single tube manometer-related question(s)

[colemanstoops]

This is probably so basic it's embarrassing but I'm missing something somewhere along the line.

Say you have a large reservoir (like a railroad tank car) hooked up ultimately to a vertical pipe, the upper end of which is open to the atmosphere. Conceptually the system is described by a single tube manometer, in that the reservoir is closed off to the atmosphere at all times. Now: let's assume that the reservoir is valved off from the "manometer" to begin with, and the valve is opened instantly. The liquid (let's say an oil with a specific gravity of ~0.8) in the reservoir drops modestly and that in the vertical pipe rises. We all know the differences in the elevation of the two liquid surfaces above/below some datum must be equal. What I'm missing/forgetting is some what to predict from the outset what that elevation difference (and thus the pressure in the reservoir) will be. It's evident that the reservoir pressure above the liquid will be subatmospheric, and the level on the reservoir side would remain higher than that on the vertical pipe side, but I'm having trouble quantifying that.

Can someone refresh me on the basics, please?

 

[Compositepro]

Your question is sure confusing so I guess you must be confused. Forget about any "datum". The pressure is equal to the difference of the two liquid levels for any kind of manometer. I assume you are simply referring to the ruler scales for reading the manometers. A U-tube manometer has a scale that goes up and down from zero in the center. This allows a reading to be taken without having to do any arithmetic (assuming the zero is in the right place). If you covered-up one leg of the U-tube so you can't see it it would look like a single tube manometer.

In a U-tube manometer the distance between zero and the one centimeter mark on the scale is actually 1/2 centimeter. In a single tube manometer the reservoir level changes very little so the distance between zero and the one centimeter mark is very slightly less than one centimeter to account for the very slight change in reservoir level.

I think this answers question, but you lost me with your comments about subatmospheric. You are to be commended for at least knowing what a manometer is. I've found that most chemistry and and engineering graduates these days do not know what it is when they see one nor do they know how to read one.

 

[LittleInch]

I had to read this several times and a diagram would have helped, but I think I've now got it.

As you say the level in the reservoir will be higher than the outer tube. The difference in level will be equal to the difference in pressure from atmosphere to the slight negative pressure equated to the head difference of the liquid which will vary with the density of the liquid.

If the reservoir is very large compared to the tube volume, the drop of the reservoir will not be significant.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[chicopee]

For this difference in elevation to take place, there is air movement across the top of the pipe, so you would expect the fluid to rise, however, since the reservoir is sealed up, you would then create a partial vacuum within the top of the fluid within the reservoir. I would not expect any difference in elevation as long as the reservoir is sealed up. The reservoir needs another opening preferably only subjected to static atmospheric air to register a difference in elevation.

 

[Compositepro]

Well,if the reservoir head space is a sealed volume, then you basically have a gas thermometer. So, what was the question?