91Ó°ÊÓ

Let $V$be the volume of the heaped coal. It is obvious that since the base is a square; let's call one of its sides $a$and the height$h$.

$h=\frac{V}{a^2}$

The rate at which the fuel must be consumed for the power plant to operate is indicated in metric tones per hour. This will be symbolized by the letter $m$. It's obvious that if we want the power plant to run for$t$hours, the amount of coal we'll need to burn is$M=mt$.

A mass $M$of a volume $V$of coal with $\mathrm{ÒÏ}$density will be

$V=\frac{M}{\mathrm{ÒÏ}}$

Therefore, we can combine our formulas to find:

$h=\frac{V}{a^2}=\frac{M}{\mathrm{ÒÏ}{a}^2}=\frac{mt}{\mathrm{ÒÏ}{a}^2}$

We know that the mass $m$required for our power plant in one hour is 300 metric tones,

localid="1649657696661" $h=\frac{3·{10}^5·24}{1500·{10}^2}=48\mathrm{m}$

The efficiency is given by,

$\mathrm{η}=\frac{P_e}{P_h}$

This is generated by multiplying the numerator and denominator by the time, where the output electric power is denoted by $P_e$and the input heating power is denoted by $P_h$. We know that when mass $M$of a substance with specific energy $c$is burned, heat is generated.

$Q=cM$

This means that we'll need to burn the same fuel at a rate of $\frac{M}{t}$to acquire heating power $P_h$.

$P_h=\frac{cM}{t}$

A mass of $300$metric tons is needed in one hour. That is, our efficiency will be

$\mathrm{η}=\frac{P_e}{P_h}=\frac{P_e}{cM/t}=\frac{P_{e}t}{cM}$

Using a numerical approach, we have

$\mathrm{η}=\frac{7.5·{10}^8·3600}{2.8·{10}^7·3·{10}^5}=32\%$