Respuesta :
Answer:
The rate at which coal burned is 111.12 kg/s.
Explanation:
Given that,
First initial temperature =750°C
First final temperature =440°C
Second initial temperature =415°C
Second final temperature =270°C
Suppose If the heat of combustion of coal is [tex]2.8×10^{7}\ J/kg[/tex], at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.
The work done by first engine is
[tex]W=eQ[/tex]
The work done by second engine is
[tex]W'=e'Q'[/tex]
[tex]W'=e'Q(1-e)[/tex]
Total out put of the plant is given by
[tex]W+W'=950\ MW[/tex]
Put the value into the formula
[tex]eQ+e'Q(1-e)=950\times10^{6}[/tex]....(I)
We need to calculate the efficiency of first engine
Using formula of efficiency
[tex]e=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e=0.65(1-\dfrac{440+273}{750+273})[/tex]
[tex]e=0.196[/tex]
We need to calculate the efficiency of second engine
Using formula of efficiency
[tex]e'=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e'=0.65(1-\dfrac{270+273}{415+273})[/tex]
[tex]e'=0.136[/tex]
Put the value of efficiency for first and second engine in the equation (I)
[tex]Q(0.196+0.136(1-0.196))=950\times10^{6}[/tex]
[tex]Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}[/tex]
[tex]Q=3111.24\times10^{6}\ W[/tex]
We need to calculate the rate at which coal burned
Using formula of rate
[tex]R=\dfrac{Q}{H_{coal}}[/tex]
[tex]R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}[/tex]
[tex]R=111.12\ kg/s[/tex]
Hence, The rate at which coal burned is 111.12 kg/s.
