hollandhvac
Mechanical
- Feb 23, 2007
- 120
Hello everybody, I am here to ask a question about battery heat dissipation. I am an HVAC engineer and I revied the heat dissipation of battery's in a battery room and noticed that our contracted had mentiond two figures. As an example I will take on of the battery room,s the first was 0,08 kW and the second was 1,786 kW this is the heat load for the same battery,s but under different conditions. Now I have questioned the contractor hhow tis is possible that there is a factor 22 between the loads and I recieved the following answer.
During float charge (normal operation) battery voltage is 2,25 V/cell. During boost charge voltage is raised to 2,4V/cell which means higher battery current than during float charge. Battery has it's internal resistance. As the internal resistance gives power looses equal to P= I^2*R this means that if current increases than power looses also increase ( battery resistance is constant). So boost charge requires higher voltage what causes higher current flowing through what produces higher power looses.
This was the answer I recieved from our contractor. The questions I now got is: is a differance in 0,15 Volt per cell sufficent to produce so much additional heat? And does the answer I recieved make sence?
Thanks for your help
During float charge (normal operation) battery voltage is 2,25 V/cell. During boost charge voltage is raised to 2,4V/cell which means higher battery current than during float charge. Battery has it's internal resistance. As the internal resistance gives power looses equal to P= I^2*R this means that if current increases than power looses also increase ( battery resistance is constant). So boost charge requires higher voltage what causes higher current flowing through what produces higher power looses.
This was the answer I recieved from our contractor. The questions I now got is: is a differance in 0,15 Volt per cell sufficent to produce so much additional heat? And does the answer I recieved make sence?
Thanks for your help