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I have an interesting mathematical/electrical dilemma and I'm not sure where to start.

I am considering buying this fridge (specs on page).

I will be hopefully buying a battery pack with it to support a long-road trip (a few months) wherein I'll keep medication in the fridge (powered by the battery pack when parked and by the car when driving). The medication needs to be keep between 2 and 8 degrees C at all times.

Here's the battery pack I'm thinking of getting:

Can someone please help me understand what I need to do in order to calculate whether or not this fridge + battery combo will work, and if so, how much time will I be able to power the fridge for.

Ideally, I need a combination of fridge and battery which will allow me to have the fridge constantly powered, assuming I sleep in the car for approx. 8 hours a night and drive for the rest of the day (although, there will be times which I'm away from the car during the day for several hours + times where I hope to have access to a wall outlet).

  • Specs:

Fridge:

  • Input voltage (AC 220-240 V
  • Input frequency 50/60 Hz
  • Rated input power (AC) 64 W
  • Rated input power (DC) 46W DC12V/50W DC24V W
  • Input voltage (DC) 12/24 V

Battery:

  • Capacity: 453Wh
  • Battery Type: Lithium-ion
  • Input Voltage: AC 110-220 V, 50/60 Hz; Solar Power: 18 V; Car Input: 12 V
  • Output Voltage: AC 110V 60 Hz; DC Output: 4 * 12V 4 A (Max.10 A); USB Output: 4 ports: 2* 5V 2.1 A and 2* 5V 1 A
  • Power Output: Rate 500 W Peak/Surge 1000 W
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    \$\begingroup\$ Hi, Joseph. Questions and answers here are supposed to stand on their own even if the links go dead. Post the relevant details into your question. In particular the voltage (V) and amps (A) of the fridge and the voltage and ampere-hours (Ah) of the battery pack. Hint: if you divide 'Ah' by 'A' you are left with 'h'. Factor in that the fridge might not run continuously if well insulated so the battery life will be longer. \$\endgroup\$ – Transistor Jul 22 '17 at 21:39
  • \$\begingroup\$ Hi Transistor,TY for the feedback, here are the specs as requested. Apologies for not mentioning earlier: Fridge: Input voltage (AC 220-240 V Input frequency 50/60 Hz Rated input power (AC) 64 W Rated input power (DC) 46W DC12V/50W DC24V W Input voltage (DC) 12/24 V Battery: Capacity: 453Wh Battery Type: Lithium-ion Input Voltage: AC 110-220V, 50/60HZ; Solar Power: 18V; Car Input: 12V Output Voltage: AC 110V 60HZ; DC Output: 4 12V 4A (Max.10A); USB Output: 4 ports: 2 5V 2.1A and 2* 5V 1A Power Output: Rate 500W Peak/Surge 1000W \$\endgroup\$ – Joseph Jeremiah Noonan Jul 22 '17 at 21:41
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    \$\begingroup\$ Check out what the experts produce at www.dulas.org.uk - they have solar direct drive refrigerators specifically designed for vaccine storage in hot countries.... \$\endgroup\$ – Solar Mike Jul 22 '17 at 21:43
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    \$\begingroup\$ Put all the info in the question rather than scattered through the comments. Again, the idea is to create good questions to generate good answers for you and future readers. Use bullet points to put them on separate lines. \$\endgroup\$ – Transistor Jul 22 '17 at 21:44
  • \$\begingroup\$ I don't think the fridge is reliable in 40'C+ ambient in sun " cooling down to 30 °C below ambient" I needs thicker insulation for lower loss and more power than 50W which can be regulated. THis would run full power and maybe not regulate well enough to satisfy 5'C +/-3 Perhaps a 2 stage fridge. \$\endgroup\$ – Sunnyskyguy EE75 Jul 22 '17 at 23:58
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There isn't enough information to answer this, because your fridge's specs only tell the maximum power draw.

I presume it draws this power when the cooler is running. But it is not running all day. So you should check for how many hours the cooler will run in a day, how much energy it will use... This depends on outside temperature, the quality of the insulation, etc... many variables, and many unknowns.

Say you have a box and you want the contents to stay below 5°C. So you install a cooling system. You also install thermal insulation.

What is the compromise? You could install a foot thick insulation, and then you'd need a tiny cooler to compensate for the losses. Or you could use just a little bit of insulation, like an inch of foam, and use a more powerful cooler.

I'm trying to tell you that your question is a little bit more complicated than you think.

Fridge manufacturers usually do not quote an average power, because it depends on ambient temperature, and also it depends on user habits, like how often you open the door and let warm air in, if you place a warm beer inside and expect the fridge to cool it, etc.

assuming I sleep in the car for approx. 8 hours a night and drive for the rest of the day

In this case, the most cost effective thing to do is not to store electricity, but to store cold. If you have a power supply available all day except during the night, then the best option is to put a few gallons of bottled water (as thermal mass) in your fridge next to your sensitive medicine. When you have power, the fridge cools the thermal mass. When you don't have power, the thermal mass absorbs the heat flowing though the insulation.

The good thing about a few gallons of cold water used as thermal mass is that it doesn't break down, doesn't need maintenance, it can't fail. Even if your car dies, your fridge dies, and everything stops working, you still have time to find someone who owns a freezer, or buy some ice, whatever. It is low-tech and foolproof.

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Just looking at the essentials:

Fridge: Rated input power (DC) 46 W, DC 12 V /50W DC24V W

Battery: Capacity: 453 Wh, Output Voltage: DC Output: 4 * 12 V, 4 A. Power Output: Rated 500 W peak

Checks:

  • 12 V fridge and 12 V power supply outlet. Pass.
  • 46 W fridge and 500 W power supply. Pass.
  • Fridge current: \$ I = \frac {P}{V} = \frac {46}{12} = 3.8 \; A \$. Socket: 4 A. Pass.
  • Run time (100% on): \$ t = \frac {Wh}{W} = \frac {500}{46} = 10 \;h \$.

Since the unit takes 7 - 8 hours to recharge it is putting about 50 W + losses. Let's say it's about 75% efficient so input power will be 50 * 4/3 = 66 W (approx). We can calculate the approximate charging current on DC as \$ I = \frac {P}{V} = \frac {66}{12} = 5.5 \; A \$.

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Worst case your fridge draws 3.8amps (46W/12V) continuously which would be 3.8Ah per hour or 92Ah (3.8*24) in a day. Depending on ambient temperature, how often you open it etc you could do a lot better than that but impossible to quantify. Your battery can supply 37.75Ah. Which would power your worst case fridge consumption for less than 10 hours (37.75A/3.8A) without a recharge from the car.

A running car could easily power the fridge and recharge the battery. I couldn't easily find how long the battery takes to recharge on DC but it says 7-8 hours from the wall so let's assume that for DC too. If you were to end the day with a 7-8 hour drive and then stop for 8 hours the battery would keep the fridge running worst case.

You did not mention alternative batteries but a 12V AGM battery could be discharged to around 20% and would recharge to around 80% off the car alternator. A regular car alternator may provide around 40A of charge. A 160Ah AGM would be quite a bit cheaper (potentially still cheaper if installed in your car, close to the alternator) and would recharge from 32Ah to 128Ah in less than 3 hours. It could then supply your fridge for 24 hours before the car would need to be run again. There are lots of lead acid choices beyond AGM but AGM is a good choice because it can take a very high charge in it's bulk charge phase and can be discharged well beyond 50%. There are also three way fridges which run off propane or battery. In your case you could run a fridge just off propane for close to a month with a 9kg bottle or between the car and propane probably even longer.

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    \$\begingroup\$ Good detailed answer. Watch your units though: "3.8amps per hour" isn't right. It's just 3.8 A continuously which in one hour would be 3.8 Ah (amps x hours, not amps / hour). Also note that SI unit convention is "V" for volt, "W" for watt. The ampere-hour isn't an official SI unit but is always written Ah. 'AH' would be ampere-henries. \$\endgroup\$ – Transistor Jul 22 '17 at 23:16

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