Joined

·
294 Posts

Vortech's site sets V3si discharge temp increase at 120 degrees over ambient temp of 70 degrees F, making a total discharge temp of 190 degrees F. Can anyone confirm similar measurements?

1 - 19 of 19 Posts

Joined

·
294 Posts

Vortech's site sets V3si discharge temp increase at 120 degrees over ambient temp of 70 degrees F, making a total discharge temp of 190 degrees F. Can anyone confirm similar measurements?

Joined

·
3,427 Posts

Most modern passenger and military aircraft are powered by gas turbine engines, also called jet engines. All types of jet engines have some parts in common. All jet engines have a compressor to increase the pressure of the incoming air. There are currently two principal compressor designs found on jet engines: the axial compressor, in which the air flows parallel to the axis of rotation, and the centrifugal compressor, in which the air is turned perpendicular to the axis of rotation. In either design, the job of the compressor is to increase the pressure of the flow. We measure the increase by the compressor pressure ratio (CPR), which is the ratio of the air total pressure pt exiting the compressor to the air pressure entering the compressor. This number is always greater than 1.0. Referring to our station numbering, the compressor entrance is station 2 and the compressor exit is station 3. The CPR is equal to pt3 divided by pt2, as shown on the slide.

CPR = pt3 / pt2 >= 1.0

To produce the increase in pressure, the compressor must perform work on the flow. In the axial compressor, cascades of small airfoils are mounted on a shaft that turns at a high rate of speed. Several rows, or stages, are usually used to produce a high CPR, with each stage producing a small pressure increase. In the centrifugal compressor, an additional pressure increase results from turning the flow radially, radiating from or converging to a common center. Since no external heat is being added to or extracted from the compressor during the pressure increase, the process is isentropic. The total temperature ratio Tt3 / Tt2 across the compressor is related to the pressure ratio by the isentropic flow equations.

Tt3 / Tt2 = (pt3 / pt2) ^((gam -1) / gam)

where gam is the ratio of specific heats.

Work must be done to turn the shaft on which the compressor is mounted. From the conservation of energy, the compressor work per mass of airflow CW is equal to the change in the specific enthalpy ht of the flow from the entrance to the exit of the compressor.

CW = ht3 - ht2

The term "specific" means per mass of airflow. The enthalpy at the entrance and exit is related to the total temperature Tt at those stations.

CW = cp * (Tt3 - Tt2)

Performing a little algebra, we arrive at the equation:

CW = [cp * Tt]2 * [CPR ^((gam -1) / gam) - 1] / nc

that relates the work required to turn the compressor to the compressor pressure ratio, the incoming total temperature, some properties of the gas, and an efficiency factor nc. The efficiency factor is included to account for the actual performance of the compressor as opposed to the ideal, isentropic performance. In an ideal world, the value of the efficiency would be 1.0; in reality, it is always less than 1.0. So additional work is needed to overcome the inefficiency of the compressor to produce a desired CPR. The work is provided by the power turbine, which is connected to the compressor by the central shaft.

Notice that the CPR is related to the total temperature ratio across the compressor. Since the CPR is always greater than 1.0 and the value of gam, the ratio of specific heats, is about 1.4 for air, the total temperature ratio is also greater than 1.0. Air heats up as it passes through the compressor. There are temperature limits on the materials of the compressor. On some engines, the temperature at the exit of the compressor becomes a design constraint, a factor limiting the engine performance. You can now use EngineSim to study the effects of different materials on engine operation.

Joined

·
3,427 Posts

Compression of a gas naturally increases its temperature.

In an attempt to model the compression of gas, there are two theoretical relationships between temperature and pressure in a volume of gas undergoing compression. Although neither of them model the real world exactly, each can be useful for analysis. A third method measures real-world results:

Isothermal - This model assumes that the compressed gas remains at a constant temperature throughout the compression or expansion process. In this cycle, internal energy is removed from the system as heat at the same rate that it is added by the mechanical work of compression. Isothermal compression or expansion more closely models real life when the compressor has a large heat exchanging surface, a small gas volume, or a long time scale (i.e., a small power level). Compressors that utilize inter-stage cooling between compression stages come closest to achieving perfect isothermal compression. However, with practical devices perfect isothermal compression is not attainable. For example, unless you have an infinite number of compression stages with corresponding intercoolers, you will never achieve perfect isothermal compression.

Adiabatic - This model assumes that no energy (heat) is transfered to or from the gas during the compression, and all supplied work is added to the internal energy of the gas, resulting in increases of temperature and pressure. Theoretical temperature rise is T2 = T1·Rc(k-1)/k, with T1 and T2 in degrees Rankine or kelvins, and k = ratio of specific heats (approximately 1.4 for air). R is the compression ratio; being the absolute outlet pressure divided by the absolute inlet pressure. The rise in air and temperature ratio means compression does not follow a simple pressure to volume ratio. This is less efficient, but quick. Adiabatic compression or expansion more closely model real life when a compressor has good insulation, a large gas volume, or a short time scale (i.e., a high power level). In practice there will always be a certain amount of heat flow out of the compressed gas. Thus, making a perfect adiabatic compressor would require perfect heat insulation of all parts of the machine. For example, even a bicycle tire pump's metal tube becomes hot as you compress the air to fill a tire.

Polytropic - This model takes into account both a rise in temperature in the gas as well as some loss of energy (heat) to the compressor's components. This assumes that heat may enter or leave the system, and that input shaft work can appear as both increased pressure (usually useful work) and increased temperature above adiabatic (usually losses due to cycle efficiency). Compression efficiency is then the ratio of temperature rise at theoretical 100 percent (adiabatic) vs. actual (polytropic).

but to answer your question, it's not 1 set number, it's a dynamic system with a lot of variables involved.

Joined

·
8,722 Posts

How's that for a lot of showing off without answering the OP's question! LOL.

--Peter

--Peter

https://www.classle.net/book/compressors-temperature-staged-compression-prime-movers

Joined

·
294 Posts

***Brownie points if you can tell me what .286 refers to? I think it's refers to a constant for gas, in this case air.

Pressure Ratio = (Ambient Pressure + Boost Pressure) / Ambient Pressure

At 9 psi, thats… (14.7+9)/14.7 = 1.61

Going back to Theoretical Discharge Temp, let's say its 25 degrees C = 77 degrees F = 298 K

(298K) * (1.61)^.286 = 341K or 150 degrees F at 9psi

That's a temperature increase of 150 - 77 = 73 degrees F

But we digress…has anyone taken actual measurements? If these calcs are right, the Vortech V3 is pretty amazing. If no one has data, I'll post mine when I get around to it. Thanks guys!

Joined

·
3,427 Posts

That was the point!How's that for a lot of showing off without answering the OP's question! LOL.

--Peter

You'd have to run a 3d contour map to see what temps are because a simple x-y plot isn't enough to really begin answering the question.

.286 = (gamma-1)/gamma

(Atmospheric Temperature) * (Pressure Ratio)^.286 = Theoretical Discharge Temp

***Brownie points if you can tell me what .286 refers to? I think it's refers to a constant for gas, in this case air.

Pressure Ratio = (Ambient Pressure + Boost Pressure) / Ambient Pressure

At 9 psi, thats… (14.7+9)/14.7 = 1.61

Going back to Theoretical Discharge Temp, let's say its 25 degrees C = 77 degrees F = 298 K

(298K) * (1.61)^.286 = 341K or 150 degrees F at 9psi

That's a temperature increase of 150 - 77 = 73 degrees F

But we digress…has anyone taken actual measurements? If these calcs are right, the Vortech V3 is pretty amazing. If no one has data, I'll post mine when I get around to it. Thanks guys!

Which you used in the isentropic pressure ratio.

To/T = (Po/P)^((gamma-1)/gamma)

The amount of times i've used that equation is not cool. I reached a point where i memorized all the isentropic flow relations because they are used daily.

How do i collect these brownie points?

The calculation you did would be fine if there was no heat transfer. But the temperature increase is much more than 73 degrees at 9 psi. I think it recall i being something around double that. My partner Jamie has hard numbers on what the V3 produces under static conditions. I'm sure he'll chime in.

-R

"My Vortech V3 discharge temps are: xxx"

and less:

"The theoretical limit of the maximum power level is fundamentally tied to the inlet temperatures possible within a closed system and for a single turbo encabulator system dual rows of inlet gas discharges must be maintained in order to reach a static vortex just outside the axial compressor wheel which serves to act as a buffer for the threshold allocation system"

Joined

·
294 Posts

I owe you a beer my friend. Bimmerfest next year?!

Uh, was that SAE or DIN?

"My Vortech V3 discharge temps are: xxx"

and less:

"The theoretical limit of the maximum power level is fundamentally tied to the inlet temperatures possible within a closed system and for a single turbo encabulator system dual rows of inlet gas discharges must be maintained in order to reach a static vortex just outside the axial compressor wheel which serves to act as a buffer for the threshold allocation system"

Regard,

Jerry

Joined

·
294 Posts

Vortech lists actual measured total discharge temp at 70F and 10.6 psig (which I think converts to 25.3psi?), as 190F.

About Roots Type Superchargers | Vortech Superchargers

This is surprising as I would imagine temps at 25psi would be way, way higher. I just got a new temp gauge so I just need to fab a fitting and I'll get some real world measurements.

I would assume that Vortech knows a little about their superchargers and how they perform so I expect that the measured temp will be close to what they say, although for some reason they don't seem to likes other types of superchargers :dunno: :rofl:

Joined

·
294 Posts

Sweet. What was average ambient temp when for these measurements?

Joined

·
294 Posts

70 F @9psi:

After warm up…

At idle: ~95F

Cruising: ~130-150F

I'll post updates if and when I take more. Will be starting a separate A2A intercooling thread relating to readings. It will involve mathematical theory for those that don't like that…you've been warned.

Joined

·
294 Posts

Indeed. More reason to like them now.

Joined

·
1,867 Posts

As Tim said, best way is to throw a probe in there and measure it, as it seems you've done!

1 - 19 of 19 Posts

Join the discussion

BMW M5 Forum and M6 Forums

M5Board is the best forum community for information on the BMW M5 E60 (V-10), E39 (V-8), E34 (straight 6), E28, F90 and F10. Discuss performance, specs, reviews and more!

Full Forum Listing
Recommended Communities

Join now to ask and comment!