7

I live close to the Rocky Mountains and drive into them quite often. I have a small naturally-aspirated four-banger that tends to lose a very noticeable amount of power getting into the higher altitudes and it doesn't help that much of the drive will be uphill with two adults and luggage.

I understand that this is due to the density of air being lower in altitude than at sea level. Less air being sucked into the engine implies less fuel to maintain the stoichiometric ratio, so you lose power to maintain the efficiency of the burn. In fact, you can roughly calculate how much horsepower you are losing using this formula:

HP Loss = (elevation in feet * 0.03 * horsepower @ sea level)/1000


The solution to this is simple: More air! Turbos compress air and shove it in along with the atmospheric pressure to create a higher density of oxygen, which is detected by the vehicle and more fuel is added, giving a boost in power. The result is that turbocharged vehicles have hardly any noticeable change in power in altitude. Since diesels often are equipped with a turbo, they rarely suffer any power loss in altitude for the same reason.


This website states that diesels have a compression ratio of 20:1 versus 8:1 for gasoline. I know that the compression ratio is unrelated to actual air intake; it is compressing the air that comes in to a specified ratio, regardless of how much air is involved. However, I imagine this might still help non-turbo diesel engines perform in higher altitudes since they tend to run lean and might change the fuel/air ratio less strictly. I think I understand why I'm off base in believing this, but I assume I am not taking everything into consideration.

My question is this: Do non-turbo diesel engines lose power comparably to non-turbo gasoline engines in higher altitude?


Edit

Considering Zaid's answer below, I threw together a quick chart to see the difference between the formula I gave above and the more accurate one Zaid so kindly provided for the percent power reduction.

Altitude (ft)     3,000   5,000   10,152   14,000
Actual Reduction  5.91%   13.46%  30.18%   40.52%
Alternate         9.00%   15.00%  30.46%   42.00%

So you can see that the simpler formula works best close to 10,000 feet and breaks down in lower altitudes. This was comparing the formulas for a gasoline engine.

  • What are the units of elevation in your formula? Feet? Meters? Other? – sixtyfootersdude Oct 9 '15 at 17:24
  • @sixtyfootersdude, sorry, they are in feet. I will edit to clarify. – Poisson Fish Oct 9 '15 at 17:51
8

Non-turbo diesels lose less power in the Rockies

At least according to the SAE J1349 standard.

(Calculations shown below).


Assumptions

Dry air pressure in the Rockies = 90 kPa   ( at 3000 ft)
           Absolute Temperature = 277.15 K ( 4 °C )

This allows us to compute the two quantities, A & B, that are used to determine the correction factors for both types of engines:

A = 99 / p_PT = 99 / 90 = 1.1
B = T_P / 298 = 277.15 / 298 = 0.9300

For gasoline engines:

Correction factor = A^1.2 * B^0.5 = 1.081

For naturally-aspirated diesel engines:

Correction factor =     A * B^0.7 = 1.045

This correction factor is used to determine the actual power made by an engine on a dyno after environmental variations are taken into consideration, it implies that the diesel engine loses about 4.3%¹ due to the altitude change compared to the 7.5%² power lost by the gasoline engine.

Since 4.3% < 7.5%, diesel wins.


The equations and explanation on how to use them are found in the Bosch Automotive Handbook.

It's nice to not have to worry about how to derive these relationships and just use the industry-wide standard for a change :D


¹ : 1 - 1 / 1.045 = 0.043 = 4.3%

² : 1 - 1 / 1.081 = 0.075 = 7.5%

  • See, this is why we should be able to do the neat things with LaTeX and MathJax that some of the other Stack Exchange sites can use. Does the book by any chance explain the reasoning behind using A^1.2*B^0.05 and A*B^0.7? Are the parameters based on empirical data, or is there some physics behind it? Are there similar equations for forced-induction engines? – Poisson Fish Oct 8 '15 at 19:56
  • @PoissonFish : Yeah, the lack of MathJax is annoying. No explanation is provided in the book regarding why A and B are related the way they are for gasoline vs diesel. The relationship is almost certainly based on empirical correlations. The gasoline correction factor doesn't change for forced-induction engines, but FI diesels have their own dedicated correction factor. Interesting question btw – Zaid Oct 8 '15 at 20:11
  • Huh, that's surprising that the correction wouldn't change for FI gasoline. As a side note, using your formulas for Leadville, CO (the highest city in US at 10,152 feet), a gasoline engine would lose 30% of its power while the NA diesel would lose 24%. All very interesting, thank you! – Poisson Fish Oct 8 '15 at 20:13
  • @PoissonFish you're welcome! – Zaid Oct 8 '15 at 20:15
1

Based on some of the understandings of throttle pates already posted (Why do heavy vehicles almost always use diesel engines?) and how/when they may apply to diesel engines - especially comments related to how diesel engines without throttle plates always ingest the maximum amount of air - for those diesel's the answer may possibly depend whether the engine in question was one of those and already ingesting more air than it actually needed (and running lean) to support effective combustion.

In that case - which suggests and earlier model diesel without sophisticated control hardware - the non turbo diesel would probably do better in high altitude even though its compression ratio would still drop.

This would probably hold true even more if the gasoline engine in the comparison really - as stated above - only had an 8:1 ratio; which is really low for a gasoline engine.

At 8:1, dropping a few more compression points would really hurt Torque; not so for the diesel at 20:1 that can still (usually) produce decent Torque at a lower engine speed that may also facilitate better cylinder filling to offset the altitude issue.

Jim.

  • 1
    I think your using your other account. I upvoted you last night on two of your questions so you could have the reputation to comment and start voting on other peoples posts. You'll want to stick with one account so you can reputation that gives you rights to do things within the site and participate in the community. Please be clear, I'm not admonishing you in any way. Just trying to be helpful. If you stick to one account it will allow any upvotes to your questions to accumulate as reputation in a single account. Cheers to you, I look forward to reading your posts in the future! – DucatiKiller Dec 30 '15 at 16:35
  • @DucatiKiller - I think I already said this, but this user is an unregistered user. Until they actually create an account, none of the upvotes you give will be of any benefit to him. You are just wasting your votes. Maybe Jim will actually create an account one of these days so he can get credit for his answers. – Pᴀᴜʟsᴛᴇʀ2 Dec 31 '15 at 7:39
  • 2
    @JimStanley - You are confusing compression ratio (CR) with cylinder pressure. CR (whether static or dynamic) doesn't change (in most cases) during the operation of the engine. The CR refers to the swept volume of the cylinder. It is a measured. Cylinder pressure can change due to atmospheric conditions. – Pᴀᴜʟsᴛᴇʀ2 Dec 31 '15 at 7:55

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