In short I'm trying to estimate the current draw in a remote weather station, to dimension its battery pack and solar panel accordingly.

The way I'm thinking of involves writing down the current draw of the system every min and then multiply this number with 3600 to get the current draw for one whole day. For now I've only tried to calculate the current draw when the station is not sending data, as I want to make sure it's correct before I proceed to when the station is transmitting.

I know the following current draws of the components:
Datalogger CR1000X: 1mA
PTB100: 8,3mA (Runs for 3 seconds every 10th second)
AT/RH: 4,5mA (Runs for 1 second every 10th second)
Modem idle use: 0,9mA

My first assumption is that the stated currents from the datasheets, are expressed in mA per second. (As Ampere is defined as charge pr. second)

With this knowledge I've made the following excel sheet to get an overview of the current draws during 1 min of operation.




I sum up the currents to get the usage per min.
Multiply the sum with 60 to get mA/h
Divide with 1000 to get A/h
And multiply with 24 to get amps drawn per day.

If all of the above is correct, does this mean that a 72Ah lead acid battery would supply the station with power for:
72/16,428 = 4,4 hours

My personal guess would be that my assumption about current being measured per second (from the datasheet) and the calculation of current drawn per hours is not correct.

What am asking is if this is the right way to do it?

Thank you.

 

The basic assumption is incorrect, current is not in units of time.
To determine the run time with a given battery you will need to calculate the current drawn by each device (milliamps=0.001Amp) and multiply by the proportion of each hour that it will run. That will give the consumption in milliamp hours per hour.
example: PTB100: 8,3mA (Runs for 3 seconds every 10th second) 8,3 x 3/10 =0.3x 8.3 milliamp hours per hour.
Then add up all of the milliamp hours to get the total consumption of milliamp hours per hour.
Then convert your battery capacity from amp hours to milliamp hours 1 amp hour=1000 milliamp hours.
Next, divide the battery capacity (milliamp Hours) by the total consumption in milliamp hoursper hour. The result will be the operating time in hours.
Example: Datalogger CR1000X: 1mA ->1ma hour per hour -> 72,000 ma hr/ 1 ma hr/hr=72000hrs

 

Don't forget also that a 72Ah battery is usually (but not always) spec'd at the 20hour rate, ie 20hours @ 3.6A. If you draw more than 3.6A it will run for less time; ie a 7.2A draw wont be 10 hours but more like 9 or less. Similarly a 1A draw is likely to be more than 72 hours, could be as much as 85 or more.

 

Don't forget also that a 72Ah battery is usually (but not always) spec'd at the 20hour rate, ie 20hours @ 3.6A. If you draw more than 3.6A it will run for less time; ie a 7.2A draw wont be 10 hours but more like 9 or less. Similarly a 1A draw is likely to be more than 72 hours, could be as much as 85 or more.

With the loads specified the load current does not exceed 20 Milliamps. So it does not seem that the high load limitation applies here.
Also, I am thinking that the TS has missed the decimal and that is the 7.2 amp-hour battery. I use those a lot. 72 amp hours is a big battery!!

 

Also note that for best lead-acid battery life, it should likely not be discharged below about 50% of its capacity.

 

The basic assumption is incorrect, current is not in units of time.
To determine the run time with a given battery you will need to calculate the current drawn by each device (milliamps=0.001Amp) and multiply by the proportion of each hour that it will run. That will give the consumption in milliamp hours per hour.
example: PTB100: 8,3mA (Runs for 3 seconds every 10th second) 8,3 x 3/10 =0.3x 8.3 milliamp hours per hour.
Then add up all of the milliamp hours to get the total consumption of milliamp hours per hour.
Then convert your battery capacity from amp hours to milliamp hours 1 amp hour=1000 milliamp hours.
Next, divide the battery capacity (milliamp Hours) by the total consumption in milliamp hoursper hour. The result will be the operating time in hours.
Example: Datalogger CR1000X: 1mA ->1ma hour per hour -> 72,000 ma hr/ 1 ma hr/hr=72000hrs

Thanks a lot for the explanation MisterBill2, it is much appreciated.
I've found an Excel template which does what you describe and expresses the "proportion of each hour that a device will run" as a duty cycle in %.
Having calculated the portion of each hour the devices are in the states "On", "Ilde" or in "standby".

(Example the radio module will only be "on/transmitting" for ~24.66 seconds in an entire day => 24.66s/86400s = 0.00028541666*100%, so a duty cycle of 0.028541666% for the tansmitting state of the RF-module)

I end up with the following:


Which to me seems more plausible.

Battery wise I intend on using 2 x G70EP VRLA batteries connected in parallel for a stated nominal capacity of 2*71Ah=142000mAh


Yes they are big and heavy batteries, trust me I've been carrying them around in the field ;-)

 

Don't forget also that a 72Ah battery is usually (but not always) spec'd at the 20hour rate, ie 20hours @ 3.6A. If you draw more than 3.6A it will run for less time; ie a 7.2A draw wont be 10 hours but more like 9 or less. Similarly a 1A draw is likely to be more than 72 hours, could be as much as 85 or more.

Thanks Irving good to know.
Even a max current draw, everything running a max at the same time. I'm not going to see a current draw close to 1A.

 

Also note that for best lead-acid battery life, it should likely not be discharged below about 50% of its capacity.

Hey crutschow,
You are right, my thinking is that the solar panel and or a small wind turbine, will ensure that the battery never does a deep discharge cycle. Meaning that I expect the batteries lifetime to be based on time rather than cycles:


Of course taken into account that the temperature deviation from 25 degree Celsius, will have a negative effect on the battery's float life.

 

The self-discharge rate of the batteries will be more than the usage rate. For SLA its around 2% - 5% a month. Lithium would be a better and lighter technology for this requirement...

But if you have solar/wind why such large batteries???

 

My question also: Why such a large battery for an application that will be supervised by an operator more frequently? For my whole day radio communications tours of duty, where the working tome will be a maximum of six hours, I have a 7.2 amp- hour battery because that is the best capacity for multiple day uses, and that give a 5X safety margin.

 

The self-discharge rate of the batteries will be more than the usage rate. For SLA its around 2% - 5% a month. Lithium would be a better and lighter technology for this requirement...

But if you have solar/wind why such large batteries???

These batteries will be placed at weather stations in the arctic regions. I want to use a VRLA battery as they are resistant against fluctuations in temperatures and especially resistant to cold temperatures like -20 Celsius or even colder, making Lithium an no go (sadly).

During the winter we have a very limited amount of solar power generated if any, meaning the batteries need to last during the cold winter months. Sadly the wind turbines we have been using, can't withstand the harsh climate and will either be blown apparat or covered in ice.

 

My question also: Why such a large battery for an application that will be supervised by an operator more frequently? For my whole day radio communications tours of duty, where the working tome will be a maximum of six hours, I have a 7.2 amp- hour battery because that is the best capacity for multiple day uses, and that give a 5X safety margin.

To give an honest anwser I don't know. I'm new at the company and my predecessors doesn't seem to have taken much of an interest in dimensioning these batteries "correctly", I assume they used "The better safe than sorry" approach.

Ideally the weather stations will not be serviced/supervised frequently, the goal is once every 2 years as it is rather expensive and difficult to to get to where they are located.

And just to be sure, does my "new" current consumption estimation look correct?

 

On a side note.
A solar controller is installed to manage the charging of the battery, adding an extra 8mA of constant consumption:

 

These batteries will be placed at weather stations in the arctic regions. I want to use a VRLA battery as they are resistant against fluctuations in temperatures and especially resistant to cold temperatures like -20 Celsius or even colder, making Lithium an no go (sadly).

During the winter we have a very limited amount of solar power generated if any, meaning the batteries need to last during the cold winter months. Sadly the wind turbines we have been using, can't withstand the harsh climate and will either be blown apparat or covered in ice.

If your solar charging is compromised in winter then you should add self-discharge as a line in your spreadsheet as its likely to be the biggest discharge value. Self-discharge is worse in hot weather.

I recently found this document on the web. You may find it helpful.

 

I'm just going to suggest actually measuring those current draws if it's not too difficult; data sheets can have errors. And, see if there are any unnecessary current drains, like power indicator LEDs.

 

Using a cold chamber to verify performance under the expected conditions is a very wise choice. Strange things can happen in extreme cold conditions.

 

If your solar charging is compromised in winter then you should add self-discharge as a line in your spreadsheet as its likely to be the biggest discharge value. Self-discharge is worse in hot weather.

I recently found this document on the web. You may find it helpful.

Good point Irving and thanks for the document. I have been reading the manual of the batteries we use and they state the following:

So as I understand it only happens under Open Circuit conditions, but in real life I assume the batteries will always loose charge with time.

When trying to determine the amount of self-discharge it gets rather difficult to get an actual number and they only state it as "Extremely low self-discharge rate".


Which also leads me to assume why my predecessors used the "Better be safe than sorry" over dimensioning approach.

 

I'm just going to suggest actually measuring those current draws if it's not too difficult; data sheets can have errors. And, see if there are any unnecessary current drains, like power indicator LEDs.

Hello bassbindevil.
Yes, I've also been doing test of the complete setup on the side, were I get a maximum current draw which is a bit higher than what the spreadsheet says is the max. (Spreadsheet estimation: ~171mA max draw)

 

Using a cold chamber to verify performance under the expected conditions is a very wise choice. Strange things can happen in extreme cold conditions.

For sure, but we sadly don't have that kind of equipment.

 

My domestic freezer gets down to -18C. I've used that to validate cold operation on the grounds that the 'real thing' will actually be encased in some heavy insulation and warmed slightly by its residual operating heat

Many moons ago I used to be part of a design team on ground-to-air comms for fighter aircraft - designing high power transmitters and sensitive receivers that worked reliably over an operating range exceeding -55C to 125+C (the former being a shutdown fighter on an arctic airstrip in a blizzard (can be colder) and the latter being the same fighter on a desert airstrip in full sunlight with the engines running - the radio bay being located in the top of the fuselage above and between the engines - not the best choice but thats where the iairframe designer put them! Fortunately we had that equipment; unfortunately we also had to do field validation!