Local pressures in turbocharger housing?

[gen x]

This is question for fluid mechanics.

Static pressure in the exhaust manifold(turbo car engine) is usually 1.2 to 2.5 times higher than the boost pressure(intake manifold pressure).Boost pressure is around 1bar guage pressure(2bar absolute).
Can the local static pressure somewhere inside a turbine housing ever be lower than atmospheric pressure, is this possible?

here some links where CFD is used:
https://www.linkedin.com/posts/comp...using-ansys-activity-6980110878163525632-AtDY

https://www.sciencedirect.com/science/article/pii/S2212540X24000221

 

[Lnewqban]

Can the local static pressure somewhere inside a turbine housing ever be lower than atmospheric pressure, is this possible?

What other reason would the air located next to the intake have in order to flow into it?
The intake of the compressor sucks air from the surroundings, while the vanes of the rotor increase its pressure via centrifugal effect.

 

[Baluncore]

Can the local static pressure somewhere inside a turbine housing ever be lower than atmospheric pressure, is this possible?

I know the Nemesis that would catch me out, would be the radial pressure gradient inside the centrifugal pump. So I must expect the lowest pressure to be hidden behind the compressor impeller, at the shaft, between the impeller and the bearing. There may be a low pressure point in a similar position, behind the turbine. Those low pressure points are ideally placed, to suck lube oil past the shaft seals, that will be damaged by worn bearings.

While pumping a high head of water up a pipeline, there is a low pressure on both sides of the impeller, at the centre. Only when the pump is turned off, does the full head of water pressure appear in the casing at those two points, which can be mechanically embarrassing. That occurs because the non-return valve will normally be on the inlet to the pump, to aid in priming, not on the output to avoid back pressure from bursting the casing.

 

[gen x]

What other reason would the air located next to the intake have in order to flow into it?
The intake of the compressor sucks air from the surroundings, while the vanes of the rotor increase its pressure via centrifugal effect.

I refer at turbine not compressor.

There may be a low pressure point in a similar position, behind the turbine. Those low pressure points are ideally placed, to suck lube oil past the shaft seals, that will be damaged by worn bearings.

Low pressure at turbine and compressor is not good, because it will suck engine oil.


@boneh3ad
@Arjan82
Can we from this below plots total pressure, velocity and density find minimum static pressure in turbine? Problem is gas velocity is very high 325m/s (1170km/h) so wil Bernoulli get correct static pressure? Is total pressure in plot, absolute or relative?

link source: https://www.sciencedirect.com/science/article/pii/S2212540X24000221

 

[Lnewqban]

I refer at turbine not compressor.

In that case, the answer to your question is no.
The local static pressure somewhere inside a turbine housing of a turbo can never be lower than atmospheric pressure.

 

[Baluncore]

The local static pressure somewhere inside a turbine housing of a turbo can never be lower than atmospheric pressure.

That is a bold statement. Is there any proof ?

If the turbine exhaust is to atmospheric pressure, then will not the centre of the flowing exhaust, have a lower pressure ?

 

[gen x]

That is a bold statement. Is there any proof ?

If the turbine exhaust is to atmospheric pressure, then will not the centre of the flowing exhaust, have a lower pressure ?

Is total pressure from my post #4 absolute or guage pressure?

 

[jack action]

In a compressible flow, "gauge" total pressure doesn't really make any sense.

With incompressible flow, total pressure is an addition:
$$p_{total_{abs}} = p_{static_{abs}}+½\rho v^2$$
So:
$$(p_{total_{gauge}} + p_{atm}) = (p_{static_{gauge}} + p_{atm}) + ½\rho v^2$$
Therefore the equation works as equally well with gauge pressure:
$$p_{total_{gauge}} = p_{static_{gauge}}+½\rho v^2$$
With compressible flow, total pressure is a ratio:
$$\frac{p_{total_{abs}}}{p_{static_{abs}}} = \left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}$$
$$\frac{(p_{total_{gauge}} + p_{atm})}{(p_{static_{gauge}} + p_{atm})} = \left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}$$
So, if one would use gauge pressures with compressible flow, it would have to come with a value for $p_{atm}$ to validate them and it would be very confusing.

Short answer: total pressures in the context of compressible flow are always represented with absolute values.

 

[gen x]

In a compressible flow, "gauge" total pressure doesn't really make any sense.

With incompressible flow, total pressure is an addition:
$$p_{total_{abs}} = p_{static_{abs}}+½\rho v^2$$
So:
$$(p_{total_{gauge}} + p_{atm}) = (p_{static_{gauge}} + p_{atm}) + ½\rho v^2$$
Therefore the equation works as equally well with gauge pressure:
$$p_{total_{gauge}} = p_{static_{gauge}}+½\rho v^2$$
With compressible flow, total pressure is a ratio:
$$\frac{p_{total_{abs}}}{p_{static_{abs}}} = \left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}$$
$$\frac{(p_{total_{gauge}} + p_{atm})}{(p_{static_{gauge}} + p_{atm})} = \left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}$$
So, if one would use gauge pressures with compressible flow, it would have to come with a value for $p_{atm}$ to validate them and it would be very confusing.

Short answer: total pressures in the context of compressible flow are always represented with absolute values.

If I use data from plot, density=0.4kg/m3, minimum total pressure(blue)= 1.06e+05Pa, velocity=325m/s, using compressible equations I get static pressure= 71000 Pa, so static pressure in that section of turbine is lower than atmospheric pressure.
Did you get the same?

 

[jack action]

From the isentropic process:
$$\frac{p}{p_t} = \left(\frac{T}{T_t}\right)^{\frac{\gamma}{\gamma - 1}}$$
and from the definition of total enthalpy $h_t$:
$$h_t = h + \frac{v^2}{2}$$
knowing that $h_t = C_pT_t$ and $h = C_pT$, we get:
$$\frac{T}{T_t} = 1 - \frac{v^2}{2C_pT_t}$$
And, since, from the ideal gas law, $p_t = \rho_t RT_t$:
$$\frac{T}{T_t} = 1 - \frac{v^2}{2C_p\frac{p_t}{R\rho_t}} = 1 - \frac{v^2}{2\frac{C_p}{R}}\frac{\rho_t}{p_t}$$
And using the heat capacity ratio relationship for an ideal gas:
$$\frac{T}{T_t} = 1 - \frac{v^2}{2\frac{\gamma}{\gamma - 1}}\frac{\rho_t}{p_t}$$
Back with our original equation:
$$\frac{p}{p_t} = \left(1 - \frac{v^2}{2\frac{\gamma}{\gamma - 1}}\frac{\rho_t}{p_t}\right)^{\frac{\gamma}{\gamma - 1}}$$
Where $\frac{\gamma}{\gamma - 1} = 3.5$ for air. So:
$$ p = 106000\ Pa \times \left(1 - \frac{(325\ m/s)^2}{2\times 3.5}\frac{0.4\ kg/m³}{106000\ Pa}\right)^{3.5} = 86336\ Pa$$

 

[gen x]

$$ p = 106000\ Pa \times \left(1 - \frac{(325\ m/s)^2}{2\times 3.5}\frac{0.4\ kg/m³}{106000\ Pa}\right)^{3.5} = 86336\ Pa$$

Isn't weird that pressure is lower than atmosperic pressure, so if you drill hole in that part of turbine housing, air will be sucked in instead blown out?

 

[jack action]

Your numbers seem to come from behind the impeller tip, where it would be normal to have a low-pressure area as the impeller turns. But even at this point, the total pressure seem to be 133 000 Pa rather than 106 000 Pa. (Which sets the static pressure a little above the atmospheric pressure.)

If you look at the numbers closer to the impeller base (where it is closer to where you are afraid oil could get in), you have velocities closer to 90 m/s, with similar density and pressure you indicated. The calculations set the static pressure closer to the atmospheric pressure.

That being said, I don't think those graphs show accurate values for positions behind the impeller wheel, where the bearing seals are.