Outlet Velocity of Steam through a Valve 3

The usual practice in this situation is to assume an isenthalpic process. T and P are known in the upstream condition. Calculate the enthalpy. Assume the same enthalpy for the downstream where P is known. Calculate T downstream.

Good luck,
Latexman

Latexman is right, use his recommendation.

Unless you use a Self-drag valve, the flow will be acoustically choked and a shock wave would occur at the valve outlet.

Thanks Latexman and Davefitz! really appreciate your help, i've got it now!

Cheers

Just to add a little consideration.

Being the pressure reduction throughout a valve an isenthalpic process (as already pointed out above), steam downstream the valve should have a temperature higher than that a saturated steam would have at the same pressure, and more precisely the temperature increase deltaT would be:

deltaT = (h1 – h2)/cp2

where:

h1 = saturated steam enthalpy at pressure P1 (upstream the valve)
h2 = saturated steam enthalpy at pressure P2 (downstream the valve)
cp2 = saturated steam specific heat at pressure P2 (downstream the valve).

So the final temperature Tf of steam downstream the valve would be

Tf = T2 + deltaT

Where:

T2 = saturated steam temperature at pressure P2

So one should have a superheated steam. This is correct for saturated dry steam.
In the real world steam has always a degree of dryness which lower than 100%. In this case the heat in excess will be used to increase the steam dryness.

ione (Mechanical)
The formula deltaT = (h1 – h2)/cp2

is for a perfect gas, constant specific heat. Yet, the entalpies you are using are for a real gas.

If you know the sat pressure at conditions 1 and 2 , then the sat temperatures and therefore delta T is easily calculated.

sailoday28,

Maybe I am wrong, but I can’t see where the problem is.

The definition of specific heat stands still for saturated steam downstream the valve at the reduced pressure P2 (I think it is undeniable). The process throughout the valve is isenthalpic (undeniable too). Now steam at pressure P2 (downstream the valve) has got an enthalpy, it wouldn’t have if it was at the saturation temperature at pressure P2 (just take a table with steam properties). So there is an excess of enthalpy (difference between enthalpy at upstream pressure P1 and enthalpy at downstream pressure P2) that is used during the transformation to increase the steam temperature.

cp2 x deltaT = h1-h2

You will get exactly the same result plotting an isenthalpic transformation from P1 to P2 on a Mollier diagram (h vs s).

I am pretty sure about what I have stated, but in this forum there many members more authoritative than me that can put a light on this matter.

ione (Mechanical
Having an equation of state and specific heat data, one may obtain expressions for enthalpy or changes in enthalpy.
integ[dH]= integ[CpdT] - integ[d(v*tau)/dtau|p]dp
for non two phase states. Where tau=1/T
Note there are 2 integrals on the right hand side
Your integration assumes the second integral on the RHS is zero and Cp is a constant giving-- delta H= Cp delta T

Typically for a throttling process, delta H is approx zero and therefore for a perfect gas deltaT is zero.

Note, once obtaining H from the integration, Cp may be obtained from DH/DT at constant pressure.

Specific heat (Cp) for a saturated liquid or sat gas is not DH/DT at constant pressure and if you have an old Keenan and Keyes C' and C" are defined as specific hts for sat liq and vapor respectifully.

Maybe I have been not too clear with index.

I should have written

deltaT = (h2 – h2’)/cp2’

where
h2 = enthalpy of steam at pressure P2 (downstream the valve) which is equal to h1 (dry steam enthalpy at pressure P1, upstream the valve)

h2’ = enthalpy of dry saturated steam at pressure P2

cp2’ = specific heat of dry saturated steam at pressure P2


Let’s make a practical example for 1 kg/hr of dry saturated steam from

P1 = 64.7 bar(g) to P2= 0.24 bar(g)

At P1, the steam enthalpy is h1 = 2777.28 kJ/kg

Downstream the valve at P2 the enthalpy is the same (isenthalpic process) h2 = 2777.28 kJ/kg

For a saturated dry steam at pressure P2 the enthalpy should be h2’ = 2685.26 kJ/kg and the specific heat cp2’ = 2.0664 kJ/(kg*K)

Downstream the valve we have a steam which is superheated and its temperature is higher than that of saturation at P2 = 0.24 bar(g) which is be T2’ = 106.069 K.

deltaT = (2777.28 – 2685.26)/2.0664 = 44.531 °C

So the temperature of the steam downstream the valve is

T2 = T2’ + deltaT = 106.069 + 44.531 = 151.221 °C

The same result can be found from a Mollier chart.

`The process is not isenthalpic since it is a flow situation where the energy , i.e. the sum of K.E and enthalpy are conserved, so the difference in the enthalpies is equal to the difference in the kinetic term, V^2/2g.
You have to use the steam tables for this and the possibility of choked flow where the mass flow rate is reduced.
In either case you can iteratively get the differences in h, and go back and get the density ratio or V ratio , etc. differences to meet the energy conservation requirement.

For an adiabatic steadystate process, zekeman gets a star.

I agree with Zekeman that the transformation should not be named isenthalpic. During the process throughout the reduction valve dynamic phenomena take place, and so in a Mollier diagram (h vs s) the transformation should not be represented with a straight horizontal line from initial to final state (isenthalpic transformation).
The transformation in a Mollier diagram could be better depicted by a broken line (first step vertical isentropic line from pressure P1 to pressure P2 and second step isobaric line at pressure P2 to the final state at enthalpy h2 = h1). The thermodynamic transformation which takes place in a pressure reduction valve is the resulting of expansion phenomena with energy transfer due to conversions of static and dynamic pressures. In a throttling valve we have an irreversible expansion with the downstream enthalpy equal to the upstream enthalpy, and this is what matters. For a DRY (ideal) steam, at the end of the process the steam is superheated (and this, from an engineering point of view makes it comparable to a perfect gas). I still consider as valid the approach I’ve described in my previous posts and also the consideration that in the real world depending on the “wetness” of steam (upstream the valve), the superheating effect is minimized.

ione. For the throttling valve, chances are that the KE upstream can be neglected. If the KE downstream ( and not heat transfer) can't be neglected, then h1= stagnation enthalpy2 at pressure2. If and only if, KE2 can be neglectged then h2=h1
And IF the gas is perfect, T2=T1.

I don’t want to be boring and repetitive or appear arrogant, since I have always something to learn, anyway this time I think I will stay with my conviction.

Just one last thing.
Please take a look at the paragraph “What use is the temperature - entropy diagram (or T - S diagram)?” from the link below: there are described the transformations that happen during throttling.
I think (or at least hope) that those guys know their job, dealing with steam from day to night.

I have found by chance on the web this calculator of steam (saturated and superheated) properties.


If you enter the following data for a superheated steam:

Pressure: 0.24 barg
Henthalpy: 2777.28 kJ/kg (the same as that of saturated steam at 64.7 barg)

You get a temperature T = 151.526 °C (practically equal to the value I’ve posted on 16 Feb 2010 3:36).

ione,

Touché!

Often as Engineers, we make judgements between what is practical ("that's close enough") and what is academically correct. It's always good to hear both sides, as we have in this thread, lest we forget to use the academically correct method when conditions warrant it.

Good luck,
Latexman

Latexman,

Thank you very much for your post.
I was not looking for approval at all costs, but I have dealt with this matter more than once and always did the way I’ve described. It was a bit sad and also frustrating to hear my approach was completely wrong. As I have already said above, I have always something to learn but I also consider some things I’ve learnt as a pillar of my (little) knowledge (especially if they've worked well till now).

Ione,

Yes, for steam it turns out that it is far from ideal and a classical throttling process ignores the effects of velocity, since it is usually low and Mollier diagrams are very convenient.

But the OP wants the velocity, not the temperature.

BTW, your example showing a decrease in temperature seems correct to me,although I don't understand why you introduced Cp*delta T , unless you have no access to a Mollier diagram and steam tables.

Can anyone help me with the calculation for the temperature of the fluid at outlet and thus help me with the velocity calculation?

Click to expand...

zekeman,

To me, it sounds like he needs help with T_out and he'll use that to calculate velocity, which he knows how to do. Just my interpretation.

Good luck,
Latexman

Zekeman,

In all my previous posts I have made considerations concerning the superheating effect on dry saturated steam due to pressure reduction, and outlined a different behaviour between real wet steam and dry steam.

I take now the opportunity to add some considerations on velocity (just describe the way I usually proceed in these cases).

The velocity of steam downstream the valve, especially with a pressure reduction as that described by the OP (from 64.7 barg to 0.27 barg,) asks for particular care. It is recommended to keep velocity below a threshold of 200 m/s. Higher velocities produce bothering noise and vibration issues.

The correlation below works (IMO) quite good.

V = 353.7* (Q* vs)/d^2

Where:

V = steam velocity downstream the valve [m/s]
Q= steam flow rate [kg/h]
vs = steam specific volume at downstream pressure P2 [m3/kg]
d = diameter of the body section valve [mm]

With the data given by the OP

Q = 235000 kg/hr (worst case)
vs = 1.54755 kg/m3 (from superheated steam table at P=0.24 barg and h=2777.28 kJ/kg)
d = 610 mm (this value must be checked as only the OP effectively knows it)

V = 353.7*235000*1.54755/372100 = 345.7 m/s (this is a too high value)

Conversely we can calculate the min diameter (dmin) of the reducing valve to ensure a max value
for the steam velocity V = 200 m/s

dmin = SQRT(353.7*Q*vs/V) ? 802 mm ? 31.5” (It seems a valve of 22” is little)