For you "formula" competent people....

[BushWakr - RIP]

Hey guys,



Check out this link and if you can, try running "our" trucks thru the process and see what we get for efficiency maps...



http://www.turbosaturns.net/articles/compressor maps_2.htm



Obviously we need to substitute our data to make it work ... .



Lets see if we can get some info that may help us with turbo sizes and boost levels etc...



Bob.

 

[Rams-n-Hogs]

As an interesting side note, whenever you solve for the engine airflow requirements you can move the decimal over to the right one place you end up with the engine's horsepower.

lbs/min of airflow

RPM N/A 5psi 7psi 10psi 12psi 15psi 18psi

2000 3. 94 5. 28 5. 83 6. 62 7. 17 7. 96 8. 75

3000 5. 91 7. 92 8. 75 9. 93 10. 76 11. 94 13. 12

4000 7. 87 10. 55 11. 65 13. 22 14. 32 15. 90 17. 47

5000 9. 84 13. 92 14. 56 16. 53 17. 91 19. 88 21. 84

6000 11. 81 15. 83 17. 48 19. 84 21. 49 23. 86 26. 22

So, this 1. 9L Saturn makes 262. 2 HP@6000rpm?



Jim

 

[formula]

Ah, more formulae Bob, eh? Let's run the numbers for the 6BT through the same steps as the saturn engine and see what pops out... ...



(CID x RPM) / 3456 = CFM



One cubic foot contains 1728 cubic inches. A 4 cycle engine completes one intake stroke per every two revolutions of the crankshaft. This is where the number 3456 comes from.



(360cid X 2500rpm) / 3456 = 260cfm @ 100% volumetric efficiency.



I really don't know what the VE of the 6BT is, so until someone posts with a number for it, the rest of these calculations may be way off. But, for fun, I'll use 85%.



260cfm x 0. 85 = 221cfm



To convert cfm to lb/min, there is a multiplication factor of 0. 069 However, this is only good for one particular set of ambient air conditons as it is dependent on barometric pressure, air temperature and moisture level. So, the water gets more muddy.



221 x 0. 069 = 15. 3 lb/min



The pressure ratio at 35psi manifold pressure is (14. 7 + 35) / 14. 7 = 3. 4 This calculation must be adjusted according to altitude as 14. 7 psi is the normal atmospheric pressure at sea level.



So, at rated speed of 2500rpm with a base mass flow of 15. 3lb/min and a manifold PR of 3. 4, it should be flowing 15. 3 x 3. 4 = 52lb/min (also works out to about 0. 39kg/sec) That move the decimal point business is obviously not going to work here... . it would give 520hp, YEAH RIGHT.



Let's not forget that in this experiment, we're comparing a gasoline engine with a diesel engine, which have quite different operating characteristics. Just for starters... ... ... a diesel engine does not have to operate at a fixed air/fuel ratio as does a gasser. More homework needs to be done to perform any of these calculations for your particular engine and atmospheric conditions.



Sean



Hummmm... ... ... . where's DieselFreak?

 

[Diesel Freak]

Originally posted by formula

So, at rated speed of 2500rpm with a base mass flow of 15. 3lb/min and a manifold PR of 3. 4, it should be flowing 15. 3 x 3. 4 = 52lb/min (also works out to about 0. 39kg/sec)

just looking at the numbers it looks pretty close.....



gotta go check some notes

 

[Diesel Freak]

. 39kg/sec at a 3. 4 PR is off the map for an HX35, but it is in the sweet spot for an HX40.



coincidence? I think not!

 

[BushWakr - RIP]

Holy crap..... look out guys... . the creature is loose again, and looking to eat stuff... . like turbo's





Pastor Bob.

 

[soslow91]

Y tables

Since were all having fun lets not forget to throw heat into the equation. We all know hot air is less dense so. .



At a pressure ratio of 3. 4 the ideal rise factor would be . 4139 x (ambient temp +460) I'll use 80 for ambient since its hot where I live =223. 506



Compressor efficiency I don't have the maps or a specific turbo in mind so I'll use 70% 223. 5/. 7=319. 29+80(ambient)=399. 29



Density ratio without intercooling= (80+460)/(399+460) or . 629 . x3. 4=2. 1386



Density ratio with a 70% efficient ic 319. 29x. 7=223. 5

399-223. 5=175. 5 (399+460)/(175+460)=1. 353

1. 353x2. 1386=2. 8935



This doesn't account for pressure loss across the intercooler

which there will be but hopefully not too much.



460 is what you add to degrees F to get degrees ? (cant remember the name) which is whats used to calculate heat rise in air (and other "Perfect Diatomic Gasses")also handy if your into hot air balloons.



Some other Y values



2/1 . 217

2. 5/1 . 296

3/1 . 366

3. 5/1 . 425

4/1 . 48

4. 5/1 . 53

5/1 . 577



So (460+F)xY= ideal rise

Ideal/compressor efficiency= actual rise

Intercooler efficiency= actual temp drop/possible temp drop



Unless I missed the point of all of this I think the KG/sec would be closer to . 33 since the density ratio is only 2. 89 but still at a 3. 4 pressure ratio also 85% volumetric efficiency is probably optomistic.





If you read this entire post and almost thought it made sense +10RWHP pendind Bob's approval but -2 points to your appeal to the opposite sex.

 

[Rams-n-Hogs]

460 is what you add to degrees F to get degrees ? (cant remember the name)

? = Kelvin



I may not understand all this formula stuff, but, what all you guys are saying is I need an HX-40???, or would I be better off using my HX-35 as the small turbo in a set of twins?



(trying to convince the wife I need a new turbo. She says we need to pay bills!!!)



Jim

 

[Diesel Freak]

Re: Y tables

Originally posted by soslow91

-2 points to your appeal to the opposite sex.

well, there you have it. . the reason I am single!

 

[formula]

Ground control to Major Tom... ... ..... zzzzzzzzz... ... ... zzzzzzzzzzzz



As for the 460 degrees, I think that's for the Rankine (sp?) scale. 0R = absolute zero. It is based on the farenheit scale, its units having the same value.



The Kelvin scale, which is more commonly used in the scientific world, is based on the Celsius scale. A temperature change of 1K is equal to a temp change of 1deg C. On the Kelvin scale, absolute zero is 0K and zero deg C is 273. 15K.



Do you really need any HX40? Look back at the calculations and you'll see that they all were made using an engine speed of 2500rpm and a manifold pressure of 35psi. So, unless you intend to drive around all the time at this speed/boost, there are other variables to consider.



Sean

 

[cerberusiam]

Jeeezzz... ... . maybe being married is a GOOD thing... ... I'd never get a date otherwise. :{

 

[BushWakr - RIP]

Would it not be correct to "solve" for a variety of RPM/boost and use the turbo combo that includes the widest range within it's efficiecy map ?? yes/no ???



Bob.

 

[Rams-n-Hogs]

The Kelvin scale, which is more commonly used in the scientific world, is based on the Celsius scale.

OOPS, that's what I get for only taking one physics course in high school, (25+ years ago). I remember just enough to make me dangerous.

Would it not be correct to "solve" for a variety of RPM/boost and use the turbo combo that includes the widest range within it's efficiecy map ?? yes/no ???

That's why I keep following all the turbo threads.



Jim

 

[hdm48]

Ok, using my last dyno run at 298. 4rwhp @ 34lbs boost, I get a VE of 103. 45. But this is "wheel" hp, not engine. I am using a stock converter, so if any of you guys know the power bleed with a stock converter the engine power could be estimated.

But, have any of you concidered PV=NRT? This is NOT constant as conditions change.

I do have turbo calculations using a given A/R ratio, turbin pitch/angle, pipe diamiter in/out,temp, ect.



Now if you REALLY want to have some fun, call turbonetics and try to get these guys to set you up with a turbo for expected mods. A lot of ummms and ahhs and "I'll have to get back to you on thats"



Oh no! Now Pastor Bob will need to get that perscription of valium refilled.



Dave

 

[formula]

Originally posted by hdm48

But, have any of you concidered PV=nRT? This is NOT constant as conditions change.

Ah, the ideal gas equation. FWIW, P = pressure, V = volume, n = number of moles of gas, R = universal gas constant, T = absolute temperature (K) Here are some of the the values for R:



0. 08206 L atm/mol K



1. 987 cal/mol K



8. 314 J/mol K



8. 314 m^3 Pa/mol K



62. 36 L torr/mol K



A change in any one of the variables will result in a change in one or more of the remaining variables. For example... ... . if pressure goes up, either volume will have to decrease, number of moles of gas will have to increase, temperature will have to increase, or some combination thereof.



But, when looking at a turbocharger system, this relationship is rather secondary. As far as the turbo is concerned, we're interested in its adiabatic efficiency. This basically means the extent to which the turbo adds energy (heat) to the thermodynamic system.



Sean

 

[jbolt]

Alright Sean now just factor in the ASV of your unladen swallow. Divide by its VE and correct for ambient air temp, barometric pressure, humidity and a 6oz pork chop and it might start to make sense.



Jay

 

[hdm48]

With the pork chop theory in mind, and the river don't rise,,,

The turbo, if it stays within it's map, will create boost as long as drive pressure is greater than VE. And temp increase will be linear as boost pressures increase above VE. Right? Or should I eat that pork chop?

Now I know from experience that if boost is greater than VE by 20%, 21% becomes redundant and is nothing more than heat and drag. I am an old top fuel guy and when going through the traps at 8500rpm we only made 50lbs boost. Any less was a power loss and any more was nothing more than heat and power loss. EGTs? if the header doesn't melt, it's ok. MPGs? well, umm, at 40+ gals per mile we won't go there.

But now I wonder, does one build a truck for racing or for pulling?

Diesel engines cannot be compaired to gas or fuel burning engines. I have noticed that over fueling is required to make big power. I do it. But why is it needed?

Can someone explain that?



Dave