--- title: "Shape factor algebra and concentric spheres" subject: "Heat Transfer" chapter: 4 key: "semester_5/heat_transfer:shape-factor-algebra-concentric-spheres" source: https://www.bctengineer.com/q/semester_5/heat_transfer/shape-factor-algebra-concentric-spheres site: BCT Engineering --- # Shape factor algebra and concentric spheres Heat Transfer · Chapter 4: Radiation Heat Transfer - Asked in: year not given, 8 marks - Video: [Shape Factor ,View Factor | Heat Transfer | Short tricks | NEGI Sir | GATE AIR - 1 (XE)](https://www.youtube.com/watch?v=_5etblbS_H4) — Unacademy GATE - ME, PI, XE - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/heat_transfer/shape-factor-algebra-concentric-spheres - Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-4 ## Question (a) Define shape factor (view factor). State the reciprocity relation, the summation rule and the superposition (additive) rule used in shape factor algebra. (b) Two concentric grey spheres have radii 0.1 m and 0.2 m. The inner sphere (emissivity 0.6) is at 600 K and the outer sphere (emissivity 0.4) is at 300 K. Find the shape factors F12, F21 and F22, and the net radiant heat exchange between the spheres. ## Short answer Shape factor F_ij is the fraction of radiation leaving surface i that directly strikes surface j. Reciprocity: A_iF_ij = A_jF_ji; summation: ΣF_ij=1; superposition adds for parts. For the concentric spheres: F12 = 1, F21 = 0.25, F22 = 0.75; net radiant heat exchange ≈ 424 W from inner to outer. ## Answer #### (a) Shape factor and algebra The shape factor $F_{ij}$ is the fraction of radiation leaving surface $i$ that strikes surface $j$ directly. It depends only on geometry and orientation. - **Reciprocity:** $A_iF_{ij} = A_jF_{ji}$. - **Summation rule:** for an enclosure of $N$ surfaces, $\sum_{j=1}^{N}F_{ij} = 1$. - **Superposition:** if surface $j$ is made of parts $j_1, j_2$, then $F_{i(j_1+j_2)} = F_{ij_1}+F_{ij_2}$ and $A_iF_{i\to j}$ adds similarly. - For a plane or convex surface, $F_{ii} = 0$; for a concave surface $F_{ii} > 0$. #### (b) Concentric spheres ```text .------------. / .------. \ | | 1 | 2 | | | T1 | T2| \ '------' / '------------' ``` Areas: $A_1 = 4\pi(0.1)^2 = 0.1257$ m$^2$, $A_2 = 4\pi(0.2)^2 = 0.5027$ m$^2$. **Shape factors.** All radiation from the inner (convex) sphere reaches the outer one, so $F_{12} = 1$ and $F_{11} = 0$. By reciprocity: $$ F_{21} = \frac{A_1}{A_2}F_{12} = \frac{0.1257}{0.5027} = 0.25 $$ By the summation rule for surface 2: $F_{22} = 1 - F_{21} = 0.75$. **Heat exchange.** For a small body enclosed by a large one: $$ Q_{12} = \frac{\sigma A_1\left(T_1^4 - T_2^4\right)}{\dfrac{1}{\varepsilon_1}+\dfrac{A_1}{A_2}\left(\dfrac{1}{\varepsilon_2}-1\right)} $$ $$ \begin{aligned} \sigma A_1(T_1^4-T_2^4) &= 5.67\times10^{-8}(0.1257)(600^4 - 300^4) = 865.7\ \text{W}\\ \text{denominator} &= \frac{1}{0.6}+0.25\left(\frac{1}{0.4}-1\right) = 1.667 + 0.375 = 2.042\\ Q_{12} &= \frac{865.7}{2.042} = 424\ \text{W} \end{aligned} $$ **Answer:** $F_{12} = 1$, $F_{21} = 0.25$, $F_{22} = 0.75$; $Q_{12} \approx 424$ W from the inner to the outer sphere. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the reciprocity relation for shape factors? A. F_ij = F_ji B. A_i F_ij = A_j F_ji C. A_i / F_ij = A_j / F_ji D. F_ij + F_ji = 1 Answer: B. Reciprocity links the shape factors in both directions through the areas. (Key point: Reciprocity relation) 2. For an enclosure of N surfaces, the summation rule states that A. the sum of all areas equals 1 B. the sum over j of F_ij equals 1 for each surface i C. F_ii is always 1 D. the sum of F_ij equals the number of surfaces Answer: B. All radiation leaving surface i must land somewhere in the enclosure, so the fractions add to 1. (Key point: Summation rule) 3. For the concentric spheres with r1 = 0.1 m and r2 = 0.2 m, which set of values is right? A. F12 = 0.25, F21 = 1, F22 = 0 B. F12 = 1, F21 = 0.5, F22 = 0.5 C. F12 = 1, F21 = 0.25, F22 = 0.75 D. F12 = 0.75, F21 = 0.25, F22 = 1 Answer: C. The inner convex sphere sees only the outer one (F12 = 1). Reciprocity gives F21 = A1/A2 = 0.25 and summation gives F22 = 0.75. (Key point: Shape factors of spheres) 4. In the heat exchange formula for a small body enclosed by a large one, the denominator is 1/e1 + (A1/A2)(1/e2 - 1). What value does it have here? A. 2.042 B. 1.667 C. 3.5 D. 1.0 Answer: A. 1/0.6 = 1.667 and 0.25 x (1/0.4 - 1) = 0.375, so the sum is 2.042, giving Q = 865.7/2.042 = 424 W. 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