Respuesta :
Answer:
(a). The height of the pile is [tex]42.708\times10^{3}\ m[/tex]
(b). The thermal efficiency is 0.143%.
Explanation:
Given that,
Mass of coal needed for one day [tex]m= 260\times1000\times24[/tex]
[tex]m=6240000\ kg[/tex]
[tex]m =6.2\times10^{6}\ kg[/tex]
We need to calculate the density of coal
Using formula of density
[tex]\rho = \dfrac{m}{V}[/tex]
Put the value into the formula
[tex]V=\dfrac{6.2\times10^{6}}{1.5}[/tex]
[tex]V=4.1\times10^{6}\ Kg/m^3[/tex]
We need to calculate the height of the pile
Using formula of volume of coal
[tex]V=L\times W\times H[/tex]
[tex]H=\dfrac{V}{L\times W}[/tex]
Where, L = length
W = width
H = height
Put the value into the formula
[tex]H=\dfrac{4.1\times10^{6}}{12\times8.0}[/tex]
[tex]H=42708.33\ m[/tex]
[tex]H=42.708\times10^{3}\ m[/tex]
(b). Thermal energy released by combustion of coal per hour is [tex]E= m x heat\ of\ combustion[/tex]
Put the value into the formula
[tex]E=260\times1000\times28\times10^{6}[/tex]
[tex]E=7.28\times10^{6}\ MJ[/tex]
Thermal power released by combustion of coal is
[tex]P=\dfrac{E}{t}[/tex]
Put the value into the formula
[tex]P=\dfrac{7.28\times10^{12}}{3600}[/tex]
[tex]P=2022.2\times10^{6}\ W[/tex]
[tex]P=2022.2\ MW[/tex]
We need to calculate the thermal efficiency
Using formula of efficiency
[tex]\eta=\dfrac{P_{e}}{P_{th}}\times100[/tex]
Put the value into the formula
[tex]\eta=\dfrac{2.9\times10^{6}}{2022.2\times10^{6}}[/tex]
[tex]\eta=0.143\%[/tex]
Hence, (a). The height of the pile is [tex]42.708\times10^{3}\ m[/tex]
(b). The thermal efficiency is 0.143%.
