2 Stroke VS. 4 Stroke heat production...

[BEN MC7]

I was curious if anyone here knows of or has compiled data regarding the BTU (or whatever label they have given it) of the different engines that we have been using in our buses.  That might be a handy chart to decide what any of us should consider when doing a re-power from an 8-71, 6-71 to an 8-92 or Series 50/60.

There is a wealth of knowledge here with all of us tinkering on our projects.  The actual numbers (instead of statements like ' alot of heat ', would benefit many doing their research.

[TomC]

Here are some observations on facts.  If you have a factory installed 8V-92TA, since they are the hardest engine to cool in the trucking industry, then you can just about install any 4 stroke engine up to 600hp with the addition of an air to air intercooler.  If you have a 8V-71N or a 6V-92TA, any 4 stroke engine up to 450hp will be cooled with the addition of the air to air intercooler.  If you have a 6V-92TA, you could install a Series 50 and remove one of the radiators and replace it with the air to air intercoooler (what Grayhound did).  If you have a 6-71, you could install a 4 stroke engine up to 400hp with the addition of a air to air intercooler.
Actual BTU outputs of engines are hard to come by since there are many differences between their usage.  For instance a 400hp 8V-92TA in a Grayhound will put out alot less heat then a 475hp 8V-92TA that Prevost would install.
Repower's that I would suggest would be Cummins ISM at 4-500hp, Cat C-12 at 470hp, Mercedes-Benz 4000 at 450hp.  All these are around the 2,000-2,200lb mark, and are much lower in height then the Series 50/60, Cummins N14 or ISX, or Cat 3406/C15 (that weigh up to 3,000lb now).  Good Luck, TomC

[Tim Strommen]

Ben,

There is a more general way to determine how much "waste heat" is being produced.  The fuel you use has a known measured energy content (in BTUs).  For Diesel #2, this number is about 140,000BTU per gallon.  Based on the power generated vs. fuel consumption – you can determine the efficiency of the combustion.  Once you have that number, it's relatively easy to calculate the amount of "waste heat" generated.

Further from that – if you know how much cooling water is going through your radiator(s) and you have an inlet outlet temperature (and a value for how much volume the radiator contains), you can determine how much heat your radiator "rejects" to the open air (and via subtraction, determine how much heat is rejected by the exhaust gasses and the engine block itself).  These calculated numbers will be very close to reality – this is the same basic way engine designers size radiators and cooling systems.

Another conversion you need: 2544.328BTU = 1HP

As an example:

If your engine is consuming 13GPH of diesel and producing only 250HP peak WOT (wide open throttle - fuel rack to the upper limit), your fuel-to-motion conversion efficiency is around 34.95%.  This means that for 13 gallons, with a BTU content of (140,000 x 13 = 1,820,000BTU) about 1.8 million BTU of energy content – only (1,820,000 x 34.95% = 636,090 is used to move the vehicle, while the balance (1,820,000 – 636,090 = 1,183,910) of about 1.2 million BTU is rejected by the radiator, exhaust gasses, and engine block to the circulating air.


Detroit Diesel may have publications that state output HP vs. fuel usage – I think the field service data book has this (I'll check it tonight).  As a general rule of thumb – the best efficiencies are around 33% for diesel engines - so knowing the fuel consumption is a key data point for sizing the cooling system.

-Tim

P.S. I just thought about something that may also help with the DDEC and MUI "V" engines...  The injectors have a "CC" injection value for them.  This is the maximum injection at WOT.  Take that number considering the maximum RPMs and the cycle type (for a 2-stroke, each cylinder will fire each rotation of the engine - while for a 4-stroke each cylinder will fire every two rotations of the engine).

Another handy conversion: 3,785.412cc = 1 gallon

Example:  6V-92 2-stroke with a .065CC injectors (note this is an example number -  I don't have my injector data book here...) running at 2100 RPM WOT would consume 6 x .065 X 2100 = 820.1726cc of fuel per minute (49,210.36CC/Hr).  So, this engine would be running at about 13GPH (and assuming the 34.95% efficiency - you'd be getting about 250HP peak...) -T