Measuring Pressure — Question 4

Question 4

A light manometer liquid can amplify a small difference

An inverted U-manometer connects two water taps at the same elevation. The upper connecting liquid is oil of density $800\,\mathrm{kg/m^3}$, immiscible with water of density $1000\,\mathrm{kg/m^3}$. The water–oil interface on the A side is $0.12\,\mathrm{m}$ above that on the B side. Use $g=10\,\mathrm{m/s^2}$.

1. Trace a pressure path and find $p_A-p_B$, including its sign.

2. For the same pressure difference, what interface separation would occur with an upper liquid of density $950\,\mathrm{kg/m^3}$? Explain the sensitivity tradeoff.

Question 4 - Solution

A light manometer liquid can amplify a small difference

(a) Derivation and calculation

Let $h=z_L-z_R>0$ and let both taps be at $z_t$. Following A to B,$$p_B=p_A-\rho_wg(z_L-z_t)+\rho_og(z_L-z_R)+\rho_wg(z_R-z_t).$$$$\boxed{p_A-p_B=(\rho_w-\rho_o)gh}.$$$$\boxed{p_A-p_B=(1000-800)(10)(.120)=240\,\mathrm{Pa}}.$$

(b) Result and interpretation

For $\rho_o^{\prime}=950\,\mathrm{kg/m^3}$,$$\boxed{h^{\prime}=240/[(1000-950)(10)]=.480\,\mathrm m}.$$$$\frac{dh}{d(\Delta p)}=\frac1{(\rho_w-\rho_o)g}.$$A smaller density contrast improves height sensitivity but requires more range and more accurate knowledge of that contrast.

Physical check

A is the higher-pressure tap; the inverted arrangement reverses the usual heavy-liquid interface sense.