Maximum performance in self-compatible thermoelectric elements

Abstract

Within the framework of a new optimization strategy based on self-compatible thermoelectric elements, the ability to reach maximum performance is discussed. For the efficiency of a thermogenerator and the coefficient of performance of a Peltier cooler, the constraint z T = k_o = const. turned out to provide a suitable criterion for judging maximum performance. In this paper k_o is calculated as an average of the temperature dependent figure of merit z T.

APPENDIX: PROOF OF INTEGRAL INEQUALITIES

Let us consider the integrals in Eqs. (8):

$$\matrix{ {\int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{1 \over T}{{\sqrt {1 + zT} - 1} \over {\sqrt {1 + zT} + 1}}dT} } \hfill & {{\rm{and}}} \hfill \cr {\int_{{T_{\rm{a}}}}^{{T_{\rm{s}}}} {{1 \over T}{{\sqrt {1 + zT} + 1} \over {\sqrt {1 + zT} - 1}}dT} } \hfill &, \hfill \cr } $$

((A1))

and their analytical solution in the case z(T)T = k_o = const.,

$$\matrix{ {{{\sqrt {1 + {k_o}} - 1} \over {\sqrt {1 + {k_o}} + 1}}\ln \left( {{T_{\rm{a}}}/{T_{\rm{s}}}} \right)} \hfill & {{\rm{and}}} \hfill & {{{\sqrt {1 + {k_o}} + 1} \over {\sqrt {1 + {k_o}} - 1}}\ln \left( {{T_{\rm{s}}}/{T_{\rm{a}}}} \right)} \hfill \cr } \quad.$$

((A2))

Let us further assume that the constant k_o is related to the average over z(T)T:

$${k_o} = {1 \over {{T_{\rm{s}}} - {T_{\rm{a}}}}}\int_{{T_{\rm{a}}}}^{{T_{\rm{s}}}} {z\left( T \right)T\;dT\quad.} $$

((A3))

The constraint z(T)T = k_o = const. can be considered as an upper/lower bound because the following integral inequalities hold for positive values T_s and T_a and any given z(T) > 0, such that z(T)T is monotonously increasing:

$$\int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{1 \over T}{{\sqrt {1 + z\left( T \right)T} - 1} \over {\sqrt {1 + z\left( T \right)T + 1} }}dT \le {{\sqrt {1 + {k_o}} - 1} \over {\sqrt {1 + {k_o}} + 1}}\ln \left( {{T_{\rm{a}}}/{T_{\rm{s}}}} \right)} \quad,$$

((A4))

and

$$\int_{{T_{\rm{a}}}}^{{T_{\rm{s}}}} {{1 \over T}{{\sqrt {1 + z\left( T \right)T} + 1} \over {\sqrt {1 + z\left( T \right)T - 1} }}dT \ge {{\sqrt {1 + {k_o}} + 1} \over {\sqrt {1 + {k_o}} - 1}}\ln \left( {{T_{\rm{s}}}/{T_{\rm{a}}}} \right)} \quad,$$

((A5))

with k_o given by Eq. (A3).

Proof: Let g be a continuous, convex (concave) function on ℝ₊; then Jensen’s inequality

$${1 \over {b - a}}\int_a^b {g\left( {y\left( x \right)} \right)dx\,_{\left( \le \right)}^ \ge g\left( {{1 \over {b - a}}\int_a^b {y\left( x \right)dx} } \right)} $$

holds for any nonnegative continuous function y: [a, b] → ℝ (see Ref. 15, p. 113). Defining

$$\matrix{ {{g_l}\left( w \right): = {{\sqrt {1 + w} - 1} \over {\sqrt {1 + w} + 1}}} \hfill & {{\rm{and}}} \hfill & {{g_u}\left( w \right): = {{\sqrt {1 + w} + 1} \over {\sqrt {1 + w} - 1}}} \hfill \cr } \quad,$$

one easily verifies that g″_l(w) < 0 for all w ≥ 0 and g″_u(w) > 0 for all w > 0. Hence, g_l is concave and g_u is a convex function on (0, ∞).

First we prove Eq. (A4). Choosing y(T) = z(T)T we obtain by means of

$$\int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{1 \over T}dT} = \ln {{{T_{\rm{a}}}} \over {{T_{\rm{s}}}}}$$

and Eq. (A3) from Jensen’s inequality the following:

$$\matrix{ {{1 \over {{T_{\rm{a}}} - {T_{\rm{s}}}}}\int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{1 \over T}dT} \int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{{\sqrt {1 + z\left( T \right)T} - 1} \over {\sqrt {1 + z\left( T \right)T} + 1}}dT} } \hfill \cr { \le {g_l}\left( {{k_o}} \right)\ln \left( {{T_{\rm{a}}}/{T_{\rm{s}}}} \right) = {{\sqrt {1 + {k_o}} - 1} \over {\sqrt {1 + {k_o}} + 1}}\ln \left( {{T_{\rm{a}}}/{T_{\rm{s}}}} \right)\quad.} \hfill \cr } $$

((A6))

As a second step, we use Chebyshev’s inequality for integrals (see, e.g., Ref. 16). Let F and G be both monotonically increasing or monotonically decreasing nonnegative functions, respectively, on an interval [a, b]; then, the inequality

$$\matrix{ {{1 \over {b - a}}\int_a^b {F\left( x \right)G\left( x \right)} \,dx} \hfill \cr { \ge {1 \over {b - a}}\int_a^b {F\left( x \right)dx{1 \over {b - a}}\int_a^b {G\left( x \right)dx} } } \hfill \cr } $$

holds. The relation reverses if one of the functions increases and the other decreases. If y(T) = z(T)T is monotonically increasing, then the composition G(T) = (g_l ○ y)(T) = g_l(z(T)T) increases as well. However, F(T) = 1/T decreases; hence, we obtain from Eq. (A6) the proposed relation (A4).

In the same way we obtain the estimate Eq. (A5) applying Jensen’s inequality on the convex function g_u and then Chebyshev’s inequality taking into account that the composition (g_u ○ y)(T) = g_u(z(T)T) now is a monotonically decreasing function if y = z(T)T increases. ▪

Remark 1: The integral on the left-hand side of Eq. (A5) cannot be bounded above by any constant depending on k_o. We give an example from the mathematical point of view. Fix some interval [T_a, T_s] with T_s–T_a = 20 (all temperatures in Kelvin), a constant c > 0, and define functions z_n(T) > 0 on [T_a, T_s] by

$${z_n}\left( T \right)T = \left\{ {\matrix{ {c{{\left( {T - {T_{\rm{a}}} + 1} \right)}^n}/{{10}^n}} \hfill & {{\rm{if}}\,{T_{\rm{a}}} \le T \le {T_{\rm{a}}} + 9} \hfill &, \hfill \cr c \hfill & {{\rm{if}}\,{T_{\rm{a}}} + 9 \le T \le {T_{\rm{s}}}} \hfill &, \hfill \cr } } \right.$$

where n ∈ ℕ. Then z_n(T)T is continuous and monotonically increasing for every fixed n and the averages k_o are uniformly bounded by c/2 ≤ k_o ≤ c for all n ∈ ℕ. However, the corresponding integrals in Eq. (A.5) are unbounded because z_n(T)T is chosen to be small for large n. To hold the average k₀ bounded below, this effect is restricted to a subinterval of [T_a, T_s]. For example, on [T_a, T_a + 4] we have 0 < z_n(T)T ≤ c(1/2)ⁿ. Then we estimate

and 0 < z_n(T)T ≤ c(1/2)ⁿ on [T_a, T_a + 4] implies that

is unbounded as n → ∞.

Remark 2: Notice that the proved inequalities Eqs. (A4) and (A5) can be used to estimate the performance parameters. The result is that the quantities (9) serve as upper bounds of the performance values given by the Eqs. (8) if the function z(T)T is monotonically increasing and k_o is given by Eq. (A3), because we get the following:

$$\matrix{ {{\rm{TEG:}}} \hfill & {1 - \exp \left( { - \int_{{T_{\rm{s}}}}^{{T_{\rm{a}}}} {{1 \over T}{{\sqrt {1 + z\left( T \right)T} - 1} \over {\sqrt {1 + z\left( T \right)T} + 1}}} dT} \right)} \hfill \cr {} \hfill & { \le 1 - {{\left( {{{{T_{\rm{s}}}} \over {{T_{\rm{a}}}}}} \right)}^{{{\sqrt {1 + {k_o}} - 1} \over {\sqrt {1 + {k_o}} + 1}}}}\quad,} \hfill \cr } $$

((A7))

$$\matrix{ {{\rm{TEC:}}} \hfill & {{{\left[ {\exp \left( {\int_{{T_{\rm{a}}}}^{{T_{\rm{s}}}} {{1 \over T}{{\sqrt {1 + z\left( T \right)T} + 1} \over {\sqrt {1 + z\left( T \right)T} - 1}}\,} dT} \right) - 1} \right]}^{ - 1}}} \hfill \cr {} \hfill & { \le {{\left[ {{{\left( {{{{T_{\rm{s}}}} \over {{T_{\rm{a}}}}}} \right)}^{{{\sqrt {1 + {k_o}} + 1} \over {\sqrt {1 + {k_o}} - 1}}}} - 1} \right]}^{ - 1}}\quad.} \hfill \cr } $$

((A8))

Equality holds if z(T) T = const. If z(T) T is decreasing, however, the above inequalities in general do not hold.