Guest Posted August 15, 2005 Report Share Posted August 15, 2005 Hi, I have an application where I need to estimate the total heat gain of an electrical room associated with the heat loss from the equipment in the room. The room will have PFC of 500kVAR for a distribution voltage of 600VAC. I am wondering what range I could expect the heat loss of the capacitors to be relative to the rated kVAR. I realise that the loss of the caps is heavily dependent on the harmonics. In this case, the electrical room will have 400hp of vfds, all with line reactors (3%) installed to help with harmonics. In a typical application, could I expect a typical heating value to be 1%, 2%, more, less? Any help you can provide would be appreciated. Thanks, -Collin Link to comment Share on other sites More sharing options...
marke Posted August 15, 2005 Report Share Posted August 15, 2005 Hello Colin I would expect that the heating from the PFC capacitors would be insignificant relative to the vfds and reactors.Are you using detuning reactors with the capacitors?I would strongly recommend that detuning reactors are used due to the presence of the drives. Unfortunately, there will be additional heat from the dtuning reactors, but the life of the pfc capacitors wil be greatly extended. Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
Guest Posted August 16, 2005 Report Share Posted August 16, 2005 Yes, detuning reactors will be used. They will add more heating, as you say, but in comparision to the reactors for the drives, the drives themselves, and other equipment in the room, it will most likely not be significant. After posting this question yesterday, I came across some manufacture's data sheets that stated a complete system with detuning reactors would not exceed 1.5W/kVAR -- with the capacitiors themselves accounting for no more than 0.5W/kVAR. In my case, this would lead to .75kW for 500kVAR (max), which is small considering the total room heating from electrical equipment will be between 50-60kW I expect. Thanks again for your help. -Collin Link to comment Share on other sites More sharing options...
marke Posted August 16, 2005 Report Share Posted August 16, 2005 Hi Colin Those numbers would certainly be in accordance with my expactation, but I would not be brave enough to put my life on them!!I would expect that the reality could be in the order of +/- 100%, but at the end of the day, in your situation, it is still insignificant relative to the heat already dissipated in the room. Have a good day,Best regards from New Zealand, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
BigMax Posted August 17, 2005 Report Share Posted August 17, 2005 Colin, I've recently been involved in the development of a complete PFC system range from the ground up. One of our primary considerations was power loss and resulting temperature rise. As you may be aware, absolute temperature is the enemy of PFC capacitors and de-rating is not applicable to them. We were therefore careful to provide sufficient forced air ventilation in order to control temperature rise. In order to do this, we had to carefuly consider and actually measure total power loss. PFC capacitors themselves are generally low loss components and I agree with your capacitor loss expectations. Detuning reactors however produce much higher losses. We tested various reactor brands and found around 15% power loss variations between them. The nett result of our development suggests losses in the order of 5Watts per kVAr. This loss figure includes capacitor losses, switchgear losses (contactors, fuse links, cables etc) and of course the predominant reactor losses. I hope this info. is helpful. Cheers! BigMax Link to comment Share on other sites More sharing options...
Guest Posted August 22, 2005 Report Share Posted August 22, 2005 Thanks Again for your help marke & BigMax. -Collin Link to comment Share on other sites More sharing options...
Guest Posted October 7, 2005 Report Share Posted October 7, 2005 DEAR SIR , HOW TO FIND HEAT LOSS CALCULATIONS AND WHATIS THE NORMSREGARDS, .GR Link to comment Share on other sites More sharing options...
marke Posted October 7, 2005 Report Share Posted October 7, 2005 Hello GRThe heat loss is dependent on the capacitors. some are in the order of 0.5W/KVAR. This information is often available on the capacitor datasheet.Harmonic currents increase the heat loss. There is also heat loss from the detuning reactors which is very dependent on the harmonic currents.Refer to the manufacturers data.Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
oskin Posted October 11, 2005 Report Share Posted October 11, 2005 I am a little late on picking up on this thread but I agree with Big MaxDO NOT underestimate the power loss with your detuning reactors.I have had experience of some brands having very high operating temperatures, 110 degree C plus core temperatures. This has completely blown my cooling calulations out the window and had to double the airflow on the boards..Be very careful. Link to comment Share on other sites More sharing options...
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