Derating power with altitude?

[FatTire]

Looking for a 4WD tractor for FEL, driveway grading, and snowblowing (72inch width, rear mount) at 9000 ft. elevation. I read once that the power loss is about 3% per 1000 ft. above sea level. If that is true, a 40 hp unit would run like 28 hp. Does this sound anywhere near correct?

How many HP should I be shooting for at this elevation? Not looking to skimp, but also not wanting to go so large that maneuverability becomes an issue...

 

[Ford tractor]

Close. I think it's roughly 3-4% per 1000ft.

Here's a calculator. You have to know temperature, humidity, absolute barometric pressure and it will tell you what you want to know more accurately.

Engine Horsepower Calculator


Winter's cold will increase the density of the air and help make it act like a lower altitude.

 

[Jerry/MT]

Looking for a 4WD tractor for FEL, driveway grading, and snowblowing (72inch width, rear mount) at 9000 ft. elevation. I read once that the power loss is about 3% per 1000 ft. above sea level. If that is true, a 40 hp unit would run like 28 hp. Does this sound anywhere near correct?

How many HP should I be shooting for at this elevation? Not looking to skimp, but also not wanting to go so large that maneuverability becomes an issue...

Idealy, horsepower for a normally aspirated engine varies inversely with ambient pressure and the square root of the ambient absolute temperature(~oRankine = 459.7 + T in oF)), i.e., HP/(pambient x Tambient) = constant. Since the engine is usually rated at "standard , sea level conditions", pambient is 14.696 psia and Tamb is 518.7 oR (59 oF). If the reference conditions at 9000 ft are 10.5 psia and std day @ 9000 ft is 26.9 oF ,

HP @9000'/ HP @sl =10.5/14.696 x(486.6/518.7)^0.5 =0.692

So a 40 hp machine would deliver a rated power of 27.7 hp at these conditions.

So by knowing the ambient pressure and the ambient temperature, you can scale the rated horsepower to any atmospheric condition.

Note that horsepower lapse rate with altitude 3% per 1000 ft would yield a hp loss of 27% and 4% would yield 36%, when the correct lapse is 30.8%.

 

[Transit]

Too much detail.

This is a problem in Density-Altitude.
Subtract 3% for every 1,000 feet above Sea Level and subtract 1% for each 10 Deg F over 60 Deg F.

At 9,000 ft. 9,000/1,000 x 0.03 = 0.27, 1.00 - 0.27 = 0.73.

40 hp x 0.73 = 29.2 hp.

At say 90 Deg F.

90 - 60 = 30. 30 x 0.01 = 0.3, 1.00 - 0.3 = 0.70.

29.2 hp x 0.70 = 20.44 hp if you consider the additional loss because of temperature.

A hp here or there is not going to make all that big a difference.

 

[LoneCowboy]

I'm at 5000 feet
on a non-turbo motor, makes a big difference
on a turbo motor, not nearly as much (or even much at all)
make sure you get a turbo motor

 

[jenkinsph]

I operate both my 110tlb and 4520 at around 8000' frequently and can notice a little difference in the 110 as it is naturally aspirated. Haven't noticed any difference in the 4520 with a turbo though. So I would agree with the comment above about getting a turbo equipped tractor.

 

[E/S]

Elevation loss is why I got a turbo charged tractor. I am at 6000' and could really tell the difference from 1000' to 6000' on my old 245dt.

E/S

 

[Jerry/MT]

Too much detail.

This is a problem in Density-Altitude.
Subtract 3% for every 1,000 feet above Sea Level and subtract 1% for each 10 Deg F over 60 Deg F.

At 9,000 ft. 9,000/1,000 x 0.03 = 0.27, 1.00 - 0.27 = 0.73.

40 hp x 0.73 = 29.2 hp.

At say 90 Deg F.

90 - 60 = 30. 30 x 0.01 = 0.3, 1.00 - 0.3 = 0.70.

29.2 hp x 0.70 = 20.44 hp if you consider the additional loss because of temperature.

A hp here or there is not going to make all that big a difference.

I thought you said the other method was too complicated.
If I subtract 1% for every 10 degrees over 60 oF, I get .0.03 so 1- 0.03 = 0.97. 0.97 X 29.2 = 28.3hp NOT 20.44 hp!

 

[Transit]

I thought you said the other method was too complicated.
If I subtract 1% for every 10 degrees over 60 oF, I get .0.03 so 1- 0.03 = 0.97. 0.97 X 29.2 = 28.3hp NOT 20.44 hp!

So lets try it again.
1% = 0.01, I always use the leading zero to indicate that the units place was not overlooked.
90-60 = 30.
0.01 x 30 = 0.3

I would show you how to use the exponent, the power of ten, but unfortunately this pc does not do well with such notation.
I will explain another way.

30 may be written as 30.00, and may also be written as 3 x 10.00

If you count each place to the left of the decimal point as a + number, 10.00 would be 1.0 and an exponent of +1.
Likewise, 0.01 would be written as 01.00 and an exponent of - 2 by shifting the decimal point to the right two places.
When multiplying, add the exponents. [+1] + [-2] = -1. therefore the value would be 0.1, hence
[3 x 10] x [1.00 x 0.01], rearranging terms, [3 x 1.00] x [10 x 0.01] is [3] x [0.1] or 0.3.
:thumbsup:


You ought to know better!!!!!!

 

[TractorRob]

...subtract 1% for each 10 Deg F over 60 Deg F...

So lets try it again.
1% = 0.01, I always use the leading zero to indicate that the units place was not overlooked.
90-60 = 30.
0.01 x 30 = 0.3
...

0.01 x 30 is 1% for each degree, you said it should be 1% for each 10 degrees.

 

[MtnViewRanch]

What ever size tractor you need to do what it is that you need to do. Spec everything out, as in how much HP is required to operate the implements that you want to use. Then look at tractors that have that HP rating that have a turbo on it and only look at tractors that have turbos on them. As long as you get a tractor that has a turbo on it that has the rated HP for what you want to do you will be a happy camper. No turbo, unhappy camper, end up buying another tractor with a turbo on it.

Oh, did I say that you should get a tractor with a turbo on it.

 

[Jerry/MT]

So lets try it again.
1% = 0.01, I always use the leading zero to indicate that the units place was not overlooked.
90-60 = 30.
0.01 x 30 = 0.3

I would show you how to use the exponent, the power of ten, but unfortunately this pc does not do well with such notation.
I will explain another way.

30 may be written as 30.00, and may also be written as 3 x 10.00

If you count each place to the left of the decimal point as a + number, 10.00 would be 1.0 and an exponent of +1.
Likewise, 0.01 would be written as 01.00 and an exponent of - 2 by shifting the decimal point to the right two places.
When multiplying, add the exponents. [+1] + [-2] = -1. therefore the value would be 0.1, hence
[3 x 10] x [1.00 x 0.01], rearranging terms, [3 x 1.00] x [10 x 0.01] is [3] x [0.1] or 0.3.
:thumbsup:


You ought to know better!!!!!!

You said 1%/10 degree F. 90F-60=30F/10=3; 3 x .01= 0.03!


Your so called "simple method" is confusing you!

I think I'll just stick to the methods I learned in engineering school. They're more accurate and I understand the principles behind them.

 

[FatTire]

OK, so in any case the rule of thumb I was using comes out "about right". Pushing a 72 inch blower through a 3 foot deep wind-compacted drift (or maybe a hard drift broken up a bit by FEL), is 40 hp in the ball park? Does Kubota even have a turbo in that range???

 

[Jerry/MT]

OK, so in any case the rule of thumb I was using comes out "about right". Pushing a 72 inch blower through a 3 foot deep wind-compacted drift (or maybe a hard drift broken up a bit by FEL), is 40 hp in the ball park? Does Kubota even have a turbo in that range???

I don't think there's a stock answer to your question. Check the specs on a variety of snowblowers. They usually specify the HP required. Then if you buy a normally aspirated engine, size it up to deal with the altitude effect. Check the specs on a turboed tractor and match that to the power requirements. The salesfolks will have to do some work to get the altitude data, but they should be able to do that for you. You don't want to find out you're be underpowered AFTER you buy the equipment. Do the research first.

 

[Don87]

Looking for a 4WD tractor for FEL, driveway grading, and snowblowing (72inch width, rear mount) at 9000 ft. elevation. I read once that the power loss is about 3% per 1000 ft. above sea level. If that is true, a 40 hp unit would run like 28 hp. Does this sound anywhere near correct?

How many HP should I be shooting for at this elevation? Not looking to skimp, but also not wanting to go so large that maneuverability becomes an issue...

Read this thread, it will help you.
http://www.tractorbynet.com/forums/owning-operating/176744-tractor-blow-lots-snow.html
The skidsteer he purchased is what you need.


Don

 

[rjkobbeman]

OK, so in any case the rule of thumb I was using comes out "about right". Pushing a 72 inch blower through a 3 foot deep wind-compacted drift (or maybe a hard drift broken up a bit by FEL), is 40 hp in the ball park? Does Kubota even have a turbo in that range???

After a quick search on Kubota's site, it appears you can't get a turbo until you get to 50 HP. However, you have quite a variety in the 50 HP range. Grand Ls, MXs and Ms. The MX5100 would probably be the lowest in price. I have a friend who has one and loves it.