The sketch shows a bubble-tube apparatus. An open tube extends to very near the bottom of a tank of oil. In operation, compressed nitrogen is supplied to the tubing at low pressure. The nitrogen bubbles from the bottom of the tube into the oil so slowly that the pressure in the gas at depth equals the oil pressure adjacent to it (at B). Use the information given.
Calculate the depth of the oil.
♦ The solution procedure is to write the pressure at a point, then write the pressure changes the point would experience were it to move in a path through the liquids and gases of the device and to return to the original point. The path we will follow begins at a point in air (A) immediately beside the gage. The pressure at point A is:
pA
The pressure inside the gage,beside the point A, is greater than the pressure at A by the gage reading. The inside pressure is:
pA + GR
Our imaginary point is now in nitrogen in the center of the gage. Were the point to move do to the bottom of the tube near the point B, its pressure would increase to become:
pA + GR + ρnitrogengo(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. Our point moves horizontally from nitrogen gas to liquid oil with the same pressure. Now we imagine the point to move upward to the surface of the oil, then from the surface upward through air to the point A, where we started and at which the pressure is pA. The extended equation is:
pA + GR + ρnitrogeng(18 m) - ρoil gD - ρatmg(18 m - D) = pA
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)(18 m - D) = 0
Consequently the depth of oil in the tank is calculated:
40,000 Pa + 229 Pa - 9,123 (Pa/m) D - 194 Pa - 10.8 (Pa/m)D = 0
Hence:
D = 4.32 m.
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. Were we to ignored the densities of gases (set them to equal zero) the answer would be 4.38 meters for an error of 1.5%.