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Specific work (We)
- Thread starter RallyeNLD
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kJ refers to the scalar work (and so energy), that is, the scalar product of two vectors: force times displacement in the same orientation or |F|.|d|.cos(teta) where teta is the angle between the two vectors.
Although J is the product of N.m, it is completely different from m.N which refers to the torque, being torque the vectorial product of force and arm and calculated as |F|.|l|.sin(teta) (if teta is zero, the torque is zero, too)
Of course you can multiply torque and angle and you'll find the work of the torque along the angle
fabio vincent
Although J is the product of N.m, it is completely different from m.N which refers to the torque, being torque the vectorial product of force and arm and calculated as |F|.|l|.sin(teta) (if teta is zero, the torque is zero, too)
Of course you can multiply torque and angle and you'll find the work of the torque along the angle
fabio vincent
- Thread starter
- #3
Thank you. I completely understand your theory, but i can't project it on the map I'm trying to read. For simplicity, here is the link to map:
The map shows the potential efficiency increase when using valvetronic. But how can I use this data to predict the efficiency increase of a certain engine, which uses torque as load?
The map shows the potential efficiency increase when using valvetronic. But how can I use this data to predict the efficiency increase of a certain engine, which uses torque as load?
GregLocock
Automotive
Just guessing. The work out from an engine is measured in J. The effective pumped volume of air into an engine is measured in m^3.
So J/m^3 is almost directly equivalent to W/(m^3/s)
Which would be a reasonable normalised measurement of efficiency.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
So J/m^3 is almost directly equivalent to W/(m^3/s)
Which would be a reasonable normalised measurement of efficiency.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
crysta1c1ear
Automotive
A Joule is the work done by pushing for a metre against a resistance of a Newton, and a (dm)^3 is a litre, so the units are identical regardless of whether there are conventions for their usage.
Speed is a word generally used for a scalar quantity and velocity is generally used for a vector quantity, ie the direction of the speed is also known. Both can be in miles per hour or metres per second or whatever. Writing units another way round doesn't change them, but there are conventions with ambigous units like 'pounds' where the same term is used for a pound of Sterling silver (£, or a British government cheque for a pound of silver which they refuse to pay up on), a pound mass (ie a number of kilograms), and the force exerted by that mass (ie a number of Newtons).
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I said before that a Joule and a Netwton Metre are the same units. If we look at what that means in practice, you need a Joule of energy to twist something through a dimensionless angle of 1 if the resistance is a Newton Metre. What is an angle of 1? It's a radian, eg a metre of cirumference per metre of radius.
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Now let's get to what I think might be the crux of your question. I'm going to make up figures so don't blame me if they are silly. Let's say a 1 litre single cylinder firing produces 628 Joules. That is 0.628 kiloJoules per decimetre cubed, 0.628 kJ/dm3.
If it is a two stroke engine, the crankshaft would have to turn through an angle of 2*pi for that one litre cylinder to fire. So the torque it could produce would be 628 Joules divided by 2*pi, ie 100 Joules, which we would more typically call a 100 Newton Metres.
If it is a four stroke engine and we are measuring the energy for a cylinder firing then the 628 Joules would be 2 rotations of the crankshaft, ie an angle of 2*(2*pi) so the torque would only be 50 N.m.
So as I see it, an engine giving 628 Joules per litre when the cylinders fire, would give 100 N.m of torque if it were a 2-stroke, and 50 N.m if it were a 4-stroke.
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You'd also need to be certain that volume meant the same in both cases, eg that you didn't have a European measuring cylinder capacity (including the bit at the top) and an American measuring displacement (swept area, excluding the bit at the top), etc.
As you can see from the volume remark, it boils down to what is being measured, rather than the units themselves, and that is something we can't really know from here, but I think Greg hit the hammer on the nail.
Speed is a word generally used for a scalar quantity and velocity is generally used for a vector quantity, ie the direction of the speed is also known. Both can be in miles per hour or metres per second or whatever. Writing units another way round doesn't change them, but there are conventions with ambigous units like 'pounds' where the same term is used for a pound of Sterling silver (£, or a British government cheque for a pound of silver which they refuse to pay up on), a pound mass (ie a number of kilograms), and the force exerted by that mass (ie a number of Newtons).
=
I said before that a Joule and a Netwton Metre are the same units. If we look at what that means in practice, you need a Joule of energy to twist something through a dimensionless angle of 1 if the resistance is a Newton Metre. What is an angle of 1? It's a radian, eg a metre of cirumference per metre of radius.
=
Now let's get to what I think might be the crux of your question. I'm going to make up figures so don't blame me if they are silly. Let's say a 1 litre single cylinder firing produces 628 Joules. That is 0.628 kiloJoules per decimetre cubed, 0.628 kJ/dm3.
If it is a two stroke engine, the crankshaft would have to turn through an angle of 2*pi for that one litre cylinder to fire. So the torque it could produce would be 628 Joules divided by 2*pi, ie 100 Joules, which we would more typically call a 100 Newton Metres.
If it is a four stroke engine and we are measuring the energy for a cylinder firing then the 628 Joules would be 2 rotations of the crankshaft, ie an angle of 2*(2*pi) so the torque would only be 50 N.m.
So as I see it, an engine giving 628 Joules per litre when the cylinders fire, would give 100 N.m of torque if it were a 2-stroke, and 50 N.m if it were a 4-stroke.
=
You'd also need to be certain that volume meant the same in both cases, eg that you didn't have a European measuring cylinder capacity (including the bit at the top) and an American measuring displacement (swept area, excluding the bit at the top), etc.
As you can see from the volume remark, it boils down to what is being measured, rather than the units themselves, and that is something we can't really know from here, but I think Greg hit the hammer on the nail.
crysta1c1ear
Automotive
It would have been better if I'd said a metre of partial circumference per metre of radius. I'm not even happy with that, but my point is not to actually define a radian, but simply to show it is dimensionless due to calcellation: metres per metre!
That is sloppy too, as I wrote part of the sentence assuming a certain engine size and generalized the other half of the sentence. Sorry.
Let me rewrite it ...
An engine giving 628 Joules per litre when the cylinders fire, would give 100 N.m of torque per litre if it were a 2-stroke, and 50 N.m per litre if it were a 4-stroke.
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