Hi All Expert,

          I have calculated Heat loss for Control panel according to following link. It is used for MM 440 drive panel. 

https://cache.industry.siemens.com/dl/files/972/22978972/att_118633/v1/pdf_power_loss_and_temperature_rise_en_v1_5.pdf

OR

https://support.industry.siemens.com/cs/document/22978972/micromaster-4-(mm4)-what-are-the-power-losses-of-micromaster-drives-and-accessories-and-how-is-temperature-rise-calculated?dti=0&dl=en&lc=es-BO

In above pdf in page no.12 its shows to formula to calculate heat load.

1.Naturally cooled cubicle

Temperature rise = Total power dissipated (W) / (5.5ii * Exposed surface of cubicle (m²)) 

2. Fan cooled cubicle

Temperature rise = (0.053iii * Total power dissipated (W)) / Air flow in cubicle (m³/min) 

Now client is asking explanation  for 5.5 and 0.053 factor.

In manual it is mention below statement 

ii factor found out experimentally

iii factor found out experimentally 

Please give some idea about this factor. So I can explain same to my customer.

 Waiting for all expert reply. It's very urgent.

Regards,

Yogendra G

JensS

This factor is found experimentally in a laboratory for a „standard cubical“

I´m afraid, but more I don´t know

 Hi JensS

Thanks for your help. Can you provide me alternative method to calculate heat load??

Waiting for your reply.

Regards,

Yogendra G

.

Hi Expert,

Is the procedure mentioned for MM4 drives applicable to the equivalent Sinamics Drives?

Thanks in Advance,

Soumava