Question

A typical coal-fired power plant burns 260 metric tons of coal every hour to generate 2.9×106 MJ of electricity. 1 metric ton= 1000 kg; 1 metric ton of coal has a volume of 1.5 m3.The heat of combustion is 28 MJ/kg. Assume that all heat is transferred from the fuel to the boiler and that all the work done in spinning the turbine is transformed into electrical energy.

Suppose the coal is piled up in a 12 m × 8.0 m room. How tall must the pile be to operate the plant for one day?

What is the power plant's thermal efficiency?

Answer 1

Answer:

(a). The height of the pile is $42.708\times10^{3}\ m$

(b). The thermal efficiency is 0.143%.

Explanation:

Given that,

Mass of coal needed for one day $m= 260\times1000\times24$

$m=6240000\ kg$

$m =6.2\times10^{6}\ kg$

We need to calculate the density of coal

Using formula of density

$\rho = \dfrac{m}{V}$

Put the value into the formula

$V=\dfrac{6.2\times10^{6}}{1.5}$

$V=4.1\times10^{6}\ Kg/m^3$

We need to calculate the height of the pile

Using formula of volume of coal

$V=L\times W\times H$

$H=\dfrac{V}{L\times W}$

Where, L = length

W = width

H = height

Put the value into the formula

$H=\dfrac{4.1\times10^{6}}{12\times8.0}$

$H=42708.33\ m$

$H=42.708\times10^{3}\ m$

(b). Thermal energy released by combustion of coal per hour is  $E= m x heat\ of\ combustion$

Put the value into the formula

$E=260\times1000\times28\times10^{6}$

$E=7.28\times10^{6}\ MJ$

Thermal power released by combustion of coal is

$P=\dfrac{E}{t}$

Put the value into the formula

$P=\dfrac{7.28\times10^{12}}{3600}$

$P=2022.2\times10^{6}\ W$

$P=2022.2\ MW$

We need to calculate the thermal efficiency

Using formula of efficiency

$\eta=\dfrac{P_{e}}{P_{th}}\times100$

Put the value into the formula

$\eta=\dfrac{2.9\times10^{6}}{2022.2\times10^{6}}$

$\eta=0.143\%$

Hence, (a). The height of the pile is $42.708\times10^{3}\ m$

(b). The thermal efficiency is 0.143%.