Modeling the Transient Response of Thermal Circuits

Abstract

Although stationary models for thermal circuits have been widely used, a direct analogy of transient responses of electric circuits to thermal systems is still difficult to establish. In this work, a thermal circuit model for transient responses is developed. The model states that each thermal object is a thermal resistance and a heat capacitor in parallel. The heat capacitor is the heat capacity of the overall material plus a correction term due to the thermal contacts of all thermal objects. The transient response of three basic thermal circuits is modeled, based on the proposed method, and validated, using the heatrapy Python package: single thermal resistance, two thermal resistances in series and two thermal resistances in parallel. A more complex model of a thermal circuit involving a heat source, a heat transfer medium and convection of heat to the surroundings is also developed and validated with data from literature of a thermal switch used in caloric cooling. The proposed method tackles computational issues introduced by the majority of numerical approaches.

1. Introduction

The analysis of heat transfer phenomena is one of the most widely discussed topic in engineering literature [1]. Precise models have been achieved for most situations, with the exception of systems involving fluids, especially when turbulence plays a role. The majority of studies in literature use numerical methods to solve the heat equation [2]

$$\rho c\frac{\partial T}{\partial t}=\nabla ·\left(k\nabla T\right),$$

(1)

where T is the temperature, t the time, k the thermal conductivity, $\rho$ the density, and c the specific heat. The heat equation can be solved analytically but for very strict conditions, which is out of the scope of the great majority of the actual conditions. An alternative technique is the lumped method where thermal elements are considered at 0 D but can only be applied for Biot numbers much below 1:

$$Bi=\frac{hL}{k}\ll 1,$$

(2)

where h is the heat transfer coefficient between the element and the environment, L is the characteristic length of the thermal element and k is the thermal conductivity of the thermal element.

A special case occurs when changes do not occur in time. Stationary thermal circuits have been used extensively to calculate temperatures and heat transfer phenomena in steady state situations [3,4]. It is one of the most simple and precise ways to study thermal processes. Nevertheless, the usual approach does not include the possibility to calculate transient responses. In fact, unlike electric circuits where the velocity of transmission of electric currents is dictated by the speed of light (300,000 ${\mathrm{ms}}^{-1}$), which is large enough to avoid collateral effects, the velocity of change in thermal circuits is extremely low, and, thus, collateral effect occur already for small characteristic lengths. As an example, the diffusion coefficient of copper, which can be seen as the equivalent to the speed of light in electric circuits, is only ~${10}^{-4}$${\mathrm{m}}^2{\mathrm{s}}^{-1}$. This makes it difficult to calculate transient responses, e.g., within the context of thermal circuits that only use thermal resistances connected to reservoirs at constant temperatures and connected between them.

The addition of heat capacitors to existing thermal circuit models was able to approximate transient responses for some applications. Bueno et al. proposed a resistance-capacity network model to investigate the energy performance of buildings [5]. Ogunsola et al. developed a time-series model using thermal capacitors for real-time thermal load estimation [6]. Panão et al. used a lumped model together with RC circuits to study a doubled skin natural and mechanical ventilated test cell [7]. More recently, Michalak and Duan et al. considered RC circuits to analyze the thermal performance of building wall and ventilation flow [8,9], and Hess et al. modeled cascaded caloric refrigeration systems based on thermal diodes or switches using heat capacitances [10]. In all situations, the thermal capacitors were connected to the ground temperature and the respective values were guessed for the specifics of each system. In this context, current tools to analyze general thermal processes lie within the scope of numerical methods, which are problematic to implement in situations where computational resources are limited.

In this work, a new method to analyze heat transient processes is proposed within the scope of thermal circuits by extending thermal elements from thermal resistances to thermal resistances and heat capacitors in parallel. A model to generalize the calculation of heat capacitor values was developed and validated for three simple thermal circuits using numerical calculations from the Python package heatrapy. The method was also validated for a more complex thermal system model using experimental data of a thermal switch from literature.

2. Thermal Circuit Model

2.1. Thermal Element

Steady state models using conventional thermal circuits only include thermal resistances, fixed temperature source and fixed heat power source components connected to each other. The application of such models accesses the temperature of each end of the thermal elements. Generally speaking, thermal resistances can be originated by thermal conductivity, thermal convection, or radiation. For thermal conductivities, the expression for the thermal resistance is [11]

$$R=\frac{L}{kA},$$

(3)

where A is the crossectional area. For the thermal convection, the expression is

$$R=\frac{1}{hA},$$

(4)

where h is the heat transfer coefficient between the thermal elements of the contact. Finally, the thermal resistance for radiation is

$$R=\frac{1}{ϵ\sigma A(T^2+T_s^2)(T+T_s)},$$

(5)

where $ϵ$ is the emissivity of the surface, $\sigma$ is the Stefan–Boltzmann constant, T is the temperature where the radiation is being emitted, and $T_s$ is the surrounding temperature.

To study the transient response, one needs to include the heat capacity and density in the model that are present in the left side of Equation (1), which is no longer zero. The heat capacity is the analogue of electric capacitor elements in electric circuits. In fact, the heat capacity is

$$C=\frac{E}{\Delta T},$$

(6)

where E is the energy that can be stored in the material when subject to a temperature difference $\Delta T$. Note that C already incorporates the total volume of the thermal element. Thus, $C=V\rho c$, where V is the volume of the thermal element and c is the specific heat.

In the proposed approach, one adds a thermal capacitor in parallel to all the thermal resistances, as shown in Figure 1. In this way, by the ohm’s law, at instant $t=0$, heat can only flow through the capacitor and not through the resistance. Then, the analysis of the transient response can be performed in the usual way.

2.2. Heat Capacity Correction

Since the ∇ operator of Equation (1) is applied to the product $k\nabla T$, and since k changes within the circuit, a correction must be performed so that the right side of Equation (1) becomes $k{\nabla }^{2}T$, which is required for applying the laws of circuits. In fact, for electric circuits, ${\nabla }^2\varphi =\frac{1}{v^2}\frac{{\partial }^2\varphi }{\partial t^2}$ [3,12], where $\varphi$ is the electric potential and v is the speed of light. The analogue for thermal circuits is

$${\nabla }^{2}T=\frac{\rho C}{k}\frac{\partial T}{\partial t}.$$

(7)

In this case, we have the first derivative of temperature on the right side. To arrive at Equation (7), one needs to insert a correction in the heat capacity. In fact, taking into account a one-dimensional model, and expanding Equation (1), one has

$$\frac{\rho C}{3}\frac{\partial T}{\partial t}=\frac{\partial k}{\partial x}\frac{\partial T}{\partial x}+k\frac{{\partial }^{2}T}{\partial x^2}.$$

(8)

Note that the division by 3 on the term on the left side is due to the fact that heat is flowing in only one direction. Equation (8) results in

$$\rho C^{\ast }\frac{\partial T}{\partial t}=k\frac{{\partial }^{2}T}{\partial x^2},$$

(9)

where

$$C^{\ast }=\frac{1}{\rho }\left(\frac{\rho C}{3}-\frac{\partial k}{\partial x}\frac{\partial T}{\partial x}\frac{\partial t}{\partial T}\right).$$

(10)

By discretizing,

$$C^{\ast }=\frac{1}{\rho }\left(\frac{\rho C}{3}-\frac{\Delta k}{\Delta x}\frac{\Delta T}{\Delta x}\frac{\Delta t}{\Delta T}\right),$$

(11)

where $\Delta x=(L_1+L_2)/2$, $\Delta k=k_2-k_1$, L is the length of the thermal object and the subscript indices represent the identity of the thermal object. $\Delta T$ can be approximated to its final steady state value $\frac{QR}{2}=\frac{Q{L}_2}{2k_2}$, and $\frac{\Delta T}{\Delta t}\sim \frac{Q}{\rho C}$. Therefore, one has

$$C^{\ast }=C\left(\frac{1}{3}-2\frac{(k_2-k_1)L}{{(L_1+L_2)}^2{k}_2}\right).$$

(12)

Since all thermal objects are in contact with each other, directly or indirectly, we have to include the corrections of all contacts in all thermal objects, so that the general corrected heat capacity of the thermal element i is

$$C^{\ast }=\gamma C,$$

(13)

where

$$\gamma =\left[\frac{1}{3}-2\sum _j\frac{|k_j-k_i|L_j}{{(L_i+L_j)}^2{k}_j}\right].$$

(14)

3. Elementary Thermal Circuits

3.1. Steady State Model

To validate the proposed approach, the steady state and transient response of the three simplest thermal circuits, which are depicted in Figure 2, were calculated. Figure 2a, is made of only one thermal resistance. Figure 2b, is made of two thermal resistances in series and Figure 2c, is made of two thermal resistances in parallel.

The steady state temperatures can be calculated in a very simple way making use of the ohm and Kirchhoff laws applied to the equivalent thermal circuits of Figure 2 (second row). In that respect, one can easily obtain the following expression for the temperature change $\Delta T$ of the thermal circuit A:

$$\Delta T=\frac{QL}{kA}.$$

(15)

By following the same approach, the two temperature changes of the thermal circuit B ($\Delta T_1=Q(R_1+R_2)$ and $\Delta T_2=Q(R_2)$) are:

$$\begin{array}{cc}\hfill \Delta T_1& =\frac{QL}{A}\left(\frac{1}{k_1}+\frac{1}{k_2}\right),\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill \Delta T_2& =\frac{QL}{A{k}_2}.\hfill \end{array}$$

(17)

Finally, for the circuit C, the first Kirchhoff law is applied to the node where the circuit splits, i.e., $Q=q_1+q_2$, while the second Kirchhoff law is applied to the closed circuit at the right side, i.e., $Q{R}_1-Q{R}_2=0$. These two equations result in the following temperature change $\Delta T$:

$$\Delta T=\frac{QL}{A(k_1+k_2)}.$$

(18)

3.2. Transient Response Model

To calculate the transient response of the three circuits, the thermal resistances were substituted by heat capacitors and thermal resistances in parallel, as proposed by the current approach. The resulted thermal circuits are depicted in the third row of Figure 2.

3.2.1. Single Thermal Resistance

By applying the second Kirchhoff law in the closed loop of the thermal element of Figure 2a, one can arrive at $R{q}_2=U/C$, where U is the energy stored in C. By differentiating both sides of this equation and by using the first Kirchhoff law applied to one node of the thermal circuit (note that $q_1=\frac{dU}{dt}$), one can easily arrive at the following simultaneous equations:

$$R\frac{d{q}_2}{dt}=\frac{q_1}{C},$$

(19)

$$q_1=Q-q_2.$$

(20)

Substituting the expression of $q_1$ of Equation (20) in Equation (19), one can obtain the following differential equation:

$$\frac{d{q}_2}{dt}+\frac{q_2}{RC}-\frac{Q}{RC}=0.$$

(21)

This differential equation is linear, ordinary and of first order, admitting solutions of the type $q_2=A{e}^{\lambda t}+B$. Substituting this function in Equation (21), one can obtain $\lambda =-\frac{1}{RC}$ and $B=Q$. Because $q_2(t=0)=0$, $A=-Q$. Therefore,

$$\begin{array}{cc}\hfill q_1(t)& =Q{e}^{-\frac{1}{RC}t},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill q_2(t)& =Q(1-e^{-\frac{1}{RC}t}).\hfill \end{array}$$

(23)

Since $\Delta T=R{q}_2$,

$$\Delta T(t)=RQ\left(1-e^{-\frac{1}{RC}t}\right).$$

(24)

Note that, for $t\to +\infty$, $\Delta T=RQ$. Therefore, the steady state is verified.

3.2.2. Series of Thermal Resistances

Let us consider the thermal circuit of Figure 2b and use the first Kirchhoff law at the nodes where the temperature is being recorded, and the second Kirchhoff law at the two thermal element loops. Hence,

$$\begin{array}{cc}\hfill Q& =q_1+q_2,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill Q& =q_3+q_4,\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill \frac{d{q}_2}{dt}& =\frac{1}{R_1{C}_1}{q}_1,\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill \frac{d{q}_4}{dt}& =\frac{1}{R_2{C}_2}{q}_3.\hfill \end{array}$$

(28)

These four simultaneous equations lead to two uncoupled differential equations:

$$\begin{array}{c}\hfill \frac{d{q}_2}{dt}+\frac{1}{R_1{C}_1}{q}_2+\frac{Q}{R_1{C}_1}=0,\end{array}$$

(29)

$$\begin{array}{c}\hfill \frac{d{q}_4}{dt}+\frac{1}{R_2{C}_2}{q}_4+\frac{Q}{R_2{C}_2}=0.\end{array}$$

(30)

These differential equations are ordinary, linear and of first order. Using the same approach as for the single resistance case, one can arrive at the following solution:

$$\begin{array}{c}\hfill q_2=Q(1-e^{-\frac{1}{R_1{C}_1}t}),\end{array}$$

(31)

$$\begin{array}{c}\hfill q_4=Q(1-e^{-\frac{1}{R_2{C}_2}t}).\end{array}$$

(32)

Therefore, the power currents $q_1$ and $q_3$ are

$$\begin{array}{c}\hfill q_1=Q{e}^{-\frac{1}{R_1{C}_1}t},\end{array}$$

(33)

$$\begin{array}{c}\hfill q_3=Q{e}^{-\frac{1}{R_2{C}_2}t}.\end{array}$$

(34)

Since the temperature changes $\Delta T_1=R_1{q}_2+R_2{q}_4$ and $\Delta T_2=R_2{q}_4$,

$$\begin{array}{cc}\hfill \Delta T_1& =R_{1}Q(1-e^{-\frac{1}{R_1{C}_1}t})+R_{2}Q(1-e^{-\frac{1}{R_2{C}_2}t}),\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill \Delta T_2& =R_{2}Q(1-e^{-\frac{1}{R_2{C}_2}t}).\hfill \end{array}$$

(36)

When $t\to +\infty$, $\Delta T_1=(R_1+R_2)Q$ and $\Delta T_2=R_{2}Q$, which match the steady states of the respective thermal circuit.

3.2.3. Thermal Resistances in Parallel

Let us consider the thermal circuit of Figure 2c. Let us also use the first Kirchhoff law at the nodes where the temperature is being recorded, and at the two nodes where the heat power is split. Let us also use the second Kirchhoff law in the thermal element loops and at the loop that connects $R_1$ and $R_2$. Hence,   

$$Q=q_1+q_2,$$

(37)

$$q_1=q_3+q_4,$$

(38)

$$q_2=q_5+q_6,$$

(39)

$$R_1{q}_3=R_2{q}_5,$$

(40)

$$\frac{d{q}_3}{dt}=\frac{1}{R_1{C}_1}{q}_4,$$

(41)

$$\frac{d{q}_5}{dt}=\frac{1}{R_2{C}_2}{q}_6.$$

(42)

Two of these differential equations are linear, ordinary, of first order and coupled. To solve this system, let us use two new variables, $u_3=\frac{d{q}_3}{dt}$ and $u_5=\frac{d{q}_5}{dt}$. By differentiating all the above equations, by using these two new variables and by doing the necessary algebra, one can decouple the differential equations and arrive at

$$\left[\begin{array}{c}\frac{d{u}_3}{dt}\\ \frac{d{u}_5}{dt}\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{R_1(C_1+C_2)}& -\frac{1}{R_1(C_1+C_2)}\\ -\frac{1}{R_2(C_1+C_2)}& -\frac{1}{R_2(C_1+C_2)}\end{array}\right]\left[\begin{array}{c}{u}_3\\ u_5\end{array}\right]$$

(43)

$$\begin{array}{cc}\hfill & =M\left[\begin{array}{c}{u}_3\\ u_5\end{array}\right],\hfill \end{array}$$

(44)

where M is a square matrix. From the nature of this last system, we know that $u_3$ and $u_5$ are linear combinations of two exponential functions with time constants given by the eigenvalues of the matrix M. In that respect, M has the trivial eigenvalue ${\lambda }_0=0$ and the eigenvalue ${\lambda }_1=-\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$, where $C^{†}=C_1+C_2$. The respective eigenvectors are

$$v_{{\lambda }_0}=\left[\begin{array}{c}1\\ -1\end{array}\right],v_{{\lambda }_1}=\left[\begin{array}{c}1\\ \frac{R_1}{R_2}\end{array}\right].$$

(45)

Therefore,

$$\left[\begin{array}{c}{u}_3\\ u_5\end{array}\right]=A\left[\begin{array}{c}1\\ 1\end{array}\right]+B{e}^{{\lambda }_{1}t}\left[\begin{array}{c}1\\ \frac{R_1}{R_2}\end{array}\right],$$

(46)

where A and B are constants. Hence, one can obtain:

$$q_4(t)=R_1{C}_{1}B{e}^{{\lambda }_{1}t}+D,$$

(47)

$$q_6(t)=R_2{C}_{2}B{e}^{{\lambda }_{1}t}+E,$$

(48)

where D and E are also constants. Since $q_4(t\to \infty )=0$, then $D=0$. The same applies to $q_6$, so that $E=0$. Since $q_3$ and $q_5$ are initially 0, (open circuit), and by the definition of thermal capacitance, we have:

$$\frac{q_4}{C_1}=\frac{Q-q_4}{C_1}.$$

(49)

Therefore, $q_4(t=0)=\frac{C_1}{C_1+C_2}Q$, which implies that $B=\frac{Q}{C^{†}{R}_1}$. Hence, one can finally arrive at   

$$\begin{array}{c}\hfill q_4=\frac{C_{1}Q}{C^{†}}{e}^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t},\end{array}$$

(50)

$$\begin{array}{c}\hfill q_6=\frac{C_{2}Q}{C^{†}}{e}^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t}.\end{array}$$

(51)

By Equation (41),

$$\begin{array}{c}\hfill q_3=\frac{1}{R_1{C}_1}\frac{C_{1}Q}{C^{†}}\left(\int e^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t}dt+D^{\prime }\right).\end{array}$$

(52)

Since $q_3(t=0)=0$, then $D^{\prime }=\frac{Q{R}_2}{R_1+R_2}$. Hence, one finally arrives at:

$$\begin{array}{c}\hfill q_3(t)=\frac{Q{R}_2}{R_1+R_2}\left[1-e^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t}\right].\end{array}$$

(53)

The same approach can be used to find $q_5$:

$$\begin{array}{c}\hfill q_5(t)=\frac{Q{R}_1}{R_1+R_2}\left[1-e^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t}\right].\end{array}$$

(54)

Since $\Delta T(t)=R_1{q}_3=R_2{q}_5$,

$$\begin{array}{c}\hfill \Delta T(t)=\frac{Q{R}_1{R}_2}{R_1+R_2}\left[1-e^{\frac{1}{C^{†}}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)t}\right].\end{array}$$

(55)

3.3. Validation

To validate the developed transient response model, the obtained analytical expressions were compared to thermal simulations of the equivalent circuits using the heatrapy Python package, which has been validated using several benchmark tests [13,14]. As a first test, the analytical expression for a copper block of 1 ${\mathrm{m}}^2$ area and 1 m long, under a heat power source of Q = 1 kWm⁻³ and ground temperature of 293 K, was used. The heatrapy simulation was obtained by using the one-dimensional class SingleObject1D, with a discretization of 10 points, $\Delta x=0.1$ m and $\Delta t=0.1$ s. Figure 3 shows the time dependency of the temperature $T_1$ of the thermal circuit of Figure 2a. It is evident that both the steady state temperature and time constant of the transient model agree, with a very high level of accuracy, with the simulated heatrapy curve.

The validation of the model for the thermal circuit with two thermal objects in series was performed for two situations: Cu as the thermal resistance in contact with the power source and Si as the thermal object in contact with the ambient; Si as the thermal resistance in contact with the power source and Cu as the thermal object in contact with the ambient. These two situations were considered to verify if the heat capacity correction is valid. Figure 4a,b show the temperature evolution when the thermal object in contact with the heat power source is Cu and Si, respectively. As in the single resistance case, the used heatrapy simulation considered a 10-point discretized model, with 1 ${\mathrm{m}}^2$ of area, 1 m long, under a heat power source of Q = 1 kWm⁻³ and a ground temperature of 293 K. One can observe again that, for the steady state, the temperatures are almost identical. Since the developed model agrees with the heatrapy simulations for the two described situations, then one can conclude that the heat capacity correction is valid.

To create a heatrapy model for the circuit with parallel thermal resistances, a two-dimensional class was used. Moreover, since power sources need to have a material in common, an additional thermal object was added. The fact that an additional Al thermal element was added does not imply that the found equation for $T(t)$ is not valid since the currents $q_1$ and $q_2$ are decoupled from the other currents. The two-dimensional heatrapy model was developed with the SingleObject2D class. The system used a 2 × 2 m square, with the left half being Al, and the second half being Cu and Si separated by a vacuum medium line. Figure 5 shows the temperature evolution of $T_1$ and $T_2$ using the developed model and the heatrapy simulations. Once again, the curves agree with a very high degree of accuracy.

4. Analytical Modeling of a Complete Thermal System

Recurrent thermal systems include heat sources and several transfer media that connects the heat sources to the surrounding air. Since this type of system appears in common engineering problems, a complete thermal system was modeled and compared with an experiment in literature. The system is depicted in Figure 6a. It consists of a heat source that delivers heat to the air and to a heat transfer medium. The heat transfer medium also rejects excess heat to the air. The temperature closest to the heat source is termed $T_{hot}$, while the temperature at the end of the heat transfer medium (in contact with the air) is termed $T_{cold}$. In the following, the steady state and transient response models will be developed.

4.1. Steady State Model

The thermal circuit used to model the complete system is depicted in Figure 6b. The thermal resistances $R_1$ and $R_3$ represent the resistance of the heat flow to the air, and the thermal resistance $R_2$ represents the resistance of the heat flow within the heat transfer medium. Using the Kirchhoff’s circuit laws applied to thermal circuits, one can write the following equations:

$$Q=q_1+q_2,$$

(56)

$$R_1{q}_1=(R_2+R_3)q_2.$$

(57)

Solving this system of equations, one can obtain the heat currents $q_1$ and $q_2$:

$$q_1=\frac{R_1}{R_1+R_2+R_3}Q,$$

(58)

$$q_2=\frac{R_2+R_3}{R_1+R_2+R_3}Q.$$

(59)

Defining the ground temperature as being $T_{amb}$, and since the temperatures $T_{cold}$ and $T_{hot}$ can be obtained by $q_2{R}_3+T_{amb}$ and $q_1{R}_1+T_{amb}$, respectively, then one arrives at

$$T_{cold}=\frac{R_1{R}_3}{R_1+R_2+R_3}Q+T_{amb},$$

(60)

$$T_{hot}=R_1\frac{R_2+R_3}{R_1+R_2+R_3}Q+T_{amb}.$$

(61)

Hence,

$$\Delta T=T_{hot}-T_{cold}=\frac{R_1{R}_2}{R_1+R_2+R_3}Q.$$

(62)

Since $R_2=\frac{L}{kA}$, $R_1=\frac{1}{h_{1}Af}$ and $R_3=\frac{1}{h_{3}A}$, where A is the sectional area, f is the fraction of area in contact with the ambient, L the thickness of the heat transfer medium, k the heat transfer medium effective thermal conductivity and h the heat transfer coefficients,

$$\Delta T=\frac{L{h}_3}{A(k{h}_3+L{h}_{1}f{h}_3+h_{1}fk)}Q.$$

(63)

4.2. Transient Response Model

To determine the transient response model, each thermal resistance of the steady-state model was converted into a thermal resistance and heat capacitor in parallel. The new circuit is depicted in Figure 6c. Using the Kirchhoff’s circuit laws, one can write the following eight simultaneous equations:

$$Q=q_1+q_2,$$

(64)

$$q_1=q_5+q_6,$$

(65)

$$q_2=q_7+q_8,$$

(66)

$$q_2=q_3+q_4,$$

(67)

$$\frac{d{q}_5}{dt}=\frac{1}{R_{1}C}{q}_6,$$

(68)

$$\frac{d{q}_3}{dt}=\frac{1}{R_{2}C}{q}_4,$$

(69)

$$\frac{d{q}_7}{dt}=\frac{1}{R_{3}C}{q}_8,$$

(70)

$$R_1{q}_5=R_2{q}_3+R_3{q}_7.$$

(71)

Since $q_2=q_7+q_8$ and $q_2=q_3+q_4$, then $q_7+q_8=q_3+q_4$. Moreover, $Q=q_1+q_2=(q_5+q_6)+(q_3+q_4)$. Let us also consider three new variables:

$$u_3=\frac{d{q}_3}{dt},$$

(72)

$$u_5=\frac{d{q}_5}{dt},$$

(73)

$$u_7=\frac{d{q}_7}{dt}.$$

(74)

Then, by differentiating the equations, we have:

$$u_3+\frac{d{q}_4}{dt}=-u_5-\frac{d{q}_6}{dt},$$

(75)

$$u_7+\frac{d{q}_8}{dt}=u_3-\frac{d{q}_4}{dt},$$

(76)

$$\frac{d{u}_5}{dt}=\frac{1}{R_{1}C}\frac{d{q}_6}{dt},$$

(77)

$$\frac{d{u}_3}{dt}=\frac{1}{R_{2}C}\frac{d{q}_4}{dt},$$

(78)

$$\frac{d{u}_7}{dt}=\frac{1}{R_{3}C}\frac{d{q}_8}{dt},$$

(79)

$$\frac{d{q}_6}{dt}=\frac{d{q}_4}{dt}+\frac{d{q}_8}{dt}.$$

(80)

Using Equations (75), (76) and (80), and grouping them into a single system, one arrives at the following system of three differential equations in the matrix notation:

$$\left[\begin{array}{c}\frac{d{u}_4}{dt}\\ \frac{d{u}_5}{dt}\\ \frac{d{u}_7}{dt}\end{array}\right]=M\left[\begin{array}{c}{u}_4\\ u_5\\ u_7\end{array}\right],$$

(81)

where

$$M=\left[\begin{array}{ccc}-\frac{2}{3C{R}_2}& -\frac{1}{3C{R}_2}& \frac{1}{3C{R}_2}\\ -\frac{1}{3C{R}_1}& -\frac{2}{3C{R}_1}& -\frac{2}{3C{R}_1}\\ \frac{1}{3C{R}_3}& -\frac{1}{3C{R}_3}& -\frac{2}{3C{R}_3}\end{array}\right].$$

(82)

These three differential equations are coupled, but they are linear and of first order. Therefore, it suffices to obtain the eigenvalues of the matrix M to calculate the time constants. In this case, there are three time constants:

$$det\left[\begin{array}{ccc}-\frac{2}{3C{R}_2}-\lambda & -\frac{1}{3C{R}_2}& \frac{1}{3C{R}_2}\\ -\frac{1}{3C{R}_1}& -\frac{2}{3C{R}_1}-\lambda & -\frac{2}{3C{R}_1}\\ \frac{1}{3C{R}_3}& -\frac{1}{3C{R}_3}& -\frac{2}{3C{R}_3}-\lambda \end{array}\right]=0.$$

(83)

By doing the appropriate calculations, one arrives at the time constants:

$${\lambda }_0=0,$$

(84)

$${\lambda }_1=-{\lambda }^{\prime }-{\lambda }^{″},$$

(85)

$${\lambda }_2=-{\lambda }^{\prime }+{\lambda }^{″},$$

(86)

where

$${\lambda }^{\prime }=-\frac{1}{3C{R}^{\ast }},$$

(87)

$${\lambda }^{″}=-\frac{1}{3C}\sqrt{\frac{1}{R^{\ast \ast }}-\frac{1}{R^{\ast \ast \ast }}},$$

(88)

$$\frac{1}{R^{\ast }}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3},$$

(89)

$$\frac{1}{R^{\ast \ast }}=\frac{1}{R_1^2}+\frac{1}{R_2^2}+\frac{1}{R_3^2},$$

(90)

$$\frac{1}{R^{\ast \ast \ast }}=\frac{1}{R_1{R}_2}+\frac{1}{R_2{R}_3}+\frac{1}{R_1{R}_3}.$$

(91)

Let us suppose that T, V, W are the eigenvectors associated with the eigenvalues ${\lambda }_0$, ${\lambda }_1$ and ${\lambda }_2$, respectively. Then, the solution is

$$\left[\begin{array}{c}{u}_3\\ u_5\\ u_7\end{array}\right]=AT+BV{e}^{{\lambda }_{1}t}+DW{e}^{{\lambda }_{2}t}.$$

(92)

The sum of two exponential functions is never equal to one exponential. However, for certain conditions, it is possible to approximate the sum to one exponential function. In fact, $A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}=A{e}^{-{\lambda }^{\prime }t}{e}^{-{\lambda }^{″}t}+B{e}^{-{\lambda }^{\prime }t}{e}^{{\lambda }^{″}t}=e^{-{\lambda }^{\prime }t}(A{e}^{-{\lambda }^{″}t}+B{e}^{{\lambda }^{″}t})$. If $t\gtrsim {\lambda }^{″}$, then $A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}\simeq e^{-{\lambda }^{\prime }t}(B{e}^{{\lambda }^{″}t})=B{e}^{-({\lambda }^{\prime }-{\lambda }^{″})t}=B{e}^{-{\lambda }_{2}t}$.

To calculate $T_{cold}$ and $T_{hot}$, one needs $q_5$ and $q_7$. Hence,

$$u_5(t)=A_5+B_5{e}^{-{\lambda }_{2}t},$$

(93)

$$u_7(t)=A_7+B_7{e}^{-{\lambda }_{2}t}.$$

(94)

Thus,

$$q_5(t)=\int u_{5}dt=E_5+A_{5}t-\frac{B_5}{{\lambda }_2}{e}^{-{\lambda }_{2}t}={\alpha }_5+{\beta }_5{e}^{-{\lambda }_{2}t},$$

(95)

$$q_7(t)=\int u_{7}dt=E_5+A_{7}t-\frac{B_7}{{\lambda }_2}{e}^{-{\lambda }_{2}t}={\alpha }_7+{\beta }_7{e}^{-{\lambda }_{2}t}.$$

(96)

Since $q_5(t=0)$, then ${\beta }_5=-{\alpha }_5$. Thus, $q_5(t)={\alpha }_5(1-e^{-{\lambda }_{2}t})$. In the same manner, because $q_5(t=+\infty )=\frac{R_2+R_3}{R_1+R_2+R_3}$, then ${\alpha }_5=\frac{R_2+R_3}{R_1+R_2+R_3}$. Therefore,

$$q_5(t)=\frac{R_2+R_3}{R_1+R_2+R_3}(1-e^{-{\lambda }_{2}t}).$$

(97)

Using the same approach,

$$q_7(t)=\frac{R_1}{R_1+R_2+R_3}(1-e^{-{\lambda }_{2}t}).$$

(98)

Hence, since $T_{hot}(t)=R_1{q}_5(t)$ and $T_{cold}(t)=R_2{q}_7(t)$, then

$$\begin{array}{cc}\hfill T_{hot}(t)& =R_{1}Q\frac{R_2+R_3}{R_1+R_2+R_3}Q(1-e^{-{\lambda }_{2}t})+T_{amb},\hfill \end{array}$$

(99)

$$\begin{array}{cc}\hfill T_{cold}(t)& =\frac{R_2{R}_{3}Q}{R_1+R_2+R_3}Q(1-e^{-{\lambda }_{2}t})+T_{amb}.\hfill \end{array}$$

(100)

Because $R_1=\frac{1}{h_{1}Af}$, $R_2=\frac{L}{kA}$ and $R_3=\frac{1}{h_{3}A}$,

$$T_{hot}(t)=\frac{Q}{h_{1}Af}\frac{h_1{h}_{3}Lf+k{h}_{1}f}{kA{h}_3+h_{1}f{h}_3+k{h}_{1}f}(1-e^{-{\lambda }_{2}t})+T_{amb},$$

(101)

$$T_{cold}(t)=\frac{QL}{A}\frac{1}{kA{h}_3+h_{1}f{h}_3+k{h}_{1}f}(1-e^{-{\lambda }_{2}t})+T_{amb},$$

(102)

where

$${\lambda }_2=-\frac{A}{3CL}\left[h_{1}fL+k+h_{3}L-\sqrt{h_1^2{f}^2{L}^2+k^2+h_3^2{L}^2-h_{1}fkL-L^2{h}_1{h}_{3}f-Lk{h}_3}\right].$$

(103)

Note that ${\lambda }_2(k)$ is a decreasing function that tends to 0 when k goes to infinity.

4.3. Validation

One thermal component that has been increasingly addressed in recent years is the thermal switch [15,16]. This component can be used in many applications, such as in caloric cooling, space applications, energy harvesting, among others [17,18,19]. Therefore, their development has been intense, and the respective modeling has been exclusively performed numerically [20,21]. Here, we use the transient response modeling developed in the last section to compare with the experiment published by Puga et al. [22]. In their experiment, a Peltier device was used as heat source in contact with a thermal switch based on a nanofluid cage. The effective thermal conductivity of the thermal switch is low if the thermal switch is inactive (magnet is at rest) or high if the thermal switch is active (magnet is approaching and put away in a specific frequency). Thus, heat is either released to the air or to the thermal switch (and then to the air), which is a good representation of the developed model. The experiment undergoes two stages: one with the thermal switch with a low effective thermal conductivity, and one with a high thermal conductivity. The switch occurs at ∼650 s.

Since one can approximate the temperature evolution to exponential functions, the experiment data were fitted with Equations (101) and (102) for the four curves: hot plate before $t=650$ s, hot plate after $t=650$ s, cold plate before $t=650$ s, and cold plate after $t=650$. Figure 7 shows a very precise agreement between the developed transient response model and the measured temperatures of the hot and cold plates. In particular, the time constants for the two curves after $t=650$ s are smaller than those before $t=650$ s. This is due to the fact that the effective thermal conductivity of the heat transfer medium (nanofluid cage) increases during the switch (at $t=650$ s), which reduces the value of ${\lambda }_2$.

5. Conclusions

The theory of circuits has been widely used in addressing steady state heat transfer problems. However, a complete and reliable model focused on transient responses is still lacking. In this work, a thermal circuit model for transient responses was developed. The model consists of thermal objects made of thermal resistances and heat capacitors in parallel. The used heat capacity values are corrected to incorporate all thermal contacts. The developed analytical models were validated for three basic thermal circuits using the heatrapy Python package: single thermal resistance, two thermal resistances in series and two thermal resistances in parallel. All the heatrapy simulations agreed with the analytical solution. The proposed model was also validated by comparing it to an experiment based on nanofluidic thermal switches. These results show that the developed model provides a reliable way to analyze simple and complex thermal systems with simplified analytical tools that can easily solve computational issues of numerical methods, such as computation time. The main cost of the proposed method is the more complex formulation of the model when compared to numerical methods. Although there is nothing hindering the use of temperature dependent thermal properties in the scope of thermal circuits, the respective validation will be performed in a future investigation.