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traindriver (Automotive)
22 Oct 09 12:50
Has anyone had experience with generating heat (to heat a fluid to 240 F) by using pressure drop through an orifice plate?  I want to heat hydraulic oil from 70F to 240F without using electric heaters, just pressure drop through a nozzle or an orifice.
Any help with general formulas would be most appreciated.
racookpe1978 (Nuclear)
22 Oct 09 17:22
Think very, very carefully about your fundemental physics, mechanical energy of the fluid stream, and compressed gas "real world" before you go further.

How are you going to compress the fluid you are expanding?  That alone will take more energy than you can get back out of the gas/fluid flowing past the orifice.  Compressing the fluid (a potentially explosive and flamable situation) will in itself increase temperature of the fluid - in this case, the oil -  with electricpower to run the compressor.

Why do you think that any expanding gas or fluid (from high pressure side of the orifice to low pressure side) is going to heat up?  (My experience with flow like that is that expanding gasses cool off.  My physics and heat transfer and fluid flow equations teach me the same thing.  I will grant you your experience - please explain your thermo.)  

Also - Be careful of the flamability of a highly pressurized liquid expanding through an orifice into a low pressure region.  In diesel and gas engines and gas turbine, that process is done by what's called a fuel injector - with good reason.   

You want to heat the flowing fluid at the orifice plate, right?  Please consider how you will do that more clearly.  

Thanks!  Robt
MortenA (Petroleum)
23 Oct 09 1:44
racookpe1978

There are two kinds of adiabatic expenasion: Isentropic (think expander) and isenthalpic (think valve). The temperature following expansion of the first will be lower than for the second - why? Because of the work performed when letting down pressure is "removed" in the expander (enthalpy is kept constant) whereas it "stays in the gas/fluid" when going through a valve (enthalpy is constant). Gas gets colder in both circumstances due to the expansion - but a liquid does not expand that much so the liquid gets varmer. How much depends on dP. It guranteed!

Best regards

Morten
Helpful Member!  Drexl (Chemical)
23 Oct 09 10:51
Hi,

If you have a closed loop (with no heat loss) with a pump and a orifice plate and insert the power P[kW] as mechanical power in the pump this will increase the temperature as:

P * t = m * cp * dT

P = Power [kW]
t = Time you run the pump [s]
m = mass of fluid [kg]
cp = specific heat [kJ/kgK]
dT = temperature increase [K]

  
Helpful Member!  GjFi (Mechanical)
23 Oct 09 11:37
Traindriver:

What is your flow rate (mass rate)?

Amount of heat to warm up the oil from 70F to 240F seems to be considerable.
Heat = Mass rate x Specific heat capacity x Temp. Difference

The heat will be generated by the power loss in the orifice.
Power loss = Heat (assuming an adiabatic process)

The pressure drop needed.
Pressure drop = Power loss / Flow rate

Regards
 
traindriver (Automotive)
23 Oct 09 14:54
Drexl and GjFi  Thanks....This is what I was looking for.  I have not worked with this stuff for 30 years so I needed a little jump start.  I know it seems odd to try and do it this way but I have a very small and limited box (regulations) to work within to get heat generated.

Thanks again
Zesti (Mechanical)
24 Oct 09 16:36

Actually, it is not the pressure drop at the orifice that is going to heat up your fluid.

The pump has an efficiency factor: the useful part of the power driving the pump is put into moving the oil flow with a specific raise in pressure.

The not so useful part is taken up by losses at the impeller, turbulence etc. Ultimately this is all heat.

So, right at the pump, just stirring the fluid if you like, there is a major source of heat going into the fluid.

If you have a tiny orifice, the pump has to work harder to get the fluid round the closed loop. This is more heat going into the fluid at the pump.

The pump is the source of the heat, not the orifice.
A more restrictive orifice will make the pump work harder, generating more heat.

Anyway, why do you not want to use an electric heater?
 
GjFi (Mechanical)
25 Oct 09 17:46
Zesti:

I agree that the energy source is the pump.

But, the pressure drop in the orifice yields a temperature rise in the direction of the flow.
There is about a 1 degF rise in temperature across any hydraulic resistance (orifice, pipe, etc.) for each 140 psi drop if the fluid has a petroleum base for which density = 0.03 lb/in^3.     

Regards
chicopee (Mechanical)
25 Oct 09 22:06
Pressure drop thru an orifice will yield a temperature drop in the expanding side of the fluid.  Any heat generated will be from the orifice skin friction but that increased value will be so much smaller than the decreased temperature value that the orifice will feel colder than warmerfrom the net heat transfer.
MortenA (Petroleum)
26 Oct 09 1:47
chicopee

You are absolutely wrong for two reasons:

1) The original poster is asking for a LIQUID - at some point gas got into this - dont ask me why

2) Where will the work go?

It may be a language barrier - but what is the meaning of ".. the orifice wil feel colder than warmerfrom the heat transfer"?

We have been over this many tile - but take a look at the treads recommended by 25362 above.

Best regards

Morten
CJKruger (Petroleum)
26 Oct 09 15:07
I know liquids are counter-intuitive (as we are used to gases) but MortenA and Drexl are correct.

For an ideal liquid through an orifice/valve, the equation is:

Cp(Tout − Tin) = V(Pin − Pout)

Clearly a temperature rise with drop in pressure.

Cilliers
www.korf.co.uk
Helpful Member!  mauricestoker (Mechanical)
27 Oct 09 8:22
I used to work for the Linde Division, cryogenics, and am familiar with the Linde process.

Whether heat increases or decreases depends on the working medium, pressure and temperature. I think you need to determine the inversion point for the medium in order to answer the question, and whether the Joule Thomson coeffient is postive or negative in the temp/pressure range you are operating.

This works for gases as well as fluids, depending on the gas. Cryogenic production would be much different otherwise.

Never considered it for oil, but it makes sense as the resovoirs will be deep under ground and under pressure. The heat gain would need to be considered. Google oil and inversion point and you will get many, many hits.
zaphod (Mechanical)
27 Oct 09 8:44
traindriver:

Trying to go from 70 to 240 just on pressure drop is possible.

GjFi is correct that it will take about 140 psi drop per deg F you want the oil temperature to rise.  (I'm assuming .45 Btu/lb deg F heat capacity and a density of about .9 specific gravity, both pretty typical of petroleum based hydraulic oils).

So you need to take a total drop of 28,000 psi to get there.

Typical hydraulic pumps are capable of around 5000 psi.  You could run a heatup loop that eventually discharges into wherever you wanted to go.

You'd be a little borderline on hydraulic oil temperature for the standard pump though.  Usually they like to keep things under 180 deg F.  You'd have to run it up to 205 deg F to get your 240 deg F.

A good equation to keep in mind:

Pressure (psi) x Flow (gpm)/1714 = HP

Good luck,
Zaphod
 
zaphod (Mechanical)
27 Oct 09 8:57
mauricestoker:
    I'm familiar with the Joule Thomson effect.  A gas will cool when depressurized because of the expansion of the gas.  I believe the expansion of hydraulic oil from pressurized to depressurized state is so little that JT effect can be safely ignored.

    I used to use hydraulic pumps to generate heat off of diesel engines to heat high pressure liquid nitrogen.  There was no JT effect on the hydraulic oil.  I always got 1 deg F for every 140 psi (roughly).

    When I google Oil and inversion point, it comes back all about water and oil mixtures.  I'm not sure I understand the relevance to this discussion.
mauricestoker (Mechanical)
27 Oct 09 11:01
You need to understand whether heat gain is from friction or from Joule Thomson effect. If you plot temperature versus pressure for solutions to Cp/Cv = 0. All area within the dome will have a positive coefficient, the line definig the graph is zero, and outside of the dome will have a negative value. If negative, throttling will cause an increase in temperature.

For the specific medium, the curve will change with temperature and pressure. In the right range, water will demonstrate the effect, increasing temperature as it flows up, and decreasing when flowing down (gravity is included as the gz term for enthalpy).

For oil I would guess it is important because depending on off-gassing and mixture, as well as the gz factor, the coefficeint may be either positve or negative, meaning the top of the well head could be freezing while the bottom is increasing in temperature.

If the JT can be ignored at the temperature and pressure you use, then the discussion is moot and heat gain is frictinal, not JT. Same can be done with water.
iainuts (Mechanical)
27 Oct 09 14:34
I think the question as posed is missleading.  There's very little temperature rise or fall across an orifice, but that's not really the issue here.  The question might be better posed, "Given a closed loop hydraulic system, can the oil be heated from 70 to 270 F by pumping the oil to pressure and dropping it across an orifice?"

The answer was already provided by Drexl above, along with an equation.  But to clarify that responce, I'd like to point out that there is a significant assumption made in deriving that equation.  That is, the amount of heat loss to atmosphere, either through the tubing wall or through a heat exchanger is small and negligable compared to the work input.  

Draw a control volume around your hydraulic circulation system.  It should encompass the hydraulic circuit, pump and orifice.  Consider also any heat exchangers involved.  

If the amount of heat lost to the environment is small compared to the work input by the pump, then the first law can be reduce to:
dU = Win
To determine work and internal energy change, use the equation provided by Drexl

If we add heat lost to the environment, the first law reduces to:
dU = Win - Qout

You may want to do a simplified analysis of heat transfer from the tubing if there's much tubing.  If there's a heat exchanger in the circuit, it will obviously need to be bypassed.
 
mauricestoker (Mechanical)
27 Oct 09 15:16
The amount of temperature cahnge, and whether it will be postive or negative, is based upon the temperature, pressure, and materials Cp and Cv values.

The equation, to include the JT coefficient, for flow throttling (to include gravity effects) can be found in Fundamentals of Classical Thermodynamics, Wylen and Sonntag. If the value is negative, temperature increases with throttling. If postive, temperature decreases with throttling.
MortenA (Petroleum)
28 Oct 09 1:23
Maurice:

Would you agree with me:

For a gas:

JT is just a "fancy" (read not chemical student) word for an adiabatic isenthalpic expansion (e.g. like in a valve, orifice aor pipeline). Here the energy statys in the fluid e.g. when pressure is reduced due to friction

The "opposite" is adiabatic isentropic expansion (e.g. like in an ideal expander). In this process the gas gets colder than in the isentropic expansion (or JT process) since work is removed from the fluid e.g. by the shaft. A vertical pipe suc as a well string is a mix because pressure is reduced due to less hydrostatic heigh and due to friction.

And then in reality the process is not adiabatic because heat IS transferred with surroundsing - but thats another story.

For a liquid:

Its in principle the same - but normally the change in volume is very small and therefore the heating from work becomes the major factor.

For a two phase system:

Condensation releases a lot of energy and the temperature drop is therefore less than expected for a gas system

Best regards

Morten
mauricestoker (Mechanical)
28 Oct 09 7:38
Morten,

I'd agree that the JT effect is not applicable to ideal gases, and represents a change in internal energy only. I'd agree that theoretic approach is given which does not match reality, and that JT effect is dependent on the material characteristics, specifically Cp and Cv, and ratio of Cp/Cv
for a set temperature and pressure range, determines whether throttling will result in increase in temperature. At STP, argon does not show the effect. At cryo temperatures, it is a major factor. For pulling water or pumping fluids over large changes in elevation , the JT effect will be noticeable (the z term is not neglected ). For water, JT effect changes with gravitational effect, will increase or decrease going up or down over large elevation change.

If an oil field were deep under ground and under large pressure, I would expect that the pressure gradient between atmospheric and field pressure would have to be controlled because of JT effect and fractionation (off-gassed products would have different JT coefficeint, may increase temperature upon throttling).

I believe the effect is primarily due to van der Waals and compressability, not directly from ideal gas law. Not all adiabatic isenthalpic expansion results in temperature increase. It depends on the JT coefficient, pressure and temperature.

I would never have considered this if I hadn't worked in a cryo plant. Feed from nitrogen goes to the oxygen condenser to go below the highest available pressure point in the inversion curve, then JT effect provides the cooling. Oxygen feeds the argon condenser in a daisy chain effect, and enough LN/LOx was always kept on site to reload the condenser any time an impurity occurred. Wish I understood it as well as the large brained mathematician and mechanical engineer they sent from Tonawanda, but it's a few billion working brain cells above my level.

Probably would be a better question for cryo or petroleum forums. I've been out of that area since Bhopal.



 

 
iainuts (Mechanical)
28 Oct 09 7:58
The discussion on isenthalpic pressure drop across an orifice is all well and good, but it isn't the pressure drop that increases the temperature from 70 to 240 F.  If heat isn't added to the system, then the only way to increase the temperature (assuming a closed loop hydraulic system) is to add work.   
mauricestoker (Mechanical)
28 Oct 09 9:12
Iainuts,

You're right, in the temp and pressure range given, the effect is frictional, and JT is of no importance. I would agree that without addition of heat, no work is performed.
I disagree with the rest, and addition of heat does not imply work is performed.

If you think in terms of energy and not work, consider a closed pressure vessel. When heat is added, no work occurs, but internal energy changes. Similar, if a paddle is installed and run by a pump, no work is performed but internal energy changes.

The rate of change in temperature divided by the rate of change in pressure is the JT coefficient. At the state point where change in pressure does not impact change in temperature, the coefficient is zero, and no JT effect takes place. When plotted against against presure and temperature (which is not normally near STP), the graph will be dome shaped. For points within the dome, throttling will cause decrease in temperature. For any point not on the graph or under the dome, throttling will result in increase in temperature.

If you don't agree, I'd recommend going to any cryo plant, or review the mechanical engineering thermo books. Most of all, this does not involve work or enthalpy, it involves internal energy and state points. If you can identify an argon plant that does not recognize this, please put me in touch as currently being an energy engineer, I'll show them how they can possibly save money.


 
iainuts (Mechanical)
28 Oct 09 9:33
Hi Maurice,
Why are you concerned with the JT affect?  What you're saying about the JT affect is all perfectly valid.  But if heat can't be added to the hydraulic fluid, the only way to increase internal energy is to have work performed on the fluid.  That can be done very easily by pressurizing the fluid with a pump and dropping the pressure across an orifice.  That's all the OP is asking about.  The only other consideration is to determine how much heat might be lost in the system as it gets hot.  And yes, I'm quite aware of cryo plants, I've worked in the cryogenic field for over 21 years now, and much of my work is process related.   
mauricestoker (Mechanical)
28 Oct 09 10:43
Maybe I missed the initial question: can you use throttling to increase temperature? I must have missed the part about deadheading the pump.

If you dead head the pump, like mentioned above, you change the internal energy. What does that have to do with the original question, can use of orifice plate or throttling device increase temperature? If you know of another condition at which throttling results in increased temperature, I would very much like the education. Judging by some of the responses above, I would gather that many people assume that you cannot increase temperature by throttling. At least one response asked for the thermo behind it, and there it is.
iainuts (Mechanical)
28 Oct 09 11:25
I think we all agree that we can't significantly increase the temperature of a fluid by throttling.  That's too narrow of a focus, we need to step back and look at the system and the process.  

Also, I'm not suggesting the pump be dead headed.  Typical hydraulic circuits have pumps, and if the hydraulic fluid is forced to flow around a circuit in which there is a pump and a restriction such as an orifice, and if there isn't any work being done by the fluid nor any heat being removed, then the hydraulic fluid will gain all that energy being put in by the pump and won't be rejecting any energy.  End result is the fluid gets hot.  The reason the fluid gets hot is because energy is added at the pump which isn't removed by the isenthalpic expansion.  So if energy isn't removed, the fluid gradually warms up from all the work being done on it.  The fluid only rises by a few degrees with each cycle, but if that heat isn't removed through either a heat exchanger or by having that pressurized fluid do work on something, then the energy remains in the fluid and it gradually gets hot.  I've seen hydraulic circuits run like this for about 15 minutes and get up to 200 F during that time.

We can apply the first law to this and find dU = Win.  The equation provided by Drexl above is the same equation as this.  It appears to me that Drexl has the same interpretation as I do regarding the intent of the OP.  

 
 
davefitz (Mechanical)
28 Oct 09 13:26
You can do it indirectly, and the Linde reference provided the hint.

One means of cooling a gas is by expanding the high pressure gas thru an isentropic expansion, otherwise called a turbine. Linde had done this in the past, and if there was no useful load for the turbine to drive, it might spin a shaft that is immersed in oil with appropriate shear devices in the oil filled space. The oil would then heat up.  As long as the oil is not then transferring heat to the environment, it would continuously heat up.

So, in this case , a source of high pressure gas would expand across a turbine, the tubine shaft would drive a high shear propellor immersed in the oil bath, and the oil would heat up.
mauricestoker (Mechanical)
29 Oct 09 7:15
Thank you, davefitz, you obviously get the point.
mjpetrag (Mechanical)
29 Oct 09 10:16
Say you have compressed liquid water at 5000 psi and 400 F.  
h1 = 381.14 Btu/lbm

You drop this across a valve to 500 psi at constant enthalpy.

T2 =  405.3 (interpolation)

DeltaT = +5.3 across valve


Now if you tried it with steam.
5000 psi @ 1000F -> 1365.5 Btu/lbm
dropping across the valve to 500 psi @ constant enthalpy
T2 = 715.3 F

DeltaT = -284.7 F

 

-Mike

25362 (Chemical)
30 Oct 09 0:29

Miller's J-T inversion curve for all fluids shows that for 1.56 reduced pressure [Pr] -the liquid water case in hand-a reduced temperature [Tr] below 0.84 would involve heating on isenthalpic expansion.

In general, for a Pr≈0 this would happen for all fluids at Tr<0.78 or Tr>5. This curve shows a maximum at Pr>11.8 beyond which all fluids would heat up on J-T expansion.

Liquid water and steam appear to comply with Miller's findings.
chicopee (Mechanical)
7 Nov 09 8:34
I admit I was incorrect MortenA. I examined several phase diagrams and there are chemicals that exhibit a temperature rise with pressure drop in the liquid region when holding enthaly constant.

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