Penguins huddling: To withstand the harsh weather of the Antarctic, emperor peng

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Penguins huddling: To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups. Assume that a penguin is a circular cylinder with a top surface area a =0.26

m^{2}

and height h = 90 cm. Let

P_{r}

be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus

N P_{r}

is the rate at which N identical, well-separated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area Na and height h, the cylinder radiates at the rate

P_{h}

. If N = 1000, by what percentage does huddling reduce the total radiation loss?

Options:

A . 33%

B . 52%

C . 13%

D . 27%

Answer: Option D
:
D
Heat loss by individula penguin =

σ e A_{p e n g u i n} (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4})

=

σ e (2 π r_{p e n g u i n} h + 2 π r_{p e n g u i n}^{2}) (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4})

Let,

σ e (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4}) = k

∴ H e a t l o s t b y 1000 p e n g u i n s r a d i a t i n g f a r a w a y f r o m e a c h o t h e r, Q_{1}

= 1000 (2 π r_{p e n g u i n} h + 2 π r_{p e n g u i n}^{2}) k

Now,

π r_{p e n g u i n}^{2} = 0.26 m^{2} o r r_{p e n g u i n} = 0.287 m

Also,

π r_{h u d d l e}^{2} = N a_{p e n g u i n} = 1000 \times 0.26 o r r_{h u d d l e} = 9.09 m

∴ H e a t

lost by hudle of [enguins,

Q_{2}

= σ e A_{h u d d l e} (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4}) = (2 p i r_{h u d d l e} h + 2 π r_{h u d d l e}^{2}) k

NOw, percentage of reduce in radiation loss is given by

\frac{Q_{2}}{Q_{1}} = (2 p i r_{h u d d l e} h + 2 π r_{h u d d l e}^{2}) \times 100 /

N (2 p i r_{p e n g u i n} h + 2 π r_{h u d d l e}^{2}) = 26.67 %

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