Convert $/Ah to $/kWh: Simple Explanation

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SUMMARY

This discussion focuses on converting the price of battery cells from $0.76/Ah to $/kWh, specifically for a 2.8-3.0 Ah battery cell at an average voltage of 3.6 V. The conversion involves multiplying amp-hours (Ah) by volts (V) to obtain kilowatt-hours (kWh), while considering the necessary factor of 1,000 for the kilo prefix. Additionally, it is emphasized that the voltage is not constant and varies based on discharge curves and specific conditions such as temperature and discharge rates.

PREREQUISITES
  • Understanding of electrical units: Amperes, Volts, and Watt-hours
  • Familiarity with battery specifications and discharge curves
  • Basic knowledge of conversion factors, particularly the kilo prefix (10^3)
  • Experience with calculating energy costs in relation to battery output
NEXT STEPS
  • Research the impact of battery discharge curves on voltage and capacity
  • Learn about calculating energy costs for different battery types
  • Explore the relationship between amp-hours and watt-hours in battery systems
  • Investigate the efficiency of DC versus AC power systems in energy conversion
USEFUL FOR

Battery engineers, electrical engineers, and anyone involved in energy storage solutions or battery cost analysis will benefit from this discussion.

km011
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It has been years since my college physics class and I'm struggling to get an answer that makes sense. I'm trying to convert the price of battery cells from $0.76/Ah (for a 2.8-3.0 Ah battery cell) assuming 3.6 V to $/kWh.

This is likely really simple, but I really want to understand how to do this.

Thanks!
 
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Amps are Coulumbs/sec (so Ah would be a measure of columbs)
Volts are Joules/Coloumb
so to get KWh (which is joules) You'd take Ah*V
 
cpscdave said:
Amps are Coulumbs/sec (so Ah would be a measure of columbs)
Volts are Joules/Coloumb
so to get KWh (which is joules) You'd take Ah*V

there's an unaccounted for "kilo" in there somewhere ?
 
Also realize that the voltage is not a constant 3.6V. Look at the discharge curves for the batteries you are considering. Also, the Ah rating is at a particular specified discharge rate only. (and temperature)
 
amp hours are used for volume of a battery. ex: 9v Duracell = 500ma/hr. which only means the battery will produce 500ma for one hour. or 100ma for 5 hrs. you had dollar signs so I can only take that as what would the cost be in dc battery output compared to alternating commercial supplied current. using 500ma is too small so figure a couple 50AH car batteries. last time I checked one kwh (AC) was about ten cents a kwh. so your 100AH batteries deliver (12 x 100) = 1200 or about 12 cents per hour. By the time the trickle charge was done you probably would have spent 3 times that amount. and that doesn't account for what bigguy said about curves, drop, etc.
 
jim hardy said:
there's an unaccounted for "kilo" in there somewhere ?

Opps there is indeed :) Only off by 10^3 that's a good day for me :D
 
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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