12 replies to this topic

[raghumech]

Hi,

Can anyone help me out with this?
Whether pressure of the fluid can be increased by reducing the velocity and temperature of it??
can it be obtained by changing the flow type?

Thanks in advance.

[thorium90]

I dont think one can increase pressure by reducing temperature if you think about the ideal gas law. But by reducing velocity, in a way yes, if you think of Bernoulli equation. Flow type i'm gonna assume laminar and turbulent? In a way yes, because you have to put more pressure which makes the flow higher to change it from laminar to turbulent. But also in a way no, since mathematically speaking no additional work is done comparing different types of flow. For example if there are lots of obstacles in the flow stream the flow would break and become turbulent which in no way increases pressure.

[kkala]

Let us consider practically incompressible flow in an horizontal pipe. Between two non distant points (friction between them ~ 0) Bernoulli equation (mechanical energy balance in Perry's fluid dynamics, p. 6-7 in 7th edition) can be expressed as p+γv²/2=constant (p=pressure, γ=density, v="average" velocity), expressing conservation of pressure energy+kinetic energy. Following is based on this equation.
1. In theory pressure can increase by reducing velocity. In practice this increase is rather small and neglected in a lot of engineering applications (but considered in special cases).
For instance if water flows at a velocity of 6 m/s (max recommended by Norsok std P-001) and this is reduced to 0.1 m/s, pressure will increase by 1/2*1000(36-0.01) kg/(ms2)=17995 N/m2=0.18 Bar, but neglecting frictional pressure drop for flow enlargement (Perry, 7th ed, p 6-16).
2. Reducing temperature would increase density and reduce velocity; but effect on pressure can practically be (close to) negligible, even for gases (the latter flowing at small pressure drop to be considered incompressible).
For instance, suppose air flow at 50 m/s, 250 oC, 2 Bara, cooled to 50 oC. Density changed from 1.292*(273/523)*(2/1.013)=1.332 kg/m3 to 1.292*(273/323)*(2/1.013)=2.156 kg/m3. Velocity changed from 50 m/s to 50*(1.332/2.156)=30.9 m/s. Pressure will increase by 1/2*1.332*50² - 1/2*2.156*30.9²= 636 Pa = 0.00636 Bar; that is by 63 mm H2O, but neglecting frictional pressure drop for cooler.
2a. Mentioned pressure change as above in gas cooling for t oC to t1 oC (t>t1) is always positive (P increase).
Density changed from γ to γ1=γ*(t+273)/(t1+273), velocity form v to v1=vγ/γ1=v*(t1+273)/(t+273); so P1-P =
=1/2*γv²-1/2γ1v1²=1/2γv²*(1-(γ1/γ)(v1/v)²)) =1/2γv²*(1-(t1+273)/(t+273)) >0 .
3. Changing flow type is result of changing velocity, consequences as above.
Comments on these views welcomed.

Edited by kkala, 22 January 2013 - 09:54 AM.

[Ajay S. Satpute]

Dear Raghu,

 

Do you have any practical issue for this query?

 

Regards.

 

Ajay

[raghumech]

Dear Raghu,

 

Do you have any practical issue for this query?

 

Regards.

 

Ajay

 

 

Yes ajay, i just want to convert the high temperature fluid to pressurized one...searching for mode to convert it.....infact its gas...

[thorium90]

In other words, you want to pressurize a hot gas stream? That sounds simple, use a gas compressor.

Frankly, without more details, I think a gas compressor will seem to fit the bill.

[raghumech]

In other words, you want to pressurize a hot gas stream? That sounds simple, use a gas compressor.

Frankly, without more details, I think a gas compressor will seem to fit the bill.

Hii..

      I have some constraints in using gas compressor, so i thought to use the temperature of gas as input to convert it as pressurized one....

 

Like i have 130*c gas....which is going to be wasted...so i want to use this energy as input to convert gas around 2bar(possibly)..

[thorium90]

So you have some gas stream at some unknown pressure and composition that is at 130C and you want cool it down to some unknown temperature and compress it from that unknown pressure up to 2 bar using some unknown technique that can cool and compress at the same time and you cant use a compressor due to some unknown constraints and you are hoping someone might be able to help you with that?

Edited by thorium90, 04 February 2013 - 04:31 AM.

[Ajay S. Satpute]

Raghu,

 

You may not be able to get rid of the compressor, but certainly as the system starts with a higher temperature, you will get even higher temperature and that heat if you can recover from compressor outlet stream and re-use elsewhere in the plant, then you will have an energy efficient system.

 

I have done a simulation in hysys for CH4 gas. PFA the snapshot.

 

As thorium90 pointed out there are many unknowns in this query, it is difficult to understand what exactly is your problem.

 

 

 

Regards.

 

Ajay

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[kkala]

1. A thought, raghumesh, would be to exploit the heat content of mentioned 130 oC gases, yet conclusion would be probably negative. If these gases are coming from combustion, they cannot be cooled below their acid dew point and a safety margin of (say) 15 oC should be added to calculated acid dew point. Besides the gases will lose some pressure due to friction, when passing from cooler.
Increasing gas stream pressure (understood not far from 1 Atma) to 2 bara only by cooling is judged not possible, see example of para 2, post no 3.
Note: In para 2 of post no 3 (by kkala) please read "Pressure will increase by 1/2*2.156*30.9^2  - 1/2*1.332*50^2 = 636 Pa" (apparent error).
2. Concerning "p+0.5*γv^2=constant" (post no 3), practical examples can be found in Physics textbooks, e.g.  venturi meter, pitot tube, Bunsen burner, ejector for vacuum, wind pressure tending to take building roof off during a strong wind, etc.

 

Editing note: Corrected from p+γv^2=constant to p+0.5*γv^2=constant.

Edited by kkala, 06 February 2013 - 05:36 AM.

[raghumech]

So you have some gas stream at some unknown pressure and composition that is at 130C and you want cool it down to some unknown temperature and compress it from that unknown pressure up to 2 bar using some unknown technique that can cool and compress at the same time and you cant use a compressor due to some unknown constraints and you are hoping someone might be able to help you with that?

 

yes ur rit..

[raghumech]

Raghu,

 

You may not be able to get rid of the compressor, but certainly as the system starts with a higher temperature, you will get even higher temperature and that heat if you can recover from compressor outlet stream and re-use elsewhere in the plant, then you will have an energy efficient system.

 

I have done a simulation in hysys for CH4 gas. PFA the snapshot.

 

As thorium90 pointed out there are many unknowns in this query, it is difficult to understand what exactly is your problem.

 

 

 

Regards.

 

Ajay

 

Thanks ajay..

[raghumech]

1. A thought, raghumesh, would be to exploit the heat content of mentioned 130 oC gases, yet conclusion would be probably negative. If these gases are coming from combustion, they cannot be cooled below their acid dew point and a safety margin of (say) 15 oC should be added to calculated acid dew point. Besides the gases will lose some pressure due to friction, when passing from cooler.
Increasing gas stream pressure (understood not far from 1 Atma) to 2 bara only by cooling is judged not possible, see example of para 2, post no 3.
Note: In para 2 of post no 3 (by kkala) please read "Pressure will increase by 1/2*2.156*30.9^2  - 1/2*1.332*50^2 = 636 Pa" (apparent error).
2. Concerning "p+γv^2=constant" (post no 3), practical examples can be found in Physics textbooks, e.g.  venturi meter, pitot tube, Bunsen burner, ejector for vacuum, wind pressure tending to take building roof off during a strong wind, etc.

 

Thanks for the info first of all...

 

Its not the gas coming from combustion, observed that heat from some other mode......actually gas will act as coolant so it observed heat...