Submitted: Tuesday, Sep 23, 2003 at 21:30

ThreadID: 7393
Views:1913
Replies:5
FollowUps:3

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I have an electric engine on my boat that is less than ½ a horsepower (314 watts). To make it useful I need to attach it to a battery so the question of the size of battery arises. The boat in question is a sailing boat, so weight is a problem. I thus need to answer the question

"What is the lightest battery I need to get full power out of my engine for a period not exceeding 30 minutes?”

I only know 3 electrical equations

Ohm’s Law V=IR

Power Law P=VI

Peukert’s Law C=InT

where

V = Voltage in Volts

I = Current in ohmns

C = Capacity in Amp/hrs

N = Peukert number

P = Power (watts)

R = Resistance in amps

T = Time in hours

Known

Volts V 12.00

Watts P 314.00

Capacity C 110.00

Peukert n 1.30

I can thus deduce

calculate amps I 26.17

calculate ohms R 0.46

Calculate time T 1.58

I think that the above is a correct answer to the question

"What is the battery I need to get full power out of my engine.”

To get the size of the battery down I played with the Capacity as this seems to be the limiting factor with weight. I also noticed that most battery web sites seem to imply that the full 12 volts may not be available so I played a little with the Volts.

I got the following results

Fixed

Peukert n 1.30 1.30 1.30 1.30

Watts P 314.00 314.00 314.00 314.00

Variable

Volts V 12.00 10.00 12.00 10.00

Capacity C 110.00 110.00 30.00 30.00

Calculated

amps I 26.17 31.40 26.17 31.40

ohms R 0.46 0.32 0.46 0.32

time T 1.58 1.25 0.43 0.34

I know nothing goes according to theory so I took the lightest battery indicated in the last column (30 amp hours) and added 30%. So I went out and bought a 40 Amp Hour battery safe in the knowledge that I could get half an hour of flat out motoring with my engine.

WRONG!

I doubt I’m getting half the power I should have.

So what am I doing wrong?

1. Is there a limitation to Peukert’s theory?

2. Is there another equation my teachers forgot to tell me when I was at school?

3. Some other reason?

This is not a theoretical question as the sailing season is fast approaching and I need an answer soon!

Any suggestions will be appreciated (but No - I will not return to a stinky old petrol outboard!)

"What is the lightest battery I need to get full power out of my engine for a period not exceeding 30 minutes?”

I only know 3 electrical equations

Ohm’s Law V=IR

Power Law P=VI

Peukert’s Law C=InT

where

V = Voltage in Volts

I = Current in ohmns

C = Capacity in Amp/hrs

N = Peukert number

P = Power (watts)

R = Resistance in amps

T = Time in hours

Known

Volts V 12.00

Watts P 314.00

Capacity C 110.00

Peukert n 1.30

I can thus deduce

calculate amps I 26.17

calculate ohms R 0.46

Calculate time T 1.58

I think that the above is a correct answer to the question

"What is the battery I need to get full power out of my engine.”

To get the size of the battery down I played with the Capacity as this seems to be the limiting factor with weight. I also noticed that most battery web sites seem to imply that the full 12 volts may not be available so I played a little with the Volts.

I got the following results

Fixed

Peukert n 1.30 1.30 1.30 1.30

Watts P 314.00 314.00 314.00 314.00

Variable

Volts V 12.00 10.00 12.00 10.00

Capacity C 110.00 110.00 30.00 30.00

Calculated

amps I 26.17 31.40 26.17 31.40

ohms R 0.46 0.32 0.46 0.32

time T 1.58 1.25 0.43 0.34

I know nothing goes according to theory so I took the lightest battery indicated in the last column (30 amp hours) and added 30%. So I went out and bought a 40 Amp Hour battery safe in the knowledge that I could get half an hour of flat out motoring with my engine.

WRONG!

I doubt I’m getting half the power I should have.

So what am I doing wrong?

1. Is there a limitation to Peukert’s theory?

2. Is there another equation my teachers forgot to tell me when I was at school?

3. Some other reason?

This is not a theoretical question as the sailing season is fast approaching and I need an answer soon!

Any suggestions will be appreciated (but No - I will not return to a stinky old petrol outboard!)

Tuesday, Sep 23, 2003 at 22:00

Dont know about the battery What has this to do with 4WD ????
AnswerID:
31826

Tuesday, Sep 23, 2003 at 22:12

he needs the 4wd to get the boat to the river loljamieson wild cat
FollowupID:
22702

Wednesday, Sep 24, 2003 at 07:55

He needs the boat to tow the 4WD OUT of the river! ROFLMAO
FollowupID:
22722

Wednesday, Sep 24, 2003 at 11:59

Thanks Guys I knew there had to be a 4WD in there somewere.
FollowupID:
22736

Tuesday, Sep 23, 2003 at 22:14

Just a few more equations you may want to knowAmps

V/R

P/V

Square root P/R

Volts

I*R

P/I

Square root P*R

Ohms

V/I

P/I square

V square/P

Watts

I square*R

V square/R

V*I

Also a wet cell battery would not be able to hold the current for that long maby try a dry cell or a dura cycle 1 battery may not even be enough

I'll ask one of the trades men at work and see what they can come up with

i'll let you know.

Paul

AnswerID:
31832

Tuesday, Sep 23, 2003 at 22:34

hasbeanI don't doubt your maths.

The issue is, of course, the power actually available from your battery.

If your motor draws 26 amps you will need a .... Hmmmm ..... I see your problem.

A 26AH deep cycle battery (in theory) would last for 1 hour at 26A but in practice would last 30mins at 26 amps before entering a discharged state. A 40AH battery should do the job.

The answer is simple - phone a battery company.

The official documentation from Century batteries says to multiply the amperage draw by the hours used and add 20%

Thus 26A x .5hrs=13AH

13AH + 20%=15.6AH

How dodgy is that?

But wait there's more

I quote

"Note that the faster a battery is discharged, the fewer amp hours it will deliver.

That is why deep cycle batteries carry an amp hour rating for three standard lengths of discharge time." (20hr, 5hr, & 2hr)

Okay

"For each discharge rate, and battery, divide the amp hour capacity by the number of hours to determine how many amps you can draw per battery over a specified time period." Huh!!

The appropriate table indcates 3 different "capacities" for the 20hr, 5hr, & 2hr discharge cycles.

The Table for the 47T Century battery (the smallest battery on my list) seems to indicate that you can draw a total of 50A out of the battery over a 20hr time frame (Slow?? or Intermittent???) but the "capacity" of the battery drops to 31A max if it is discharged over 2 hrs (fast discharge). Over 5hrs of discharge the "capacity" is 39A.

Could it be that continuous (fast??) discharge of your battery at 26A has caused a loss in discharge "capacity"? It still doesn't make sense does it?

Sorry that's the best I can do.

I suppose that while I was spending 30 mins discharging my brain, someone else has hit upon the right answer for you, before I can hit the "submit" button.

Oh well

Cheers

oskar

AnswerID:
31836

Tuesday, Sep 23, 2003 at 22:37

hasbean.There are a few variables, the arithmatic is ok, the 314w power of the motor may be the output power not the input power, a good motor is about 60% efficient so the input could be 500w, the second variable is the charge of the battery, measuring the voltage is a very poor way of measuring the charge, the normal charger is set low to reduce the risk of over charging, the most reliable way to measure charge is to calculate the time x amps = ah capacity, whatever happens don't let the battery boil.

Eric.

AnswerID:
31837

Wednesday, Sep 24, 2003 at 08:23

Based on Erics 500 wats (worst case) input power that would work out at 500 W/12 V = 42 amps continuous current, best case at 320 W, would be 27 amps. This means for a one hour life you would need a 42 A/H battery, or for your "not exceeding 1/2 hour" would be a 21 A/H battery worst case. Note that both of these assumptions are from fully charged to dead flat, not a nice thing to do to any battery, and also that the normal way A/H's are specified are at a 10 or 20 hour rate so what you would be looking for would be specified at a higher rating than what you would think that you need (about +30%).At an educated guesstimate try for about a 25-30 A/H Odyssey (PC925), or an Orbital marine type. If you are in a smallish yacht, avoid the "sloppies", because if you capsize or broach you are likely to end up with acid in your bilges and no battery (and at a guess a very looong row to shore....).

AnswerID:
31854