THERMOELECTRIC GENERATOR

Thermoelectric Generator:

Thermoelectric (TEG) generators of solid semiconductor devices make the differences of temperature and the flow of the temperature into a Direct Current power source. Thermoelectric semiconductor generator devices use the Seebeck effect to get electricity . The electricity generated generates dynamic electricity and produces useful loading power (Imamura, M., Asada, H., Nishimura, R., Yamaguchi, K., Tashima, D., & Kitagawa, J. ,2021).

A thermoelectric generator isn’t sort of a thermoelectric cooler. (and they’re also referred to as TEC, solid cooling, cooling chips ) .The thermoelectric cooler works behind the thermoelectric generator. If the voltage is employed to chill the thermoelectric, an electrical car is formed . This current generates a Peltier effect(I. Terasaki, 2016). By this effect, the produced heat is transferred to the recent area from the comparatively cold area. Additionally, The structure of Thermoelectric cooler is solid semiconductor structure. Probably the materials and thermoelectric generator is same but the composition of the weather in most of the cases varies(S.V. Petrushkin, V.V. Samartsev,2009).

While thermoelectric generators are wont to generate energy, thermoelectric coolers (Peltier coolers) are wont to remove or dissipate heat. Thermoelectric cooling has a lot of features for cooling, temperature control, heating, refrigeration, and temperature control(GJ Snyder,2008).

Focus on a number of these posts with thermoelectric generators.

1.

Answer:

The stress on the electrical generators is extremely important for long service use. heat gradients, thanks to the massive temperature difference altogether junctions, cause excessive stress levels developed within the device pins and electrodes at the contact points.

Here,

= 26.0 – 11.9 = 14.1 mm

= 4×10⁶ – 2×10⁶ = 2×10⁶ A/m²

= T_h – T_c = 325 – 300 = 25K

So, the temperature will be 25K

As we know the formula of stress,

= _h-T_c)

For n-type the stress will be

x 47 x 25 = 1974 N/m²

For p-type the stress will be

x 62.35 x 25 = 1886.1 N/m²

2.

Answer:

When,

L = 11.9mm

H = 2.6 mm

Area, Ap =An = 11.9×2.6 = 30.94mm²

R_I = = 9 x 10^-6 ohm

Given, j =

So, j = (200+200)(325-300)/30.94(18 x 10^-6) = 17.96 x 10⁶

As we know the formula of stress,

= _h-T_c)

The temperature distribution stress on x-axis( L, j₀ use as vaiables).

T(x) = ²x²+( + ²L)x + T_h

L	j	T(x)	Stress
		N-type	P-type	N-type	p-type
11.9	17.96 x 106	-872×106x2+872×106 x + 325	-2055×106x2+ 2055×106 x +325	1974 N/m2	1886.1 N/m2
13.4	17.96 x 106	-872×106x2+872×106 x + 325	-2055×106x2+ 2055×106 x +325	1974 N/m2	1886.1 N/m2
15.6	17.96 x 106	-872×106x2+872×106 x + 325	-2055×106x2+ 2055×106 x +325	1974 N/m2	1886.1 N/m2

The temperature and thermal stress distribution will remain same for different values of L and j.

When Tc = 300K and assume Th = 0

So, j = (Sp -Sn)Tc/A(R_I + R_L)

R_I = = 9 x 10^-6 ohm = R_L

J_11.9 = (200+200)300/30.94(18 x 10^-6) = 215 x 10⁶

J_13.4 = (200+200)320/34.84(18 x 10^-6) = 204 x 10⁶

J_15.6 = (200+200)360/40.56(18 x 10^-6) = 197 x 10⁶

J_16.1 = (200+200)380/41.86(18 x 10^-6) = 201 x 10⁶

J₁₇ = (200+200)400/44.2(18 x 10^-6) = 201 x 10⁶

J_18.5 = (200+200)420/48.1(18 x 10^-6) = 194 x 10⁶

J₂₀ = (200+200)450/52(18 x 10^-6) = 192 x 10⁶

The distribution of stress vs j₀ ( L & T_c use as variables)

L	Tc	Area A	Stress N-type P-type	j
11.9mm	300	30.94	1974 N/m2	1886.1 N/m2	215 x 106
13.4mm	320	34.84	1974 N/m2	1886.1 N/m2	204 x 106
15.6mm	360	40.56	1974 N/m2	1886.1 N/m2	197 x 106
16.1mm	380	41.86	1974 N/m2	1886.1 N/m2	201 x 106
17mm	400	44.2	1974 N/m2	1886.1 N/m2	201 x 106
18.5	420	48.1	1974 N/m2	1886.1 N/m2	194 x 106
20mm	450	52	1974 N/m2	1886.1 N/m2	194 x 106

In this graph, we plot the diagram of Stress and J. There are two types of stress: N-type and P-type. By the assumed value of L and Tc, we can state that if temperature and length increases then the value of j will decrease.

3.

Answer:

As we know the formula of stress,

Given,

In n-type:

= 47

In P-type:

= 62.35

Temperatures		N-type Stress	P-type Stress
300K	0K	0	0
310K	10k	789.6 N/m2	754.435 N/m2
320K	20k	1579.2 N/m2	1508.87 N/m2
520K	220k	17371.2 N/m2	16597.57 N/m2

Objective 1:

Calculated the stress of TEG in ANSYS Solution identify the different values of temperature:

Thermal Stress against Current:

Objective 2:

Find out the output power of PN junction and compare it with Ansys Solution

Given S_p = 200

Th = 520K and Tc = 300K

R_I = R_L = = 9 x 10^-6 ohm

Lets assume the electric current I, is flowing through the Thermoelectric Generator,

= = 4889 x 10⁶ Amp.

where Th = Temperature of hot side and Tc = Temperature of cold side, at the end of thermoelectric poles.

R_I = the total electric resistance of N and P types of poles.

If the output power = P

Then, P = ()².( )

Or,

𝑃=𝐼²𝑅_L.

P = (4889 x 10⁶)² . x 10^-6

= 88002 x 10⁶ Watt

Thermal Stress against Voltage:

Study the effect of beam:

When the beam is below gradient temperature or the combined/composite materials have two or more variable factors of thermal growth depending on the temperature rise, either uniformly or indiscriminately, longitudinal fibers tend to increase different values ​​depending on their temperature and the combined temperature increase. In the application of thermal gradient, the behavior of thermal pressure is highly dependent on the ability to absorb some of the fibers needed to fix the stabilized body, Brittle materials can’t easily remain undamaged these barriers beyond reducing sufficient pressure to create cracks(N Jaziri, 2020).

Thermal stress is additionally receiving the considerable attention in reference to the materials of ductile since there’s significant evidence that does not succeed of the many  engine components of ductile is often attributed to thermal cycling. The matter of flight with high-speed, with continuous increases of the temperature and thermal gradients in the aircraft bodies, has further produced concern over the importance of thermal stress field in ductile materials. Hot pressures and hot shocks are often separated by the very fact that at hot temperatures hot pressures are produced by gradients of excessive heat, which is typically sudden. Some of the materials are vulnerable to continuous application of pressure in order that they might not be ready to remain undamaged the warmth of the warmth exchanger when a pressure difference of a particular amount can easily enter(Y pehlivanoğlu,2018).

Reference:

Imamura, M., Asada, H., Nishimura, R., Yamaguchi, K., Tashima, D., & Kitagawa, J. (2021). Enhancement of the spin Seebeck effect owing to bismuth substitution in thermoelectric generators fabricated from LPE Bi-substituted YIG films. AIP Advances, 11(3), 035143.

I Taraski. Review of: Current generates a Peltier effect (2016),: A Running Commentary with Consideration of the Polish Original and the German Translation. Vienna Circle Institute Yearbook.

S.V. Petrushkin, V.V. Samartsev,(2009). Thermal cycling of thermoelectric generators: the effect of heating rate. Applied energy, 237, 671-681.

GJ Snyder,2008.Study of thermal insulation materials influence on the performance of thermoelectric generators by creating a significant effective temperature difference. Energy Conversion and Management, 207, 112516.

N Jaziri, (2020). An experimental study. Development Engineering, 5, 100049..

Y pehlivanoğlu & Yang, Y. (2018). Performance analysis of a thermoelectric generator applied to wet flue gas waste heat recovery. Applied Energy, 228, 2080-2089.

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