It's another way to skin the cat.

 

Less hot piping under the hood, easier installation, no exhaust backpressure, quicker response. . .

 

I guess I'm having a hard time grasping the concept of the only energy in the exhaust gas is heat.

So, if the exhaust gas was cooled to ambient temperature before entering the turbo, exhaust flow through the turbine would not spin it?

Or, if you just had a heat source at the inlet of the turbine, it would spin with no flow?

I can understand that higher heat equals higher energy, but flow has to occur for any work to be done.

Highlander- If you disconnected the discharge piping of the turbo or just ran an exhaust housing, hell yeah you would lose a ton of power. But that's because your trying to look at the exhaust and compressor side of the turbo as seperate entities, when it really has to be looked at as a system.

And all of the exhaust energy doesn't come directly from the piston 'pushing' it out. Your exhaust valve opens approx 50bbdc, and alot of your exhaust energy comes from blowdown before there are any pumping losses.

Once boost is created there is exhaust backpressure and intake (boost) pressure. Once this happens the turbo is in 'closed loop'. It's like a revolving door. More boost equals more exhaust which equals more boost which equals even more exhaust.

Someone also stated that some turbo cams have 'zero' overlap. As in zero seat to seat overlap. I can't think of any situation where that would be ideal, or even good.

 

INTMD8, I sent you and e-mail a day or two ago. Did you get it?

I don't have time now to explain in detail, but this may help. The power any turbine produces is:

P=mdot*(h1-h2)*n

h1 and h2 are internal energy of the exhaust gas before and after the turbine, respectively. 'h' is a function of temperature ONLY.

n is the efficiency of the turbine. Similar to compressor efficiency, only backwards.

mdot is mass flow rate through the turbine.

So, you have to have mass flow rate to spin the wheel as well as heat.

If the exhaust was cooled to ambient before it went into the turbine, either one of two things would occur. 1. the exhaust would be at ambient pressure also and would, therefore, no longer possess the ability to do work, or 2. the exhaust would cool to sub-ambient as it passes through the turbine.

Think about this: Nitrous Oxide is at ambient in the bottle, but when you de-pressure it, it drops in temperature dramatically.

Mike

 

I'll have to read that post again when I feel like thinking


And no, I didn't get your email. Can you resend?

intmd8@sbcglobal.net

 

DO you have an email where you can confirm this... I mean I knew the pistons pushed the backpressure.. but EVERYONE and every article that I have read only talks about the temp differential... I know gas will travel through pressure differentials but... THere is so much power that can have.. literally pistons is what moves the turbine through backpressure... If we didn't have backpressure there wouls be very little boost.

Mike.. let me know if these strikes you at correct...

W=PÄV right?

Work is what move the turbine... a way of proving that piston is the one that pushes DIRECTLY the turbine would be to put it this way...

Heat will only increase pressure..... so more heat more work done... the pressure differential before the turbine and after the turbine only helps in the movement... which accounts to a small percentage...

But.. after the explosion the chamber is filled with exhaust gases and there is pressure there since pressure is what pushes the pistons...

So as soon as the exhaust valve opens there is an initial volume of gas from backpressure and what was inside the combustion...

The ÄV comes from the piston going up... in this case it would be 775mL of ÄV.

so you have the pressure and you have the change in ÄV... so you have work going on the turbine... am I correct or just fantasizing?

 

In order to keep from confusing the issue, we have to split it up into 2 parts: 1) the turbine and 2) the piston on exhaust stroke.

We can do this because they are only attached by a pipe and what leaves one retains its properties and goes to the other.

Don't try to analyze them both at the same time because they are two totally different processes.

1. The Turbine. A very simple Thermodynamic energy converter. The turbine converts heat and mass flowrate into shaft power (turbine shaft, not crankshaft). A turbocharger turbine might produce 35 hp, which is applied to the compressor wheel and compresses air. When you draw a box around the turbine and perform the Thermo calc's, you find that it takes around a 100 deg. F temperature drop of the exhaust gas to produce 35 hp. At 1200 degrees, the 100 degree temp drop equates to around 4 psi pressure drop. The cooler the exhaust gas, the greater the pressure drop must be. For instance, if the exhaust were 800 degrees (a la STS, perhaps), the pressure drop might be 8 psi. This 4 or 8 psi is added as exhaust backpressure.

Now that we know the affect a turbo has on backpressure, we can look at the piston and try to determine how this affects the engine.

2. The piston on exhaust stroke. This is a bit more complicated than the turbine because there are two distinct parts of the exhaust cycle. 1) Exhaust valve opens 50 degrees BBDC (still on the power stroke), releasing high pressure gas into the header and 2) piston travels up, pushing the remaining exhaust gas out. It is my opinion that added backpressure doesn't affect the 1st part of this since the pressure in the cylinder is very high (maybe 200 psi) when compared to the low pressure (maybe 20 psi) that the turbocharger causes. In fact, the added backpressure might help us here since we are still on the power stroke. However, during the exhaust stroke, the piston is pushing against an extra 4 or 8 psi. Using W=P*A*stroke/12in/ft*4(strokes per revolution)*5000 rpm/60sec/min/550ftlb/s, I get that it takes 13 hp to make 6 psi extra backpressure. Note that this DOES NOT include the affects of cam overlap, which is probably the biggest contributor to lost power from backpressure. You see, when both valves are open (and ALL cams have some overlap), the high pressure exhaust forces its way into the lower pressure intake port, diluting the intake charge. Diluted intake charge = more lost power.

Clear as mud?

Mike

 

so basically.. as a rule of thumb you aresaying is taht it takes 13hp/6psi.. so 2.2hp loss/psi of backpressure in just pushing?

so if you get 20 psi back pressure for 12psi boost you are loosing only 44hp? Just to the backpressure alone for the piston pushing up?? i would tend to think its a little bit more than that...

Hey mike did you solve your stuff? let me know highlander@caribe.net

if I can help out I will

 

If "theoretically" Boost pressure is the same as exhaust back pressure, would the effects of backpressure be eliminated once under boost assuming equal intake and exhaust duration in the cam during the exhaust and intake strokes? We are looking at what the piston has to push up against during the exhaust stroke, but we are also forgetting that the piston is also being pushed DOWN during the intake stroke by positive pressure. I know the cam would have some to do with this as the exhaust valve opens BBDC and the intake valve opens BTDC so part of the cam event doesn't actually happen during their respective intake/exhaust stroke.

 

According to engineermike's equation, without massflow work is not possible, and without a differential in temperature work is not possible on a turbine. If you were to take a bottle of compressed gas at room temp and hook it to the turbine inlet you would first have differential pressure due to bottle pressure and atmospheric pressure, thus with a differential press you have flow. Also think that when you decompress a gas it gets cold, because it it loosing potential energy from being under pressure. Due to conservation of energy the energy of the gas in the bottle must go somewhere. The temp of the compressed air would most likely be less than ambient. We could measure that and massflow and figure out Work.

 

Because of the harmful and hard-to-calculate effects of overlap with backpressure, I'm saying that it is a MINIMUM of 13 hp per 6 psi backpressure. This all assumes 1000 cfm flow, 10 psi boost, 350 cid, 5000 rpm, etc. . . Other combo's will be different.

I'm solving this stuff using fundamental Thermodynamics and Physics equations out of textbooks.

Mike

 

You're absolutely correct.

This is the case for superchargers also.

 

I dont think the push down vs boost will help in the same way as the push up vs backpressure gives disadvantages. Remember that pressure is being exherted in both sides of the engine, intake and exhaust with the main disadvantage that in the exhaust stroke you are 1) working agasint MORE backpressure than boost and that the backpressure is against the natural movemen of the piston. On the intake side. specially at 5000rpm the push for the boost is not much since basically the piston creates a vacuum anyways due to its speed down that is completely filled by boost.. so I dont belive in the thing that boosted applications have a positive effect in the intake stroke...

The negative side of boost is with the intake valves...

 

I haven't had much time to post online for a while but I couldn't help but stick my head in here

That refers to internal energy (U), which is what thermodynamics mainly focuses on. The 1st law is Q-W=dKE-dPE-dU

That is Heat (Q) minus Work (W) is always equal to the change in Kinetic energy (dKE) minus change in Potential energy (dPE) minus the change in internal energy (dU). Put shortly, energy is conserved!

Of course this energy starts as chemical (hydrocarbons) which is then transferred to thermal (combustion). The exhaust gasses have mass and they have velocity, therefore they have Kinetic energy... which cannot be ignored. I'm sure this gets even more confusing as the KE contributes to U with its interaction in the system, and U contributes to KE as a function of pressure in the system... now it becomes symantecs as to when and where it happens and what you consider driving the turbine or what your "system" is. It's all just energy so you might as well say it's gasoline doing the job In the end both the KE and U are interchangable since it's the work done that matters.

All turbines operate on the principle of differing pressure. If mass has velocity and strikes the turbine it imparts energy into it, and also create heat. Heat in itself creates pressure which creates velocity when there's a differential available. Both can happen at the same time.

Temperature determines the sensible internal energy but there's also the latent internal energy which is a function of the state of matter. It's also more of THE definition, not a "function of."

This doesn't help any

 

What hapenned to work?