I'm not 100% sure if this is the right site for this, since this has nothing to do with electronic design. But hey, it's fun to take on some math problems once and then.
Let's ask ourselves this:
- How much total heating capacity is required for one whole hour?
- For every kWh we need 14.5 MJ, how many kWh is there in 300 MWh?
Number of kWh in 300 MWh is = \$\frac{300×10^6Wh}{10^3Wh} = 300×10^3\$
Total heating capacity requred for one hour = \$14.5 ×300×10^3\$ MJ = \$14.5×300\$ GJ.
Now we know how much energy this power plant requires for one full hour in order to deliver 300 MW.
Let's start with coal and see how much that would cost. Coal can give 29 MJ / kg. How many kg's of coal do we need to deliver \$14.5×300\$ GJ?
Well that would be equal to x kg = \$\frac{14.5×300×10^9}{29×10^6}=\frac{14.5×300×10^3}{29}\$ kg.
The price per ton (1000 kg) is 35€. So how much would that cost?
x€ = \$\frac{14.5×300×10^3}{29}×\frac{35}{10^3}=\frac{35×14.5×300}{29}=5250\$ €
Now we're done with coal, let's calculate the gas price. Gas can give 36 MJ / \$m^3\$
How many \$m^3\$ of gas do we need to deliver \$14.5×300\$ GJ?
Well that would be equal to x \$m^3\$ = \$\frac{14.5×300×10^9}{36×10^6}=\frac{14.5×300×10^3}{36} m^3\$.
The price per 1000 \$m^3\$ is 90€. So how much would that cost?
x€ = \$\frac{14.5×300×10^3}{36}×\frac{90}{10^3}=\frac{90×14.5×300}{36}=10875\$ €
So the amount of money you have to pay to supply the power plant with enough coal for one hour is 5250€. And the amount of money you have to pay to supply the power plant with enough gas for one hour is 10875€.