Modeling Coolant Flow in Lumped Parameter Thermal Network

Transcription

1 Modeling Coolant Flow in Lumped arameter Thermal Network Tapani Jokinen Aalto University (Finland) Abstract The paper deals with modelling a coolant flow in thermal networks It is shown that the coolant flow can be modelled by heat flow controlled temperature sources Normal circuit analysis programs can be used to solve this kind of networks It is also shown how the thermal network with coolant flow is solved if a circuit analysis program is not available I INTRODUCTION The lumped parameter thermal model has been used for a long time in calculating the temperature rises in electric machines; eg Soderberg used thermal networks for temperature calculations of turbinegenerators in 93 [] Hak made a noted contribution to thermal networks in the 5's eg [] [3] [4] [5] Several researchers have made their doctoral theses on thermal modeling and networks eg Roberts [6] Kylander [7] Kolondzovski [8] Hong [9] Nategh [] Alexandrova [] and olikarpova [] Development of a thermal network model can be divided into four parts: Forming the thermal network for the machine Determining the thermal resistances 3 Determining the losses and their distribution in the machine 4 Modeling the coolant flow through the machine All these items are important in the calculation of temperature distribution The last item coolant flow can be modeled in different ways The simplest way to model the coolant flow is to assume that the temperature of the coolant is constant and equal to its mean value That gives sufficiently good results if the temperature rise of the coolant is small as it normally is in totally enclosed fancooled motors If the temperature rise of the coolant is high as in motors having opencircuit cooling the constant temperature approximation is not sufficient We can estimate the temperature rise of the coolant in different parts of the motor [3 5] After solving the thermal network we know the heat flow distribution and we can recalculate the temperature rise of the coolant correct our estimation and solve again the network to obtain a more accurate result The most accurate way to consider the coolant flow is to handle the heat flow equations and the coolant flow equations at the same time [ 3] The system equations of thermal networks with passive components are linear and the system matrix is symmetrical This results in the property of reciprocity ie the temperature rise of any part A per watt in part B is same as the temperature rise of part B per watt in part A The equations describing the temperature rises of the coolant in different motor parts are also linear but they do not have the properties of symmetry and reciprocity This is the reason the coolant flow cannot be modeled by a passive electrical network In this paper it is shown that the coolant flow can be modeled by heat flow controlled temperature sources in the thermal network The circuit analysis programs such as Spice Saber or Aplac can be used to solve thermal networks with heat flow controlled sources The heat flow controlled temperature source is described by a current controlled voltage source in the program Examples to form thermal networks are presented too II METHOD OF ANALYSIS Let us examine the cooling of the stator of an open motor (Fig ) The coolant flow q enters one of the end winding regions The losses ew absorbed from the end winding and the friction losses ρ in the end winding region warm up the coolant The temperature rise is R ( ) ew ρ end q ew ρ ρcpq where ρ is the density and c p the specific heat capacity of the coolant q the coolant flow and the term R q ρc q () p R q has the dimension of the thermal resistance [K/W] It is assumed that the mass flow ρq does not depend on the temperature of the coolant The temperature rise of node ( ) in Fig can be assumed to be the average temperature rise in the end winding region According to () we get Fig Temperature rise of the coolant in an opencircuit machine () 8

2 end Rq ( e w ρ) (3) The losses ys absorbed from the stator yoke warm up the coolant by the amount of R R ( ) end end 3end end 3end end 3 q ys q ew ρ 3 (9) ys ρc q end end p Substituting end from () and using the term () we get R ( ) R end q ew ρ q ys (5) The temperature rise of node ( ) in Fig is the average temperature rise of the coolant over the stator yoke end end Rq( ew ρ) Rqys (6) Analogously we get for node 3 (4) where the heatflowcontrolled temperature sources are R q ( ew ρ ys ) () 3 R q ( ys ew ρ ) () The equivalent network satisfying (3) (8) and (9) is shown in Fig 3 The rule for writing the temperature source equations is now: Rule : The temperature source between two coolant flow nodes m and n is equal to the sum of losses absorbed by the coolant in the nodes m and n multiplied by R q R ( ) R R ( ) (7) 3 q ew ρ q ys q ew ρ Equations (3) (6) and (7) can be interpreted as heatflowcontrolled temperature sources; for instance for the source there are two controlling heat flows ew ρ and ys The thermal network in Fig matches (3) (6) and (7) Fig Interpretation of the coolant as heatflowcontrolled temperature sources and 3 The rule for writing the temperature source equations is formulated as follows: Rule : The temperature source connected between a coolant flow node and earth is equal to the sum of two products The first is R q multiplied by the losses absorbed by the coolant before the coolant flow node and the second is R q multiplied by the losses absorbed in the coolant flow node under consideration According to Fig the temperature rises and 3 can also be written in the form Fig 3 Interpretation of the coolant as heatflowcontrolled temperature sources and 3 III EXAMLE The total enclosed fancooled induction motor in which there is also an inner coolant flow (Fig 4) is presented as an example of how to form the coolant flow part of the thermal network The outer and inner coolant flows are qo and qi The friction losses in the winding end regions and in the outer fan are ρ ρ and ρ3 respectively The losses transferred from the nondrive end winding region and from the stator core to the outer coolant flow are 6 and s3 The losses transferred from the driveend winding region to the ambient are 4 It is assumed that the outer coolant flow does not cool the bearing shield in the drive end The losses from the stator and rotor core to the inner coolant flow are s5 and r7 The losses from the stator and rotor end windings to the inner coolant flow are ews6 ews4 ewr6 ewr4 The thermal network will be formed only for the coolant flow The thermal network inside and between the stator and rotor cores and windings is not presented That part of the network can be formed according to the literature eg [4] and [5] R ( ) R end end end end q ew ρ q ys (8) Fig 4 Total enclosed fan cooled induction motor with outer and inner coolant cycles 9

3 The thermal network of the coolant flow is presented in Fig 5 The thermal resistances over the bearing shields are R 6 and R 4 The coolant flow is modelled according to rule and Fig 3 The heat flow controlled sources are: θ R qo ρ3 () θ R qo ( ρ3 6 ) (3) θ 3 R qo ( 6 s3 ) (4) θ 45 R qi ( ρ4 ewr4 ews4 4 s5 ) (5) θ 56 R qi ( s5 ρ ewr6 ews6 6 ) (6) θ 67 R qi ( ρ ewr6 ews6 6 r7 ) (7) where R qo R qi ρcpqo ρcpqi (8) (9) Note the correct signs of the heat flows in the equations for and 67 The heat flows 6 and 4 have a negative sign because they flow in a direction opposite to the other heat flows in nodes 4 and 6 (Fig 5) Note also that the equivalent circuit representing the inner coolant flow is not a closed loop but an open loop because a voltage source between nodes 7 and 4 would shortcircuit the circuit representing the inner coolant flow and the heat flow would be infinite If we write the heatflowcontrolled temperature source between nodes 7 and 4 according to Rule we find that the source is a linear combination of the temperature sources and 67 We may use any node of the coolant flow as a starting point; in Fig5 node 4 has been chosen The end point is the last node in the coolant cycle before the cycle closes node 7 ρ θ ρ3 θ 6 R 6 ews6 6 ewr6 θ3 θ 56 s5 s3 5 θ r7 θ 45 ews4 ewr4 4 R4 ρ Fig 5 Thermal network of the coolant flow for the TEFC motor presented in Fig 4 IV SOLUTION OF EQUIVALENT CIRCUIT Let us examine the solution of an equivalent circuit (Fig 6) without a circuit analysis program To decrease the number of equations only a simple stator circuit is considered in Fig 6 The machine has an opencircuit cooling There are 9 nodes in the circuit The line connecting the nodes 7 8 and 9 indicates the circulation of the cooling air The losses are following: Fey iron losses in yoke Fed iron losses in teeth Cuu resistive losses in slots resistive losses in end winding ρ and ρ friction losses in end winding spaces The nodal point method known from circuit theory is used in solving the circuit The circuit equations written in matrix form are 4 Fig 6 Thermal circuit of the stator of an opencircuit cooling machine 3

4 Fey Fed Cuu () The conductance G nm refers to the conductance between nodes n and m For instance! " On the diagonal of the matrix in () there is the sum of the conductances which are connected to the node under consideration Everywhere else there are the conductances between the nodes with a minus sign For instance the three conductances G 43 G 45 and G 46 are connected to node 4 and their sum G 43 G 45 G 46 is on the diagonal On the same row G 43 is in the third column G 45 in the fifth column and G 46 in the sixth column Between the other nodes node 4 does not have a connection and these elements are zero in the matrix Equation () can be written in short form as #$%#&%' #(% # ) * % '#(% #% ' #% #) * % () where Fey Fed #(% Cuu and () where ρ and ρ are the friction losses in the end winding spaces and R q the thermal resistance () In matrix form q 8 9 q q q q q or in short q ρ q ρ (7) q ρ q ρ #& * %#: e %#) * %; ρ (8) The unknown heat flows are solved from (8) #) * %#: e % > #& * % #: e % > ;&? (9) Substituting (9) for [Φ e ] in () we obtain #$%#&%' #(% #% ' #: e % > #& * % #: e % > ;&? (3) and #: e % > A#&%' #(% #% #: e % > A (3) ;&? from which the solution for the temperature rises in 9 nodes are obtained #&%@#$%' > #: e % > A #: e % > A (3) ;&? # ) * % (3) In Equation () there are eleven unknown temperature rises and three unknown heat flows Ф 57 Ф 8 and Ф 69 or [Ф e ] thus we need three more equations These are obtained from the temperature rise of the coolant According to (3) (6) and (7) the temperature rises in nodes 7 8 and 9 are q ( ρ ) (4) q 5 ρ 6 q (5) q 5 ρ 6 q q 5 ρ 6 (6) V CONCLUSION The paper deals with modeling a coolant flow in lumped parameter thermal networks There are two modeling methods In the first method heat flow controlled temperature sources are connected between coolant flow nodes and earth In the second method heat flow controlled temperature sources are connected between the coolant flow nodes The heat flow controlled temperature sources depend on the method used Simplest means to solve a thermal network is to use a circuit analysis program If this kind of a program is not available we have to form the circuit equations ourselves The paper describes also how the circuit equations are built and solved 3

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