Measuring Pressure — Question 8

Question 8

Two calibration points reveal both offset and gain

A pressure sensor obeys $S=S_0+kp_g$, where $S$ is in volts and gauge pressure $p_g$ is in kilopascals. Against a reference manometer it reads $0.40\,\mathrm{V}$ at zero gauge pressure and $2.80\,\mathrm{V}$ at $12.0\,\mathrm{kPa}$.

1. Infer $S_0$ and $k$, then find the gauge pressure for a reading of $1.90\,\mathrm{V}$.

2. At a different site atmospheric pressure is $86\,\mathrm{kPa}$. The sensor still reads $1.90\,\mathrm{V}$ while referenced to the local atmosphere. Find absolute pressure and explain why the original atmospheric pressure is irrelevant.

Question 8 - Solution

Two calibration points reveal both offset and gain

(a) Derivation and calculation

$$S=S_0+kp_g,\quad\boxed{S_0=.400\,\mathrm V}.$$$$\boxed{k=(2.80-.40)/12.0=.200\,\mathrm{V/kPa}}.$$$$\boxed{p_g=(S-S_0)/k=(1.90-.40)/.200=7.50\,\mathrm{kPa}}.$$

(b) Result and interpretation

$$p_{\mathrm{abs}}=p_{\mathrm{atm,local}}+p_g\Rightarrow\boxed{p_{\mathrm{abs}}=86.0+7.50=93.5\,\mathrm{kPa}}.$$The gauge reference is the atmosphere at its current reference port. Calibration determines gain and electrical offset, not a permanent atmospheric pressure.

Physical check

Using $S/k$ without subtracting $S_0$ would introduce a spurious 2.00 kPa.