Question

What will be shape factor $F_{22}$ of a very long concentric cylinders as shown , indicated areas are cross-section area of cylinders?

• A. 0.72
• B. 0.8
• C. 0.6
• D. 0.3

Answer 1

The correct option is C 0.6

$F_{11}=0$ $[\therefore F_{12}=1]$
$F_{21}=\frac{A_1}{A_2}$
$F_{21}=\frac{\pi d_{1}l}{\pi d_{2}l}=\frac{d_1}{d_2}$
$\text{A=curved surface area, a= cross-sectional area}$
$\frac{a_1}{a_2}={\left(\frac{d_1}{d_2}\right)}^2$
${\left(\frac{100\pi }{625\pi }\right)}^{1/2}=\frac{d_1}{d_2}$
$\frac{d_1}{d_2}=0.4$
$F_{21}+F_{22}=1$
$F_{22}=1-0.4=0.6$
$F_{22}=0.6$