Penguin huddling. To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups (Figure). Assume that a penguin is a circular cylinder with a top surface area $\mathit{a}\mathbf{=}\mathbf{0}\mathbf{.34}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$and height $\mathit{h}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.1}\mathbf{}\mathbf{\text{m}}$ . Let ${\mathit{P}}_{\mathbf{r}}\mathbf{}$ be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus $\mathit{N}{\mathit{P}}_{\mathbf{r}}\mathbf{}$ 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 and height , the cylinder radiates at the rate ${\mathit{P}}_{\mathbf{h}}$. If $\mathit{N}\mathbf{=}\mathbf{}\mathbf{1000}$ , (a) what is the value of the fraction ${\mathit{P}}_{\mathbf{h}}\mathbf{/}\mathit{N}{\mathit{P}}_{\mathbf{r}}$and (b) by what percentage does huddling reduce the total radiation loss?

1. The top surface area of the penguin, $a=0.34{\text{m}}^{\text{2}}$
2. The height of the penguin, $h=1.1\text{m}$
3. Te rate at which individual penguin radiates energy is $P_r$
4. The rate at which N penguins radiate energy is $N{P}_r$
5. After huddling, the rate at penguins radiate energy is $P_h$ with top surface area, Na and height, h .
6. Number of penguins, $N=1000$

The process by which heat is transferred from the hotter end to the colder end of the object is known as conduction. Heat spontaneously flows from a hotter to a colder body. For example, heat is conducted from the hotplate of an induction bottom to the bottom of a saucepan in contact with it. Thus, here also the penguins due to the increasingly harsh and colder temperature conditions of the region tend to create a huddle to have the proper transfer of the heat from one another keeping themselves warm.

Formulae:

The rate of conduction energy by the body, $P_{rad}=\frac{kA(T_H-T_C)}{L}$ …(i)

Where,K is the thermal conductivity of the material, A is the surface area of radiation,$T_H$is the temperature at the hotter end,$T_L$is the temperature at the colder end, L is the length of the conduction.

The top surface area of cylinder,$a=\pi r^2$ …(ii)

Where, $r$is the radius of the surface.

The lateral surface area of cylinder,$l=2\pi rh$ …(iii)

Where, r is the radius of the base surface and h is the height of the cylinder.

The required total surface area of the radiated energy by the penguin can be given as the sum of the top surface area and the lateral surface area.

So, the surface area of an individual penguin can be given using equations (ii) and (iii) as follows:

$A=\pi r^2+2\pi rh$

Since, here the value of radius is unknown, we can write the above equation using formula of equation (ii) as:

$A=a+2h\sqrt{a\pi }\text{(}\because \text{Radiusofonepengiun,r=}\sqrt{\frac{a}{\pi }}\text{)}$ …(iv)

Now, the surface area of N penguins can be given using equations (ii) and (iii) and the given data as follows:

$A_h=Na+2h\sqrt{Na\pi }\text{(}\because \text{RadiusofNpengiuns,r=}\sqrt{\frac{Na}{\pi }}\text{)}$ …(v)

Again, comparing the values in equation (i), it can be seen that,

$P\alpha A\text{(forT=constant)}$

Thus, the value of the fractioncan be given as:

$\begin{array}{c}\frac{P_h}{N{P}_r}=\frac{A_h}{NA}\\ =\frac{Na+2h\sqrt{Na\pi }}{N(a+2h\sqrt{a\pi })}\text{(fromequations(iv)and(v))}\\ \text{=}\frac{1+2h\sqrt{\frac{\pi }{Na}}}{1+2h\sqrt{\frac{\pi }{a}}}\\ =\frac{1+2\left(1.1\text{\hspace{0.17em}m}\right)\sqrt{\frac{\pi }{1000\left(0.34{\text{\hspace{0.17em}m}}^2\right)}}}{1+2\left(1.1\text{\hspace{0.17em}m}\right)\sqrt{\frac{\pi }{\left(0.34{\text{\hspace{0.17em}m}}^2\right)}}}\text{(substitutingthegivenvalues)}\\ =0.16\end{array}$

Hence, the value of the required fraction is $0.16$

From the calculations of part (a), we got the ratio of powers radiated by the penguins, thus, the total radiation loss is reduction by the value,

$\begin{array}{c}\text{Loss}=1-0.16\\ =0.84\\ \text{Loss}\%=84\%\end{array}$

Hence, the percentage of reduction in the radiation is $84\%$