Math, asked by manjunathasnrh, 9 months ago

applications of annular fins​

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Answered by Anonymous
8

In thermal engineering, an annular fin is a specific type of fin used in heat transfer that varies, radially, in cross-sectional area. ... Annular fins are often used to increase the heat exchange in liquid–gas heat exchanger systems.

Answered by Anonymous
8

\huge\bold{\underline{\underline{{AnsWer:-}}}}

In thermal engineering, an annular fin is a specific type of fin used in heat transfer that varies, radially, in cross-sectional area. Adding an annular fin to an object increases the amount of surface area in contact with the surrounding fluid, which increases the convective heat transfer between the object and surrounding fluid. Because surface area increases as length from the object increases, an annular fin transfers more heat than a similar pin fin at any given length. Annular fins are often used to increase the heat exchange in liquid–gas heat exchanger systems.

The maximum possible heat loss from an annular fin occurs when the fin is isothermal. This ensures that the temperature difference between the fin and the surrounding fluid is maximized at every point along the fin, increasing heat transfer by convection, and ultimately heat loss Q:

{\displaystyle Q=k\,\left(4\pi r_{1}t\right)\left(T_{b}-T_{e}\right)\beta \left[C_{2}\,K_{1}\left(\beta r_{1}\right)-C_{1}\,I_{1}\left(\beta r_{1}\right)\right].}

Q = k\, \left( 4 \pi r_1 t \right) \left( T_b - T_e \right)

\beta \left[ C_2\, K_1\left( \beta r_1 \right) - C_1\, I_1\left( \beta r_1 \right) \right].

The efficiency ηf of an annular fin is the ratio of its heat loss to the heat loss of a similar isothermal fin:

{\displaystyle \eta _{f}={\frac {\displaystyle {\frac {2r_{1}}{\beta }}\,K_{1}\left(\beta r_{1}\right)\,I_{1}\left(\beta r_{2}\right)-I_{1}\left(\beta r_{1}\right)\,K_{1}\left(\beta r_{2}\right)}{r_{2}^{2}-r_{1}^{2}\,K_{0}\left(\beta r_{1}\right)\,I_{1}\left(\beta r_{2}\right)+I_{0}\left(\beta r_{1}\right)\,K_{1}\left(\beta r_{2}\right)}}.}

\eta_f =

\frac{ \displaystyle \frac{2 r_1}{\beta}\, K_1\left( \beta r_1 \right)\, I_1\left( \beta r_2 \right)

- I_1\left( \beta r_1 \right)\, K_1\left( \beta r_2 \right)}

{ r_2^2

- r_1^2\, K_0\left( \beta r_1 \right)\, I_1\left( \beta r_2 \right)

+ I_0\left( \beta r_1 \right)\, K_1\left( \beta r_2 \right)}.

\huge{\boxed{\rm{\red{Note:-}}}} Mark @Paromita94 as brainliest

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