# Tricky pressure question

1. Aug 30, 2004

### suffian

The problem is stated as follows:

"The following is quoted from a letter. It is the practice of carpenters hereabouts, when laying out and leveling up the foundations of relatively long buildings, to use a garden hose filled with water, into the ends of the hose being thrust glass tubes 10 to 12 inches long. The theory is that water, seeking a common level, will be of the same height in both the tubes and thus effect a level. Now the question rises as to what will happen if a bubble of air is left in the hose. Our greybeards contend the air will not affect the reading from one end to another. Others say that it will cause important innaccuracies. Can you give a relatively simple answer to this question, together with an explanation?"

[Diagram of air bubble above an obstruction such that water is higher than obstruction on each side of air bubble]

Okay, my guess is it can cause incorrect results. The way i see it, two conditions must be satisfied for correct results. When the surface is level, the heights in each tube must be equal. Also, when the surface is uneven the heights in each tube must be unequal. I think I can disprove the first condition.

Suppose the surface is indeed level. Also, let h1 and h2 be the the height of the water in each tube and let x1 and x2 be the rise in water level on the appropriate side near the air bubble. It seems reasonable to me to assume that the pressure over the air bubble is uniform. Then, the external pressure on each side of the air bubble should be equal. Hence, we have (with d = density of water), dg( h1 - x1 ) = dg( h2 - x2 ) or h1 - h2 = x1 - x2. Since the air bubble could be of any size, we might let x1 and x2 be unequal, then the heights in each tube are unequal even though the surface is level and static equilibrium seems to be satisfied (thus incorrect results).

I know it looks like the problem is solved, but i'm not entirely sure, and i couldn't find the answer. Is this correct?

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