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Question 11.8: A Peltier generator is made up of two thermoelectric element...

A Peltier generator is made up of two thermoelectric elements connected in series (Fig. 11.1). One side of the Peltier generator is maintained at temperature T^+  and the other temperature T^− . The electric current I generated by the Peltier generator flows through the thermoelectric materials labelled 1 and 2. The plate which is heated up to temperature T^+ connects electrically these two materials but is not electrically available to the user.
Its electric potential is V^+ . The other ends of the thermoelectric materials are on the cold side, at a temperature T^− . They are connected to the electric leads of the device. A load resistance R_0 is connected to these leads. The voltage V designates the electric potential difference between the leads.
Analyse the operation of this generator using the electric charge and heat transport
equations,

j_{q1} = −σ_1 ε_1 ∇ T_1 − σ_1∇ \varphi _1        and      j_{Q1} = −κ_1∇ T_1 + T_1 ε_1 j_{q1}

j_{q2} = −σ_2 ε_2 ∇ T_2 − σ_2∇ \varphi _2        and      j_{Q2} = −κ_2∇ T_2 + T_2 ε_2 j_{q2} .

The thermoelectric materials 1 and 2 have a length d and a cross-section surface area A, which can be written as,

d = \int_{0}^{d}{dr} .\hat{x}                                             A = \int_{S}{dS} .\hat{x}

where \hat{x} is a unit vector oriented clockwise along the electric current density j_q , and the infinitesimal length and surface vectors dr and dS are oriented in the same direction. The temperature difference between the hot and cold ends is given by,

\Delta T= T^+ - T^- = \int_{0}^{d}{dr} .\nabla T_1 = \int_{0}^{d}{dr}.(-\nabla T_2 )

Likewise, the electric potential differences \Delta \varphi _1 and \Delta \varphi _2 between the hot and cold ends are written as,

\Delta \varphi _1 = V^+ = \int_{0}^{d}{dr}.\nabla \varphi _1 .

\Delta \varphi _2 = V^+ − V = \int_{0}^{d}{dr}.(-\nabla \varphi _2)

The electric charge conservation implies that the electric current densities are the same in each material, i.e. j_{q1} = j_{q2} . The electric current I flowing through materials 1 and 2 is the integral of the electric current densities j_{q1} and j_{q2} over the cross-section area A,

I = \int_{S} {j_{q1}} .dS =\int_{S} {j_{q2}} .dS

According the relation (10.104), the thermal powers P_{Q1} and P_{Q2} are the integrals of the heat current densities j_{Q1} and j_{Q2} flowing through materials 1 and 2 over the cross-section area A,

τ ≡ T\frac{∂ε}{∂T} .       (11.104)

P_{Q1} =  \int_{S} {(-j_{Q1)}}.dS                                       P_{Q2} =  \int_{S} {j_{Q2}}.dS

Determine:
a) the thermal power \acute{P}_Q applied on the hot side of the device when no electric current flows through the device.
b) the effective electric resistance R of the two thermoelectric materials when the temperatures are equal, i.e. T^+ = T^− , and no electric current flows through the resistance R_0 , i.e. when R_0 = ∞. Instead, an electric current flows through the thermoelectric materials.
c) the electric current I as a function of the temperature difference ΔT.

d) the thermodynamic efficiency of the generator defined as,

η =\frac{R_0 I^2}{P_Q}

where here, P_Q is the thermal power at the hot side when the electric current is flowing through the device. Show that the optimum load resistance is given by

\frac{R_0}{R} = \sqrt{1+\zeta }

where \zeta is a dimensionless parameter given by,

\zeta = \frac{T^+ (ε_1 − ε_2)^2}{(κ_1 + κ_2) \Bigl(\frac{1}{\sigma _1} + \frac{1}{\sigma _2} \Bigr) }
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a) When no electric current flows through the device, i.e. j_{q1} = j_{q2} = 0 , in order to determine the thermal power \acute{a}_Q, we integrate the heat transport equations over the volume V. The integrals over the volume are the product of integrals over the crosssection area A times integrals over the length d of the thermoelectric materials,

\int_{S}{(-\acute{j}_{Q1} )}.dS \int_{0}^{d}{dr} .\hat{x} = \kappa _1\int_{0}^{d}{dr} .\nabla T_1 \int_{S}{dS}.\hat{x}

 

\int_{S}{\acute{j}_{Q2}}.dS \int_{0}^{d}{dr} .\hat{x} = \kappa _2\int_{0}^{d}{dr} .(-\nabla T_2 )\int_{S}{dS}.\hat{x}

which reduce to,

\acute{P}_{Q1} = \kappa _1 \frac{A}{d} ΔT

 

\acute{P}_{Q2} = \kappa _2 \frac{A}{d} ΔT

Thus, the total thermal power is given by,

\acute{P}_{Q} = \acute{P}_{Q1} +\acute{P}_{Q2} = (κ_1 + κ_2)\frac{A}{d}ΔT

b) When the temperatures of the hot and cold sources are equal, i.e. T^+ = T^−, the temperature gradients vanish, i.e. ∇T_1 = ∇T_2, which implies that there is no thermoelectric effect. The integrals of the electric charge transport equations over the volume are the product of integrals over the cross-section area A times integrals over the length d of the thermoelectric materials,

\int_{S}{(-{j}_{q1} )}.dS \int_{0}^{d}{dr} .\hat{x} = \sigma _1\int_{0}^{d}{dr} .\nabla \varphi _1 \int_{S}{dS}.\hat{x}

 

\int_{S}{(-{j}_{q2} )}.dS \int_{0}^{d}{dr} .\hat{x} = -\sigma _2\int_{0}^{d}{dr} .(-\nabla \varphi _2) \int_{S}{dS}.\hat{x}

Since, the load resistance is infinite, i.e. R_0 = ∞, the electric current used for the measurement flows in the opposite direction, I→−I under the condition that V^+ < V.
In this case, the electric charge transport equations integrated over the volume reduce to,

I = σ_1 \frac{A}{d}Δ\varphi _1 = σ_1 \frac{A}{d} V^+

 

I = – σ_2 \frac{A}{d}Δ\varphi _2 = – σ_1 \frac{A}{d} (V^+ – V)

The potential difference across the thermoelectric materials 1 and 2 connected in series is given by,

V = Δ\varphi _1 – Δ\varphi _2 = I \frac{A}{d} \Bigl(\frac{1}{\sigma _1} +\frac{1}{\sigma _2} \Bigr) =I \frac{2 d}{A} \frac{1}{σ}

where σ is the effective conductivity of both thermoelectric materials. Since the effective electric resistivity ρ is the inverse of the effective electric conductivity ρ,

ρ = \frac{1}{σ} = \frac{1}{2 }\Bigl(\frac{1}{\sigma _1} +\frac{1}{\sigma _2} \Bigr)

Thus, the potential difference across both thermoelectric materials is written as,

V = ρ \frac{2 d}{A} I= R I

where 2 d is the effective length of both materials of length d each, section area A, connected in series, and R is their effective resistance. Hence,

R = ρ\frac{2 d}{A} = \frac{1 }{2} \Bigl(\frac{1}{\sigma _1} +\frac{1}{\sigma _2} \Bigr) \frac{2 d}{A}

c) The integrals of the electric charge transport equations over the volume are the product of integrals over the cross-section area A times integrals over the length d of the thermoelectric materials,

\int_{S}{j_{q1}}.dS \int_{0}^{d}{dr} .\hat{x} = -\sigma _1 \varepsilon _1\int_{0}^{d}{dr} .\nabla T_1 \int_{S}{dS} .\hat{x} – \sigma _1 \int_{0}^{d}{dr} .\nabla \varphi _1 \int_{S}{dS}. \hat{x}

 

\int_{S}{j_{q2}}.dS \int_{0}^{d}{dr} .\hat{x} = \sigma _2 \varepsilon _2\int_{0}^{d}{dr} .(-\nabla T_2) \int_{S}{dS} .\hat{x} + \sigma _2 \int_{0}^{d}{dr} .(-\nabla \varphi _2) \int_{S}{dS}. \hat{x}

The electric charge transport equations integrated over the volume reduce to,

I = −σ_1 ε_1 \frac{A}{d} ΔT − σ_1 \frac{A}{d}V^+

 

I = σ_2 ε_2 \frac{A}{d} ΔT − σ_2 \frac{A}{d}(V^+ – V)

Ohm’s law for the load resistance is written as,

V = R_0 I

In view of this relation that charaterises the electric properties of the load, the charge transport equations can be recast as,

V^+ = −\frac{1}{\sigma _1}\Bigl(\frac{d}{A}I – σ_1 ε_1 ΔT \Bigr)

 

V^+ = \frac{1}{\sigma _2}\Bigl( \Bigl(\frac{d}{A} + σ_2 R_0\Bigr)I − σ_2 ε_2 ΔT\Bigr)

which implies that the electric current is given by,

I = \frac{ε_2  ε_1}{\Bigl(\frac{1}{\sigma _1} +\frac{1}{\sigma _2} \Bigr) \frac{d}{A} +R_0} ΔT = \frac{ε_2  ε_1}{R + R_0} ΔT

This result for the current I is consistent with the current obtained in the analysis of the Seebeck loop (§ 11.6.2), which is equivalent to a Peltier generator (Fig. 11.1) in which the load resistance is set at zero, i.e. R_0 = 0.

d) In order to determine the thermal power P_Q entering through the hot plate at temperature T^+ , we integrate the heat transport equations over the volume V. The integrals over the volume are the product of integrals over the cross-section area A times integrals over the length d of the thermoelectric materials,

\int_{S}{(-j_{Q1})}.dS \int_{0}^{d}{dr} .\hat{x} =\kappa _1 \int_{0}^{d}{dr} .\nabla T_1 \int_{S}{dS} .\hat{x} – T^+ \varepsilon _1 \int_{S}{j_{q1}}.dS \int_{0}^{d}{dr} .\hat{x}

 

\int_{S}{j_{Q2})}.dS \int_{0}^{d}{dr} .\hat{x} =\kappa _2 \int_{0}^{d}{dr} .(-\nabla T_2) \int_{S}{dS} .\hat{x} + T^+ \varepsilon _2 \int_{S}{j_{q2}}.dS \int_{0}^{d}{dr} .\hat{x}

which reduce to,

P_{Q1} = κ_1 \frac{A}{d} ΔT − T^+ ε_1 I

 

P_{Q2} = κ_2 \frac{A}{d} ΔT + T^+ ε_2 I

and implies that,

P_Q = P_{Q1} + P_{Q2} = (κ_1 + κ_2) \frac{A}{d} ΔT + T^+ (ε_2 − ε_1) I

Therefore, the efficiency η for any load resistance R_0 is given by,

η = \frac{R_0 I^2}{P_Q} = \frac{R_0 \frac{(ε_2 − ε_1)^2 ΔT^2}{(R + R_0)^2}}{(κ_1 + κ_2) \frac{A}{d} ΔT + T^+ (ε_2 − ε_1)^2 \frac{ΔT}{R + R_0}}

which can be recast as,

η = \frac{ΔT}{T^+} \frac{\frac{R_0}{R}}{\frac{(κ_1 + κ_2)}{T^+ (ε_2 − ε_1)^2} \Bigl(\frac{1}{\sigma _1} +\frac{1}{\sigma _2} \Bigr) \Bigl(1+\frac{R_0}{R}\Bigr)^2 + \Bigl(1+\frac{R_0}{R}\Bigr) }

Using the definition of the coefficient  \zeta > 0, the temperature difference ΔT = T^+ − T^− > 0 and the ratio r = 1 + R_0/R > 0 , the efficiency is reduced to,

η = \Bigl(1 -\frac{T^−}{T^+} \Bigr) \frac{r − 1}{\zeta ^{-1} r^2 +r}

To find the optimal load resistance, we have to optimise the efficiency η with respect to the ratio r,

\frac{dη}{dr} = \Bigl(1 -\frac{T^−}{T^+} \Bigr) \frac {\zeta ^{-1} r^2 +r -(r-1) \bigl(\zeta ^{-1} r^2 +r\bigr)}{\bigl(\zeta ^{-1} r^2 +r\bigr)^2} =0

which implies that,

r^2 − 2 r − \zeta = 0

Thus, the optimum ratio r > 0, is given by,

r = 1+\sqrt{1+\zeta }

Hence, for an optimal resistance load, the efficiency is given by,

\eta = \Bigl(1- \frac{T^-}{T^+} \Bigr)\frac{\zeta \sqrt{1+\zeta } }{(1+\sqrt{1+\zeta } )^2 +\zeta (1+\sqrt{1+\zeta } )} \leq 1-\frac{T^-}{T^+}

In the limit \zeta → ∞ , the efficiency of the Peltier generator η tends towards the Carnot efficiency η_C (7.46),

 

lim_{\zeta → ∞}η =1- \frac{T^-}{T^+}= η_C

 

η_C = \frac{T^+ -T^-}{T^+} = 1- \frac{T^-}{T^+}<1                 (7.46)

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