Evaluation of the Heater Influence in Determining Thermal Conductivity by the Constant Heat Flow Method
Abstract Generated abstract
This paper estimates the error introduced by neglecting heater thickness and thermal inertia in determining thermal conductivity by the steady heat flow method. The authors model the heater as a thin plate adjacent to a semi-infinite specimen, formulate the coupled heat conduction problem, and use a Laplace transform approximation for small heater thickness to derive a correction to the standard thermal conductivity expression. For typical Plexiglas experimental parameters, the calculated correction is about 0.0032 W/m deg, approximately two percent of the measured value, indicating that heater effects can be small but measurable under the stated conditions.
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EVALUATING THE EFFECT OF THE HEATER IN DETERMINING THERMAL CONDUCTIVITY BY THE STEADY HEAT FLOW METHOD
V. N. Shmandina, L. T. Fuks
In calculating thermal conductivity by the steady heat flow method [1], the expression used is
\[ \lambda=\frac{2q\sqrt{a\tau}}{\vartheta}\operatorname{ierfc}\frac{x}{2\sqrt{a\tau}}, \tag{1} \]
obtained for the case of heating a semi-infinite body.
Expression (1) contains the following quantities:
\(\lambda\) — coefficient of thermal conductivity,
\(q\) — heat-flow density,
\(a\) — coefficient of thermal diffusivity,
\(\tau\) — time,
\(x\) — distance from the heater to the temperature-measurement point,
\(\vartheta\) — temperature excess at the point \(x\) over the time \(\tau\).
When manufacturing a flat heater for a rod, it is necessary to make it of such a design that, as far as possible, it does not affect the formation of the temperature field of the object under investigation. We made the heater from uniformly wound constantan wire of diameter \(d=0.1\text{–}0.2\) mm. To impart mechanical strength, the flat winding was pasted on both sides with paper using BF-4 glue. The finished heater had a thickness of 0.28 mm. In other cases, other low-inertia designs are used, for example heaters made of a shaped cutout of metal foil.
In the calculated dependence (1), the thickness and inertia of the heater are not taken into account.
The present work is an attempt to estimate the error arising from neglecting these conditions.
Let heater 1, Fig. 1, be a plate of small thickness \(b\), and let a constant heat flow \(q\) act at its left boundary. Then the temperature at the point \(x\) of the investigated specimen 2 can be found as the result of solving the system:
\[ \frac{\partial^2\vartheta_1(x,\tau)}{\partial x^2} = \frac{1}{a_1}\, \frac{\partial\vartheta_1(x,\tau)}{\partial\tau}, \qquad -b\leq x\leq 0, \tag{2} \]
\[ \frac{\partial^2\vartheta_2(x,\tau)}{\partial x^2} = \frac{1}{a_2}\, \frac{\partial\vartheta_2(x,\tau)}{\partial\tau}; \qquad 0\leq x\leq \infty, \tag{3} \]
\[ \vartheta_1(x,0)=\vartheta_2(x,0)=0, \tag{4},(5) \]
\[ -\lambda_1 \frac{\partial \vartheta_1(-b,\tau)}{\partial x}=q, \tag{6} \]
\[ \vartheta_1(0,\tau)=\vartheta_2(0,\tau), \tag{7} \]
\[ \lambda_1 \frac{\partial \vartheta_1(0,\tau)}{\partial x} = \lambda_2 \frac{\partial \vartheta_2(0,\tau)}{\partial x}, \tag{8} \]
\[ \frac{\partial \vartheta_2(\infty,\tau)}{\partial x}=0. \tag{9} \]
Here the subscript 1 denotes quantities pertaining to the heater, and the subscript 2 quantities characterizing the specimen. The remaining notation was explained earlier.
Fig. 1
The solution of the system (2)—(9) is given in [2, 3].
If we introduce the notation
\[ M=\frac{1-\dfrac{\lambda_2}{\lambda_1}\sqrt{\dfrac{a_1}{a_2}}} {1+\dfrac{\lambda_2}{\lambda_1}\sqrt{\dfrac{a_1}{a_2}}}, \]
then, for the temperature of the specimen, the solution will have the form
\[ \vartheta_2= \frac{2q\sqrt{a_2\tau}(1-M)}{\lambda_2} \sum_{n=1}^{\infty} M^{n-1}\operatorname{ierfc} \left[ \frac{ \dfrac{x}{\sqrt{a_2}}+ \dfrac{(2n-1)b}{\sqrt{a_1}} } {2\sqrt{\tau}} \right]. \tag{10} \]
It is difficult to use this solution, since the series (10) converges very slowly. We shall give an approximate estimate of the influence of the heater on the determination of the thermal-conductivity coefficient of the specimen.
Applying the integral Laplace transform, we write the transform of the specimen temperature in the form
\[ \vartheta_{2L}(x,s)= \frac{ q\exp\left(-\sqrt{\dfrac{s}{a_2}}\,x\right) } { \lambda_1 s \sqrt{\dfrac{s}{a_1}} \left[ \operatorname{sh}\left(\sqrt{\dfrac{s}{a_1}}\,b\right) + N\operatorname{ch}\left(\sqrt{\dfrac{s}{a_1}}\,b\right) \right] }, \tag{11} \]
where
\[ N=\frac{\lambda_2}{\lambda_1}\sqrt{\frac{a_1}{a_2}}. \]
For a small value of the heater thickness, as a first approximation one may set
\[ \operatorname{sh}\left(\sqrt{\frac{s}{a_1}}\,b\right) \approx \sqrt{\frac{s}{a_1}}\,b, \]
\[ \operatorname{ch}\left(\sqrt{\frac{s}{a_1}}\,b\right) \approx 1. \]
Then equation (11) is transformed as follows:
\[ \vartheta_{2L}(x,s)=\frac{q a_1}{\lambda_1 b}\, \frac{\exp\left(-\sqrt{\frac{s}{a_2}}\,x\right)} {s\sqrt{s}\left(\sqrt{s}+f\right)}. \tag{12} \]
Here
\[ f=\frac{N\sqrt{a_1}}{b}. \]
Passing from the image to the original gives
\[ \vartheta_2=\frac{q a_1}{b\lambda_1}\left[ \frac{2}{f}\sqrt{\frac{\tau}{\pi}}\exp\left(-\frac{x^2}{4a_2\tau}\right) -\frac{1+f\frac{x}{\sqrt{a_2}}}{f^2}\, \operatorname{erfc}\frac{x}{2\sqrt{a_2\tau}} +\frac{1}{f^2}\exp\left(f\frac{x}{\sqrt{a_2}}+f^2\tau\right) \operatorname{erfc}\left(\frac{x}{2\sqrt{a_2\tau}}+f\sqrt{\tau}\right) \right]. \tag{13} \]
Using the fact that, in our experiments, the quantity
\[ \frac{x}{2\sqrt{a_2\tau}}+f\sqrt{\tau}>10, \]
and that for large values of the argument the complementary probability integral is expressed with sufficient accuracy by
\[ \operatorname{erfc}x\simeq \frac{1}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x}, \tag{14} \]
after some transformations we obtain
\[ \vartheta_2=\frac{q}{\lambda_2}\left( 2\sqrt{a_2\tau}\,\operatorname{ierfc}\frac{x}{2\sqrt{a_2\tau}} -b\frac{\lambda_1}{\lambda_2}\frac{a_2}{a_1}\, \operatorname{erfc}\frac{x}{2\sqrt{a_2\tau}} \right). \tag{15} \]
Hence the value of the thermal-conductivity coefficient determined with allowance for the thermal inertia and thickness of the heater will be equal to
\[ \lambda_2^*=\lambda_2-\Delta\lambda =\lambda_2-\frac{2q\sqrt{a_2\tau}}{\vartheta_2}\, \operatorname{ierfc}\frac{x}{2\sqrt{a_2\tau}} -\frac{bq\,c\rho_1}{\vartheta_2\,c\rho_2}\, \operatorname{erfc}\frac{x}{2\sqrt{a_2\tau}}. \tag{16} \]
Here \(\lambda_2\) is determined from expression (1).
Calculation of the correction \(\Delta\lambda\) from relation (16) for an experiment with Plexiglas at values
\[ q=800\ \text{W}/\text{m}^2,\quad \tau=60\ \text{s},\quad a_2=10^{-7}\ \text{m}^2/\text{s}, \]
\[ b=4\cdot10^{-5}\ \text{m},\quad x=3\cdot10^{-3}\ \text{m},\quad \lambda_2=0.18\ \text{W}/\text{m}\cdot\text{deg}, \]
\[ \vartheta_2=3.3^\circ\text{C},\quad c\rho_1=1.5\cdot10^6\ \text{J}/\text{m}^3\text{deg}, \]
gave the correction \(\Delta\lambda=0.0032\ \text{W}/\text{m}\cdot\text{deg}\).
In this case \(\lambda_2\) according to formula (1) will be equal to
\[ \lambda_2=0.18\ \text{W}/\text{m}\cdot\text{deg}. \]
Thus, the correction is a quantity of the order of two percent.
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