Diferential pressure measurment 3

Its true that the electrical analogy breaks down when you include the weight of liquids.

I'd vent the air, or you'd have to add the air column weight to the liquid column weight, or neglect the air weight and only measure the liquid column height... in all cases, it complictates the readings.

Regardless of gauge elevation, and at no flow, the differential pressure applied to the gauge will be zero, PROVIDED both gauge tubes are full of water. If there is air in one or both gauge tubes, leading to a difference in water level between the tubes, this difference in water level will be reflected like for like in the differential pressure reading.

Why leave an "air cushion" anyway?

Brian

You will have about 3.24% error in your dp reading due to the elevational difference. However, this is standard and you can consider as the starting point. It is always better to bleed air as the cushioning effect is useless and causes inaccuracy of the measurement.

Thank you all for taking the time to explain this.

My thinking was to cushion the pumping pulsations that I noticed on my pressure gauge. Not needed at all. Just a dumb idea.

It seems to be not as simple as using the initial reading as a starting point, as the fluid levels will change with flow. I see that now!

I'll just bleed the air out. A little air between the last fitting and the gauge should not be significant?

The proper way to do this for both gauges is an orifice ?

I wish I had more head room. It looks like a closed vertical loop from one side to the other would self level and show foh visually in the tube. I'm looking at 15 foh dp.

Now, to highjack my own thread with another question that will once again show my ignorance of fluid flow.

At one time I would winterize my marine v8 motor by disconnecting the water line feeding the exhause and block.

Insert a funnel.
Hold it a foot or so higher than any part of the (open to atmosphere) water jackets.
Pour in antifreeze until it displaced any standing water and ran out the normal way.

This of course did NOT work without adding conciderably more pressure to force fluid thru the maze.

I honestly do not understand what type of chamber internally would prevent the higher level of fluid from flowing thru and seeking the lower level.

Air lock is a term I've heard, but thought it was only related to pumping.

It sounds like you have two pressure gages, not a diff pressure gage. If so, as you start up the readings of two pressure gages at equal levels should be the same, but as flow in the system increases, the system curve head losses will depress the individual readings, but the difference in the readings of the two gages is what you are interested in and that difference should remain proportional to flow in the exchanger at all times.

Air or vapor lock can happen in any "gooseneck" configuration. Even with a higher gravity head driving the fluid, it can take awhile to clear in small diameter lines, especially if the internals rise up to almost as high as the fluid entry point, (which is why most liquid cooled engines come equipped with water pumps), as theoretically if the levels could be arranged nicely, an engine could be (at least partially) cooled by heat driven circulation only.

tnx again,

Actually it IS a dif pressure gauge. Although one side has a tee going to a gauge to measure the static pressure.

I just bled the lines and all is fine now. I was just totally suprised that air in the lines would change things.

So, you are saying that in the motor situation, it will eventually fill under any "open" condition, as long as the fill level is higher than any internal dam? That was my gut feeling, but it just didn't seem to work that way.

I can't visualize any type of trap that would not fill and pass fluid.

In this case the fill line and funnel would just remain full.

Almost... BUT, There are some cases where it might remain locked, unless considerable pressure can be applied. In a series of goosenecks /see diagram below/ and in a start-up flow scenario with liquid at the bottoms of the "U"s and a little push starts that liquid going into the up columns and the air that was trapped in the inverted "U" starts moving into the downcolumns, the head or pressure equivalent at the liquid filling column must be the sum of the weights of the liquid in all liquid filled up-columns... to get them all moving, then once they start moving you must add more pressure to compensate for liquid frictional losses.

The diagram illustrates that with atmospheric pressure at the end of the goosenecks, how each water column adds pressure (0.4333 psi/ft) as it moves higher. (I said the weight of this (special) air in each column is zero and this special air is also not compressible.



As you can see, the additional pressure to lift the liquid columns trapped between vapor-locked goosenecks can get relatively high. And we still haven't cleared the invert elevations of the goosnecks.

Oops, looks like I forgot to move the last liquid column, but I suppose you get the idea. I'll correct it in a bit.

It might also be worth mentioning that if the liquid was gasoline and the pressure at any one of the tops of the goosneck was less than about 8.5 psia at 60ºF, or probably 20 psia or more if its temperature was high, a vapor pocket would appear. At that point in time, the vapor would weigh approximately zero and you would be looking at something almost like the start-up case above where all locked heads must be added together, except you would have some flow going already, but in any case an immediate increase in inlet pressure would be needed to keep things moving along.