Fluids at Rest — Question 6

Question 6

A sealed headspace adds an offset to every depth

A closed tank contains a static liquid of density $900\,\mathrm{kg/m^3}$. Absolute pressure sensors at depths $0.20$ and $0.80\,\mathrm{m}$ below the liquid surface read $112.0$ and $117.292\,\mathrm{kPa}$. Take $g=9.80\,\mathrm{m/s^2}$.

1. Check whether the difference is consistent with hydrostatics and infer the headspace pressure.

2. If the headspace pressure rises by $8.0\,\mathrm{kPa}$ with liquid depth unchanged, find both new readings and explain what happens to their difference.

Question 6 - Solution

A sealed headspace adds an offset to every depth

(a) Derivation and calculation

$$\Delta p=117292-112000=5292\,\mathrm{Pa}.$$$$\rho g\Delta h=(900)(9.80)(.80-.20)=5292\,\mathrm{Pa}.$$$$\boxed{p_s=p_1-\rho gh_1=112000-1764=110236\,\mathrm{Pa}}.$$

(b) Result and interpretation

An 8.0 kPa surface-pressure increment gives$$\boxed{p_1^{\prime}=120.0\,\mathrm{kPa}},\qquad\boxed{p_2^{\prime}=125.292\,\mathrm{kPa}}.$$$$p_2^{\prime}-p_1^{\prime}=5.292\,\mathrm{kPa},\quad\frac{dp^{\prime}}{dh}=\rho g.$$The common offset changes absolute pressure but leaves the depth gradient unchanged.

Physical check

Extrapolation from either sensor produces the same headspace pressure.