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Erschienen in:

Open Access 01.02.2025

Simple thermodynamic model of thermostats for a liquid into a tank: an analytical approach

verfasst von: Robert Salazar, Felipe Deaza, José Zamudio, Leonardo García

Erschienen in: Journal of Engineering Mathematics | Ausgabe 1/2025

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Abstract

In this manuscript, we propose a simplified mathematical model based on the heat transfer laws to predict the temperature profiles of a liquid controlled by a simple thermostat. The model result in a set of linear ordinary differential equations ODEs with forcing which turn on and off at a priori unknown times \({\mathcal {T}}_M=\left\{ \zeta _0,\zeta _1,\ldots ,\zeta _M\right\} \). The pth switch-time \(\zeta _p\in {\mathcal {T}}_p\) is calculated from the zeros of a function \({\mathcal {Q}}(\chi )={\mathcal {Q}}(\chi ;\zeta _1,\ldots ,\zeta _{p-1})\) coming from analytical solutions of the system depending on the previous times \(\zeta _1,\ldots ,\zeta _{p-1}\). The mathematical problem can be solved by using standard techniques for solving ODEs once \({\mathcal {T}}_M\) is calculated by M-successive iterations of the conditional expression \({\mathcal {Q}}(\chi =\zeta _p)=0\) and the Newton–Raphson method. We provide analytical expressions for the temperature as a function of time and \({\mathcal {T}}_M\) considering direct (DC) and alternate (AC) feeding voltages. We solve the system using this numerical-analytical approach and compare it with the results of the 4th Runge–Kutta method finding a good agreement between both methods.
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1 Introduction

In this work, we present a simple mathematical model of a thermostat to control fluid temperature in thermal contact with the environment. With no thermostat, the fluid temperature T(t) will approach the room’s temperature \(T_{\text {env}}\) as \(t\longrightarrow \infty \) to reach the thermal equilibrium. In this scenario, a thermostat is a device that sets a target temperature \(T_{\text {target}} > T_{\text {env}}\) and transfers heat to approach the fluid temperature T(t) to the target.
In practice, thermostats are used to control the temperature of systems. A typical example corresponds to the control temperature in ferment tanks of non-distilled alcoholic beverages, e.g., beer [16] and wine [79]. Thermostats are also important for feeding processes as [10], bioreactors [11], reproduction of bacteria in lactose [12], biofuel production, among others [1318] including micro-controllers and analogical components.
Thermostatted system models are essential in thermodynamics and statistical mechanics for regulating temperature in physical and simulated systems, facilitating the study of equilibrium and non-equilibrium processes [1921]. Classical models like the Lorentz gas and Ehrenfest gas introduced thermostatted dynamics to explore thermal fluctuations, entropy, and equilibrium behavior [22, 23]. These systems allow simulations to transition from Newtonian dynamics (N, V, E) to canonical ensembles (N, V, T) and other configurations, such as constant pressure or enthalpy [24]. Modern advancements incorporate technologies like on-off valves and fuzzy controllers [25, 26] to enhance performance in nonlinear systems [25, 26].
In the modeling of many particle systems via Molecular Dynamics MD, thermostats also play an important role. Even when the thermostats in MD simulations are algorithms programmed in the numerical method, the MD’s thermostats as the real ones define a target temperature \(T_{\text {target}}\) which corresponds to the thermodynamic average of the kinetic energy of the system in the equilibrium. Examples of them in MD are the Nose-Hoover, Langevin, and Andersen thermostats, which properly reproduce the system’s thermal fluctuations in the thermal equilibrium [27, 28].
In the current work, we study a simplified thermostat model based on Newton’s cooling law. For simplicity, it is assumed that the room’s temperature is constant. Even when the model is linear, the analytic study of the system is not necessarily trivial since the forcing depends on the fluid temperature. That term turns on/off depending on the condition where T reaches or not \(T_{\text {target}}\) which occurs at a given set of unknown switch times \({\mathcal {T}}_M=\left\{ \zeta _0,\zeta _1,\ldots ,\zeta _M\right\} \). We shall show that it is possible to build the temperature profiles as a function of the time t from analytical expressions including explicitly the switch times up to time t, this is the solution in the interval \(t\in [\zeta _o, \zeta _M]\). The system studied in this document is essentially an on/off switch controller. These controllers are common in many conventional systems [2932]. We aim to provide solutions for a simplified on/off temperature controller that enables the understanding of its dynamical evolution by using analytical tools.
Previous studies on bioreactors with on-off control model thermal processes by using the analytic solutions of Newton’s law of cooling [33]. In that study the temperature profiles of the step response during heating and cooling are modeled mathematically by using the exact exponential solutions of the Newton’s law of cooling at least for DC voltage fed. The Newton’s cooling law can be also used to derive models for Proportional Integral Derivative (PID) temperature controlling [34].
The document is organized as follows: The model based on Newton’s cooling law is presented in Sect. 2. The next section is devoted to the analytical treatment. In Sect. 3.1 is briefly described the homogeneous solution of the system, this is, with no voltage source. The solution with (direct current) DC and (alternate current) AC voltage are presented in Sects. 3.3 and 3.4, respectively. A brief analysis of the limit cycles emerging on the system as a function of the voltage frequency \(\omega _o\) is presented in Sect. 3.4.1. Section 4 presents a comparison between the analytical expressions and the temperature profiles calculated via the 4th Order Runge–Kutta Method (RK4). A conclusion section is at the end of the manuscript.

2 A theoretical simplified model

The first approach that we are going to consider consists of modeling heat flows using Newton’s cooling law [35]. For instance, for the case fluid at temperature T in thermal contact with the environment at temperature \(T_{\text {env}}\), the law reads:
$$\begin{aligned} \frac{\textrm{d}Q}{\textrm{d}t} = - \kappa _1 S_1 (T-T_{\text {env}}), \end{aligned}$$
here, \(\textrm{d}Q/\textrm{d}t={\dot{Q}}\) is the heat flux, \(\kappa _1\) is the heat transfer coefficient, \(S_1\) is the heat transfer area, and the minus indicates lose of fluid energy \(\textrm{d}Q/\textrm{d}t<0\) if \(T>T_{\text {env}}\). Since phase changes in the fluid are not considered, heat flows will only involve temperature changes depending on the fluid’s specific heat \(c_f\):
$$\begin{aligned} c_f = \frac{1}{m_f} \frac{\textrm{d}Q}{\textrm{d}T}, \end{aligned}$$
then
$$\begin{aligned} \frac{\textrm{d}T}{\textrm{d}t} = - \frac{T-T_{\text {env}}}{\tau _1} \quad \text{ with } \quad \tau _1=\frac{c_f m_f}{\kappa _1 S_1}. \end{aligned}$$
The solution of that equation is the standard exponential decaying \(T(t)=T_{\text {env}}+(T(0)-T_{\text {env}})\exp (-t/\tau _2)\) where \(\tau _1\) is a characteristic time constant. Let us now introduce a resistance completely immersed into the fluid. The temperature of the resistance is \(T'\), and it is fed via a voltage V(t). Since the fluid plays the role of the environment for the resistance, then
$$\begin{aligned} \frac{\textrm{d}Q'}{\textrm{d}t} = - \kappa _2 S_2 (T'-T) + {\dot{q}}, \end{aligned}$$
where \(\kappa _2\) is the heat transfer coefficient resistance/fluid, \(S_2\) the effective area of heat transfer, and \({\dot{q}}\) is the power of the voltage source. The heat transfer can be written as follows \({\dot{Q}}'=c_R m_R {\dot{T}}'\) with \(C_R=c_R m_R\) a heat capacity associated to the resistance. As a result, the dynamic equations for the system including the thermostat are
$$\begin{aligned} \frac{\textrm{d}T}{\textrm{d}t}&= - \frac{T-T_{\text {env}}}{\tau _1} - \frac{T-T'}{\tau _2}, \end{aligned}$$
(1)
$$\begin{aligned} \frac{\textrm{d}T'}{\textrm{d}t}&= - \frac{T'-T}{\tau _3} + \frac{1}{m_R c_R} {\dot{q}}(T,T_{\text {target}},t), \end{aligned}$$
(2)
with \(\tau _2=m_f c_f / (\kappa _2 S_2)\) and \(\tau _3=m_R c_R / (\kappa _2 S_2)\). These expressions Eqs. (1) and (2) should be solved for the fluid and resistance temperature, T(t) and \(T'(t)\), respectively.
We shall assume that target temperature of the thermostat is \(T_{\text {target}}\), and the electric current thought the resistance is activated via a relay switch circuit. Thus, the heating term \({\dot{q}}(T,T_{\text {target}},t)\) is a function defined as follows:
$$\begin{aligned} {\dot{q}}(T,T_{\text {target}},t):= {\left\{ \begin{array}{ll} \dot{{\mathcal {W}}}(t) & \text {if}\; T(t) < T_{\text {target}}, \\ 0 & \text {otherwise}, \end{array}\right. } \end{aligned}$$
(3)
where \(\dot{{\mathcal {W}}}(t)\) is the power supply of the heat source \(\dot{{\mathcal {W}}}(t)=V(t)^2/R\) with R the resistance.
Even when the equations governing the fluid and the heat source temperatures involve a system of coupled first-order ordinary differential equations, solving this system analytically is not straightforward because the function \({\dot{q}}(T,T_{\text {target}},t)\) depends not only on the type of power supply used (direct current, harmonic alternating current, etc.) but also on a conditional statement that includes the fluid temperature T(t). Nevertheless, the linearity of the system enables to deal with the unknown switch times, say \(\left\{ \zeta _0,\zeta _1,\ldots ,\zeta _\infty \right\} \), implicit in the piece-wise function in Eq. (3) by finding analytic expression for the temperatures as it will be shown in the next section.
The governing equations of the system Eqs. (1) and (2) correspond to a liquid at temperature T(t) in thermal contact with the environment at temperature \(T_{\text {env}}\) with a heat source of temperature \(T'(t)\) completely immersed in the fluid. The heat source turns on/off depending on the conditions defined in Eq. (3). These equations determine the dynamic evolution of the temperature profiles of an ideal system controlled by a thermostat. In practice, these thermostats are devices that regulate the energy of the heat source by turning on or off the circuit. There are multiple ways to design a thermostat among them there are the Microcontroller-Based Designs (e.g., Arduino, ESP32, or STM32) these systems often incorporate digital temperature sensors like the DS18B20 or analog sensors like thermistors for accurate measurement. For instance, a standard circuit responsible for controlling the fluid temperature can be managed by a ESP32 [36] microcontroller as it is shown in Fig. 1. This device operates the optocoupler, which in turn activates or deactivates the relay when the temperature falls below the specified target temperature \(T_{\text {target}}\). Additionally, the microcontroller handles communication with three DS18B20 temperature sensors, which monitor the temperature of the fluid T(t), the heater \(T'(t)\), and the surrounding environment \(T_{\text {env}}\).

3 Analytical treatment

In this section, we shall solve the set of equations Eqs. (1), (2) with the source \({\dot{q}}(T,T_{\text {target}},t)\) given by Eq. (3). It is advisable to combine the first-order ODEs Eqs. (1) and (2) to eliminate one the temperature of the resistance. To this, we can write Eq. (1) as follows:
$$\begin{aligned} \frac{\textrm{d}T}{\textrm{d}t} = -\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}\right) T(t) + \frac{1}{\tau _2}T'(t) + \frac{T_{\text {env}}}{\tau _1}, \end{aligned}$$
taking a first derivative with respect to time
$$\begin{aligned} \frac{\textrm{d}^2T}{\textrm{d}t^2} = -\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}\right) \frac{\textrm{d}T}{\textrm{d}t} + \frac{1}{\tau _2}\frac{\textrm{d}T'}{\textrm{d}t}, \end{aligned}$$
combining the previous equation with Eq. (2) we have
$$\begin{aligned} \frac{\textrm{d}^2T}{\textrm{d}t^2} = -\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}\right) \frac{\textrm{d}T}{\textrm{d}t} + \frac{1}{\tau _2}\left[ - \frac{T'-T}{\tau _3} + \frac{1}{m_R c_R} {\dot{q}}(T,T_{\text {target}},t)\right] , \end{aligned}$$
this is
$$\begin{aligned} \frac{d^2T}{dt^2} = -\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}\right) \frac{\textrm{d}T}{\textrm{d}t} + \frac{1}{\tau _2\tau _3}T - \frac{1}{\tau _2\tau _3}T' + \frac{1}{\tau _2} U(T,T_{\text {target}},t), \end{aligned}$$
(4)
where we have defined
$$\begin{aligned} U(T,T_{\text {target}},t):= \frac{1}{m_R c_R} {\dot{q}}(T,T_{\text {target}},t). \end{aligned}$$
(5)
We can isolate \(T'(t)\) from Eq. (1) as follows
$$\begin{aligned} T'(t) = \tau _2 \frac{\textrm{d}T}{\textrm{d}t} + \left( 1+\frac{\tau _1}{\tau _2}\right) T(t) - \frac{\tau _2}{\tau _1} T_{\text {env}}. \end{aligned}$$
(6)
Replacing Eq. (6) in Eq. (4) and arranging the terms, we obtain
$$\begin{aligned} \frac{\textrm{d}^2T}{\textrm{d}t^2} + \left( \frac{1}{\tau _1}+\frac{1}{\tau _2}+\frac{1}{\tau _3}\right) \frac{\textrm{d}T}{\textrm{d}t} + \frac{1}{\tau _1\tau _3} T - \frac{T_{\text {env}}}{\tau _1\tau _3} = \frac{1}{\tau _2}U(T,T_{\text {target}},t). \end{aligned}$$
The factors including the inverse of \(\tau _1,\tau _1\), and \(\tau _3\) have units of angular frequency. Thus, we may define for convenience
$$\begin{aligned} \omega _A:= \frac{1}{2}\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}+\frac{1}{\tau _3}\right) \quad \text{ and }\quad \omega _B = \frac{1}{\sqrt{\tau _1\tau _3}}, \end{aligned}$$
(7)
and the following shifted temperature
$$\begin{aligned} \Theta (t):= T(t) - T_{\text {env}}, \end{aligned}$$
(8)
then
$$\begin{aligned} \frac{d^2\Theta }{dt^2} + 2\omega _A \frac{\textrm{d}\Theta }{\textrm{d}t} + \omega _B^2 \Theta&= \frac{1}{\tau _2}U(T,T_{\text {target}},t) \\&= {\left\{ \begin{array}{ll} {\mathcal {U}}(t)/\tau _2 = \frac{\dot{{\mathcal {W}}}(t)}{\tau _2 m_R C_R} & \text {if}\; T(t) < T_{\text {target}} \\ 0 & \text {otherwise}. \end{array}\right. } \end{aligned}$$
The left-hand side of the is a linear second-order ordinary differential equation which resembles a classical damped harmonic oscillator. On the other hand, the term \(U(T, T_{\text {target}},t)\) can be interpreted as a forcing term. However, we have to remark that \(U(T, T_{\text {target}},t)\) depends explicitly on T(t), in other words, U depends on the dynamical evolution of the system, a priori unknown. The piece-wise condition on \(U(T, T_{\text {target}},t)\) is the way we introduce the relay device into the mathematical model. That device allows or denies the flux of current to the resistance at specific times, say
$$\begin{aligned} {\mathcal {T}} = \left\{ \zeta _0,\zeta _1,\ldots ,\zeta _M \right\} \quad \text{ with }\quad \zeta _0<\zeta _1<\cdots <\zeta _M, \end{aligned}$$
and \(M\in {\mathbb {N}}^{+}\). To write more propriety \(U(T, T_{\text {target}},t)\), we can write
$$\begin{aligned} U(T, T_{\text {target}},t) = {\mathcal {U}}(t)\sum _{(n=0,2,\ldots ,\infty )}\Pi _{\zeta _n,\zeta _{n+1}}(t), \end{aligned}$$
with \(\Pi _{t_1,t_2}(t)\) the boxcar function defined as
$$\begin{aligned} \Pi _{t_1,t_2}(t) = H(t-t_1)-H(t-t_2), \end{aligned}$$
and H(t) the Heaviside step function
$$\begin{aligned} H(t):= 0 \quad {{\textbf {if}}} \quad t<0 \quad {{\textbf {otherwise}}} \quad 1. \end{aligned}$$
Thus, the problem is to solve
$$\begin{aligned} \frac{d^2\Theta }{dt^2} + 2\omega _A \frac{\textrm{d}\Theta }{\textrm{d}t} + \omega _B^2 \Theta = \frac{{\mathcal {U}}(t)}{\tau _2} \sum _{(n=0,2,\ldots ,\infty )}\Pi _{\zeta _n,\zeta _{n+1}}(t), \end{aligned}$$
(9)
including the unknowns \(\left\{ \zeta _n\right\} _{n=0,\ldots \infty }\). The second-order ordinary equation Eq. (9) can be solved by using the Laplace Transform:
$$\begin{aligned} F(s) = {\mathcal {L}}\left\{ \Theta (t) \right\} := \int _{0}^\infty \Theta (t) \exp (-s t) dt. \end{aligned}$$

3.1 The homogeneous solution

Let be \(\theta (t)\) the solution for \({\mathcal {U}}(t)\) is equal to zero (the homogeneous solution), then
$$\begin{aligned} \frac{\textrm{d}^2\theta }{\textrm{d}t^2} + 2\omega _A \frac{\textrm{d}\theta }{\textrm{d}t} + \omega _B^2 \theta = 0, \end{aligned}$$
(10)
the Laplace transform of each term is
$$\begin{aligned} & {\mathcal {L}}\left\{ \frac{\textrm{d}^2\theta }{\textrm{d}t^2} \right\} = s^2 {\mathcal {L}}\left\{ \theta \right\} - s\theta (0^{-}) - {{\dot{\theta }}}(0^-) = s^2 G(s) - s\theta _o - {{\dot{\theta }}}_o\\ & {\mathcal {L}}\left\{ \frac{\textrm{d}\theta }{\textrm{d}t} \right\} = s {\mathcal {L}}\left\{ \theta \right\} - \theta (0^{-}) = s G(s) - s\theta _o, \end{aligned}$$
where the dots denote derivatives with respect the time. Hence, the Laplace transform of Eq. (10) takes the form
$$\begin{aligned} {\mathcal {L}}\left\{ \frac{\textrm{d}^2\theta }{\textrm{d}t^2} + 2\omega _A \frac{\textrm{d}\theta }{\textrm{d}t} + \omega _B^2 \theta \right\} = (s^2 + 2\omega _A s + \omega _B^2)G(s) - (s+2\omega _A)\theta _o - {\dot{\theta }}_o = 0, \end{aligned}$$
therefore,
$$\begin{aligned} G(s) = \frac{(s+2\omega _A)\theta _o + {\dot{\theta }}_o}{s^2 + 2\omega _A s + \omega _B^2}. \end{aligned}$$
One can write \(s^2 + 2\omega _A s + \omega _B^2 = (s + \omega _A+\omega )(s + \omega _A - \omega )\) for a given \(\omega \). It is not difficult to show that \(\omega = \sqrt{\omega _A^2-\omega _B^2}\). Hence, \(s^2 + 2\omega _A s + \omega _B^2 = (s+\omega _+)(s+\omega _-)\) with
$$\begin{aligned} \omega _{\pm } = \omega _A \pm \sqrt{\omega _A^2-\omega _B^2}. \end{aligned}$$
Then one can propose to write
$$\begin{aligned} G(s) = \frac{(s+2\omega _A)\theta _o + {\dot{\theta }}_o}{(s+\omega _+)(s+\omega _-)}:= \frac{K_+^{h}}{(s+\omega _+)} + \frac{K_-^{h}}{(s+\omega _-)}, \end{aligned}$$
where the coefficients \(K_+^{h}, K_-^{h}\) can be found straightforwardly by using partial fractions. They are given by
$$\begin{aligned} K_-^{h} = \frac{(2\omega _A-\omega _-)\theta _o + {\dot{\theta }}_o}{\omega _+ - \omega _-}\quad \text{ and }\quad K_{+}^{h} = \theta _o - K_-^{h}=-\frac{(2\omega _A-\omega _+)\theta _o + {\dot{\theta }}_o}{\omega _+ - \omega _-}, \end{aligned}$$
then the homogeneous solution is
$$\begin{aligned} \theta (t) = {\mathcal {L}}^{-1}\left\{ G(s) \right\} = K_{+}^{h} {\mathcal {L}}^{-1}\left\{ \frac{1}{s+\omega _{+}} \right\} +K_{-}^{h} {\mathcal {L}}^{-1}\left\{ \frac{1}{s+\omega _{-}} \right\} , \end{aligned}$$
this is
$$\begin{aligned} \theta (t) = K_{+}^{h} \exp (-\omega _{+}t) + K_{-}^{h} \exp (-\omega _{-}t). \end{aligned}$$
According to Eq. (8), we have \(\theta (t) = T(t) - T_{\textrm{env}}\) if there is no source of current feeding the circuit, thus
$$\begin{aligned} T(t)&= T_{\textrm{env}} + \frac{(2\omega _A-\omega _-)(T_o-T_{\textrm{env}}) + {\dot{T}}_o}{\omega _+ - \omega _-} \exp (-\omega _{-}t) \nonumber \\&\quad - \frac{(2\omega _A-\omega _+)(T_o-T_{\textrm{env}}) + {\dot{T}}_o}{\omega _+ - \omega _-} \exp (-\omega _{+}t). \end{aligned}$$
(11)
In the previous equation, the initial condition on \({\dot{T}}_o\) can be evaluated from the starting on the fluid and temperatures \(T(0)=T_o\) and \(T'(0) = T'_o\) as follows:
$$\begin{aligned} {\dot{T}}_o = -\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}\right) T_o + \frac{1}{\tau _2}T'_o + \frac{T_{\text {env}}}{\tau _1}, \end{aligned}$$
where we have used Eq. (1). The resistance temperature \(T'(t)\) is computed straightforwardly from Eqs. (11) and (6). The typical behavior of the analytical homogeneous solution with no source is shown in Fig. 2.
Damped and overedamped solutions take place for real values of \(\omega _{\pm } = \omega _A \pm \sqrt{\omega _A^2-\omega _B^2}\). On the other hand, oscillatory solutions require \(\omega _A^2<\omega _B^2\) for complex values of \(\omega _{\pm }\). One can expand explicitly the discriminant \(\sqrt{\omega _A^2-\omega _B^2}\) by using the definitions given in Eq. (7) as follows:
$$\begin{aligned} \omega _A^2-\omega _B^2&= \frac{1}{4}\left( \frac{1}{\tau _1}+\frac{1}{\tau _2}+\frac{1}{\tau _3}\right) ^2 - \frac{1}{\tau _1\tau _3}\\&= \frac{1}{4\tau _1^2} + \frac{1}{4\tau _2^2} + \frac{1}{4\tau _3^2} + \frac{1}{2\tau _1\tau _2} - \frac{1}{2\tau _1\tau _3} + \frac{1}{2\tau _2\tau _3}\\&= \frac{1}{4}\left( \frac{1}{\tau _1^2} + \frac{1}{\tau _3^2} - \frac{2}{\tau _1\tau _3}\right) + \frac{1}{4\tau _2^2} + \frac{1}{2\tau _1\tau _2} + \frac{1}{2\tau _2\tau _3}\\&= \frac{1}{4}\left( \frac{1}{\tau _1} - \frac{1}{\tau _3}\right) ^2 + \frac{1}{2\tau _2}\left( \frac{1}{2\tau _2}+\frac{1}{\tau _1}+\frac{1}{\tau _3}\right) > 0, \end{aligned}$$
since \(\tau _1, \tau _2\), and \(\tau _2\) are positive reals. As a result, \(\omega _{\pm }\in {\mathbb {R}}\) is always real and the homogeneous problem does not have oscillatory solutions. Additionally, the term \(\omega _{+}-\omega _{-}\) given by
$$\begin{aligned} \omega _{+}-\omega _{-} = 2 \sqrt{\omega _A^2-\omega _B^2} = \sqrt{\left( \frac{1}{\tau _1} - \frac{1}{\tau _3}\right) ^2 + \frac{2}{\tau _2}\left( \frac{1}{2\tau _2}+\frac{1}{\tau _1}+\frac{1}{\tau _3}\right) }, \end{aligned}$$
is also a positive real number.

3.2 Non-homogeneous solution

If we take the Laplace transform of equation Eq. (9), we obtain
$$\begin{aligned} (s^2 + 2\omega _A s + \omega _B^2)F(s) - (s+2\omega _A)\Theta _o - {\dot{\Theta }}_o = {\mathcal {L}}\left\{ \frac{{\mathcal {U}}(t)}{\tau _2} \sum _{(n=0,2,\ldots ,\infty )}\Pi _{\zeta _n,\zeta _{n+1}}(t)\right\} , \end{aligned}$$
with F(s) the Laplace transform of the fluid’s shifted temperature \(\Theta (t)\), hence,
$$\begin{aligned} F(s) = \frac{(s+2\omega _A)\Theta _o + {\dot{\Theta }}_o}{(s+\omega _+)(s+\omega _-)} + \sum _{(n=0,2,\ldots ,\infty )}\frac{1}{(s^2 + 2\omega _A s + \omega _B^2)}{\mathcal {L}}\left\{ \frac{{\mathcal {U}}(t)}{\tau _2} \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\} . \end{aligned}$$
We can follow the same procedure of previous section concerning the homogeneous solution for the first term to write
$$\begin{aligned} \Theta (t)&= \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) \\&\quad + \sum _{(n=0,2,\ldots ,\infty )} {\mathcal {L}}^{-1} \left\{ \frac{1}{(s^2 + 2\omega _A s + \omega _B^2)}{\mathcal {L}}\left\{ \frac{{\mathcal {U}}(t)}{\tau _2} \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\} \right\} , \end{aligned}$$
with \(K_{\pm }\) given by
$$\begin{aligned} K_{\pm } = \mp \frac{(2\omega _A-\omega _{\pm })\Theta _o + {\dot{\Theta }}_o}{\omega _+ - \omega _-} = \mp \frac{(2\omega _A-\omega _{\pm })(T_o-T_{\text {env}}) + {\dot{T}}_o}{\omega _+ - \omega _-}. \end{aligned}$$
Then the fluid’s temperature takes the form:
$$\begin{aligned} T(t)&= T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) \nonumber \\&\quad + \sum _{(n=0,2,\ldots ,\infty )} {\mathcal {L}}^{-1} \left\{ \frac{1}{(s+\omega _+)(s+\omega _-)}{\mathcal {L}}\left\{ \frac{{\mathcal {U}}(t)}{\tau _2} \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\} \right\} . \end{aligned}$$
(12)
At this point, we study separately two cases regarding the fed of the resistance: direct current (constant voltage), and alternate current (harmonic voltage).

3.3 DC voltage source

This case corresponds to \({\mathcal {U}}(t) = {\mathcal {U}}_o = \dot{{\mathcal {W}}}_o/m_R C_R\) with \(\dot{{\mathcal {W}}}_o = V_o i_o\) the power, \(V_o\) the voltage and \(i_o\) the current. The Laplace transform of the source terms is
$$\begin{aligned} {\mathcal {L}}\left\{ {\mathcal {U}}(t) \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\}&= {\mathcal {L}}\left\{ {\mathcal {U}}(t) \left[ H(t-\zeta _n)-H(t-\zeta _{n+1})\right] \right\} \\&= {\mathcal {L}}\left\{ {\mathcal {U}}(t)\right\} _{t\longrightarrow t+\zeta _n} \exp \left( -s \zeta _n \right) \\&\quad - {\mathcal {L}}\left\{ {\mathcal {U}}(t)\right\} _{t\longrightarrow t+\zeta _{n+1}}\exp \left( -s \zeta _{n+1} \right) \\&=\frac{{\mathcal {U}}_o}{s}\left[ \exp \left( -s \zeta _{n} \right) -\exp \left( -s \zeta _{n+1} \right) \right] , \end{aligned}$$
then
$$\begin{aligned} T(t) = T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) + \frac{1}{\tau _2}\sum _{(n=0,2,\ldots ,\infty )} {\mathcal {L}}^{-1} \left\{ {\mathcal {F}}_n(s) - {\mathcal {F}}_{n+1}(s) \right\} , \end{aligned}$$
with
$$\begin{aligned} {\mathcal {F}}_m(s):= \frac{\exp (-s \zeta _m)}{s(s+\omega _+)(s+\omega _-)}\quad (m=1,\ldots ,\infty ). \end{aligned}$$
We can use partial fractions to write
$$\begin{aligned} {\mathcal {F}}_m(s):= \frac{\exp (-s \zeta _m)}{s(s+\omega _+)(s+\omega _-)} = \left( \frac{P}{s} + \frac{P_+}{s+\omega _+} + \frac{P_-}{s+\omega _-}\right) \exp (-s \zeta _m). \end{aligned}$$
This implies
$$\begin{aligned} P(s+\omega _+)(s+\omega _-) + P_{+}s(s+\omega _-) + P_{-}s(s+\omega _+) = 1, \end{aligned}$$
or equivalently
$$\begin{aligned} P+P_{+}+P_{-}&= 0\\ (\omega _{+} + \omega _{-}) P + \omega _{-}P_{+} + \omega _{+}P_{-}&= 0 \\ P \omega _{+}\omega _{-}&= 1. \end{aligned}$$
Solving for \(P, P_{\pm }\) one obtain
$$\begin{aligned} P_{\pm } = \mp \frac{1}{\omega _{\pm }(\omega _{+}-\omega _{-})} \quad \text{ and }\quad P = \frac{1}{\omega _+\omega _-}. \end{aligned}$$
The inverse Laplace transform of \({\mathcal {F}}_m(s)\) is
$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ {\mathcal {F}}_m(s)\right\}&= {\mathcal {L}}^{-1}\left\{ \frac{\exp (-s \zeta _m)}{s(s+\omega _+)(s+\omega _-)} \right\} \\&= H(t-\zeta _m) {\mathcal {L}}^{-1}\left\{ \frac{1}{s(s+\omega _+)(s+\omega _-)} \right\} _{t\longrightarrow t-\zeta _m}\\&=H(t-\zeta _m)f_1(t-\zeta _m), \end{aligned}$$
with
$$\begin{aligned} f_1(t)&= {\mathcal {L}}^{-1}\left\{ \frac{1}{s(s+\omega _+)(s+\omega _-)} \right\} \\&=P{\mathcal {L}}^{-1}\left\{ \frac{1}{s} \right\} + P_+{\mathcal {L}}^{-1}\left\{ \frac{1}{s+\omega _+} \right\} + P_-{\mathcal {L}}^{-1}\left\{ \frac{1}{s+\omega _-} \right\} , \end{aligned}$$
this is
$$\begin{aligned} f_1(t):= f(t)H(t) \quad \text{ with }\quad f(t) = \left[ \frac{1}{\omega _+\omega _-} + \sum _{\sigma \in \left\{ +,-\right\} } P_{\sigma } \exp ( - \omega _{\sigma }t)\right] . \end{aligned}$$
(13)
Therefore,
$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ {\mathcal {F}}_m(s)\right\}&= \left[ \frac{1}{\omega _+\omega _-} + \sum _{\sigma \in \left\{ +,-\right\} } P_{\sigma } \exp [- \omega _{\sigma }(t-\zeta _m)]\right] H(t-\zeta _m)\\&= f(t-\zeta _m)H^2(t-\zeta _m). \end{aligned}$$
As a result, the Fluid’s temperature takes the form:
$$\begin{aligned} \boxed { T(t) = T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) + \frac{1}{\tau _2}\sum _{n \in 2{\mathbb {N}}^0} \sum _{m\in \left\{ n,n+1\right\} }(-1)^m \left[ \frac{1}{\omega _+\omega _-} + \sum _{\sigma \in \left\{ +,-\right\} } P_{\sigma } \exp [- \omega _{\sigma }(t-\zeta _m)]\right] H^2(t-\zeta _m) }, \nonumber \\ \end{aligned}$$
(14)
with
$$\begin{aligned} 0 = \zeta _o< \zeta _1< \zeta _2 \ldots \zeta _{n}< \cdots \zeta _{n-1}< \cdots <\zeta _{\infty }, \end{aligned}$$
and under the condition
$$\begin{aligned} T(\zeta _n) = T_{\text {target}} \quad \forall \quad n \in {\mathbb {N}}^0. \end{aligned}$$
(15)
Now we have to compute the times in \(\left\{ \zeta _n \right\} _{n=0,1,\ldots ,\infty }\). We know the first one \(\zeta _o = 0\) assuming that voltage turn on at \(t=0\), this the case \(T(0) < T_{\text {env}}\). The condition in Eq. (15) for \(t=\zeta _1\) is
$$\begin{aligned}&\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _1) + \frac{1}{\tau _2}\sum _{n \in 2{\mathbb {N}}^0} \sum _{m\in \left\{ n,n+1\right\} }(-1)^m\\&\quad \times \left[ \frac{1}{\omega _+\omega _-} + \sum _{\sigma \in \left\{ +,-\right\} } P_{\sigma } \exp [- \omega _{\sigma }(\zeta _1-\zeta _m)]\right] H^2(\zeta _1-\zeta _m) = T_{\text {target}}-T_{\text {env}}, \end{aligned}$$
since the \(\zeta _n's\) are ordered then we know that
$$\begin{aligned} H(\zeta _1-\zeta _m) = 1 \quad {\textbf {if}}\quad m \in \left\{ 0, 1\right\} \quad {\textbf {else}}\quad 0. \end{aligned}$$
Hence,
$$\begin{aligned}&\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _1) + \frac{1}{\tau _2}\sum _{n=0}^\infty \left\{ \sum _{m\in \left\{ 2n,2n+1\right\} } f(\zeta _1-\zeta _m)H^2(\zeta _1-\zeta _m)\right\} \\&\qquad = T_{\text {target}}-T_{\text {env}} \\&\qquad \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _1) \\&\quad \qquad + \frac{1}{\tau _2}\sum _{n=0}^\infty \left\{ f(\zeta _1-\zeta _{2n})H^2(\zeta _1-\zeta _{2n})-f(\zeta _1-\zeta _{2n+1})H^2(\zeta _1-\zeta _{2n+1})\right\} \\&\qquad = T_{\text {target}}-T_{\text {env}}\\&\qquad \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _1) + \frac{1}{\tau _2} \left[ f(\zeta _1-\zeta _0) - f(\zeta _1-\zeta _1)\right] = T_{\text {target}}-T_{\text {env}}\\&\qquad \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _1) + \frac{1}{\tau _2} \left[ f(\zeta _1) - f(0)\right] = T_{\text {target}}-T_{\text {env}}. \end{aligned}$$
The previous relationship enables to compute \(\zeta _1\) from \(\zeta _0=0\). Similarly, we can employ
$$\begin{aligned} H(\zeta _2-\zeta _m) = 1 \quad {\textbf {if}}\quad m \in \left\{ 0, 1,2\right\} \quad {\textbf {else}}\quad 0, \end{aligned}$$
to write for \(t=\zeta _2\)
$$\begin{aligned}&\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _2) \\&\quad \qquad + \frac{1}{\tau _2}\sum _{n=0}^\infty \left\{ f(\zeta _2-\zeta _{2n})H^2(\zeta _2-\zeta _{2n})-f(\zeta _2-\zeta _{2n+1})H^2(\zeta _2-\zeta _{2n+1})\right\} \\&\qquad = T_{\text {target}}-T_{\text {env}}\\&\qquad \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _2) + \frac{1}{\tau _2} \left[ f(\zeta _2-\zeta _0) + f(\zeta _2-\zeta _2) - f(\zeta _2-\zeta _1)\right] \\&\qquad = T_{\text {target}}-T_{\text {env}}\\&\qquad \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _2) + \frac{1}{\tau _2} \left[ f(\zeta _2) + f(0) - f(\zeta _2-\zeta _1)\right] = T_{\text {target}}-T_{\text {env}}. \end{aligned}$$
The last expression enables to calculate \(\zeta _2\) from the previous times \(\zeta _1\) and \(\zeta _0\). This procedure can be generalized for a given \(\zeta _p\). Again we can use
$$\begin{aligned} H(\zeta _p-\zeta _m) = 1 \quad {\textbf {if}}\quad m \in \left\{ 0, 1,2,\ldots ,p\right\} \quad {\textbf {else}}\quad 0, \end{aligned}$$
for any positive integer p such that \(p\le m\). Then at \(t=\zeta _p\)
$$\begin{aligned}&\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _p)\\&\qquad + \frac{1}{\tau _2}\sum _{n=0}^\infty \left\{ f(\zeta _p-\zeta _{2n})H^2(\zeta _p-\zeta _{2n})-f(\zeta _p-\zeta _{2n+1})H^2(\zeta _2-\zeta _{2n+1})\right\} \\&\quad = T_{\text {target}}-T_{\text {env}}\\&\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \zeta _p) \\&\qquad + \frac{1}{\tau _2} \left[ \sum _{n=0}^{\lfloor p/2 \rfloor } f(\zeta _p-\zeta _{2n}) - \sum _{n=0}^{\lfloor (p-1)/2 \rfloor } f(\zeta _p-\zeta _{2n+1})\right] = T_{\text {target}}-T_{\text {env}}, \end{aligned}$$
with \(\lfloor z \rfloor \) the floor function. We can put together in a single sum the sum of the even and odd terms as follows:
$$\begin{aligned} \sum _{n=0}^{\lfloor p/2 \rfloor } f(\zeta _p-\zeta _{2n}) - \sum _{n=0}^{\lfloor (p-1)/2 \rfloor } f(\zeta _p-\zeta _{2n+1}) = (-1)^p f(0) + \sum _{n=0}^{p-1} (-1)^n f(\zeta _p-\zeta _n). \end{aligned}$$
Let be \(\chi =\zeta _p\) then we can define the following function:
$$\begin{aligned} \boxed { {\mathcal {G}}(\chi ;\zeta _{p-1},\ldots ,\zeta _1) = T_{\text {env}}-T_{\text {target}}+\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \chi ) + \frac{1}{\tau _2} \left[ (-1)^p f(0) + \sum _{n=0}^{p-1} (-1)^n f(\chi -\zeta _n)\right] }. \nonumber \\ \end{aligned}$$
(16)
The zeros of this function, this is, the values of \(\chi \) that satisfies
$$\begin{aligned} {\mathcal {G}}(\chi ;\zeta _{p-1},\ldots ,\zeta _1) = 0, \end{aligned}$$
are precisely \(\zeta _1,\zeta _2,\ldots ,\zeta _p\). Of course, we are interested in the last one \(\chi =\zeta _p\) if we know the previous zeros \(\zeta _1,\zeta _2,\ldots ,\zeta _{p-1}\). Then an strategy is to use recursively Eq. (16) to find \(\chi =\zeta _2\) from \(\zeta _1\), later \(\chi =\zeta _3\) from \(\zeta _2, \zeta _1\), after this one compute \(\chi =\zeta _4\) from \(\zeta _3, \zeta _2,\zeta _1\) and so on. An example of code in Wolfram Mathematica to compute the first M switch-times \(\zeta _1,\dots ,\zeta _M\) is shown in Listing 1. Once the switch times are determined, one can use Eq. (14) to evaluate the fluid’s temperature.
A plot of the voltage source and fluids’ temperature for the following numerical values \((\tau _1,\tau _2,\tau _3)=(3,3,4)\), \(T_{\text {target}},T_{\text {env}},T_o,T'_o = (25,20,10,20)\), \(V_o=120\) and \(R=3000\) is shown in Fig. 3. The profiles of that figure were computed by using Eq. (14) truncating the sum up to the first M even numbers, this is
$$\begin{aligned} \sum _{n \in 2{\mathbb {N}}^0} \longrightarrow \sum _{n \in (0,2,4,\ldots ,2M)}, \end{aligned}$$
setting \(M=80\), which corresponds to the number of times that the voltage source turned on & off, or conversely the number of box-cars functions in the plot of \({\dot{q}}(t)\) (see Fig. 3-bottom).
We observe that the fluid’s temperature T(t) (blue solid line) oscillates around target temperature \(T_{\text {target}}\) approaching asymptotically to that value. A zoom of the behavior of T(t) in the range \(t\in [\zeta _10,\zeta _{2M+1}]\) is illustrated in Fig. 4-top. We remark the fact that Eq. (14) includes exponential of real arguments which are by definition non-oscillatory individually. However, these exponential functions collectively turning on at the right times \(\left\{ \zeta _m\right\} _{0<m<\infty }\) generate a oscillatory profile.
The time derivative of the fluid’s temperature is1
$$\begin{aligned} {\dot{T}}(t)&= -\sum _{\sigma \in \left\{ +,-\right\} } \omega _\sigma K_{\sigma } \exp ( - \omega _{\sigma }t) - \frac{1}{\tau _2}\sum _{n \in 2{\mathbb {N}}^0} \sum _{m\in \left\{ n,n+1\right\} }(-1)^m \nonumber \\&\quad \times \left[ \sum _{\sigma \in \left\{ +,-\right\} }\omega _\sigma P_{\sigma } \exp [- \omega _{\sigma }(t-\zeta _m)]\right] H^2(t-\zeta _m). \end{aligned}$$
(17)
This function is continuous in \(t\in [0,\infty )\) including the switch times \(\left\{ \zeta _m\right\} _{m=1,\ldots \infty }\). The temperature of the resistance is computed by replacing T(t) (given by Eq. (14)) and \({\dot{T}}\) (given by Eq. (17)) in Eq. (6). The temporal average of the resistance temperature is
$$\begin{aligned} \langle T'(t) \rangle _{t_1,t_2} = \tau _2 \langle {\dot{T}} \rangle _{t_1,t_2} + \left( 1+\frac{\tau _1}{\tau _2}\right) \langle T(t) \rangle _{t_1,t_2} - \frac{\tau _2}{\tau _1} T_{\text {env}}, \end{aligned}$$
(18)
with
$$\begin{aligned} \langle T'(t) \rangle _{t_1,t_2}:= \frac{1}{t_2-t_1} \int _{t_1}^{t_2} T'(t) dt. \end{aligned}$$
The solution Eq. (14)) tends to \(T_{\text {target}}\) as \(t\longrightarrow \infty \) and \(\langle {\dot{T}} \rangle _{\zeta _1,t}\longrightarrow 0\) as t goes to infinity. The resistance temperature tends to the following value:
$$\begin{aligned} T'_{\text {lim}} = \lim _{t \rightarrow \infty }\langle T'(t) \rangle _{\zeta _1,t} = \left( 1+\frac{\tau _1}{\tau _2}\right) T_{\text {target}} - \frac{\tau _2}{\tau _1} T_{\text {env}}. \end{aligned}$$
(19)
A plot of the resistance temperature \(T'(t)\) is shown in Fig. 4-bottom. One can observe that \(T'(t)\) approaches to \(T'_{\text {lim}}\) (given by Eq. (19)) as \(t\longleftrightarrow \infty \). Even when \(T'(t)\) is a continuous function in \(t\in [0,\infty )\), its times derivative \({\dot{T}}'(t)\) is not continuous since \(T'(t)\) is not smooth at switch-times \(\left\{ \zeta _m\right\} _{m=1,\ldots \infty }\) implying jumps of \({\dot{T}}'(t)\) at that times.

3.4 AC voltage source

This second case includes a harmonic source term \({\mathcal {U}}(t) = \dot{{\mathcal {W}}}(t)/m_R C_R = {\mathcal {U}}_o\cos ^2(\omega _o t)\) with \(\dot{{\mathcal {W}}}_o = V_o^2 \cos ^2(\omega _o t) / R\) the power, \(V_o\) the voltage amplitude, and \(\omega _o\) the angular frequency of the voltage source. Of course, if we take \(\omega _o \longrightarrow 0\), we recover the case studied in the previous section. The Laplace transform of the source terms is
$$\begin{aligned} {\mathcal {L}}\left\{ {\mathcal {U}}(t) \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\}&= {\mathcal {L}}\left\{ \sum _{k\in \{n, n+1\}}{\mathcal {U}}(t) (-1)^k H(t-\zeta _k) \right\} \\&= \sum _{k\in \{n, n+1\}} (-1)^k{\mathcal {L}}\left\{ {\mathcal {U}}(t)\right\} _{t\longrightarrow t+\zeta _k} \exp \left( -s \zeta _k \right) , \end{aligned}$$
assuming that \(n\in 2{\mathbb {N}}^0\). Now,
$$\begin{aligned} {\mathcal {L}}\left\{ {\mathcal {U}}(t)\right\} _{t\longrightarrow t+\zeta _n}&= {\mathcal {L}}\left\{ {\mathcal {U}}_o\cos ^2[\omega _o (t + \zeta _n)] \right\} \\&= \frac{{\mathcal {U}}_o}{2} {\mathcal {L}}\left\{ \exp [{\varvec{i}}\omega _o (t + \zeta _n)] + \exp [-{\varvec{i}}\omega _o (t + \zeta _n)] \right\} \\&= \frac{{\mathcal {U}}_o}{2}\left[ \frac{1}{s} + \left( \frac{s}{s^2+4\omega _o^2}\right) \cos (2\omega _o \zeta _k) - \left( \frac{2\omega _o}{s^2+4\omega _o^2}\right) \sin (2\omega _o \zeta _k)\right] , \end{aligned}$$
hence,
$$\begin{aligned} {\mathcal {L}}\left\{ {\mathcal {U}}(t) \Pi _{\zeta _n,\zeta _{n+1}}(t)\right\}&= \frac{{\mathcal {U}}_o}{2}\sum _{k\in \{n, n+1\}} (-1)^k \left[ \frac{1}{s} + \left( \frac{s}{s^2+4\omega _o^2}\right) \cos (2\omega _o \zeta _k)\right. \\&\left. \quad - \left( \frac{2\omega _o}{s^2+4\omega _o^2}\right) \sin (2\omega _o \zeta _k)\right] \exp (-s \zeta _k). \end{aligned}$$
Replacing the Laplace transform of the source term in Eq. (12), we obtain
$$\begin{aligned} T(t)&= T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t)+ \sum _{(n=0,2,\ldots ,\infty )} {\mathcal {L}}^{-1}\\&\quad \times \left\{ \frac{1}{(s+\omega _+)(s+\omega _-)}\sum _{k\in \{n, n+1\}} (-1)^k \frac{{\mathcal {U}}_o}{2\tau _2}\right. \\&\left. \quad \times \left[ \frac{1}{s} + \left( \frac{s}{s^2+4\omega _o^2}\right) \cos (2\omega _o \zeta _k) - \left( \frac{2\omega _o}{s^2+4\omega _o^2}\right) \sin (2\omega _o \zeta _k)\right] \exp (-s \zeta _k)\right\} , \end{aligned}$$
arranging the terms
$$\begin{aligned} T(t)&= T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) + \frac{{\mathcal {U}}_o}{2\tau _2} \sum _{n\in 2{\mathbb {N}}^0} \sum _{k\in \{n, n+1\}} (-1)^k \\&\quad \times \left[ {\mathcal {L}}^{-1} \left\{ \frac{\exp (-s \zeta _k) }{s(s+\omega _+)(s+\omega _-)} \right\} \right. \\&\quad + \left. {\mathcal {L}}^{-1} \left\{ \frac{s \exp (-s \zeta _k) }{(s^2+4\omega _o^2)(s+\omega _+)(s+\omega _-)} \right\} \cos (2\omega _o \zeta _k) \right. \\&\left. \quad - {\mathcal {L}}^{-1} \left\{ \frac{2\omega _o \exp (-s \zeta _k) }{(s^2+4\omega _o^2)(s+\omega _+)(s+\omega _-)} \right\} \sin (2\omega _o \zeta _k) \right] , \end{aligned}$$
and using the inverse Laplace transform shifting property \({\mathcal {L}}^{-1}\left\{ \exp (-a s)F(s)\right\} = {\mathcal {L}}^{-1}\left\{ F(s)\right\} _{t \longrightarrow t-a} H(t-a)\) the fluid’s temperature takes the form
$$\begin{aligned} T(t)&= T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) + \frac{{\mathcal {U}}_o}{2\tau _2}\sum _{n\in 2{\mathbb {N}}^0} \sum _{k\in \{n, n+1\}} (-1)^k H(t-\zeta _k) \\&\quad \times \left[ {\mathcal {L}}^{-1} \left\{ \frac{1}{s(s+\omega _+)(s+\omega _-)} \right\} _{t \longrightarrow t-\zeta _k} \right. \\&\quad + \left. {\mathcal {L}}^{-1} \left\{ \frac{s}{(s^2+4\omega _o^2)(s+\omega _+)(s+\omega _-)} \right\} _{t \longrightarrow t-\zeta _k} \cos (2\omega _o \zeta _k) \right. \\&\left. \quad - {\mathcal {L}}^{-1} \left\{ \frac{2\omega _o}{(s^2+4\omega _o^2)(s+\omega _+)(s+\omega _-)} \right\} _{t \longrightarrow t-\zeta _k} \sin (2\omega _o \zeta _k) \right] . \end{aligned}$$
The following inverse Laplace transform
$$\begin{aligned} {\mathcal {L}}^{-1} \left\{ \frac{1}{s(s+\omega _+)(s+\omega _-)} \right\} _{t \longrightarrow t-\zeta _k}, \end{aligned}$$
was already calculated in the previous section regarding DC voltage source, and it corresponds to the function \(f_1(t)\) given by Eq. (13). The other functions are
https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_Equ85_HTML.png
The inverse Laplace transforms can be computed straightforwardly again with partial fractions as we have done with \(f_1(t)\). As a result, the fluid’s temperature including an AC thermostat is
https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_Equ86_HTML.png
with
https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_Equ87_HTML.png
and
https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_Equ88_HTML.png
Here, it is convenient to define
https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_Equ89_HTML.png
to write more compactly
$$\begin{aligned} \boxed { T(t) = T_{\text {env}} + \sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma }t) + \frac{{\mathcal {U}}_o}{2\tau _2}\sum _{n\in 2{\mathbb {N}}^0} \sum _{m\in \{n, n+1\}} (-1)^m H^2(t-\zeta _m) {\mathcal {Q}}_m(t-\zeta _m;\omega _o) }. \nonumber \\ \end{aligned}$$
(20)
One can replace the functions f, https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq162_HTML.gif and https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq163_HTML.gif to write explicitly the function \({\mathcal {Q}}_{m}(t)\) as follows:
$$\begin{aligned} {\mathcal {Q}}_m(t,\omega _o)&= \frac{1}{\omega _+\omega _-}+\frac{1}{\omega _{+}-\omega _{-}}\\&\quad \times \sum _{\sigma \in \left\{ +,-\right\} } \left[ \frac{ 2\omega _o\sin (2\omega _o \zeta _m)+\omega _\sigma \cos (2\omega _o\zeta _m)}{(4\omega _o^2 + \omega _{\sigma }^2)}+\frac{1}{\omega _\sigma } \right] \sigma \exp (-\omega _{\sigma }t)\\&\quad +\frac{(\omega _{+}\omega _{-}-4\omega _o^2)\cos [2\omega _o (t+\zeta _m)] + 2\omega _o(\omega _{+}+\omega _{-})\sin [2\omega _o (t+\zeta _m)]}{(4\omega _o^2+\omega _{+}^2)(4\omega _o^2+\omega _{-}^2)}. \end{aligned}$$
Note that the solution for fluid’s temperature regarding AC voltage is mathematically similar to the one found in the case of DC voltage. Essentially one may do the following identification
$$\begin{aligned} f(t-\zeta _m) \longleftrightarrow \frac{1}{2}Q_m(t-\zeta _m;\omega _o), \end{aligned}$$
between Eqs. (14) and (20) to go from the expression for DC to AC voltage. Then there is no need to repeat all the procedures to find iterative conditions to calculate \(\zeta _1,\ldots ,\zeta _p\). It is more convenient to start from Eq. (16) valid for DC voltage and applying the change f by \(\frac{1}{2}Q_m\) to write the expression
$$\begin{aligned} \boxed { {\mathcal {G}}(\chi ;\zeta _{p-1},\ldots ,\zeta _1) = T_{\text {env}}-T_{\text {target}}+\sum _{\sigma \in \left\{ +,-\right\} } K_{\sigma } \exp ( - \omega _{\sigma } \chi ) + \frac{1}{2\tau _2} \left[ (-1)^p Q_o(0;\omega _o) + \sum _{n=0}^{p-1} (-1)^n Q_n(\chi -\zeta _n;\omega _o)\right] }, \nonumber \\ \end{aligned}$$
(21)
which enables with the condition \({\mathcal {G}}(\chi ;\zeta _{p-1},\ldots ,\zeta _1) = 0\) to compute the switching times when a harmonic AC voltage of angular frequency \(\omega _o\) is considered.
We illustrate the fluid’s temperature solution of Eq. (20) setting the parameters as follows: \((\tau _1,\tau _2,\tau _3)=(3,3,4)\), \(T_{\text {target}},T_{\text {env}},T_o,T'_o = (25,20,10,20)\), \(V_o=120\), \(R=3000\) and \(\omega =4\omega _B\).

3.4.1 The effect of the voltage frequency

In the previous subsections, we obtained analytic expression to compute the fluid and resistance temperature as a function of time including a harmonic voltage source of frequency \(\omega _o\). We can observe in Fig. 5 that T(t) and \(T'(t)\) starts in a transitory state in the interval \(t\in (0,\zeta _{2m}]\) including few switch-times (m around 2). After a few additional switch times (typically m around 5), the fluid and resistance temperatures become periodic functions oscillating around \(T_{\text {target}}\) and \(T'_{\text {lim}}\), respectively.
These periodic functions emerging at \(t \longrightarrow \infty \) become in the \(T T'\)-plane limit cycles depending on the value of the voltage angular frequency \(\omega _o\) as shown in Fig. 6. In general, these cycles are closed trajectories or https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq184_HTML.gif -orbits turning https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq185_HTML.gif times around the point \((T_{\text {target}},T'_{\text {lim}})\) (yellow open points in Fig. 6) corresponding to the target temperature and the limit resistance temperature given by Eq. (19). If the voltage frequency is too large compared with the natural frequency \(\omega _o\gg \omega _B\) then the system remains in an orbit of period one, a 1-orbit. This is illustrated in Fig. 6 for \(\omega _o=50\omega _B\). If the frequency \(\omega _o\) is decreased at some point the period of the orbit doubles its value as it is shown for the cases \(\omega _o = 10\omega _B\) and \(\omega _o = \omega _B\) in Fig. 6 where temperatures are in 2-orbit.
The period of the orbits increases rapidly when the frequency \(\omega _o\) becomes smaller with respect \(\omega _B\) as it is shown in Fig. 7 for \(\omega _o=0.75\omega _B\), \(\omega _o=0.5\omega _B\), and \(\omega _o=0.15\omega _B\) regarding a 3-orbit, 9-orbit, and 94-orbit, respectively. One can observe that period https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq198_HTML.gif of the https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq199_HTML.gif -orbit corresponds exactly to the number of times that the voltage source turned on (or turned off) along a period of T(t) or \(T'(t)\). Graphically, this can be observed by counting the shaded horizontal stripes in T vs t plots in Figs. 6 and 7. Mathematically, these cycle limit solutions are given by Eq. (20) for \(t \gg 1\), this is
$$\begin{aligned} T_{\textrm{cycle}}(t) = T_{\text {env}} + \frac{{\mathcal {U}}_o}{2\tau _2}\sum _{n\in 2{\mathbb {N}}^0} \sum _{m\in \{n, n+1\}} (-1)^m H^2(t-\zeta _m) {\mathcal {Q}}_m(t-\zeta _m;\omega _o), \end{aligned}$$
where the exponential part coming from the homogeneous solution of the problem is negligible.

4 Numerical comparison

In the previous sections, we presented an analytical approach to calculate temperature profiles of a system controlled by a simple on-off circuit. These solutions are invaluable for providing deep insights into the underlying physics of the system. However, the correctness of Eq. (20) must be rigorously validated.
To ensure the validity of our solution, we compare it with results obtained from numerical simulations using the fourth-order Runge–Kutta (RK4) method. As independent of the analytical derivation, numerical methods offer a complementary approach to verify accuracy and robustness. Moreover, this comparison highlights the reliability of our analytical solution across the parameter regimes considered. By demonstrating strong agreement between the two methods, we establish confidence in the applicability of our approach to the broader study of temperature profiles of simple on-off systems.
In this section, we shall perform a brief numerical comparison of the results obtained with the Eq. (20) and the 4th-order Runge–Kutta method (RK4).
That numerical method is generally used to generate numerical solutions ODEs, e.g., dynamical governing equations of classical systems including chaotic ones [37, 38]. We have numerically solved the system defined by Eqs. (1) and (2) with source according to Eq. (3) implementing a standard algorithm of the RK4 in Python. The temperature profiles calculated with the RK4 method and the Eq. (20) are shown in Fig. 8. The angular frequency of the voltage source was set as \(\omega _o = 0.5\omega _B\) and the other parameters and starting conditions are the same set in Fig. 5. In Fig. 8 we observe that numerical calculations of the RK4 method (white and yellow symbols for T and \(T'\)) are in excellent agreement with the temperature profiles (solid red and blue lines) calculated with Eq. (20).
The error in fluid temperature profiles between the solution derived from Eq. (20) and the one obtained via RK4 method is shown in Fig. 9. This error is analyzed as a function of the maximum number of time steps used in the RK4 method. If the number of temporal steps is insufficient, the numerical error in the RK4 method can lead to divergent solutions of the ODEs, as illustrated in Fig. 9-top, particularly for high-frequency source switching. This divergence can be mitigated by increasing the maximum number of time steps in the RK4 method. In Fig. 9, the standard deviation \(\sigma = \sqrt{\langle \delta T^2 \rangle }\), where \(\delta T = T^{(\textrm{RK}4)}(t) - T(t)\), quantifies the fluctuations between the numerical temperature \(T^{(\textrm{RK}4)}(t)\) and the analytical temperature T(t) from Eq. (20). The results indicate that \(\sigma \) approaches zero as the maximum number of RK4 steps is increased demonstrating improved agreement between the RK4 method and the analytical solution. This is the case of results presented in Fig. 8 where 300 maximum steps are enough to have a good agreement between both solutions.

5 Limitations

We remark that Eqs. (1) and (2) model the temperature profiles of an idealized version of real thermostats, such as the one illustrated in Fig. 1. There are considerations that must be taken into account to improve the model. Among these considerations is the fact that the fluid’s temperature is not necessarily uniform, as assumed in the model. Since the heat source is completely immersed in the liquid, the fluid in the regions around the resistance could have higher temperatures than regions of the fluid far from the heater. This results in internal temperature gradients and convection effects, which are assumed to be irrelevant in our model but not necessarily in real devices controlled by standard on/off circuits.
Another limitation regards Newton’s law of cooling, which was used to derive Eqs. (1) and (2). This law requires uniform fluid temperature and sufficiently small temperature differences between the heater/fluid and the fluid/environment. These assumptions are not necessarily fulfilled since, depending on the system parameters and initial conditions, the heater temperature can increase significantly (e.g., when using a high-voltage source). In such scenarios, a refined version of Newton’s cooling law, including higher-order temperature differences, must be implemented. Another important assumption of the model is that the environmental temperature is constant, which implies a steady heat transfer between the fluid and the environment. If the environment temperature varies drastically, the strategy of connecting homogeneous solutions of Eqs. (1) and (2) will fail and other considerations would be required to refine the model.

6 Conclusion

In this manuscript, we proposed a simple model to describe thermostats to control the temperature of a fluid. Even when the dynamic equations of the systems are ordinary and linear, they need numerical methods to solve them since the corresponding fed depends on the fluid’s temperature evolution. The system behaves as a damped and driven harmonic oscillator including force term depending on temperature. We showed that the evolution of the homogeneous solution involves pure real parameters \(\omega _{\pm }\in {\mathbb {R}}\) which result in exponential profiles of fluid and resistance temperatures. The forcing term depends on the fluids temperature profile T(t) and more precisely the history of switch-times \(\left\{ \zeta _n\right\} _{n\in {\mathbb {N}}^0}={\mathcal {T}}\) where T(t) reaches the target temperature \(T_{\text {target}}\) of the thermostat. The p-th switch-time \(\zeta _p\) is zero of a function \({\mathcal {G}}(\chi ;\zeta _{p-1},\ldots ,\zeta _1)\) depending on the previous times and satisfying the condition \(\zeta _m > \zeta _{m-1} \forall m=1,\ldots ,p\). This enables us to compute the time \(\zeta _p\) from successive p-iterations of the condition \({\mathcal {G}}(\chi =\zeta _p;\zeta _{p-1},\ldots ,\zeta _1)=0\), as well as, to evaluate the analytic expression of the fluid and resistance temperature.
Typically, in the current model, the fluid and resistance temperature profiles start in a transitional region where the thermostat is still working to approach T(t) to \(T_{\text {target}}\). Once that state ends, the fluid’s temperature can have one of two behaviors depending on the type of voltage used to feed the resistance. On one hand, a DC voltage source results in an oscillatory but asymptotic approach of the fluid’s temperature to the target temperature. On the other, an AC harmonic voltage results in oscillatory solutions of fluid’s temperature \(T_{\textrm{cycle}}(t)\) around the target temperature with no asymptotic approach which depends on the AC frequency \(\omega _o\). For high voltage frequencies \(\omega _o \gg \omega _B\) the system trajectory in the \(TT'\)-plane is a 1-orbit turning around the point \((T_{\text {target}},T'_{\text {lim}})\) with \(T'_{\text {lim}}\) given by Eq. (19). The period of the https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq225_HTML.gif -orbit corresponds to the number of times that voltage source turns on during a period of \(T_{\textrm{cycle}}(t)\). The period https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq227_HTML.gif increases as \(\omega _o\) is decreased recovering the non-periodic DC solution in the limit \(\omega _o \longrightarrow 0\) where https://static-content.springer.com/image/art%3A10.1007%2Fs10665-024-10416-5/MediaObjects/10665_2024_10416_IEq230_HTML.gif goes to infinity.

Acknowledgements

This work was partially supported by Vicerrectoría de investigación, Universidad ECCI.

Declarations

Conflict of interest

The authors declare no competing interests.
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Anhänge

Appendix: Code

Fußnoten
1
This calculus includes the derivatives of a squared Heaviside function term:
$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t}H^2(t-\zeta _m) = 2 H(t-\zeta _m) {\dot{H}}(t-\zeta _m) = 2 H(t-\zeta _m) \delta (t-\zeta _m) = 0, \end{aligned}$$
with \({\dot{H}} = \delta (t)\) the Dirac-delta function. That time derivative is zero since \(H(t)\delta (t)\) is zero for \(t \ne 0\), and
$$\begin{aligned} \lim _{t \rightarrow 0} H(t)\delta (t) = 0. \end{aligned}$$
 
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Metadaten
Titel
Simple thermodynamic model of thermostats for a liquid into a tank: an analytical approach
verfasst von
Robert Salazar
Felipe Deaza
José Zamudio
Leonardo García
Publikationsdatum
01.02.2025
Verlag
Springer Netherlands
Erschienen in
Journal of Engineering Mathematics / Ausgabe 1/2025
Print ISSN: 0022-0833
Elektronische ISSN: 1573-2703
DOI
https://doi.org/10.1007/s10665-024-10416-5

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