Question

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown liquid is 1.5 m and the depth of the oil(specificweight=8.5kN/m3)floating on top is 5.0 m. A pressure gage connected to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid?

Answer 1

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$

where $h_1$ and $h_2$ are the height of the column of oil and the unkown liquid, respectively. Writing for $\gamma_{unk}$, we have

$\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.$

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.