Bubble-Tube Depth Gage

The sketch shows a bubble-tube apparatus. A tube, open at the bottom, extends to very near the bottom of a tank of oil. In operation, compressed nitrogen from a cylinder flows slowly through the tubing at low pressure. The nitrogen bubbles from the bottom of the tube into the oil so slowly that the pressure of the gas in the bottom of the tube equals the pressure of the oil adjacent to it (at B). Use the information given.

Calculate the depth of the oil.
♦ For solution, a little experience is required. The procedure is to assume you know the pressure at a point in a fluid. Next choose a second point and to the first pressure, add the change of pressure between the original and second point. Continue this process until, pasing through the fluids, yo return to the original point. The resulting equation will contain the answer.

The path we will follow begins at point (A) in air, immediately beside and outside the gage. We designate the pressure at point A as:

p_A

The pressure inside the gage, immediately beside the point A, is greater than the pressure at A by the gage reading. Thus the inside pressure is:

p_A + GR

Now, "where we are," is in nitrogen in the center of the gage. The next point of interest is in nitrogen, at the bottom of the tube near the point B. The pressure there is greater than at the center of the gage by the amount: ρ_nitrogeng_o(18m). Consequently the pressure is:

p_A + GR + ρ_nitrogeng_o(18 m)

But the pressure in nitrogen inside the tube at its bottom (where it bubbles out) equals the pressure in the oil adjacent to it. So the above pressure is the same as the pressure in the oil at the depth B.

We do two steps now. We consider a point above B, at the surface of the oil in the tank. The pressure at that point will be "less" by: ρ_oilgD. Next moving up through air to the point A, the pressure will decrease by: ρ_atmg(18m - D). But the pressure at point A, where we started and at which the pressure is p_A. The extended equation becomes:

p_A + GR + ρ_nitrogeng(18m) - ρ_oilgD - ρ_atmg(18m - D) = p_A

Since we began at A and returned to A, the equation reduces. With the numbers inserted, it becomes:

40,000 Pa + 1.3 kg/m³(9.81 m/s²)(18 m)

- 930 kg/m³ (9.81 m/s²)D - 1.1 kg/m³(9.81m/s²)(18m - D) = 0

Consequently the depth of oil in the tank is calculated:

150,000 Pa + 229 Pa - 9,123(Pa/m)D - 194 Pa - 10.8(Pa/m)D = 0

Hence:

D = 16.4m.

Notice that the dimension of the depth, "m" for meters, arrives in a natural way with the numerical answer.

Our calculation included the effect of the column of nitrogen gas. That piece is small.