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DCS Fw 190 D-9 Flight Manual


SimFreak

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Water-methanol injection system in Jumo is not much different than systems in other engines, US, Brit, Russian; engines for planes, sport cars and whatnot. Basic concept - cooling the charge to avoid pre-ignition and increase its density is always the same - all these engines operate by the same laws of thermodynamics after all ;). So all comments above are more or less correct and say the same thing.

 

I agree however, that the quoted sentence in the manual is poorly worded, because:

a) it doesn't really explain how and why injecting water in reciprocating engine works;

b) it's confusing by suggesting that cooling any piston engine in any way somehow allows it to "suck" more air and mysteriously produce more power. If a reader of the manual is not familiar with history and details of high-performance piston engines, his natural reaction will be - "WTF???".

 

Nobody asks to include a long, detailed MIT-style thesis on the subject in the manual, but if only "more air" was replaced by "more dense air" and "cooler engine" was replaced by "cooler charge / mixture", with another two or three explaining sentences thrown in, that part of the text would be much more understandable.

 

Thank you. We are on the same page. (no pun intended)

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b)cooling any piston engine in any way somehow allows it to "suck" more air and mysteriously produce more power.

 

Cooling an engine does allow it to produce more power. As HP = Thermal efficiency * fuel flow * BTU released by fuel combustion/ 42.4 ( 1HP = 42 BTU per min)

 

A cooler engine is running at a greater mass air flow per unit of time, compared to a less thermal efficient engine. As a larger vacuum is created on the more powerful down stroke of a more thermal efficient engine.

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Cooling an engine does allow it to produce more power. As HP = Thermal efficiency * fuel flow * BTU released by fuel combustion/ 42.4 ( 1HP = 42 BTU per min)

 

A cooler engine is running at a greater mass air flow per unit of time, compared to a less thermal efficient engine. As a larger vacuum is created on the more powerful down stroke of a more thermal efficient engine.

 

The problem with statement above is - the efficiency of the engine gets lower with cooling, as less input heat from the burning fuel gets converted to mechanical work. Thermal efficiency You quoted above equals, for any heat engine, 1-(heat expelled)/(heat absorbed from fuel). Most of the heat expelled goes away with exhaust, some of it with cooling and the rest with mechanical friction and imperfections of combustion process. If we lived in Narnia, we could have exhaust gases leaving system at absolute zero temperature and "neveroverheating" pistons and cylinders (thus none cooling required whatsoever) :D. Sadly, real world ain't working this way.

 

I agree we strive for as high mass flow rate as we can, but cooling the engine doesn't have much to do with it. Sure, If I run preheated air through some cool intake system, some of the unwanted temperature will get carried away, but that's NOT what MW50 (or any water injection system for that matter) was invented for, while discussed sentence in the draft manual suggests otherwise. Crucially important factor here is cooler, denser air/fuel mixture itself and this is what's mostly responsible for higher power output.

i7 9700K @ stock speed, single GTX1070, 32 gigs of RAM, TH Warthog, MFG Crosswind, Win10.

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Browsing the manual, I was very surprised to see 2 groups of switches under covers on the cockpit's right hand side. On D-9 the rear group of switches was deleted, and there is actually some proof for that. My question to ED's developers would be: what documents do you have that show on D-9 there are 2 groups of switches under covers? I would be very curious to see it.

 

Also I was surprised by several switches there. For example, the Außenbord switch. The last Fw 190 version to have this switch was A-5, it was deleted on all newer versions.

 

So the question is: where are those switches? Or is there a master switch for eletric circuits somewhere?:helpsmilie:

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Question for me is also handle on left console pos.1 (MW-50 to Fuel Handle of water-methanol tank) and switch pos.9 (Radio self-destruction button) I never seen it before. What is the source for this????

 

Both controls appear in a Fw 190 D-9 technical manual (Lehrmittel) from January 1945.

 

If the control "MW-50 to Fuel Handle of water-methanol tank" really looks like that, I haven't yet seen any evidence, and probably nobody really knows.

 

"Radio self-destruction button" - in reality that is IFF (FuG 25a) destruction button, not radio. I think it is safe to say that on the real plane they did not put for this a switch similar with the Netzausschalter (electric-kill switch)...

 

Nobody would put a big, accessible switch like that that could be pressed inadvertently and destroy the plane's IFF.

 

http://forums.eagle.ru/showpost.php?p=2008632&postcount=12

Here I pointed that the switch was most likely under a metal guard, and I also indicated the switch type according to Fw 190 A-8 tech documentation, I am reasonably sure they didn't changed it on D-9, after all is the same IFF type. Search the switch type on http://www.deutscheluftwaffe.de/ , actually it is a simple toggle switch.

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The problem with statement above is - the efficiency of the engine gets lower with cooling, as less input heat from the burning fuel gets converted to mechanical work. Thermal efficiency You quoted above equals, for any heat engine, 1-(heat expelled)/(heat absorbed from fuel). Most of the heat expelled goes away with exhaust, some of it with cooling and the rest with mechanical friction and imperfections of combustion process. If we lived in Narnia, we could have exhaust gases leaving system at absolute zero temperature and "neveroverheating" pistons and cylinders (thus none cooling required whatsoever) biggrin.gif. Sadly, real world ain't working this way.

 

You're interpreting the equation incorrectly.

 

Firstly, an internal combustion engine doesn't convert heat to mechanical energy. It's the increased pressure, as the fuel air mixture expands during combustion, that drives the engine. Thats where the potential energy of the fuel is converted to the mechanical energy which drives the pistons. In an internal combustion engine, heat is wasted energy, as it's energy that's not being converted to mechanical energy.

 

Secondly, increased thermal efficiency doesn't lower the energy potential of the fuel, which is what BTU is denoting in this case. BTU is a measure of energy potential of the fuel. A pound of gas burned releases about 19,000 BTU. 1 horsepower = 42.4 BTU's per minute or 2545 BTU's per hour.

 

Cooling the engine improves thermal efficiency, TE. Meaning less of the energy potential is wasted as heat, as HP= TE * Fuel Flow (Pound per hour) * BTU's of the fuel/2545.

 

So lets look at the Jumo 213 a-1, using the data posted by Yo-Yo

 

Peak HP at sea level is 1770, fuel flow is 1054.58 lbs per hour.

 

1770 HP = TE * 1054.58 * 19,000(BTU's of 1 lbs of gas, aka the energy potential of fuel)/ 2545 (BTU per HP per Hour)

 

1770HP = TE * 1054.58 * 7.46

 

Solve for TE, 1770/1054.58*7.46

 

.225 TE

 

1770 HP = .225 * 1054.58 * 7.46

 

So what does this say? It says that 77.5 percent of the potential energy of the gas is wasted. If the engine was 100% efficient in converting the potential energy of the gas it would produce 7867.166 HP.

 

So "the mysteriously generated 100 hp", generated by cooling, is very easy to find. Via a very minor increase in the thermal efficiency of the engine. Which is provided by the MW50.

 

To get an extra 100 hp out of the engine all we need to do is improve the the thermal efficiency of the engine by .013

1870 = X *1054.58* 7.46

X =.238

.238 - .225 = .013

 

 

I agree we strive for as high mass flow rate as we can, but cooling the engine doesn't have much to do with it. Sure, If I run preheated air through some cool intake system, some of the unwanted temperature will get carried away, but that's NOT what MW50 (or any water injection system for that matter) was invented for, while discussed sentence in the draft manual suggests otherwise. Crucially important factor here is cooler, denser air/fuel mixture itself and this is what's mostly responsible for higher power output.

 

When we talk about a "denser air" we're discussing Volumetric Efficiency (VE) of the engine. That is amount of air the engine ingests compared to the theoretical maximum.

 

While greater Volumetric efficiency does provide greater power. They don't necessarily provide as much horsepower per unit of efficiency as TE gains do. VE gains also run into issues of limiting returns. The compressibility of air and fluid dynamics limit how much air you can stuff into a cylinder, and makes it exponentially more energy expensive to create linear increases in density.

 

Lets look at the 213-A again, this time in terms of VE. Lets see how much more VE is necessary to derive the extra 330 HP gained from the MW 50 system. MW50 is reported to boost HP to 2100, compared to max 1770 HP.

 

Here we calculate the required VE necessary to generate the HP at the RPM

 

V-213 A Specs

RPM 3250 @ max power

2136 inch Displacement

Brake Specific Fuel Consumption of the 213-a is .59

BSFC= Fuel flow/HP

1054.58 (fuel flow)/1770(horsepower)=.59

 

REQUIRED VE = ( 9411 x HP x BSFC ) / (DISPLACEMENT x RPM)

 

At max power, 3250 RPM

1.41=(9411*1770*.59) / (2136 * 3250)

 

VE is over 1 meaning with have to compress the air, ie supercharge.

 

At Max RPM 3250, with MW 50

1.67=(9411*2100*.59) / (2136 * 3250)

 

Meaning, you need an 26% increase in volumetric efficiency, to generate the extra 330 horsepower output, when the MW50 is active in the 213. That number is so large that it's likely that at least 1/3 to 1/2 of the increased HP output, due to MW50, is a result of the smaller thermal efficiency gains. Which provide more HP per increase in efficiency than VE gains. In the case of this engine a TE efficiency increase of 1% yields 76hp compared to VE increase of 1% which yields a increase of 12hp.


Edited by Curly
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I was working off the Data posted in this thread. http://forums.eagle.ru/showthread.php?t=114094. I think you translated the data there. I think he dismissed the 1900 in that thread as experimental http://forums.eagle.ru/showpost.php?p=1878676&postcount=36

 

 

The manual also states "The powerplant consists of a Jumo engine that delivers approximately 1,776 horse power at 3,250 RPM. This could be further increased to 2,240 horse power by the use of MW-50 water-methanol injection. Maximum emergency power in level flight was 1,600 horse power at 3,250 RPM."

 

Also of interest is once you have the required VE you can compute the required manifold pressure to achieve it.

 

VE * 29.92 = Manifold absolute pressure in inches of mercury, HG.

So for our 1770 Hp 213-a with a VE of 1.41 we need 42 HG of manifold pressure, or 1.45 ATA.

 

As 1.41*29.92= 42.18

Convert HG to ATA 42.18/28.95 = 1.45 ATA

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  • ED Team
Engine Temp – I would assume they are referring to the intake temp and thus the temp of the charge in the cylinder prior to compression and ignition cycle. By adding the water into the intake you are effectively cooling the pre charge of air and also increasing its density. This means you drive the engine harder (more rpm’s) without premature detonation of the charge. This was the German answer to not having as much access to higher octane fuel supplies that the allies had.

 

It could be considered as German answer to Merlin's inter- and aftercooler. MW-50 does the same work and does not required heat exchangers, radiator, powerful pump, piping, etc. Only a tank and minor piping.

Ніщо так сильно не ранить мозок, як уламки скла від розбитих рожевих окулярів

There is nothing so hurtful for the brain as splinters of broken rose-coloured spectacles.

Ничто так сильно не ранит мозг, как осколки стекла от разбитых розовых очков (С) Me

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You're interpreting the equation incorrectly.

 

Firstly, an internal combustion engine doesn't convert heat to mechanical energy. It's the increased pressure, as the fuel air mixture expands during combustion, that drives the engine. Thats where the potential energy of the fuel is converted to the mechanical energy which drives the pistons. In an internal combustion engine, heat is wasted energy, as it's energy that's not being converted to mechanical energy.

 

Secondly, increased thermal efficiency doesn't lower the energy potential of the fuel, which is what BTU is denoting in this case. BTU is a measure of energy potential of the fuel. A pound of gas burned releases about 19,000 BTU. 1 horsepower = 42.4 BTU's per minute or 2545 BTU's per hour.

 

Cooling the engine improves thermal efficiency, TE. Meaning less of the energy potential is wasted as heat, as HP= TE * Fuel Flow (Pound per hour) * BTU's of the fuel/2545.

 

So lets look at the Jumo 213 a-1, using the data posted by Yo-Yo

 

Peak HP at sea level is 1770, fuel flow is 1054.58 lbs per hour.

 

1770 HP = TE * 1054.58 * 19,000(BTU's of 1 lbs of gas, aka the energy potential of fuel)/ 2545 (BTU per HP per Hour)

 

1770HP = TE * 1054.58 * 7.46

 

Solve for TE, 1770/1054.58*7.46

 

.225 TE

 

1770 HP = .225 * 1054.58 * 7.46

 

So what does this say? It says that 77.5 percent of the potential energy of the gas is wasted. If the engine was 100% efficient in converting the potential energy of the gas it would produce 7867.166 HP.

 

So "the mysteriously generated 100 hp", generated by cooling, is very easy to find. Via a very minor increase in the thermal efficiency of the engine. Which is provided by the MW50.

 

To get an extra 100 hp out of the engine all we need to do is improve the the thermal efficiency of the engine by .013

1870 = X *1054.58* 7.46

X =.238

.238 - .225 = .013

 

 

 

 

When we talk about a "denser air" we're discussing Volumetric Efficiency (VE) of the engine. That is amount of air the engine ingests compared to the theoretical maximum.

 

While greater Volumetric efficiency does provide greater power. They don't necessarily provide as much horsepower per unit of efficiency as TE gains do. VE gains also run into issues of limiting returns. The compressibility of air and fluid dynamics limit how much air you can stuff into a cylinder, and makes it exponentially more energy expensive to create linear increases in density.

 

Lets look at the 213-A again, this time in terms of VE. Lets see how much more VE is necessary to derive the extra 330 HP gained from the MW 50 system. MW50 is reported to boost HP to 2100, compared to max 1770 HP.

 

Here we calculate the required VE necessary to generate the HP at the RPM

 

V-213 A Specs

RPM 3250 @ max power

2136 inch Displacement

Brake Specific Fuel Consumption of the 213-a is .59

BSFC= Fuel flow/HP

1054.58 (fuel flow)/1770(horsepower)=.59

 

REQUIRED VE = ( 9411 x HP x BSFC ) / (DISPLACEMENT x RPM)

 

At max power, 3250 RPM

1.41=(9411*1770*.59) / (2136 * 3250)

 

VE is over 1 meaning with have to compress the air, ie supercharge.

 

At Max RPM 3250, with MW 50

1.67=(9411*2100*.59) / (2136 * 3250)

 

Meaning, you need an 26% increase in volumetric efficiency, to generate the extra 330 horsepower output, when the MW50 is active in the 213. That number is so large that it's likely that at least 1/3 to 1/2 of the increased HP output, due to MW50, is a result of the smaller thermal efficiency gains. Which provide more HP per increase in efficiency than VE gains. In the case of this engine a TE efficiency increase of 1% yields 76hp compared to VE increase of 1% which yields a increase of 12hp.

 

I think that the simple calculations referring to effectiveness do not explain the things... because fuel flow itself and thus BSFC can vary in wide range but the engine power has only minor changes.

Ніщо так сильно не ранить мозок, як уламки скла від розбитих рожевих окулярів

There is nothing so hurtful for the brain as splinters of broken rose-coloured spectacles.

Ничто так сильно не ранит мозг, как осколки стекла от разбитых розовых очков (С) Me

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Page 27. "Turning the system on immediately increases engine power by almost 100 HP due to the fact that a cooler engine can pull in more air. At the same time, turning on the MW-50 enables much higher supercharger boost levels."

 

This doesn't seem quite right to me. Please do correct me if I'm wrong but "engine cooling" by this method would hardly be "immediate". If anything, I'd think the cooling effect on the air, making it more dense, would cause the immediate power increase. I'm always willing to learn something new, though.

I found it misleading too, and someone may end up with a wrong assumption.

While the engine cooling should be quite immediate it's the air cooled by the water-methanol mixture that allows a better chamber filling and thus an increased power output, especially in a supercharged engine without an intercooler.

Then with cooler air and the help of the remaining water that reach the chamber you can increase the boost pressure without encountering preignition achieving even a greater power increase.

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Curly, and where does the "increased pressure" in cylinders come from? From the heat being one of products of combustion. It seems that we're just using word "heat" in different meanings. I use it in a more general sense, as a form of thermal energy, just as it's often written in descriptions of engine cycles (input heat, output heat), while You use the word in more detailed context of energy wasted. Hence our mutual misunderstanding. Maybe it's the language thing (I'm not a native English speaker) - where I learned about thermodynamics, the "general" context of this word was used more often, so I tend to do it as well.

 

I'm not sure, however, how I'm interpreting the equation incorrectly - it's the very basic definition of thermal efficiency of any heat engine. Sometimes written with Q (heat), sometimes with T (absolute temperature) but no matter how one looks at it - the smaller expelled heat is - the higher thermal efficiency one gets. I think heat lost by cooling counts in here as well, might be wrong though.

 

While we're at it, It's quite possible we both mean something different when using a word "cooling" as well. My take on the issue is: since water has great heat absorption properties, not only it increases the charge density, but it absorbs tremendous amounts of heat during combustion process and, amongst other things like increased mean effective pressure, decreases amount of heat transferred to cylinder heads (thus increasing heat efficiency as You pointed out). So it is sort of a "internal cooling", but not what average guy understands when someone says "engine runs cooler". The infamous sentence in the draft manual doesn't attempt to explain the whole phenomenon better, which in my opinion might confuse the reader and that's what all the fuss is about.

i7 9700K @ stock speed, single GTX1070, 32 gigs of RAM, TH Warthog, MFG Crosswind, Win10.

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I think that the simple calculations referring to effectiveness do not explain the things... because fuel flow itself and thus BSFC can vary in wide range but the engine power has only minor changes.

BSFC gets larger as HP declines, it gets closer to 1. As HP declines and BSFC gets closer to 1, the numerator in the VE calculation, stays roughly the same. So you end up with similar VE through out the range.

 

Lets compare the VE of the Jumo 213-A at sea level Vs 4km, and 7km using the previous referenced data.

 

VE sea level max power.= 1.41

1.41=(9411*1770*.59) / (2136 * 3250)

1054.58 (fuel flow)/1770(horsepower)=.59

 

V-213a spec @ 4000 meters

1600 HP,

3250 RPM

fuel flow at 296 G/HP/H (1044.10 Lbs/per hour)

BSFC = .6525

(1044.10/1600)

 

VE=(9411*1600*.6525)/(2136*3250)

9825084 6942000

VE @ 4000 meters = 1.415

 

only a difference of .005 in VE, despite a loss of 170 hp and an increase in BSFC.

 

 

213-A @ 7000 meters

Fuel flow 322 G HP H (1043 lbs per hour)

1470 HP

3250 Rpm

BSFC=.709

1043.53/1470

 

VE=(9411*1470*.709)/(2136*3250)

 

9808426.52/6942000

 

VE @ 7000 meters = 1.41

VE is pretty much constant through the altitude range.

 

Fuel flows are calculated in Grams per horsepower per hour(G/hp H). What's happening is fuel flow is at maximum but horsepower declines. Which is why G/HP/H goes up as HP declines. G/HP/H has an inverse relation to HP. As HP decrease G/HP/H goes up, gets closer to 1, as the denominator is decreasing.

 

As alt goes up available the amount of O2 per unit of atmosphere declines. Which is why you need higher ATA's to get less power at higher altitudes. You need more atmospheres of pressure at altitude to provide the same amount of oxgeyen at sea level. We're losing air in the fuel air mixture, that's were the power loss derives from. To keep the power output constant you would need to boost to 2 ATA at 4000 meters, roughly 60 inHG.

 

To me it looks like the bench data on the 213-a was derived from these equations, especially for the first set of power, fuel consumption and Manifold pressure charts. The TE and VE equations are often were engineers start off when designing or evaluating an engine's power requirements.


Edited by Curly
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Curly, and where does the "increased pressure" in cylinders come from? From the heat being one of products of combustion. It seems that we're just using word "heat" in different meanings. I use it in a more general sense, as a form of thermal energy, just as it's often written in descriptions of engine cycles (input heat, output heat), while You use the word in more detailed context of energy wasted. Hence our mutual misunderstanding. Maybe it's the language thing (I'm not a native English speaker) - where I learned about thermodynamics, the "general" context of this word was used more often, so I tend to do it as well.

 

I'm not sure, however, how I'm interpreting the equation incorrectly - it's the very basic definition of thermal efficiency of any heat engine. Sometimes written with Q (heat), sometimes with T (absolute temperature) but no matter how one looks at it - the smaller expelled heat is - the higher thermal efficiency one gets. I think heat lost by cooling counts in here as well, might be wrong though.

It's not the increased temperature of the cylinder heads (heat) that creates the motion of the piston. The actual physical force, that moves the piston, is the expansion of fuel air mixture, as it expands when it combusts. A pressure wave smacking into the piston head.

 

You can heat the pistons all you want and you will not get more power. Unless that heat is accompanied by either a great volume or a more rapid expansion of the fuel air mixture. Put a piston on a stove in a vacuum, turn it to high, let me know when it moves.

 

Heat is a form of work and units of measure of both can be converted into each other. A British Termal Unit is defined as the heat energy required to raise the temperature of one pound of pure water by one degree F, and is equivalent to 778 foot-pounds of work / energy. One horsepower (33,000 ft-lbs per minute) is the equivalent of 42.4 BTU's per minute or 2545 BTU's per hour

 

33,000 / 778 = 42.4165

 

42.4165 × 60 = 2544.98

 

The thermal efficiency of a heat engine is the percentage of heat energy that is transformed into work. TE= work / heat(energy). If you do less work for more heat(energy), your have a lower TE.

 

Look at the Jumo 213-A. It only captures .22 of the potential work that fuel is capable of outputting, 1770 HP = TE * 1054.58 * 19,000, If it was capable of converting all the BTU's into work it would output 7867.166 HP. Since we know the energy output of the fuel is constant, we can then say the engine is only 22 percent efficient in capturing the potential energy of the fuel. Thus we say the TE of 213-a .225 or TE=.225.

 

While we're at it, It's quite possible we both mean something different when using a word "cooling" as well. My take on the issue is: since water has great heat absorption properties, not only it increases the charge density, but it absorbs tremendous amounts of heat during combustion process and, amongst other things like increased mean effective pressure, decreases amount of heat transferred to cylinder heads (thus increasing heat efficiency as You pointed out). So it is sort of a "internal cooling", but not what average guy understands when someone says "engine runs cooler". The infamous sentence in the draft manual doesn't attempt to explain the whole phenomenon better, which in my opinion might confuse the reader and that's what all the fuss is about.

The block of the engine is physically cooler when MW 50 is introduced to the cylinders. As it's being sprayed with a water coolant mixture. As show by the previous calculations, it's likely a large percentage of the increase power output is due to the better thermal efficiency.

 

Don't get me worng, I'm an not saying the majority of the increased power output isn't due to VE enchantments provided by the MW 50. I'm saying that the 100 hp quoted in the manual due to cooling (TE gains) is reasonable. The portion of the sentence that is somewhat misleading is; "due to the fact that a cooler engine can pull in more air." For the sentence to be more succinct it should read; "Turning the system on immediately increases engine power by almost 100 HP due to increases in the thermal efficiency of the engine." It should then say "The system provides as much as 440 additional horsepower, over max output, due to gains in airflow and thermal efficiency.

 

 

More power through cooler air is a result of Charles' Law. Which states Volume is proportional to temperature. A cooler manifold environment means you can get more air in the same area for the same amount pressure, ATA in the case of Fw190 measurement. This means your restoring the parts of oxygen per unit of air back to sea level conditions, or greater, depending on how much you cool the air.

 

The anti detonation effects derived from mw50 are due to the increased air in the mixture. Which prevents early detonation. You're essentially diluting pure gas with air so it doesn't ignite before top dead center. The increased horsepower benefits of the anti detonation effects are reaped in the form of more aggressive the ignition timings, closer to top dead center. Which results in a higher compression ratio and hence a higher VE. The brawn in the MW50 is really nothing with out the brains of the MBG control unit.


Edited by Curly
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It could be considered as German answer to Merlin's inter- and aftercooler. MW-50 does the same work and does not required heat exchangers, radiator, powerful pump, piping, etc. Only a tank and minor piping.

 

Yo-Yo - you are absolutely correct, my statement was written in a hurry and wasn't well thought out. It's all trying to achieve the same results just in different ways (such as the early altitude compensating carbs where the Entente powers used the vacuum approach which leaned the fuel out as the aircraft gained height, whilst the Central powers used additional air to thin the mixture out to achieve a similar result)


Edited by Oesau
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  • ED Team
BSFC gets larger as HP declines, it gets closer to 1. As HP declines and BSFC gets closer to 1, the numerator in the VE calculation, stays roughly the same. So you end up with similar VE through out the range.

 

 

 

To me it looks like the bench data on the 213-a was derived from these equations, especially for the first set of power, fuel consumption and Manifold pressure charts. The TE and VE equations are often were engineers start off when designing or evaluating an engine's power requirements.

 

What about the case if, for example, fuel flow at the certain altitude is changed for some reason? I will operate with ALPHA parameter that is Alpha = (Air mass flow/Fuel mass flow)/14.9, where 14.9 is full combustion ratio. The engine will provide almost constant power if you vary fuel flow getting Alpha from 0.7 to 1. BSFC will change according the fuel flow.

Ніщо так сильно не ранить мозок, як уламки скла від розбитих рожевих окулярів

There is nothing so hurtful for the brain as splinters of broken rose-coloured spectacles.

Ничто так сильно не ранит мозг, как осколки стекла от разбитых розовых очков (С) Me

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What about the case if, for example, fuel flow at the certain altitude is changed for some reason? I will operate with ALPHA parameter that is Alpha = (Air mass flow/Fuel mass flow)/14.9, where 14.9 is full combustion ratio. The engine will provide almost constant power if you vary fuel flow getting Alpha from 0.7 to 1. BSFC will change according the fuel flow.

 

The mixture is leaning out, it should be outputting less power.

 

.7= (10.4/1)/14.9

.7= (20.8/2)/14.9

.8 = (10.4/.087)/14.9

.8= (20.8/1.74)/14.9

 

You're taking, roughly, 13 % of the fuel out from the mix; With every decline of .1 Alpha. You should see a drop in power. There is a lot less gas in the cylinders, combustion will be weaker. Unless you're starting out overly rich, than leaning would actually increase power.

 

Or does your model derive power out based on the difference of Alpha from 14.9?

 

Stoichiometry, 14.9, fuel air mixture, isn't the point of peak power in the fuel air mixture. It's just the point where all of the air input with the gas is also consumed. Peak power typically occurs somewhat richer than stoichiometry, down in the 12's (12.5 is typically assumed as best power mixture). While peak efficiency tends to be closer to stoichiometry but in the 15's.

 

Mixture controls and a throttle that controls airflow are useful to a piston engine pilot because they allow one to control the power band. So that they are always maximizing the VE of their craft.

 

A throttle which controls fuel flow is more appropriate for a jet, because the dynamics of the engine fuel flow has a very direct relation with power output.

 

"Alpha = (Air mass flow/Fuel mass flow)/14.9" This is an equation to calculate the difference between current mix and stoichiometry. It's not a good equation to derive power out put of a piston engine from. As you're not taking into account the varying power output achieved by mixture settings and what's realistically achievable by the engine and the induction system. If you keep fuel flow constant in the equation, then rises in mass flow reach such high levels, that one is not operating in a realistic band of manifold pressures. Again, a 13 percent increase in mass airflow is need to keep the power constant. If the volume of cylinders stays constant, at some point your either likely to exceed the engine's capacity to sustain those MP, or be at such a place where the equipment need to compress the air to those level, would no longer be efficient. IE running a 4 horsepower supercharger to gain 2 horsepower.

 

.7= (10.4/1)/14.9

.8= (11.92/1)/14.9

 

Further more, we can calculate the mass of the air based on the fuel flow. At a fuel flow of 1055lbs per hour (from the Jumo) assuming the same mixture here of 14.9

Air flow = 14.9 * fuel flow

10,972= 14.9 * 1055 (both sides in Lbs per Hour)

 

7= (10.4/1)/14.9

10,972/1055

Convert to Standard cubic feet per min

Mass flow = Fuel flow(pph) * 3.246

3418 Scfm =1055 * 3.246

 

Air mass has rise with alpha, you hit a series of limiting returns at this point. If you wanted peak power, you'd be better off enriching the mixture. As your Mass air flow would be come realistically achievable.

 

 

Before auto mix systems pilots tended to run to rich, because they thought it would give them more power. Though they easily could end up with less power than their planes we're capable of producing, trying to dump gas into a craft who's beyond the VE to use it. This also lead to maintenance nightmares for ground crews. As running rich would tend to cause issues with plug fouling, ect.


Edited by Curly
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From DCS manual, page 29:

 

The tanks are daisy-chained and fed into one another.

 

No, they don’t fed into one another. Front and rear fuselage tanks fed into the engine’s main pump.

 

Page 31:

 

When drop tanks are used, the Fuel Selector Switch should be set to “Hinten”. The Fuel Contents Gauge will continue to display full for as long as the drop tanks continue to feed the rear and in turn the forward tanks. Once the drop tanks are emptied, the fuel quantity in the rear tank begins to decrease.

 

Unfortunately things are not that simple. First, the rear tank doesn’t feed the forward tank. It can’t. As I said above, both fuselage tanks can only feed the engine pump. Actually there is no piping whatsoever between the front and rear fuselage tanks. This is true for any Fw 190 ever built, from A-1 to D-9, as the technical manuals clearly show.

 

Second, the fuel quantity in the rear tank won’t show full, then begin to decrease after the drop tank is empty. That is not how it works.

 

The pipe that feeds from the drop tank to the rear tank actually connects to a special limiting valve, mounted in the rear tank. If the plane carries a drop tank, that limiting valve will only open when the rear tank content drops below 240 liters.

 

So, if a drop tanks is used, fuel consumption order would be like that:

At first, no fuel is consumed from the drop tank, because the limiting valve is closed. So in the beginning fuel will be consumed from the rear tank, until its level drops to 240 liters. Only then, the limiting valve will open and allow fuel from the drop tank to feed in to the rear tank. When the drop tank is empty, the fuel level in the rear tank will drop below 240 – this is the indication that the drop tank is empty.

 

Since the fuel system from D-9 is practically identical with A-8, I suggest ED to check Fw 190 A documentation from http://www.luftfahrt-archiv-hafner.de/

 

In the docs there is a manual, Bedienungsvorschrift-Fl with pilot instructions for all Fw 190 A version, from A-1 to A-8. How the fuel selector and pumps switches should be used is clearly described. The entire part about fuel system usage by the pilot covers several pages, so what’s in DCS manual is only scratching the surface.

 

Very short summary about fuel system use from the instructions – when a drop tanks is attached, in the beginning the forward fuselage tank fuel pump shouldn’t be switched on. Fuel from the forward tank will still feed the engine pump (on condition the fuel selector is on “Auf”, that is how it should be), but it will feed only due to gravity, so at a much slower rate than the rear tank, with its pump on. Necessary for correct center of gravity position. When the drop tank is empty, switch its pump off, and switch on the forward tank fuel pump. Next, when the white warning light for rear tank 10 liters is on, close the rear tank with the fuel selector, rear tank pump off, switch indicator to forward tank.

 

But what if there is no drop tank, the forward tank pump is on, so most likely the forward tank 90-100 liter level light will go on first? Close the forward tank with the fuel selector and empty the rear tank, then close the rear tank, rear pump off. As a general rule, once a fuel tank is empty, it must be closed with the fuel selector, to avoid air from the tank reaching the engine’s main pump.

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  • ED Team
page 24:

The english translation of the german terms should probably be put between parenthesis:

 

page 25, paragraph 2:

The blade pitch angle is not "coarse" or "fine", it is "larger" or "smaller" (or is "bigger" and "smaller" better?), isn't it?

Disregard this, "Coarse" and "Fine" seem to be the better terms (http://forums.eagle.ru/showpost.php?p=2087064&postcount=9).

 

page 32:

55-liyrt --> 55-liters

And in the next paragraph some temperatures are mentioned. It should probably be stated that these are in degree Celsius.

 

page 39, paragraph 2:

Y-Verfarhen --> Y-Verfahren

... But the whole sentence is somewhat difficult to understand. "Verfahren" means "method". So, while having no clue what this "Y-method" actually is, I can only guess ... maybe something like this would be better:

 

page 44:

The reddish/brownish colour of some instruments (air speed indicator, variometer) on that page, or rather in the cockpit itself (as seen in other screenshots on the forums as well) looks odd to me. Hrm ... I don't want to be an ass, but was this the source for that: http://www.nationalmuseum.af.mil/shared/media/photodb/photos/061114-F-1234P-002.jpg ? Note the manually scribbled km markings at the speed indicator and compare those to the markings of the DCS Fw 190 ... Well, erm, that reddish/brownish colour ... that isn't rust in our plane, isn't it? :D

 

page 62:

behalter --> Behälter

 

page 63:

1. MotorBediengerät --> Motorbediengerät

2. The handle is labled "Notzug für Bedien-Getr.", abbreviated for "Notzug für Bedien-Getriebe". Getriebe = gear box or gear mechanism while Gerät = device. I suppose, the gear mechanism ("Bedien-Getr.") is just a part of the engine control unit ("Motorbediengerät"). Maybe this needs a bit more detailing?

 

page 72:

The translation of "Anlassen" as "Cruise" seems odd to me. I know the term "Anlassen" only as the process of starting an engine (i.e. "start"). In this context, I would assume that the other lever position labelled "Start" means "take off".

The throttle positions would then be: "Off" - "Engine start" - "Climb" - "Take off" ... that would make more sense, no?

 

page 74:

The hyphenation of the switch labels is odd ... well, wrong. And therefore the description text is also incorrect. The hyphen belongs at the end of the first line. The labels should read therefore

Kopf-

lastig

and

Schwanz-


lastig


and the discription should not include the hyphen and thus just read

But I can not completely rule out that such labeling back then wasn't used...

Note also it is "...lastig", not "...lasting".



 

page 75:

Stabilier --> Stabilizer

 

page 78, chapter "Communications - Homing Switch"

I think, this chapter could be detailed a bit more in general. The meaning and actual use of the settings "Ft" and "Abst" are not clear (to me). The implications of the two different frequencies for sending and receiving for setting I (and II?) are also quite foggy (again, to me), the meaning of the comment in the red box totally escapes me...

All in all, personally, I would appreciate some more details of how the radio works and how it is (meant) to be used.

 

page 81:

backer --> breaker

 

page 88:

pomp --> pump :D

 

page 89:

Warnup --> Warmup

 

page 90:

Taxing --> Taxiing

Thanks!

Will correct.

Единственный урок, который можно извлечь из истории, состоит в том, что люди не извлекают из истории никаких уроков. (С) Джордж Бернард Шоу

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