Question 3 of 7: First-Order Thermometer Dynamics

Question 3: First-Order Thermometer Dynamics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) Given. A first-order thermometer exchanging heat with its surroundings by convection, with mass $m$, specific heat $c$, surface area $A$, and a convective heat-transfer coefficient $h$ at the sensor/fluid interface.

Find. The parameters making up the time constant $\tau$, and a check that their combination reduces to time units.

Approach. Write the lumped-capacitance energy balance for the sensing element and read off $\tau$ as the coefficient multiplying $d\theta/dt$, then substitute SI base units for each parameter.

1. Energy balance. Heat entering the sensing mass by convection equals the rate of change of its stored thermal energy: $$hA(T_{fluid}-T_{sensor})=mc\,\dfrac{dT_{sensor}}{dt}.$$ Rearranging into standard first-order form, $\tau\,dT_{sensor}/dt+T_{sensor}=T_{fluid}$, identifies $$\boxed{\tau=\dfrac{mc}{hA}}$$ — the time constant is built from the sensing element's thermal mass ($mc$, its heat-storage capacity) divided by its convective heat-loss rate per degree ($hA$).
2. Unit check. Substituting SI base units for each quantity ($m$ in $\text{kg}$, $c$ in $\text{J}\,\text{kg}^{-1}\text{K}^{-1}$, $h$ in $\text{W}\,\text{m}^{-2}\text{K}^{-1}$, $A$ in $\text{m}^2$): $$[\tau]=\dfrac{\text{kg}\cdot\text{J}\,\text{kg}^{-1}\text{K}^{-1}} {\text{W}\,\text{m}^{-2}\text{K}^{-1}\cdot\text{m}^2} =\dfrac{\text{J}\,\text{K}^{-1}}{\text{W}\,\text{K}^{-1}} =\dfrac{\text{J}}{\text{W}}=\dfrac{\text{J}}{\text{J/s}}=\boxed{\text{s}}.$$ The kilograms and both powers of metres cancel completely, and $\text{J/W}$ reduces to seconds since a watt is a joule per second — confirming $\tau=mc/(hA)$ is dimensionally a time, as it must be.

b) Immerse the thermometer in a bath at a different, steady temperature (a step change) and record its output continuously as it responds. For a true first-order system the response is a single decaying exponential, and $\tau$ is read off directly as the time taken to reach $63.2\%$ of the total step change (equivalently, the time axis intercept of the tangent drawn at $t=0$). Fitting the whole recorded curve to an exponential (rather than reading one point) gives a more robust estimate and also confirms whether the system is genuinely first order.

c) Given. A first-order thermometer of time constant $\tau$ immersed in a liquid whose true temperature rises at a constant rate $r=dT_{true}/dt$ (a ramp input).

Find. The steady-state dynamic (lag) error between the indicated and true temperature.

Approach. Solve the first-order response to a ramp input and identify its steady-state error term.

1. Ramp response. For $\tau\,dT_{ind}/dt+T_{ind}=T_{true}=rt$, after the initial transient dies out the indicated temperature settles into tracking the ramp at a constant time (and hence temperature) lag: $$T_{ind}(t)\to r(t-\tau)=T_{true}(t)-r\tau.$$
2. Steady-state error. $$e_{ss}=T_{true}-T_{ind}=\boxed{r\,\tau}$$ — a first-order sensor under a steadily rising input reads a constant temperature deficit equal to its time constant multiplied by the true rate of rise; it never catches up while the ramp continues, and the faster the liquid heats or the sluggier (larger $\tau$) the thermometer, the larger this lag error.

d) The time constant $\tau=mc/(hA)$ is reduced by attacking either term: shrink the thermal mass $mc$ (a smaller bulb, thinner sheath, less fill fluid), or increase the convective coupling $hA$ (higher fluid velocity past the sensor, forced stirring/circulation, or a larger, thinner-walled surface giving more area per unit mass). A thinner sheath is usually the most practical lever, since it reduces $m$ while simultaneously increasing the surface-to-volume ratio that raises $hA$.

e) A radiation (infrared) pyrometer measures temperature without contact by focusing the object's emitted thermal (blackbody) radiation onto a photodetector and relating the measured radiant power to temperature through the Stefan-Boltzmann relation, $q=\varepsilon\sigma T^4$. Because the reading depends on the object's surface emissivity $\varepsilon$, an accurate non-contact reading requires either a known/assumed emissivity for the target material or a dual-wavelength (ratio) pyrometer design that cancels the emissivity dependence.