Question

assume a 1000 - megawatt power plant has an overall efficiency of 33% and that on average it operates at 50% of capacity in a days operation. using coal that has an energy content of 13,000 btu per pound, what is the daily supply amounts of coal in tons?. (remember 1 btu = 1055 joules and 1 watt = 1 joule/second.)

Explanation:

Step1: Calculate the actual power output per day

The power - plant capacity is $P_{capacity}=1000$ MW. It operates at 50% of capacity, so the actual power output $P = 0.5\times1000\times10^{6}$ W. In one day ($t = 24\times3600$ s), the energy output $E_{out}=P\times t=0.5\times1000\times10^{6}\times24\times3600$ J.

Step2: Calculate the required input energy

The overall efficiency $\eta = 0.33$. We know that $\eta=\frac{E_{out}}{E_{in}}$, so the required input energy $E_{in}=\frac{E_{out}}{\eta}=\frac{0.5\times1000\times10^{6}\times24\times3600}{0.33}$ J.

Step3: Convert the energy content of coal to joules per pound

The energy content of coal is $13000$ Btu per pound. Since $1$ Btu = $1055$ J, the energy content of coal in joules per pound is $13000\times1055$ J/lb.

Step4: Calculate the mass of coal in pounds

Let the mass of coal be $m$ (in pounds). We know that $E_{in}=m\times(13000\times1055)$. So, $m=\frac{E_{in}}{13000\times1055}=\frac{\frac{0.5\times1000\times10^{6}\times24\times3600}{0.33}}{13000\times1055}$ lb.

Step5: Convert the mass of coal from pounds to tons

Since $1$ ton = $2000$ lb, the mass of coal in tons $M=\frac{m}{2000}$.

First, calculate $E_{out}=0.5\times1000\times10^{6}\times24\times3600 = 4.32\times10^{13}$ J.
Then, $E_{in}=\frac{4.32\times10^{13}}{0.33}\approx1.309\times10^{14}$ J.
The energy content of coal per pound in joules is $13000\times1055 = 1.3715\times10^{7}$ J/lb.
$m=\frac{1.309\times10^{14}}{1.3715\times10^{7}}\approx9.54\times10^{6}$ lb.
$M=\frac{9.54\times10^{6}}{2000}=4770$ tons.

Answer:

4770