Steam Pressure Measurement Using a Siphon Tube 2

[dchansig]

I’ve come across an issue, which I have never seen an exact calculation for. I need to take a pressure measurement off a steam line at ~330 deg F. I plan on using an uninsulated ¼” stainless steel siphon tube to connect the large steam line to a pressure instrument. The instrument is rated up to 250 deg F. I plan on using a pigtail or “U” shape to isolate the steam. I know there is going to be a heat loss in the siphon tube, but I don’t really know how to calculate how much will be transferred into the ambient or what my temp is going to be at the pressure instrument. The issue I keep having is there is no flow thru this siphon tube. Can someone point me in the right direction?

 

[steamdog]

The siphon tube provides a water (condensate) seal between the gauge and the steam line and this is what protects the gauge from the high temperatures. I do not know of a calculation to figure out the heat loss with this water seal. However, one good piece of advice is to mount the gauge on the side of the pipe rather than on top, this minimizes radiant heat. You can also install a 'tee' near the 'pigtail' and fill it up with water to prevent that first blast of 100 psig steam from overtemping your transmitter.

 

[DRWeig]

Here's a start (see attached).

I have written a more complete article on this topic but I can't seem to find it anywhere...

If I do, I'll post it later.

Good on ya,

Goober Dave

 

[DRWeig]

Let me try to attach the file again, sorry!

Goober Dave

 

[DRWeig]

Other tips:

If the steam line isn't insulated, come off the top of the steam line, then elbow horizontally to get your siphon and transmitter (and double-block-n-bleed valves hopefully) off the top of the pipe. Shield your transmitter from radiation if the steam line isn't insulated also.

Good on ya again,

Goober Dave

 

[dchansig]

DEWeig,

Thanks for the great info. Where did you pull this table from? I need to reference an accepted national standard, and if I were to see the cals that produced this information, that would be great.

Thanks,

dchansig

 

[DRWeig]

Hi dchansig,

I don't think you'll find a national standard on this topic, unless there's something out there for the power plant industry.

I'll try to dig out the originator of the chart, but it might be next week before I get time... Meantime, you can try your Google skills. The research and calculations were originally done by a firm named Data Instruments -- a long, long, long time ago.

Another option is to hang on until some of the heat transfer forum aces pitch in. There are some folks here with real depth on the subject.

Good on ya,

Goober Dave

 

[zekeman]

If you post the steam pressure it is fairly simple to do this analytically.

 

[dchansig]

90lbs

 

[zekeman]

First close off the tube of length L allowing the steam to condense to solid water inside the tube.

Now for the steady state solution we have a tube rejecting heat to the ambient at temperature Tamb so we can write the second order differential equation

K*Pi/4(D0^2-Di^2)*d^2T/dx^2-h*PiD0*(T-Tamb)=0

K*(D0^2-Di^2)/4D0*d^2T/dx^2-h*(T-Tamb)=0

where

K = thermal conductivity SS =10 BTU/hr-Ft-deg F
airlyD0, Di outside and inside dia of tube
T= temperature of tube and water at position x

The solution is

T-Tamb=(T0-Tamb)*(cosh(a(L-x))/cosh(aL)

where it is seen that T0 is temperature at x=0
a=square root(4hD0/K(D0^2-Di^2)=sqrt(60)=7.74
and the second boundary at x= L is satisfied since the solution shows that
dT/dx=0 at x=L

Now find the length L corresponding T to 200 degrees. Substituting in the solution equation

200-100=(330-100)cosh(0)/cosh(aL)

Cosh(0)=1
cosh(aL)=230/100=2.3
aL=1.47
L=1.47/a=1.47/7.74 ft=12*1.47/7.74=2.27 inches

I deliberately used h=1.5 BTU/hr-ft^2-deg F for the film coefficient like the author of the paper cited bl DRWeig.
Remarkably he doesn't include radiation which is at least as big as convection and would result in an even smaller length of tubing. Also, when I tested my values against his for various temperatures, my lengths were consistently smaller.

So on 2 grounds that authors results are conservative.

Fairly simple?

 

[DRWeig]

zekeman,

Thanks for the discourse -- you've unfogged my (not-as-young-as-it-used-to-be) mind.

Best to you,

Goober Dave