Curly

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.

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

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.

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.

Yes, you're essentially misting the cylinders with engine coolant. Are you asking whether a change in fuel air density/mass air flow or thermal efficiency has a greater effect on power output? You also have to take into account you're essentially spraying alcohol into the cylinders which is going to up power output. Then the steam in the cylinders, created by the combustion fuel in the MW50 environment, also raises the pressure in the cylinder and creates increased power. What do you mean by "more immediate"? Basic video on water methanol systems Research paper on effects of water injection and fuel mass ratio. http://omicsonline.org/scientific-reports/JAME-SR-591.pdf The methanol has multiple effects, besides acting as an anti freeze. A. It's actually a better coolant than water alone. Due to it's lower freezing and higher boiling point. B. It acts as a fuel additive and raises the octane of the fuel. Methanol is alcohol. You partially answered your question with your first statement. A. The water is mixed with methanol so that the boiling point is higher. Combine with ambient temperatures at 10,000 ft, + were temps are likely be between 12 and -30, and we have a reduced likelihood of boiling. B. The 50 MW cools intake temps quickly and further reduces temps to below the boiling point. C. Water will conduct heat as long as there is a temperature difference between it and another surface. So as long as the cylinder is hotter than water, the cylinder will transfer energy to the water, whether the water is in a liquid or aerosol state.

The cooling is a secondary effect to water methanol systems, it is almost immediate. After the methanol evaporates in the intake, reducing intake air temps and increasing the density. The leftover water in the MW-50 (it's 50% water, 50% methanol) is injected into the cylinders, which reduces temperatures in the cylinders and cylinder walls. This reduces the chance of pre detonation. Thus allowing for the engine to be operated at a higher pressure. Thus we have a cooler engine that is pulling in more air, operating at a higher A.T.A. So the sentence is correct. The engine outputs more HP at the same ATA when the MW50 is injected. Subsequently the engine can also be operated at higher ATA too, do to the reduction in risk of pre detonation. The danger from a water Methanol systems is corrosion leading to component failure and raising cylinder pressures above operating limits. Both of which can result in catastrophic failure.

1.2.7 Here are quick few, but it seems to be most of the mission that come with game. East Georgia: Group Tank AD is only supposed to activate once the bridge is destroyed. The SA 9 Streala 1 is present before hand. The group Patrol is supposed to activate only after, group supply is dead. It is active at the start. In The Weeds: Shilka Groups 1,2,3 are supposed to have a 50/50 chance of spawning. They all spawn because they are not set to late activation. Also it looks like there maybe a bug in the trigger condition for the group 3 trigger. The trigger title indicates that it is supposed to apply to group 3 but it applies to group 4. The SA9 groups 1,2 and 3 are also supposed to have 50/50 chance to spawn as well but all are present. Kahshuri Gap: Russian AD 4,5 and 6, RU tank 2-1, 2-2 and 2-3 along with some of the german units are supposed to activate after a trigger. They all just go at the start. Pretty much any mission that is supposed to have group activate after a condition, fails to do so, all the units just go active from the start.

Many of the stock missions make use of triggers which use the Activate Group Command. Though none of the units to be activated have a late start selected. Which means all the units spawn at once. For a lot of the missions to become playable, you have to use the mission editor and make them late starts. This seems to be effecting all of the solo missions and the instant action missions too.