Have you ever come up with a neat idea? An idea for some really useful sort of device? And then you find out that it is, at least for now, a physical impossibility?
I had just such a neat idea this week. At least, I thought it was a neat idea. You see, for the past few weeks, I’ve been traveling around the country interviewing for various different jobs. One of these jobs is with John Deere in Waterloo, Iowa. Now for those of you who don’t know, it gets cold up in Iowa. Their average January temperature is a mere 16°F. I was imagining myself leaving work after a long day, going out into the cold, and then shivering in a very cold Jeep Liberty (my car) for the drive home. So I’ve been thinking, wouldn’t it be nice if I could somehow heat my car while it’s outside?
Well, I don’t want to run a heater using my car battery, for obvious reasons. However, I have this nifty 90W solar panel that I’m not really using at the moment (although it is acting as a glorified cellphone charger, and has been put to good use in the past). Perhaps I could mount this to the top of my Jeep and use it to run a small heater. My only question – would 90W be enough power to effect a substantial (20-30°F) temperature rise?
Initially, my gut feeling said no. For one thing, from experience I know that my 90W photovoltaic (PV) panel will never produce 90W if mounted horizontally on the roof of my car. Even on a perfectly clear and sunny winter day, it won’t be getting strong, direct sunlight. So perhaps I could expect 60W at best, but on a cloudy day I might see less than half of that. But more importantly, on a sunny day, wouldn’t I already be trapping more than 90W via my windows and the greenhouse effect? Because if so, this effect has never produced a particularly warm interior.
Of course there’s only one way to answer these questions. Science! Particularly, the study of thermodynamics. I’ve taken a couple of courses on the subject, but I still had to refer to my book for this formula, derived from the first law of thermodynamics:
This says that the rate of change of an object’s temperature (dT/dt) is equal to the power (P) absorbed or released by that object divided by its mass and its specific heat. So for my first experiment, I decided to heat the interior of my Jeep by idling the engine and turning the heat to full-blast. I then hopped out and remotely measured the interior temperature during cooling. How did I accomplish this? I temporarily re-purposed my Doom Box, which now contains both a LM34 temperature sensor and an XBee wireless transceiver. Here’s a snapshot of the latest PCB revision, powered by an ATMega644P MCU:
So here’s what the temperatures looked like during my first test. The yellow vertical line below indicates when the engine was shutdown and cooling began:
I should note that the Jeep’s temperature measurement was filtered by a 40 second (eight sample) moving average filter implemented within LabVIEW. Also, the internal temperature sensor was placed on the floor in front of the back seats. Exterior temperature measurements were taken manually using a digital thermometer and are represented by a linear trend-line shown in red above.
So during this test, my jeep was outdoors, parked in the shade, from 3-5PM. You’ll notice that during this time the exterior temperature dropped by about ten degrees as the sun started to set. However, the temperature inside the Jeep dropped slightly faster, as you would expect. There was also very little wind during this test.
Now to apply the formula! The drop in temperature between t = 4000s and t = 6000s is easily determined from the graph: 4.7°F or 2.6°C. The trick is in determining values for mass and specific heat. For this first test I decided to simply approximate the mass of air and metal within the car. I calculated that the Jeep contained about 4kg of air at a cp of 1005J/kg-K as well as about 20kg of steel (e.g. in the seats and wheel) at a cp of 490J/kg-K (it’s pretty interesting to note that air has a higher specific heat than most metals). I decided to ignore small pieces like plastic and seat stuffing, as well as everything under the hood (an assumption which proved later to be somewhat stupid).
So plugging all of that into the equation above yields a power loss of just about 18W.
Well now, eighteen watts is pretty surprising. I wasn’t expecting losses to be that low for an interior-exterior temperature difference of 25°F. This called for a second round of testing. This time, I moved my Jeep into an insulated, closed garage. Doing this allowed me to get rid of temperature variations due to solar radiation and wind. It also prevented rapid changes in exterior temperature. I then inserted a controllable heat source (an old laptop) into the car, located a few feet away from my interior temperature sensor. The laptop was configured in such a way as to draw about 46W via a 120VAC inverter. In theory, this level of power should have raised the Jeep’s internal temperature by about 12°F per hour. Did it? Well, I think the graph below speaks for itself:
Yep, that’s definitely a failure. It turns out that an input of 46W could only raise the car’s interior temperature to about five degrees (F) above ambient. So the bottom line is my 90W solar panel isn’t going to have much of an effect.
So what went wrong with that first test? Why did I calculate the need for only 18W? Two reasons. First, I didn’t account for all the residual heat in the engine compartment. This helped keep the Jeep’s interior temperature higher than it would have been otherwise. I likely also underestimated both the mass and heat capacity of the car’s interior. In my defense, it’s not exactly an easy thing to determine. In thermodynamics classes, you get problems about isolated, well-defined blocks of aluminum and ideal, constant heaters. In reality, you get an ill-defined mixture of materials and variable heat sources.
Just out of curiosity, I used the first few minutes of my second test to make a better guess at the Jeep’s mass and specific heat. In five minutes (300s), the temperature rose by about 0.7°F or 0.4°C. Assuming this rise came only from heat input by the laptop, the term mcp = 34,500J/K. That’s actually not too far from my initial guess for the first test, which worked out to 13,800J/K. What this likely means then is that in my initial outside test, the Jeep was losing about 45W instead of just 18W. But I’m also guessing that a lot of heat was still coming in from the engine compartment.
Anyway, I hope these results aren’t too confusing. I’m not making any guarantees about the correctness or validity of my methods or assumptions. What I’m really trying to say here is that you’ll need a lot more than 90W of PV panels to heat your car’s interior. That, I can say with confidence. I’d love to hear comments on this, particularly if you’re an expert in thermodynamics! Where have I gone wrong here? 🙂