The image presents a derivation of the common expressions for radiative heat transfer rates between two surfaces, based on their geometrical configurations. Below is a transcription of the content for an educational context:---**Radiative Heat Transfer Rate Derivations****a) Infinite Parallel Plates:**- *Diagram Description:* Two parallel plates, identified as \( A_1, T_1, \varepsilon_1 \) and \( A_2, T_2, \varepsilon_2 \).- *Equation:* \[ q_{12} = \frac{A\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1} \]**b) Infinitely Long Concentric Cylinders:**- *Diagram Description:* Two concentric cylinders with radii \( r_1 \) and \( r_2 \).- *Equation:* \[ q_{12} = \frac{A_1\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2} \left(\frac{r_1}{r_2}\right)} \]**c) Concentric Spheres:**- *Diagram Description:* Two concentric spheres with radii \( r_1 \) and \( r_2 \).- *Equation:* \[ q_{12} = \frac{A_1\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2} \left(\frac{r_1}{r_2}\right)^2} \]**Explanation:**- \( A \) or \( A_1 \) are the areas of the surfaces.- \( T_1 \) and \( T_2 \) represent the temperatures of the first and second surfaces, respectively.- \( \varepsilon_1 \) and \( \varepsilon_2 \) are the emissivities of the surfaces.- \( \sigma \) is the Stefan-Boltzmann constant.- The expressions calculate the rate \( q_{12

Consider heat transfer by radiation takes place between two opaque and non black bodies (say body-1…

Verify (i.e., derive) the common expressions for the radiative heat transfer rate between two surfaces below. a) Infinite parallel plates: A₁, T₁, E1 c) Concentric spheres: A2, T2, E2 b) Infinitely long concentric cylinders: 12 ۲۱ 12 912 912 912 Ar(T4 – Tâ) 1 1 + E1 E2 A₁ (T₁-T₂) 1-82 (G) + E1 E2 A₁ (T₁ – T₂) +1=52 (12) ² 1 E1

Verify (i.e., derive) the common expressions for the radiative heat transfer rate between two surfaces below. a) Infinite parallel plates: A₁, T₁, E1 c) Concentric spheres: A2, T2, E2 b) Infinitely long concentric cylinders: 12 ۲۱ 12 912 912 912 Ar(T4 – Tâ) 1 1 + E1 E2 A₁ (T₁-T₂) 1-82 (G) + E1 E2 A₁ (T₁ – T₂) +1=52 (12) ² 1 E1

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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You have stated that the convective area is πr^2. Please explain this. πr^2 is the area for a circle, not a sphere. It should be 4πr^2 as this is equal to the surface area of the sphere as you have done correctly in the radiative area. A conv= 4πr^2*3*293-292 = 0.377 A rad = 4πr^2*0.2*5.67x10^-8*(292^4-291^4) = 0.144 I have tried to multiply the convective area by 4πr^2 instead of πr^2 but it is giving me an incorrect total heat transfer rate of 0.377+0.144 = 0.521 the correct answer is supposed to be 0.236