Thursday, 31 January 2013

Engine - Turbos

Engines can be classified into "Naturally Aspirated" (NA) and "Forced Induction" (FI). An NA engine draws air at essentially atmospheric pressure into the cylinders during the intake stroke. An FI engine compresses the air going into the cylinders so that a greater quantity of fuel can be added and the explosion can be made larger, thus generating more torque and horsepower. The F10 M5 is an FI engine.

There are two main types of FI engines.  One is called "Supercharged", where the air compressor is run by fan belts driven by the crankshaft.

Superchargers are often found in American muscle cars such as the 640 HP 2013 Corvette ZR1.

The other is called "Turbocharged". This engine type uses exhaust gasses to spin a turbine which in turns spins a compressor that compresses the fresh air headed for the cylinders.

This is what is used in the F10 M5 and is the essential difference between the NA engine in my old E60 545i making 325 HP, and the FI engine F10 M5 making 560 HP. The engines are very similar otherwise: both 8 cylinders, 4.4L of displacement, 4 valves per cylinder, 10:1 compression ratio, fuel injection, variable valve timing and lift. All the other major differences are to keep up with the increased output (better cooling, better lubrication, higher-strength components, and so on).

There are in fact two turbos in the F10 M5, each feeding its own cylinder bank, but getting exhaust from cylinders on both banks. Each turbo is of the twin-scroll design, meaning the exhaust is fed into two separate streams hitting different parts of the turbine wheel. The main advantage is that the two separate streams do not interfere with one another as much. In particular, you don't want an exhaust pulse from cylinder A to partially hit the turbine, and partially run back up a pipe to cylinder B's exhaust thereby putting unneeded back pressure on the exhaust.

The image below is a cut away of the turbo in a Subaru WRX STi that shows the important components.
The exhaust side is on the right, and we see it is of the twin scroll design because of the two separate inlet channels and the corresponding shape of the turbine veins. Exhaust exits through the centre  The compressor side is on the left. Air is drawn in from the centre, compressed by the wheel, and pushed out through the single channel. In the middle is the bearing assembly. Oil is fed in from the top, directed to the ball bearings, and flows out through the bottom. The extra channels in the body towards the turbine side are for water cooling.

The spinning materials are all very strong and lightweight to allow it to spin up fast and without a lot of internal resistance. Turbos can spin at up to 200,000 RPM. The turbos are manufactured by Honewell Turbo Technologies, formerly Garrett who were pioneers in turbo technologies from California and who powered the very first production turbocharged car, the Oldsmobile Jetfire Rocket from the early 60's.

In the M5, the maximum boost pressure above atmospheric pressure (called "gauge pressure") is 1.5 bars (1 bar is about 1 atmosphere = 14 pounds per square inch - psi). Therefore the air is compressed to 21 psi gauge, or 35 psi absolute at sea level. The actual amount of boost changes as the exhaust flows change and as the so-called wastegate valve is manipulated by the DME (Digital Motor Electronics computer). The wastegate allows exhaust to bypass the turbine wheels in order to control boost pressure.

The M5 turbos are shown in the image below. The exhaust end is facing us, and the exhaust manifold is shown feeding the exhaust to the turbines. The compressors are on the far side, taking air in from the larger pipes in the centre, and directing it out through the smaller ones on top.

A key design issue with turbos is minimizing "turbo lag", the time between when you apply your foot to the gas and when full torque is available. My old turbocharged Volvo S70 T5 had lag. You would push the accelerator and get regular non-turbocharged acceleration, and then about a second later the acceleration would kick up a notch, this time turbo-powered! While a delightful surprise, it did mean you were bit slow off the mark when trying to accelerate hard.

Most of BMW and M's intellectual energy went into solving this issue which is not at all present in naturally aspirated engines and the reason BMW stayed away from turbos for so long. They did a lot of work on this for the N63 engine in the 550i, more still for the S63 in X5 M, and even more still for the S63top in the M5. in fact, most of the development effort in the S63top was to decrease turbo lag over the S63, not particularly to add horsepower. The proof is in the pudding, and there is no noticeable turbo lag on the car.

One reason for lag is because the exhaust gasses have a ways to flow before hitting the turbo. Normally, V engines intake in the middle of the V and exhaust towards the outside of the V. In the M5 motor, these are switched so as to minimize the distance from exhaust ports to turbos that can therefore be nestled on the inside of the V. These exceptionally short paths are shown clearly in the above illustration.

Another cause of turbo lag is the inertia of the turbine and compressor. To minimize this, two smaller turbos are used instead of one larger one, and the spinning materials are all exceptionally lightweight.

The turbo on the M5 is shown in more detail below.
The exhaust turbine (1) takes in gasses from the bottom, turns them 90 degrees (thus spinning the turbine) and shunts them to the exhaust system through the large diameter passages on the left. The compressor (3) is turned via the sealed bearing seat (2). Charge air is taken in from the centre and directed out on top. The vacuum operated wastegate valve (5) and actuator (4) is used to divert exhaust gasses so that they do not hit the turbine in order to control speed and boost.

The following illustration shows the intake airflow.

Air is taken in from the front of the car through the intake snorkel (1) into the intake silencers and air filters (2) past an air mass sensor (3). Cleaned air from the crankcase ventilation system is connected at (4). The turbo then pressurizes the air and sends it to a charge air cooler (6 - discussed in more detail below) and past a charge air pressure sensor (7). The air passes through a throttle valve (8) into the intake manifold (9) and from there into the cylinders. A charge air pressure and temperature sensor is located at (10). A connection from the tank vent valve is made at (11) for dirty gasses to be recycled back into the intake.

Most turbos have a blow-off valve which is used to dump pressure from the charge air system when the accelerator is suddenly released. Normally, this causes the compressed air to suddenly have no place to go, and the surge in pressure can damage the system (called "pump"). However, the combination of the design of the compressor vanes, Valvetronic and computer control make the system insensitive to pump, and hence no blow-off valve is needed. 

To help further reduce turbo lag, when in "Sport" engine mode the wastegate is kept closed longer. This leads to increased fuel consumption as the exhaust is backed up, but increases response as the turbos are kept better spooled up in case they are needed.

One consequence of the compression of the air is that it heats up to over 130 C. To allow more fuel to be mixed in and keep the compression ratio in the cylinders up, the charge air is cooled by two very large water-cooled charge air coolers back to around 50 C. These are especially large to promote better cooling, to de-throttle the intake air, and to reduce the distance to promote faster throttle response. The charge air coolers are the two prominent metal boxes at the front of the engine.

The charge air coolers are water cooled from a low temperature cooling circuit separate from the engine cooling circuit. The same liquid cooling circuit is shared with the DME (Digital Motor Electronics) computers which are sitting on the diagonal at the back near the firewall, one per cylinder bank, with the master DME on driver's right.

The turbos themselves can get very hot with the >1000 C exhaust temperatures, and are also water-cooled from the main engine cooling circuit (the metal pipes sitting on top of the intake system are for coolant). A particular issue with turbos is "heat soak" after the car is turned off. The exhaust manifold is still very hot, and the heat soaks into the turbine bearings, damaging them over time. To prevent this, the cooling circuit and main fan is kept going even after the engine is turned off until the temperature of the coolant reaches a reasonable value.

Due to the standard V8 crankshaft considerations, the design of the crankshaft is a crossplane design (see my blog post V8 Crankshaft). However, this inevitably leads to a less than ideal cylinder firing order, which is responsible for the "burble" you hear in a typical V8. 

The problem with the V8 burble is that the exhaust pulses on each cylinder bank are irregular. This tends to throttle the exhaust flow. In the case of a turbocharger, the irregular pulses lead to a pulsating acceleration and deceleration of the exhaust turbines which is inefficient. To counteract this problem, BMW designed an ingenious (and patented) "crossbank exhaust manifold".
The illustration above gives the official BMW cylinder numbering convention for V8s, and gives the cylinder firing order in brackets. Note how cylinders 8 and 6 (on the same bank) fire one after another, as do cylinders 2 and 1 The crossplane crankshaft forces this firing order (including the two adjacent cylinders 1 and 2 firing one right after the other which causes heat buildup issues on the cylinder wall they share).

However, despite the uneven firing order, the design of the crossbank exhaust manifold ensures that the turbos each get alternate exhaust pulses, and within each turbo, the pulses alternate between the two scrolls.

An overview of the entire system air intake and exhaust is diagrammed below. The front of the car is on top.
The air flows into the intake silencers (4), pass the air mass sensors (5) and into the turbos (6). The air is compressed and sent to the charge air coolers (3) where they pass a temperature and pressure sensor (2), through a throttle (1) and into the intake manifold where the pressure is again monitored (13). From there, the intake valves lift and draw the air into the cylinders on the intake stroke, mix fuel in on the compression stroke and then combust, driving the pistons down for the power stroke. 

On the exhaust stroke the exhaust valves open and the returning piston drives the hot exhaust gasses out into the exhaust manifold where they are crossed over and enter into the turbos (6). A wastegate valve (7) operated by an electro-pneumatic pressure converter (8) is used to slow down the turbines when necessary. The exhaust gasses then flow out to the exhaust through the catalytic converters (11), which includes a control oxygen sensor (9) and a monitoring oxygen sensor (10).

All of this is controlled by two Digital Motor Electronics (DME) computers (12), one controlling each cylinder bank with the one on the right of the diagram being the master.

The wastegate valves on the turbos are operated by vacuum pressure. Electronically controlled electromagnets (solenoids) open and close under computer control. This allows vacuum pressure or atmospheric pressure onto one side of the wastegate control, which moves the wastegate valve open and closed. Because of the turbo-charged nature of the engine, vacuum is sometimes in short supply as compared to a naturally aspirated car which operates with the intake manifold in a constant state of vacuum.

Therefore there is an electric pump (1) that generates the vacuum, and it does not at all rely upon the intake manifold vacuum (or lack of it).

The vacuum system generates and stores energy in a vacuum reservoir (6) to power the power brakes (via connection 2) and the wastegate valves (4) on the turbos (via the electro-pneumatic switch 5).

BMW took quite some to move to turbocharged engines, but the move eventually became inevitable as the only way of increasing output while adding low end torque and controlling emissions and fuel economy. However, BMW was keen to ensure that the driving experience remained consistent with an NA engine, and so went to great lengths, especially in the M5, to eliminate turbo lag.

Tuesday, 29 January 2013

Engine Thermodynamics

The study of engines is really the study of thermodynamics, the study of the movement of heat. The four-stroke engine in the M5 is an example of a heat engine. The explosion of the fuel releases heat as part of a chemical reaction, and that heat energy is converted into a different kind of energy: work - the application of a force over a distance that moves the car.

A very useful tool in understanding how a heat engine functions is the p-V diagram. This diagram plots pressure versus volume inside the cylinder. The pressure in question is the pressure above the piston, the volume in question is the volume above the piston.

Area in the p-V diagram corresponds to energy (work). Pressure is a kind of potential energy. If the result of a pressure is to move a piston so as to change the volume by dV (in cubic inches) and the pressure by dp (in pounds per square inches), the product dp*dV has units of pound-inches, or force over a distance, which is work energy.

Bearing in mind that work corresponds to area in the p-V diagram, the following p-V diagram is for an idealized four-stroke (Otto cycle) engine. "Idealized" means
  • a completely free-flowing intake and exhaust,
  • instantaneous combustion of all the fuel,
  • instantaneous opening and closing of valves,
  • no momentum to the gasses,
  • no gain or loss of heat (called an adiabtic process in thermodynamics).
Along the vertical axis is the gas pressure inside the cylinder (p). Along the horizontal axis is the volume of the cylinder above the piston, varied as the piston moves back and forth (V). The four-stroke Otto cycle follows the path 1-2-3-4-5-6-1-2-3-... and so on, as shown below.

Assuming completely free-flowing intake and exhaust, the pressure in the cylinder is at atmospheric pressure during both the intake stroke (1→2) and the exhaust stroke (6→1). During the intake stroke the volume of the cylinder increases because the piston moves. During the compression stroke (2→3), the valves are closed and the pressure rises as the gas is squeezed into a smaller volume. We'll explain the curved shape between 2 and 3 (and 4 and 5) in a moment.

At 3 the spark is ignited, the fuel is instantly combusted, and heat is released. The released heat instantaneously increases the pressure in the cylinder according to the Ideal Gas Law

pV = nRT

This law says that for any gas, air included (we can ignore the fuel mixed in for now), pressure (p) times volume (V) for a given quantity of air molecules (n) are both directly proportionate to temperature (expressed in degrees Kelvin: Celsius degrees above absolute zero which is -273.15 Celsius). R is a constant of proportionality called Avogadro's constant.

So when T instantly rises as a result of the explosion  the piston has had no time to move, hence V is unchanged, the amount of air in there (n) is also unchanged, so it is pressure (p) that must rise instantaneously. Hence the rise from 3 to 4.

From 4 to 5 is the power stroke during which  work is done to the piston head (Force over a distance). During this time the volume increases as the piston moves down and the pressure also decreases.

Between 5 and 6 the exhaust valves are instantaneously opened and the pressure immediately equalizes to atmospheric pressure (due to the completely free-flowing exhaust). Because of the pressure drop, heat is also lost to the flow of exhaust (but not as much heat is lost as was gained previously, the rest having been converted to work).

Finally the exhaust is pushed out through the exhaust stroke from 6 to 1, and the cycle repeats.

The reason for the curved shape between 2-3 and 4-5 is as follows.

One might naively assume by looking at the Ideal Gas Law above that p and V should vary in a proportionate way. That is, if V goes down by a factor of 10 (as it does in the M5), then p should go up by a factor of 10. But while we see volume going down by a lot, pressure does not rise as much as this. This is because temperate is not constant as air is squeezed. In fact, assuming no heat is gained or lost through the piston walls, then

T = C * p^(2/f+2)

The quantity f is the number of degrees of movement freedom of the gas molecules. Air is mainly made up of N2 and O2, which are both two-atom molecules. They have 5 degrees of freedom (the pair of atoms can move around in 3 dimensions, and the two atoms can move in 2 additional dimensions relative to one another and still not break, making f = 5 for air). Thus temperature (in Kelvin) rises in proportion to the 7th root of pressure squared.

We can then plug this into the Ideal Gas Law to get that

p = C / V^(7/5)

This relation between pressure and volume gives us the curved shape between 2-3 and 4-5.

The crucial part is that from 2 to 3 work is done by the piston, and from 4 to 5, work is done to the piston. The net work applied to the crankshaft in one complete cycle is proportional to the area outlined by 3,4,5,6.

This is the theory of the Otto engine. In practice, none of the idealized assumptions are valid, and so a more realistic p-V diagram for an Otto cycle is shown below.

The more important deviations from ideal are as follows.
  1. During the compression stroke heat is being generated and is being lost through the cylinder walls to the cooling system. 
  2. Towards the end of the compression stroke the spark is fired, because in reality combustion takes time and before it is finished the piston has already hit Top-Dead-Center and is returning while the pressure continues to rise due to the combustion.
  3. During the power stroke more heat is lost to the surroundings, and hence the shape is not the idealized adiabatic one.
  4. Because the valves do not move instantaneously, the exhaust valve must start opening while the piston is still moving down as it must be mostly open before the exhaust stroke happens or even more work will be required to move the piston back.
  5. All of the above change the shape of the work loop, making it smaller than in the ideal situation.
  6. During the exhaust stroke the air is backed up by the exhaust system and the piston must work against that backup. Hence we see the pressure in the cylinder is higher that atmospheric.
  7. On the intake stroke there is always a partial vacuum in the intake manifold, and hence work must be done against that vacuum.
  8. At cylinder Bottom-Dead-Centre the closing of the intake valve takes some time.
  9. The area enclosed by the lower loop is net negative work called "pumping losses". This offsets the positive work done in the upper loop with the net work being the difference between the two.
Actual pumping losses from a multi-cylinder engine are shown below, with each line representing a different cylinder. As you can see, the situation is actually much more complicated than the simple illustrations show!

Pumping losses are the main controllable factor in an engine design. The pumping losses are high both at low engine speed when the incoming air is restricted, thus affecting fuel economy; and at high engine speed when the air cannot go in or be pushed out fast enough, thus affecting both fuel economy and performance.

At high engine speed, pumping losses can be reduced by "de-throttling" the air going into the cylinders allowing the engine to "breathe" more freely. This is done at the air intake, air filter, throttle, intake manifold, and intake valves.

High engine speed pumping losses can also be minimized by freeing up the outgoing flow of exhaust gasses through the exhaust manifold, catalytic converter, muffler, and exhaust pipes.

This freeing up of the airflow helps keep the air pressure the same during intake and exhaust strokes, thus minimizing the area within the pumping loss loop and maximizing the area of the power loop.

Notice how on the intake stroke we are drawing air in at "Intake Manifold Pressure", which is lower than atmospheric pressure. This is because we cannot feed air in fast enough in general for the cylinders and a partial vacuum is created in the intake manifold. On the exhaust stroke, the pressure is slightly above atmospheric pressure, as we are trying to push the exhaust gasses out through the exhaust system which gets "backed up" a little due to the relatively narrow valves, pipes, catalytic converts, mufflers, and tail pipes.

In the normal Otto cycle, the area surrounded by 5,6,7 below is wasted work that is blown out the exhaust valves (heated and pressurized gasses). 

In other words, had the piston stroke been longer, we could have used that work, but as it is, we had to dump it all out the exhaust. It is possible to recoup some of that work and put it to good use. This is the job of the turbochargers. The M5 has turbos that are spun using some of the otherwise wasted energy in the exhaust gasses. The turbines then spin compressors, which are used to compress the air going into the cylinders. The compressed air has more Oxygen and can therefore support a larger explosion.

As it relates to the p-V diagram, the effect of the turbo is to raise the intake manifold pressure well above atmospheric. However, that comes at the expense of also having to raise the exhaust manifold pressure, as there is a pushback from the turbos. The net of all of this is that everything in the p-V diagram is just raised upwards, with just a bit of net loss coming from the less than 100% efficiency of the turbocharger. In other words, even though the piston has to work harder at pushing exhaust out and this robs power, the pre-compressed air entering the cylinder adds back almost the same amount of work to the piston.

However, any small net losses here are more than made up for by the fact that the pressurized air can support a larger quantity of fuel, and hence the rise between 3 and 4 can be much greater as a result with a dramatic increase in net positive work overall.

Sunday, 27 January 2013

V8 Crankshaft

There are two common crankshaft design for V8 engines: flatplane and crossplane. Most road car V8's, including the M5, use a crossplane crankshaft design.
The crossplane is on the left, and the flatplane on the right.

In the crossplane design, opposing cylinders in the V8 attach at all four points of the compass. In the flatplane design, they attach to two-points of the compass, each repeated once.

The pistons attach as shown below for the 90 degree V8 configuration (this happens to be flatplane below - you can tell because the pistons on either end are moving in phase).
The important thing in crankshaft design is to make sure there is no imbalance that leads to a shaking car. The piston heads, pins, and connecting rods are all heavy, and they move very quickly. In a single-piston engine, this would cause vibrations.

Referring to the diagram above, as the crank goes around its circle, the piston moves straight up and down, causing an up and down shaking. As well, the crank pin and a component of the connecting rod are thrown side-to-side, causing a side-to-side vibration. All crankshafts have some counterweights to counteract this side-to-side action of their own rotating parts (in other words, the crank pin and the big end of the connecting rod need to be balanced by a weight on the other side). But the rotating counterweight can never counter the motion of a piston, pin, and small end of the connecting rod traveling up and down.

Thus the point of having multiple cylinders and clever crankshaft designs is to make all of these shakes cancel out, leaving a silky smooth engine with no vibrations at all.

In the flatplane design, for every piston moving down there is an opposing piston moving up to cancel the vibrations. As well, this is symmetrical about the middle of the crankshaft, so neither is there any net torque on the crank.

However, this is not true of the crossplane V8. While the ups and down all cancel each other out, there is a net back and forth rotation about the middle of the crankshaft.

Fortunately, these can be cancelled out by the strategic placement of counter-weights on the crankshaft. While all crankshafts have counterweights to balance their own internal rotating weight, putting in place a large enough weight to also balance the pistons and rods is unique to the V8.
Is is actually a fortunate accident of a 90 degrees V8 engine that these weights do the job. Normally, one could not expect a rotating weight to balance an up and down motion. However, in a 90 degree V8 where two pistons are connected to one crank pin, the weight does double-duty for both pistons at once, and can cancel things out properly.
Once these weights are in place, the crossplane V8 also has first order balance, as does the flatplane, but at the expense of the added rotating weight which means the car cannot rev as high or as fast as the flatplane.

So if they are both the same, but the crossplane is heaver, why ever use the crossplane? The answer is that while neither arrangement has any first-order imbalance, the flatplane has a second order imbalance, whereas the crossplane does not. I'll now explain where this second-order imbalance comes from.
How the up-and-down location of the piston head (at P) relates to the crank angle A from TDC (Top Dead Centre) is actually quite complex, and is given by the formula above, solving for quantity x (l and r are constants of the engine).

One implication of the formula is that the way the piston moves with the crank angle is slightly different at the top of the motion than at the bottom.

In the graph above, looking at the blue plot, I am plotting the piston position in inches around the midpoint of its motion. Notice the shape of the top hump is considerably broader than the shape of the bottom one. To make this clearer, the red plot is the blue one 180 degrees back and flipped over so you can directly compare peak to trough. The green line plots the difference between the red and blue plots. If they overlaid perfectly, it would be flat. It is clearly not flat, and this is where second-order imbalance comes from.

Another way of looking at the red and blue plots above is that I am plotting the position of the two pistons whose motion is opposing one another. This is made clearer by flipping over one of the lines.

Imagine the blue is the motion of the cylinder on one end of the crankshaft, while the red is the opposing motion of a cylinder on the other end of the crankshaft. They almost cancel each other out (no first-order imbalance so no extreme shaking), but not quite as shown by the wavy green line which is the engine shaking up and down as a result of the second-order imbalance. It is called second-order because the frequency of the vibration is twice the crankshaft rotational frequency (note how the small wavy line waves twice as fast as the larger piston ones).

The flatplane V8 has no first-order imbalance, but it does have a second order imbalance that cannot be practically corrected. This is why flat-plane designs can only be used with small light V8 engines found in some sportscars with smaller displacements, shorter strokes, and lighter piston heads and rods (such as F1 racing cars, Ferrari V8s, and the Lotus Esprit V8).
Lotus Esprit flatplane V8
Even given the lightweight engine construction, they will tend to shake you up a bit, but folks say they enjoy it!

With the crossplane design, the second order imbalances actually cancel each other out completely, so it runs a lot smoother as a result, even in larger and heavier engines (such as the M5's relative to those other sportscars). This property of the crossplane crankshaft was only discovered in the 1920's (by Cadillac and Peerless at the same time - they share the patent), and before then all V8s were flatplane, and shook!

Other than the extra weights, there is another problem with the crossplane relative to the flatplane, and that is the piston firing order.

The ideal firing order for any "V" engine is to alternate cylinders firing on first one bank and then the other. This would spread out the heat and smooth out the exhaust pulses which leads to better exhaust flow. The flatplane firing order is like this. Very predictable right bank, left bank, right bank, and so on.

However the crossplane V8 does not have the ideal firing order. There must always be a cylinder fired on the same side at least twice, once per bank, and at least once in the firing order adjacent cylinders will fire in sequence, which leads to excess heat issues on the wall between those adjacent cylinder that must be designed around with extra cooling. The numbering and firing order (in brackets) for the cylinders in the M5 is as follows.


There are therefore heat issues in the wall between cylinders 1 and 2, and both 1,2 and 8,6 are on the same side leading to uneven exhaust pulses. With the crossplane design this is completely unavoidable.

This firing order also gives a V8 its distinctive "burbling" sound owing to the uneven exhaust pulses coming from the right and left cylinder banks each of which is typically fed through to its own tail pipe: ti-ta-tii-ta-ti-taa. Other than the sound, this leads to a slowdown in the exhaust which backs things up and robs power and fuel efficiency.

The S63 engine, however, uses a cross-bank exhaust manifold that take exhaust pulses from both cylinders banks and feeds them to the turbos, and then out the exhaust for a very smooth exhaust flow.

While the crossing over of cylinder banks to exhausts is not unheard of, it requires a complex exhaust manifold, as the exhaust valves are typically on the outside of the V, and crossing them over early is a messy business. Some of the problem can be alleviated with an H or X exhaust that crosses the two exhaust banks over half-way to the tailpipes. This alleviates some of the problem but is still not ideal.

In the S63 engine, the exhaust valves are flipped to the inside of the V, and all exhaust directed to the turbos. The exhaust pulses are crossed over right inside the V before they hit the turbos, all in the space of a foot or so, and the smoothed exhaust outputs of the turbos can each go to its own tail pipe.

So in the M5 you do not hear the V8 rumble nearly as much as a result of this clever design, and it solves all the efficiency problems associated with the crossplane V8 exhaust.

Saturday, 26 January 2013

Engine - Pistons & Crank

The crankshaft is of the V8 cross-plane design and is made of forged steel (alloy C38 = 99% iron, 0.18% carbon, 0.17% silicon, 0.44% manganese, 0.20% copper, 0.03% sulphur) with a hardened surface layer and 6 balance weights.

It is supported in the crankcase by the five bearings shown above. The outer ones are lead-free three-material bearings (three metals with a plastic coating) and the main thrust bearing is at the centre and is a bi-material bearing (two metals with a plastic coating).  The "plastic" coating is a high strength polymer composite consisting of a polymer chain matrix filled with particles of solid lubricants (molybdenum disulfide and graphite).

The connecting rod is forged from sintered steel.
Sintering is a process by which powdered metals are formed under pressure and then carefully heated to fuse them together. This adds strength.  The connecting rods are then fracture-split at the big hole to allow for a better connection after the bearings are inserted and bolted down.
This is done by carefully introducing an imperfection at the desired fracture-split points, and then an abrupt lateral sheering force (i.e., a big hammer) is applied to fracture the rods at the big hole. The ragged surface has more area to support a better connection once the two halves are bolted together again around the crankshaft.

The small end has a formed hole that allows for a more even force distribution on the power stroke (see stress diagram below).

The piston heads are manufactured by the German company Mahle who have been manufacturing pistons since 1920.
They are made of lightweight cast aluminium with an iron particle reinforced synthetic resin coating to allow them to slide across the Alusil cylinders.

The design of the piston head is very carefully considered to work with the Valvetronic III (BMW's variable valve lift technology), the valves and the way they open, and direct high-pressure fuel injection into the cylinder. Computer models were run to design the shape so as to best promote the complete dispersion and combustion of the fuel.

All of the components we have just discussed are beefed up versions for the M series engine, as they have to deal with the considerably higher force (well in excess of 7000 lb) being directed onto the piston heads, down the connecting rods, and onto the crankshaft in support of the > 500 ft-lb of torque the M5 generates at the crank (the base N63 engine generates only 440 ft-lb, so everything can be weaker and hence lighter for better fuel economy).

Friday, 25 January 2013

Engine - Block

One of BMW's core competencies is manufacturing the engines. In particular they create their own engine blocks.

The engine block of the S63top is made from a die-cast aluminum alloy called Alusil. Alusil (Al17Si4CuMg) is an aluminum-silicon alloy which contains approximately 78% aluminum and 17% silicon. These aluminum-silicon alloys, when etched, will expose a very hard silicon precipitate. The silicon surface is porous enough to hold oil, and is an excellent bearing surface (i.e., for tightening things down onto).

The block weighs 220 lb, it has 8 cylinders longitudinally arranged in a 90 degree V. Each cylinder has a bore (diameter) of 8.83cm and a stroke (piston travel) of 8.90cm, having an overall effective displacement of 4.4L.

The cylinder heads (labelled 3 below) is a new M development that contains integrated air ducts for crankcase ventilation (clearing out the exhaust gasses that blow past the pistons into the crankcase).

The parts of the block are as follows.
  1. Cylinder head covers of very light heat-resistant plastic.
  2. Cylinder head cover gaskets help to keep the seal between the cylinder head covers and cylinder heads air tight.
  3. Cylinder heads into which the valves, camshafts, spark plugs, and fuel injectors are installed.
  4. Cylinder head gaskets to keep the seal between the block and the heads air tight.
  5. Crankcase that holds the crankshaft and cylinders.
  6. Upper oil pan gasket.
  7. Upper oil pan where the oil pump is located.
  8. Lower oil pan gaset.
  9. Lower oil pan that seals the bottom of the oil pan.
  10. Timing chain guides for guiding the two independent timing chains that connect the crankshaft to the camshafts.
  11. Crankcase cover.

Thursday, 24 January 2013

Power Losses

When Horsepower or Torque are quoted for a car, you have to be careful what it is that is being quoted.

Because of power losses between the cylinders spinning the crankshaft and the wheels, power is going to be different depending on where you measure it and how you measure it. 

One method of measuring the power at the wheels is by using a wheel dynamometer. This is what tuners do to see how effective their performance upgrades are. The fully assembled car is placed is such a way that the rear wheels spin a cylinder of a given mass. From the cylinder’s acceleration, the torque can be measured and the power derived. This is called wheel horsepower.

Here are some fellows from iND Tuning putting a brand new stock F10 M5 on a wheel dyno in order to get a baseline to see if the new bits and pieces they will install do what they hope them to. Their results are as follows.

This graph shows peak engine Horsepower as measured at the wheels to be 527 HP. This is shy of BMW's stated claim of 560 HP. Is BMW cheating us? Far from it. This is wheel horsepower  Wheel horsepower is usually down by at least 13% from crank horsepower, which is what manufacturers quote. So if there really is a 13% loss before it gets to the wheels, that means the engine must be pushing out peak horsepower of 605 HP and peak torque of 547 ft-lb at the crank! Similar results have been reproduced by others, and this is fairly consistent across BMW's new turbocharged engines. We don't know why BMW is understating power...
The way car manufacturers have all agreed to quote their numbers is by using the SAE (Society for Automobile Engineering) standard J1349 which specifies a method of measurements called SAE Net, and that they have a certified SAE person standing by, overseeing the test, and signing off on the results. They are not allowed to report anything more than this, but evidently can report less.

SAE Net is where the engine is removed from the car, and completely detached from its transmission.  The standard requires that the car’s actual air filter and muffler be attached. Without these, it is called SAE Gross. Also attached and draining power is whatever else is required to make the engine run (for example, oil pump, coolant pump, alternator, and so on).  The torque is then measured at the crankshaft.

The photo above shows the BMW E60 M5 engine being dyno tested by the tuner Dinan. It is going full out, so its exhaust headers are glowing purple with the heat!

This method of measuring results in what is called Crank Horespower, Shaft Horsepower, or Brake Horsepower, because the device used to measure the torque at the crank is based on an invention by Gaspard de Prony in 1821 called the de Prony Brake. The device that measures the power is called an Engine Dyno as opposed to the Wheel Dyno we discussed first.

What causes the difference between crank and wheel horsepower are power losses in the transmission. Anything along that path that bends, generates heat, or makes noise is sucking power. So do the tires themselves deform, heat up, and make squealing noise, sucking power before the road sees it.

Horsepower is Torque

Recapping from Horsepower - Part 2, the force pushing the car forwards (F) comes from the engine torque (T=500 ft-lb) multiplied by the gear ratio (gr=15) to get wheel torque (T*gr=7500 ft-lb). The wheel torque is then converted to pushing force (F=6667lb) at the tire contact spot, which pushes with a force that is larger the shorter the "lever arm" (the tire radius, tr=13.5"=1.125').
  • F = T*gr/tr
    • (see my little tutorial on Math if you are freaked out by this equation!)
Power (P) is Work (W) per unit time, and Work is Force (F) applied over a distance (d).
  • P = W / t
  • W = F * d
  • P = F * d / t
But a distance per unit time (d/t) is speed (S), so another way of saying this is that Power (P) is a Force (F) consistently applied at Speed (S)
  • P = F * S
So if I'm able to continue applying a force of say 1167 lb to the wheels while travelling at 180 mph, that corresponds to 560 HP.
  • 180 mph = 180 * 5280 feet per hour = 180*5280/60 feet per minute
  • Power = 1167 lb * (180*5280/60) ft/min =18,485,280 ft-lb/min
  • 1 HP = 33,000 ft-lb/min
  • Power = (18,485,280/33,000) HP = 560 HP
If we can apply that 1167 lb while still moving that quickly, well that's power my friend! At that speed, the air starts pushing back with that same force of 1167 lb, and that defines the top speed you can go, when the force pushing you forward is the same as the force holding you back, then you won't accelerate or decelerate, you'll just strain against the wind at top speed.

Another way of looking at it is say I am in a tiny old Renault Cinq (my first car, thank you very much!) and I have it running at 60 mph on the highway.

If the most force I can muster in top gear with the engine running so fast is say 500 lb, then my little car is only generating 80 HP. Say I saved my pennies and installed a sporty new exhaust which cranked my horsepower to 100 HP at those same engine revs. Then at that same 60 mph speed I can now generate 620 lb of force, which is an excess of 120 lb over the 500 lb of air resistance that holds me back at that speed, which means I can speed up again past 60 mph, maybe even all the way to 70 mph!

So power and top speed are very related. Power is torque delivered at speed.

So how is speed (S) related to engine RPMs? Speed is distance per unit time. RPMs is rotations per unit time. So we have the time part, but how do rotations relate to distance?

If the engine rotates at a certain RPM, then the tires rotate slower due to the gear ratio, gr times slower in fact. So say the M5 is making 3000 RPMs at the crankshaft. In first gear that translates into 1/15th of that at the wheels, so the wheels are only spinning 200 times per minute.

The distance travelled each minute is the number of times the wheels rotate around each minute, times the perimeter around the edge of the tires.

In the picture above, if the wheels rotate around once, the car would go the distance indicated, equivalent to once around the perimeter of the tires.
  • Speed = (wheel rotations per minute) * (tire perimeter)
The wheel rotations are RPM/gr. The tire perimeter is PI (3.14) times the tire diameter, which is twice the tire radius, so PI*2*tr. Bringing this and the other two relations all together we get
  • S = (RPM/gr) * (PI*2*tr)
  • F = T*gr/tr
  • P = F * S
So power is proportional to force and speed. Force goes up with gear ratio and down with tire radius. Speed goes down with gear ratio and up with tire radius in exactly the same proportions. What that means is that the tire radius and the gear ratio both get cancelled out, leaving
  • P = T * RPM * PI*2
Very cool! If T is in ft-lb and RPM in revolutions per minute, then P is in units of ft-lb/min. Recall that Watt's draft horse could do 33,000 ft-lb/min, so the P above needs to get divided by 33,000 to get to Horsepower.
  • P = T * RPM * 2*PI/33000 
Or, more simply,
  • P = T * RPM / 5252 
So we can see that the torque and power curves are really sort of the same thing. If we know the torque at any RPM, we know the power of the car at that RPM as well. Remember that power is force at speed. Torque is "circular force" and RPMs are "circular speed", so it's natural that power is circular force times circular speed, as we showed above.

Math and Units

Now is a good time to take a little side trip to discuss math and units, as I'll use be using some math in this blog from time to time (and already have a bit in an earlier post).

Sometimes I will say X is proportional to Y. For example, for a boingy spring, force (F) is proportional to how far the spring is pressed (d).

That means that as the quantity 'd' (distance the spring is pressed) gets bigger, the force (F) gets bigger by the same factor. So if d doubles, the force doubles. If d is 5 times greater, the force is 5 times greater. In math, this is stated as
  • F = C*d
Here '*' is the symbol used to represent multiplication (common in computer languages because the times symbol 'x' is too easily confused with the letter 'x'). "C" is a constant of proportionality. It's some fixed number, such as 12 for example. Really, all that the C is useful for is converting from one unit to another. There is always a way of defining a new unit such that the C goes away.

For example, energy (E) is force (F) applied over a distance (d)
  • E = F*d
This means that energy is proportionate both to force and to distance. If force doubles, energy doubles. If distance doubles, energy doubles. If they both double, energy doubles and doubles again, so it is 4 times greater. If we want to express Energy in food calories, force in pounds, and distance in inches, this becomes
  • E(in food calories) = 0.000027 * F(in pounds force) * d(in inches)
Useful if you are lifting weights to lose weight!

However, if we use SI (System International) units, where the base units are Kilograms, Meters, and Seconds, then energy is expressed in Joules, force is expressed in Newtons, and distance is expressed in Meters, and in this case the constant goes away, getting us back to
  • E(in kg) = F(in N) * d(in m)
No constant is necessary.

The moral of the story is that the constant can always be done away with if we define the units appropriately. So say I really like pounds and inches, then I can define a new unit of energy called the Dan, and tell you that 1 Dan is defined as 1 lb*in, such that
  • E(in Dan) = F(in lb) * d(in in)
Now are you are left with is the problem of converting from Dans into more familiar energy units, but that was what the constant was for:
  • 1 Food Calorie = 0.000027 Dans
Units are never fundamentally important to understanding something, though, they are good for calculating things, so we won't worry about them too much in this blog.

We also sometimes say that X is inversely proportional to Y,  so that if Y doubles, X halves. In math,
  • X = 1/Y
where the symbol '/' represents division.

There are other, higher order relationships as well where something varies as the square or cube of something else. For example, the force is gravity is proportional to the mass of each of the two things, and inversely proportional to the distance squared. With appropriate choice of units,
  • F = m1 * m2 / d^2
The '^' symbol means "to the power of", so d^2 means d squared, or d*d. So if the distance doubles, the force falls off by a factor of 4.

Raising something to the power of a negative number is like dividing by that thing, so this can be re-stated as
  • F = m1 * m2 * d^(-2)
There we go. That's the math we will need for the time being.

Wednesday, 23 January 2013

Horsepower - Part 3

The most frequently looked at performance charactersitic of an engine is its torque and power curves. These are shown below for my old E60 545i and my new F10 M5

The vertical units are either ft-lb or HP depending on if you are looking at a torque or a power curve. The units along the bottom are engine revolutions per minute.

Where the curves stop is called the "redline" because the car's tachometer has a red line painted there. In the old days, if you took a car past its redline it would just blow up. Nowadays, the computers cut the gas and don't let you blow up the engine (it would cost BMW a lot in warranty replacements otherwise). The redline of my old car was 6500 RPM, which is fairly standard. The redline on the M5 is 7200, which is very high for a street car, but not at all high for a race car (F1 cars rev up to 18,000 RPM's  - but they use up 4 engines every racing season - my M5 has to last me ten years).

The torque curves really reflect the engine design. The 545i has a definite plateau between 3500 and 4000 RPM, and drops off on either side of that. The turbocharged M5 is engineered to have a broad flat torque curve all the way from 1500 - 5750 RPM.

In terms of driving, in the E60 if you want to accelerate you need to get the car up to around 3600 RPM, which is high enough to scare dogs and little children (they tend to turn around, look at you, and think "what an idiot"). By contrast, in the M5 you feel the torque available to you whatever your RPMs, in whatever gear you happen to be in.

When accelerating hard, you generally want to run the car out to its redline in every gear. On the other hand, you might think that shifting earlier, before the torque drops off would be better? You would be wrong. Even though the torque curve starts dropping off, its made up for by the torque division penalty you would have to pay to shift to the next gear up.

For example, if we accelerate the M5 hard out to its 7200 redline in 2nd gear, we are down to making "only" 400 ft-lb of engine torque which turns into 3200 ft-lb at the wheels due to the 8:1 gear ratio. If we then shift to 3rd gear, the engine gets dragged down to 4500 RPM (because the new gear ratio is 5:1, and the inertia of the speeding M5 is such that the wheels are not going to slow, so the engine must).

You might think, "Awesome! back up to maximum torque at 500 ft-lb, now we should get a burst of acceleration!" Nope, because the wheel torque suddenly drops to 500 ft-lb * 5 = 2500 ft-lb from the 3200 ft-lb it was at only moments ago (when it was at the "bad" part of the torque curve, but we were at 8:1 torque multiplication). And there was no point shifting any earlier than redline, we just would have taken that torque multiplication penalty sooner and enjoyed less torque from an earlier point than holding out to redline.

The curve that is actually telling you this story is the power curve. The rule is that you should shift early only if doing so gets you more power in the new gear than in the old, otherwise, hang on. Because the M5 dips so little towards the end, this situation does not really come up at all, as is the case in most good cars where the gear ratios are well-matched to the engine.

Some people swear by the torque curve, and say it is all important. Others swear by the power curve. Both camps holding any such opinions are just plain dumb, because the two curves represent exactly the same thing. One can be fully derived from the other. They cannot be varied independently at all.

  • Power (in HP) = Torque (in ft-lb) * RPM / 5252
What they really mean is that some things are easier to see in the power curve, and others in the torque curve, and they value the things that they see in each of those curves. Some people value lots of "low end grunt". A high flat torque curve hilites that. Others value the ability to go very fast. The heights of the power curve hilite that. Yet others value a high redline, but only if the power curve does not dip too much (or you would always be shifting before hitting that redline and it may as well be earlier).

In the case of the M5, it has a relatively high redline of 7200 RPM, and it holds its power right up to it. When compared to a car with a lesser redline it means the BMW will be torquing along at high multiplication factor long after the other car has had to shift up and lost its accelerating power to the new gear ratio, even if its engine torque is higher (goodbye Cadillac CTS-V!).

Horsepower - Part 2

In the previous blog post we discussed Mass, Acceleration, and Force, marching towards the notion of Horespower. In this post we will discuss Work, Power, and Torque.


Energy in the form of Work means exerting a force over a distance. For example, lifting a 4300 pound M5 5 feet straight up expends 21,500 ft-lb of energy (5 ft * 4300 lb = 21,500 ft-lb).

See how I casually mixed mass and force? The force required to lift a 4300 Pound (mass) M5 at the surface of the earth is 4300 Pounds (force). If I was lifting that same object the same distance on the moon, I would only need 4300/6 = 717 Pounds (force), and hence I would only have done 5ft*717lb = 3585 ft-lb of work on my 4300 Pound (mass) M5 instead of 21,500 ft-lb.

Now I can hear some of you clever car folks saying, "hold on, isn't foot-pounds the unit of torque?" It is indeed, and this isn't due to the confusion of force and mass, it’s due to the fact that horsepower and torque are very closely related, as we’ll soon see, so stay tuned.


This brings us to Power, as in Horsepower.

Suppose I did 66,000 ft-lb of work on my M5 by pushing it up a slight incline with a force of 100 lb over a distance of 660 ft. My energy expended was 66,000 ft-lb, whether it took me two minutes to push it that far or ten minutes.

However, if that was accomplished in ten minutes, the Power I used was 66,000 ft-lb / 10 min = 6600 ft-lb/min. If it was accomplished in two minutes, the Power would have been 66,000 ft-lb / 2 min = 33,000 ft-lb/min.

Doing the same work in one fifth the time means I need five times the power to get it done.


The unit "ft-lb/min" is a unit of power. So is the "Horesepower" a unit of Power.

A standard Horsepower was a unit that was introduced as a marketing term by the Scottish engineer James Watt in 1782.

He used the unit to compare the power of his new-fangled steam engines
with the power of the draft horses they were intended to replace.
He measured the power output of a few horses, and defined his unit of a Horsepower to be his best estimate of the maximum power that a draft horse could generate over a reasonably long period of time. He settled on using a standard definition of 33,000 ft-lb/min. 

So pushing my M5 up a slight incline by using 100 lb of force over a distance of 660 ft in 2 minutes means I had to use 1 Horespower to do it. In other words, that’s about how far and how fast a big draft horse could do it at a sustainable pace.


Torque is a measure of twisting force. Say I want to unfasten a wheel nut. I get out a tire iron that is 1 ft long and I step down on it applying 100 lb of force. I am applying a torque of 100 ft-lb to that nut. That’s how much twisting power I am applying.

Say the nut does not move. I can stand on the tire iron with both feet and apply 200 lb, still at 1 ft from it, and the torque is now 200 ft-lb.

Alternatively, I can go to my tools and get a longer tire iron, say one that is 3 ft long. If I now step down on that tire iron with the original 100 lb at the end of it, I am applying 300 ft-lb of torque.

The longer the tire iron, the more leverage I get, and the more torque I am applying throughout the tire iron.
So torque is a measure of "twisting force" and is a force applied using leverage a certain distance away from the centre of rotation.

In the diagram above, both hands are applying the exact same torque to the wrench:
  • 1 ft * 20 lb = 20 ft-lb
  • 2 ft * 10 lb = 20 ft-lb
In either case the wrench "sees" the same torque, all along its length (including at the nut).  In fact, if hand A above were pushing upwards with a force of 20 lb, then hand B would need to pull down with a force of only 10 lb to resist it. One hand is putting 20 ft-lb into the wrench, and the other is exactly resisting that with his own 20 ft-lb of torque, applied in a different manner.

So Torque is really just a “leverage-independent” way of stating force, and happens to have the same units as Work for a very good reason.

Engine Torque

Car engines generate Torque. That’s their job. Recall that a piston moves because there is pressure applied to its surface as a result of the gasses super-heating when ignited. An average gas pressure in the cylinder on the pwoer stroke is 800 psi (pounds per square inch). A cylinder with a bore of 3.5in has an area of pi*(3.5/2)^2 = 9.6 square inches. That means there is 800*9.6=7700 lb of force being applied to each cylinder during the power stroke.
How far the crank moves right to left is the same as the stroke of the engine (how far it moves up and down), which is 8.83 cm. Half of that is 0.145 ft. If the piston applies that force on average 0.145/2 ft away from the crank, that means the engine’s torque is approximately 0.145/2 ft * 7700 lb, or about 560 ft-lb. That’s likely about the torque that the M5’s cylinders generate (I made up some of the number to have it come out right, but they are all in the right ballpark).

Wheel Torque

The engine torque is transferred to torque at the wheels through the gearbox

and the differential (or "final drive").
We will discuss these in considerably more detail in a later blog post. For now, suffice it to say that the engine torque is transferred and transformed by means of gears to the wheels.

Here's how gears work.
Imagine gear A is attached to the engine crankshaft, and gear B is what turns the wheels. We see that A is going twice as fast as B. This is because B has twice as many teeth as A. The force is transferred at the point where gear A meshes with gear B. If the torque on A is 500 ft-lb, and gear A is 1ft in diameter, then the force pushing on the teeth of gear B is 500 ft-lb / 0.5 ft = 1000 lb. Because gear B has twice the teeth of gear A, it must have twice the perimeter, and therefore twice the radius. So gear B is 2 ft across. The force of 1000 lb is therefore applied 1 ft from the centre of gear B, therefore the torque on gear B is 1000 ft-lb, or double the torque on gear A.

That's the general rule of gears, no matter how complicated is the gear train. If the gear slows down the spinning by a factor of X, then it multiplies the torque by that same factor of X.

In 1st gear on the M5, every 15 turns of the engine crankshaft turns the wheels once. Therefore the torque is multiplied by a factor of 15 when in first gear. That means that if the torque generated by the pistons onto the crankshaft is 500 ft-lb, the the torque transmitted to the wheels is 500*15 = 7500 ft-lb, evenly divided between the two rear wheels, when in 1st gear (I am ignoring any frictional losses along the way that wind up as noise and heat for the time being, but will come back to that later).

The rear wheels and tires together have a radius of about 13.5 inches, or 1.125 feet, therefore the force exerted by the rubber on the road is 7500ft-lb/1.125ft = 6667 lb of force driving the car forward.

Given the mass of the car is about 4300 lb, then if the tires were perfectly sticky it should accelerate forwards at 6667/4300 = 1.55 g’s, which means it is speeding up by 34 mph each second, which would get it to 60 mph in under just 2s. Woo hoo!

Unfortunately, the M5 is traction limited off the line, as discussed in the previous blog post, and takes about 4s to get to that speed. Boo hoo!

A good all wheel drive with the same torque could theoretically hit sub 3s. The next iteration of the Mercedes-Bens E63 AMG is slated to have 4 wheel drive, so we’ll soon see if this will be the case!

Wasted Torque?

Is all that torque wasted? It certainly is in 1st gear when starting from a standstill. But the exact same engine torque is available in all gears.

For example, in fourth gear the ratio is about 4 to 1, so the torque at the wheels is 500*4 = 2000 ft-lb which means we are no longer traction limited, but rather engine torque limited. Still, the 2000 ft-lb should accelerate us at around 0.4g’s at highway speeds, meaning that we can increase our speed by about 10 mph in a second and zip past that Caddy!