A Peltier generator is made up of two thermoelectric elements connected in series (Fig. 11.13). 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 R₀ 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,

Fig 11.3

The thermoelectric materials 1 and 2 have a length d and a cross-section surface area A, which can be written as,where ˆx is a unit vector oriented clockwise along the electric current density j_q, and the infinitesimal length and surface vectors d_r and d_S are oriented in the same direction. The temperature difference between the hot and cold ends is given by,A Peltier generator has a load represented by the resistance R₀ connected to its leads. V is the voltage between the leads. The electric bridge at V+ is not accessible to the user. The regions marked 1 and 2 represent the two thermoelectric materials. The regions marked T+ and T− are the hot and cold sides of the device.Likewise, the electric potential differences Δφ 1 and Δφ 2 between the hot and cold ends are written as,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,According to 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,

Determine

a) the thermal power 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 throughthe resistance R₀, i.e. when R₀ = ∞. 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,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 bywhere ζ is a dimensionless parameter given by [188],

Transcribed Image Text:

jg₁=01&₁ VT₁-01 1 91 jq₂ = -0282 VT2 - 0242 92 and and jo₁ = jo₂ = k₁ VT₁+T ₁ €1jq₁ €2jq₂ K₂VT₂ + 12