Measuring Pressure — Question 3

Question 3

Tap elevation must be included in a differential reading

Two water-filled pressure lines connect taps A and B to a mercury U-manometer. Tap B is $0.50\,\mathrm{m}$ above A. The mercury interface on the B side is $0.10\,\mathrm{m}$ above the A-side interface. Use $\rho_w=1000$, $\rho_m=13600\,\mathrm{kg/m^3}$ and $g=10\,\mathrm{m/s^2}$. All lines are static.

1. Derive and evaluate $p_A-p_B$ using upward coordinate $z$.

2. If the two taps instead sample one connected static water reservoir, what interface-height difference should occur, despite their unequal tap pressures?

Question 3 - Solution

Tap elevation must be included in a differential reading

(a) Derivation and calculation

$$p_L=p_A+\rho_wg(z_A-z_L),\quad p_R=p_B+\rho_wg(z_B-z_R).$$$$p_L-p_R=\rho_mg(z_R-z_L),\quad h=z_R-z_L.$$$$\boxed{p_A-p_B=\rho_wg(z_B-z_A)+(\rho_m-\rho_w)gh}.$$

(b) Result and interpretation

$$\boxed{p_A-p_B=1000(10)(.50)+12600(10)(.10)=17600\,\mathrm{Pa}}.$$For two taps in one static water reservoir,$$p_A-p_B=\rho_wg(z_B-z_A)\Rightarrow(\rho_m-\rho_w)gh=0\Rightarrow\boxed{h=0}.$$

Physical check

The manometer isolates the elevation-corrected pressure difference, which vanishes in a single static water body.