How Do I Calculate Current Density from a 0.51 h Discharge Rate?

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SUMMARY

The discussion focuses on calculating current density from a specified discharge rate of 0.51 hours. The user derives current density using the formula involving 1.1 microAh per (cm² * micrometer), resulting in a calculated value of 15,840 A/m². This value is deemed reasonable by participants, as it corresponds to a current of approximately 80 mA through 10 AWG wire, which is within safe limits according to NEC standards. The relationship between the discharge rate and current density is clarified through the conversion of units and understanding of the discharge steps.

PREREQUISITES
  • Understanding of current density calculations
  • Familiarity with microampere-hour (microAh) units
  • Knowledge of finite element analysis (FEA) simulations
  • Basic electrical engineering principles, particularly regarding wire ratings
NEXT STEPS
  • Research "Current Density Calculation Methods" for detailed formulas and examples
  • Explore "Finite Element Analysis (FEA) Simulation Setup" for initial conditions and parameters
  • Learn about "NEC Wire Ratings" to understand safe current limits for different wire gauges
  • Investigate "Discharge Rate Implications in Battery Technology" for broader context on current rates
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This discussion is beneficial for electrical engineers, researchers in battery technology, and anyone involved in finite element analysis simulations requiring precise current density calculations.

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I am trying to set up intial conditions for an FEA simulation and I need to know the current density. I first need to get the current which is specified as a"constant current of 0.51 h rate." This is a disharge rate. I assumed that the h meant hour. Can I derive the currnet form that information?


Put another way, the paper said that the discharge occurs in steps of 1.1 (microAh) / ((cm^2)*(micrometer)) for 25 seconds and then maintianting an open circuit for 75 seconds. I am assuming this is just saying the 0.51 h rate another way. Can someone explain how I relate the two?

I used the 1.1 (microAh) / [(cm^2)*(micrometer)] to calculate the currnet density in A/m^2 by

[1.1 microAhr / ((10^-2)^2*(10^-6)) ] * [depth=10*10^-3] * [3600sec/hr * 1/25sec.]

= 15840 A/m^2...this seemed kind of high, but I'm not an EE.

Any insights would be appreciated
 
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Looks about right to me at a cursory glance. 15,840 A/m^2 would only put about 80 mA through, say, 10 AWG wire which is rated by the NEC for use on 30 A circuits. So, not all that high, really.
 

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