HEATING The temperature of a machine rises when it is run under steady load condition starting loading conditions. The temperature at first increases at a rate determined by power wasted. As the temperature rises, the active part of the machine dissipates heat partly by conduction, partly by radiation and in most cases, largely by means of air cooling. Therefore as the temperature rises, its rate of increase falls off owing to better heat dissipating conditions .the temperature -time curve is exponential in nature.
The temperature of any part of a machine, not only depends on the heat produced in itself but also on heat produced in other parts. This is because there is always a heat flow from one part to another. For example, the heat produced in the part of the winding embedded in the slot flows partially through the insulation to the laminations partially to the end windings. Electrical machines are not homogeneous bodies .their parts are made up of different materials like copper, iron and insulation.
these materials having different thermal resistivity’s and due to this, it is rather difficult to calculate the temperature part of the machine .the results obtained from such a theory are applicable to a certain degree, to the different parts of the machine as a whole. Q = power loss or heat developed, J/s or W G = weight of the active parts of machine, kg H = specific heat J/kg-? S = cooling surface, m2 ? = specific heat dissipation W/m2 – oC C = 1/? = cooling co-efficient oC – m2/W ? = temperature rise at any time t, oC ?_m = final steady temperature raise while heating ,? ?_n = final steady temperature raise while cooling ,? ?_i = initial temperature rise over ambient medium,? T_h = heating time constant ,s T_c = cooling time constant ,s t = time, s. 126.96.36.199 HEATING: Considering the conditions at any time t from start, heat energy developed in the body during an infinitely small time dt, = heat energy developed per second x dt = Qdt —- (1.26) If during this period dt the temperature of the body raises by d?, the heat energy stored in the body = weight of the body x specific heat x difference in temperature = G h d? —- (1.27) If in the process of heating, the temperature of the surface rises by ? over the ambient medium, at the instant considered, the heat energy dissipated by the body into the ambient medium due to radiation, conduction and convection = specific heat dissipation x surface x temperature raise x time = ? x S x ? x dt=S ? ? dt —- (1.28) As the heat developed in the machine is equal to the stored in the parts plus the heat dissipated, we have from equations 1.26, 1.27 & 1.28 Q dt=G h d?+S ? ? dt —- (1.29) dt(Q-S ? ? ) =G h d? dt=d?/(Q/Gh-S?/Gh ?) —- (1.30) Solving the differential equation 1.30 t=-Gh/S? ?log ?_e (Q/Gh-S?/Gh ?)+K —- (1.31) where K is the constant of integration, The value of K is found by applying the boundary condition, when t = 0, we have ? = ?i . Putting this in eqn.1.31 0 = – Gh/S? ?log?_e (Q/Gh-S?/Gh ?_i )+K or K=Gh/S? ?log?_e (Q/Gh-S?/Gh ?_i) Substituting this value of K in eqn. 1.31 t= – Gh/S? ?log?_e (Q/Gh-S?/Gh ?)+Gh/S? ?log?_e (Q/Gh-S?/Gh ?_i) = – Gh/S? ?log ?_e ( Q/(S? ) -?)/(Q/S? – ?_i ) —- (1.32) The machine reaches a final steady temperature rise ?_m when t = ?.
Under this condition there is no further temperature rise and the rates of heat production and dissipation are equal. This means d?=0 or Ghd?=0 when the machine attains final steady temperature rise. From eqn. 1.29, for ??=??_m, Q dt=S ? ?_m dt ?_m=Q/S? —- (1.33) ?_m=Q/S?=Q_C/S —- (1.34) Putting ?_m=Q/S? in eqn. 1.32, we have t= Gh/S? ?log?_e (?_m -?)/(?_m -?_i ) —- (1.35) The term Gh/S? has the dimensions of time and is called the heating time constant Th. ? T_h=Gh/S? —- (1.36) Putting this value, T_h=Gh/S? in eqn.
1.35 ?t= T?_h ?log?_e (?_m -?)/(?_m -?_i ) —- (1.37) or (?_m -?)/(?_m -?_i )=e^((-t)?T_h ) or ?0=??_m (1-e^((-t)?T_h ) )+?_i e^((-t)?T_h ) —- (1.38) If the machine starts from cold conditions, ?_i=0 (No temperature rise over the ambient medium) ??=??_m (1-e^((-t)?T_h ) ) —- (1.39) Equation 1.39 is the temperature rise with time. The temperature rise curve is exponential in nature as shown in fig.1.11. Heating time constant: Consider equation 1.39, ??=??_m (1-e^((-t)?T_h ) ) Putting t=T_h in above expression, we have, ??=??_m (1-e^(-1) )=0.632?_m Thus we can define the heating time as the time taken by the machine to attain 0.632?_m of its final steady temperature rise. The heating time constant of a machine is the index of time taken by the machine to attain its final steady temperature rise. Fig.1.11 Heating curve Considering the relationshipT_h=Gh/S? we conclude that the time constant is inversely proportional to ? (specific heat dissipation).
? has a large value for well ventilated machines and thus the heating constant is small. The value of heating time constant is large for poor ventilated machines. Since, the volume of machine and hence its weight increases in proportion to the third power, and the surface area in proportion to the second power of its linear dimensions, the heating time constant of a machine increases at first power of linear dimensions, because of this, large sized machines have large heating time constants. The heating time constant of a well ventilated induction motor of about 20 kW rating may be out of order of minutes while that of large or totally enclosed machines may reach several hours or even days. The heating time constant of conventional electrical machines is usually within the range of ½ to 3-4 hours.
Final Steady temperature rise: Considering an expression ?_m=Q/S?, it is clear, other things being equal, the final steady temperature rise is directly proportional to the losses. It is also evident that the final steady temperature rise is inversely proportional to surface area and specific heat dissipation. Thus for the same loss, the machine would attain a higher temperature rise if its dissipating surface is small or, if its ventilation is poor. Cooling The equation for cooling can be derived by considering eqn.1.29 Q dt = Gh d? + S?? dt The solution for the above equation is: t= Gh/S? log_e (Q/Gh-S?/Gh ?_i )+K The value of K is obtained by putting boundary conditions, When t = 0, ? = ?i.
From this we have, 0= Gh/S? log_e (Q/Gh-S?/Gh ?_i )+K Or K= Gh/S? ?log?_e (Q/Gh-S?/Gh ?_i ) Substituting this value of K and proceeding as in the case of heating, we get, ??=??_n (1-e^((-t)?T_c ) )+?_i e^((-t)?T_c ) —- (1.40) Where ?_(n )= Q/S? —- (1.41) = Q_c/S —- (1.42) and T_c=Gh/S? —- (1.43) The value of ? under cooling conditions is usually different from that under heating conditions and so the heating and cooling time constants of a machine may have different values. If the machine is shut down, no heat is produced and its final steady temperature rises when cooling is zero or ?n = 0. Under these conditions eqn. 1.40 thus reduces to ?= ?_i e^((-t)?T_c ) —- (1.44) fig.1.12 Cooling curve It is clear from eqn.1.44, that the cooling curve is also exponential in nature as shown in fig.1.12. Eqn.1.44 is applicable to machines which are shut down while eqn.1.40 is applied to machines allowed to cool owing to partial removal of load.
Cooling Time Constant: Consider the relation, ?= ?_i ? e?^((-t)?T_c ) Putting t = Tc, we have ?= ?_i ? e?^(-1)=0.368 ?_i Thus we can define the cooling time constant as the time taken by the machine for its temperature rise to fall to 0.368 of its initial value. As stated earlier the cooling time constant may be different from the heating time constant for the same machine, as the ventilation conditions in the two cases may not be the same. The cooling time constant is usually larger owing to poorer ventilation conditions when the machine cools. In self cooled motors the cooling time constant is about 2-3 times greater than the heating time constant because cooling conditions are worse at stand still. Final Steady temperature rise: Considering an expression ?_n=Q/S?, the final steady temperature rise when the machine is cooling is directly proportional to the losses. When the machine is shut down, Q = 0 and so the final temperature rise is zero, i.e.
the machine finally comes to ambient temperature conditions.