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It is a far from simple explanation. The relationship in the engine is the core of any engine development as it affect many other things.
Thnere are two definitions you need to get sorted:
Energy: The size of the bang
Power: The size of the bang time the number of bangs per minuite
Torque: The amount of twist turning the wheel.
Quote:
Originally Posted by swarwick
;D ;D ;D
I used to do my degree work in the pub.... numbed the pain!!!
The bit that's really going to get interesting is when you ask what's the difference in power v torque between a V twin and a Parallel twin and a Horizontally Opposed twin and does it make a difference where the cylinders stick out in relation to the crankshaft, gearbox and clutch i.e is the V twin a Ducati or a Moto Guzzi configuration............
Or don't think about it since each engine configuration has it's plusses and minus, pro's and con's etc etc, not everyones cuppa tea etc etc etc.
A 90º V-configuration does have perfect primary balance in memory serves me correctly (hence Hardly-Worthits vibrating like road drills with their 45º angle, and Aprilia Mille twins needing a balance shaft for their 60º twin). A conventional in-line 4 has perfect secondary balance, the "Rossi-inspired" cross-plane crank R1s require a balance shaft, something that explains why that engine is about 5kg heavier than the previous generation one... ::)
Here's a suggestion - buy a book and read it!
If i remember correctly, the best balanced engine possible is a 90deg v12.
The issue with balance is that the piston does not follow simple harmonic motion, rather a very spiky and uneven version of. So static balance can be achieved fairly easily, but dynamic balance is harder.
There are books out ther, but they do assume a certian amount of knowledge.
Yeah i'll leave it then. ;DQuote:
Originally Posted by Jon_W
Ok. To finally answer the question.
Be prepared, this could get messy.....
Right, to basics. The reciprocating (piston) engine is classed as a non-flow heat engine, and so is subject to the rules of the Non Flow Energy Equation (NFEE).
This states that Q+W=deltaE which can be shortened to Q+W=U(2)-U(1)
Where
Q=Heat transfer
W=Work (energy)
U = Internal energy
So far so good.... now to add in the "fun"
The idealised petrol engine follows the four stroke Otto cycle. The Pressure - Volume (P-V) diagran illustrates the cycle.
http://upload.wikimedia.org/wikipedi...lo_4tempos.png
A = isobaric (constant pressure) expansion (inlet stroke)
B = adiobatic (constant Q) compression (compression stroke)
C = isochloric compression (constant volume - spark) and adiobatic expansion (power stroke)
D = isochloric expansion (exhaust) and isobaric compression (exhaust stroke)
For the purposes of working out the work done by the engine the isobaric processes can be discounted. But for working out the total output the losses due to friction on these strokes bust be re-inserted.
Lets look at the curved lines of Adiabatic expansion:
So PV^gamma = K where gamma is the aidiabatic index, typically 5/3 for air.
Adiabatic processes assume no heat transfer to or from the working fluid. Hence Q = 0 So going back to the NFEE we can say that:
W = U(2) - U(1) or dW = dU but dW is a partial differential based on Q so this is where it gets complex! I'll not go into the detail but...
We can deine U = anRT
a = (alpha) = 1/2 the number of degrees of freedom
n = amount of fluid in moles
R = universal gas constant (don't ask!!)
T = Temperature
to find the change in U dU = anRdT
We are assuming here that the gasses are ideal and behave. This is a ver close approximation to reality, so we can apply the Ideal gas equation of:
nRT = PV
so dU = a d(PV)
so W = integal(between V(1) and V(2) of P.dV
Inserting the aidiabatic PV^gamma gives us W = integal(between V(1) and V(2) of P(1) (V(1)/V).dV
This becomes a hugely complex equation whic I cannot right in text!!! But the above shows us that the work done in an adiabatic process is described by the area under the curve.
No, we have two adiabatic processes, B and C on the graph. C is an expansion on the power stroke, and B is a compression done on the compression stroke. So the power gioven by each explosion is given as:
W = W(C) - W(B)
or the area between the curves.
So on to the isochloric processes.
With these we can say that as there is no change in V, the piston is not moving, so W = 0
Back to the NFEE, Q = U(2)-U(1) or dQ = dU (dQ is again a partial differential linked to W).
I won't go any further again, but you can see that at this moment all the heat (Q) is added by the ignition (look at the vertical line C) and removed in the exhaust (the vertical line D)
So, these lines represent what goes in and what comes out the exhaust. This is the first stage in determining the efficiency of the cycle.
So, we now have an idea of the work done each cycle, we can determine the Power as described in the previous post.
So to Torque.
Torque is a moment of force described by M=Fd
M = Moment (Torque)
F = Force
d = distance
Now, force can be calculated as Pressure X Area, and we have equations describing pressure. So all we need to do is to work out the mean pressure in the cylinder during expansion and minus it from the mean pressure during compression.
For this PV^gamma=K can be used.
We know V from the piston positon and Gamma so P=K/V^gamma. From that we can work out the average Pressure.
We can now differentiate this to get a figure for the average rate pressure change.
Simples.
What this shows is the physical link between torque and power within the conbustion process. It is far from simple (the above is a gross simplification, in reality ignition and exhaust are not at constant volume, and the exhaust and inlet strokes are no constant pressure, but the sums for that are more than I can write here.... let alone explain in a "simple" format....
Hope this finally answers the question. :)
I think the moderators should lock this thread. Inflicting such pain on the membership should stop... I'm sure there must be swearing going on in here as well ;D ;)
How much fluid is in a mole?Quote:
Originally Posted by Jon_W
Wonder if this little guy can help?
http://upload.wikimedia.org/wikipedi...sAquaticus.jpg
If you squeeze them hard enough you'll find out!! ;D ;D ;D
Stu