My wife has planned a cross country trip with her elderly father in his 2003 Volkswagen Golf (diesel) which gets 50 miles per gallon which he is quite proud of.
He is upset because my wife wants to bring along a Koolatron 12V cooler which draws 4.3 amps. Never mind that she’s paying for the fuel.
Can anyone translate the power usage to fuel consumption for me?
Like what is this going to cost for 5000 miles in terms of fuel volume or more specifically in miles lost per gallon?
The cooler apparently uses around 50 watts. A quick google suggests that a car’s alternator efficiency would be around 50%. That means the engine is going to have to put out an additional 100 watts to power the cooler. A quick google suggests that cars need about 20hp (15kw) to cruise. That’s 15,000 watts. The additional 100 watts equals about 0.6%
Over 5000 miles, at 50 mpg, the car should use 100 gallons. With the cooler on, maybe an extra half gallon.
That should be in the ballpark.
While I agree with these numbers, I’d like to point out that they only apply while the engine is running. When the cooler continues running with the engine off (drawing power from the battery), the engine will have to re-charge the battery accordingly later. If they keep the engine running just for the cooler, the cost will be even higher, although I doubt that The Flying Dutchman’s father-in-law would allow that.
Does this gentleman also have a fit if anyone wants to turn on the radio because it will ruin the battery?
Princhester’s analysis is valid for a diesel, but ignores the fact that most non-hybrid vehicles only cruise at less than 50% available engine power.
In addition to limiting air intake, the throttle valve causes a gasoline engine to waste a fair amount of power as pumping losses. (it has to compress the low intake manifold pressure up to the exhaust system back pressure). This is why cars with big engines get worse gas milage, even if they are light and streamlined.
So adding a small incrimental load uses a tad more gas, but also moves the engine into a more efficient operating condition, ofsetting much of the additional load…a lot of the “extra” power is power that would have been wasted anyhow.
Note that Diesels do not suffer increased pumping losses at reduced load. (no throttle valve to cause intake manifold vacuum)
The cooler will use less gas than making one extra stop to buy ice for an ice chest.
I don’t follow you here. Are you saying that at the start of the exhaust stroke, the pressure inside the combustion chamber is at intake manifold pressure level? That can’t be right. After combustion, the pressure should be much higher because the temperature is. Right?
I’m not saying you’re incorrect; I just don’t get your meaning, I suppose. I do agree there are losses due to the energy it takes to pump the exhaust gases through the backpressure.
My explaination turned into such a thread hijack that I going to start a new thread in MPSIMS. Gimme a minute.
One can also make this calculation using an assumed brake specific fuel consumption of ~0.5 lbs/hr per horsepower (cite), and assuming an average speed of 60 mph, get an additional consumption of 0.8 gallons due to the increased electrical load (you would use more fuel, say ~1 gallon total, if the average speed while operating the cooler were only 50 mph). So, a SWAG of 1/2 to 1 gallon is probably in the ballpark for a 5000 mile trip - which is in the noise compared to the total fuel consumption IMHO. Also, this analysis doesn’t address whether that 4.3A number were peak draw, steady-state, or dependent on the cooler-to-ambient temperature difference. Hope that helps. 
These arethermoelectric coolers. You set the current, which is constant, so as to give the performance you want at various temperature differences.
While the current draw will be constant when the cooling element is on, the element may well have sufficient cooling capacity to maintain the set temperature with a less than (and potentially much less than) 100% duty cycle.