Determination of the coefficient of thermal effusivity εε for a semi-confined rod

The article provides methods and formulas for calculating the coefficient of thermal effusivity ε (coefficient of heat accumulation) of solid bodies using known solutions of direct heat conduction problem. The work describes a method by which the final formulas are simplified to algebraic equations using the relative temperature M θ of heating semi-confined rod from the end. The temperature T c at the beginning of the rod is measured — x 0 and time — τ 0 from the start of heating. For calculation, the amount of heat Q accumulated during heating by a heater with a constant temperature at the end of the rod is used for the entire heating time. The technique is physically implemented during the experiment by bringing the ends of two semi-confined rods into contact. A flat, low-inertia electric heater is placed between the rods. The amount of heat ∆ Q used in the calculations is half of the total heat of the heater for the entire heating time.


Introduction
The article provides methods and formulas for calculating the coefficient of thermal effusivity of solid bodies.The work describes a method by which the final formulas are simplified to algebraic equations using the relative temperature M θ by heating semi- confined rod at a constant temperature from the end.
Finding the coefficients of thermal conductivity λ , thermal effusivity ε, thermal diffusivity a of materi- als, and heat capacity c is a difficult task.Thus, it is an inverse task of partial differential thermal conductivity.
The heat capacity is often determined by calorimetry.
According to the definition in [1] , clause 2.17, the coeffi-cient of thermal effusivity is «A value numerically equal to the square root of the product of thermal conductivity, density, and specific heat capacity».

Overview of methods
Several papers have been devoted to the task of solving inverse heat conduction tasks [2][3][4][5][6][7][8] .Some of the developed methods can be attributed to thermometric bridge schemes.In [7] , the task is solved by Laplace's operational method.Such calculations are chosen to simplify the calculation formulas and simplify the physical experiment.
The inverse task is solved using the formula for solving the direct task.According to the accepted classification [7] , the task is solved under boundary conditions of the first kind, in which the surface temperature is set as a function of time.A simpler case is considered when an electronic circuit keeps the temperature of the rod end constant from the beginning of heating and during the heating process [4] .
Monograph [2] presents a solution for the direct task of thermal conductivity of a semi-confined rod with thermal insulation of the side surface.
The statement of the task [2] is the following.At all points, the temperature of the semi-confined rod has a certain value given by the function: Fig. 1. ( [2] Fig. 4. 3).Temperature distribution in a semiconfined rod In Fig. 1 q is the heat flux density q [ ]= m W 2 ; f x ( ) is the initial temperature of the rod; Т x ( ) ,τ is the temperature from the beginning of heating near the end of the rod next to the heater.
Then, the formulas for cooling the rod are obtained by replacing the variables.At the initial moment, the end of the rod ( x 0 0 = ) takes the temper- ature T c = const , which is kept constant during the heat exchange process.In work [2] , the cooling of the rod is considered, and the final formula is obtained by replacing the variable T with the formulas for heating the rod.The differential equation is: where The boundary conditions [2] for rod cooling are as follows: perature on length, which is constant and does not depend on x).
Initially, for simplicity, let us assume [1] that T c = 0 .Also, to further simplify the calculation, let us assume that T c = 0 at the point x = 0.
If the end of the rod is not maintained at 0 ˚C, but at temperature T T c ( ) , 0 τ = = const, then by intro- ducing a new variable β = − T T c , the task is reduced to the previous one, since: .
In the monograph [2] , the task is solved by the Laplace's operational method.Paper [2] discusses the task of cooling a rod in detail.
Later, in [2] , the final calculation formula was converted into a criterion form.The ratio θ is the relative ( The final formula for calculating the core cooling temperature [2] is as follows: where Fo a x X = τ 2 [2] .
The ratio a x ⋅τ 2 is a dimensionless homochrony number for thermal conductivity processes and is called the Fourier criterion for the x coordinate.

Calculation of the coefficient of thermal effusivity ε by the author's method
For the case of rod heating, the author of this article is proposed to use the solution of the direct task according to formula (5) for the temperature value in the case of cooling the rod [2] and then use the formulas to determine the amount of heat accumulated by the rod over a specific period of time ∆τ .The difference in the author's solution is that the process is considered not when the rod is cooled but when it is heated.Hence, the calculation formulas have been changed.At heating T T x c > ( ) ,τ and T T c > 0 .Therefore, formula (3) for the temperature θ is replaced by formula (6) for the temperature M θ : The author's boundary conditions for core heating are as follows: Thus: To calculate the temperature during heating, it is not necessary to solve equation ( 1) again.
For heating, the formula ( 6) is used, which determines the relative temperature M θ .
In the special equipment, where the rod of the test material is fixed, a heater regulated at a constant temperature T c is located at one end of the rod.At the point x 0 a thermocouple is attached to the end of the rod.The side surface of the rod is thermally insulated.
Let us determine the amount of heat received [2] when its end is heated in time ∆ − ( ) τ τ 0 1 .The heat dQ passing through the unit area is equal: It is known [2] that the derivative in x of the func- , (which is in quadratic brackets in formula ( 8)) is equal: At x = 0 the exponential function is equal to one [2] , we obtain the heat flux density q, [ ] ) The amount of heat ∆Q, perceived during the time interval ∆τ , found by integrating in the range from 0 to τ by the rod, area which is equal to S, will be equal to: Hence, the coefficient of thermal effusivity of the body ε is equal to: (12)

Physical implementation of the method
The method is physically implemented by bringing the ends of two identical semi-confined rods into close contact.A flat electric heater is placed between the rods.The amount of heat used in the calculations is calculated as half of the total heat Q of the heater for the entire heat- ing period.All the heat Q that released by the heater will be distributed equally to the left and right rods.
The amount of heat released by the heater in time τ is equal: To use the method by formula (10), it is necessary to find the value of ∆Q.Mathematically, this model is implemented as follows.Fig. 1 shows a mathematical model where the rod is extended into the negative part of the graph to the left of point 0. Experimentally, this is achieved by bringing a similar rod with its end in contact with the end of the main rod to the right of point 0.Moreover, the mathematical model of heating the rod to the left and right of point 0 remains identical because the properties and dimensions of the rods are identical.The amount of heat generated by the heater will be evenly distributed to both parts of the folded rod.

Conclusions
For the method, the formulas are used to calculate the temperature T in the direct task.Then, the well-known formulas for heat flux density and accumulated heat ∆Q are used.From formula (12), the thermal effusivity coefficient ε is found.The tempera ture at a specific point of the rod -x 0 and time -τ 0 from the beginning of heating is measured.
For the calculation, the relative excess temperature is used, which is calculated from the constant temperature at the end of the rod at all times during heating.
length, which is constant and does not depend on x).The end of the rod is maintained at a certain tem-

γ
is the density of the material, [ ] γ = kg /m 3 , the coefficient of thermal effusivity of the body ε is by definition equal to λ

1 . 2 . 3 . 2 . 4 .
The first step: measure the temperature The second step: use the formula (6) to find the relative excess temperature M θ .Third step: obtain the value of the amount of heat received by the rod ∆ = Q Q The fourth step: find the coefficient of thermal effusivity (heat accumulation coefficient) -ε by formula (10).