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For a 300 MW power plant, 14.5 MJ / kWh of heating capacity must be provided. For this you can use coal with a heating value (hW) of 29 MJ / kg and a price (P) of 35 € / t or gas with a calorific value of 36 MJ / m3 and one Price of 90 € / 1,000 m3.

What is the hourly cost (KB) for these fuels?

This is a question from school. I dont understand how 14.5 MJ / kWh could be possible? I can shorten this to 14.5 k / h. This make no sense?

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  • \$\begingroup\$ Please read this thoroughly and understand it, especially the part about units cancelling each other. Dimensional Analysis is one of the things that can help you most in a career. en.wikipedia.org/wiki/Dimensional_analysis \$\endgroup\$ – Voltage Spike Oct 28 '17 at 5:53
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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€.

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For a 300 MW power plant, 14.5 MJ/kWh of heating capacity must be provided.

  • 1 kWh = 3600 kWs = 3600 kJ = 3.6 MJ (since there are 3,600 s/h).
  • If your power plant is 100% efficient then it would require 3.6 MJ of heat to produce 1 kWh of electricity.
  • If your power plant is 40% efficient it will require \$ \frac {3.6}{0.40} = 9 \ \mathrm {MJ/kWh} \$. That is, you will have to supply 9 MJ of energy in to get 1 kWh (3.6 MJ) out.
  • Since you have been given 14.5 MJ/kWh we can work out the efficiency of your power plant as \$ \frac {output \ power}{required \ heat} = \frac {3.6}{14.5} = 0.25 = 25\% \$.

I was expecting an efficiency of about 45% for a modern power plant so this is poor but maybe not all that unusual.

Can you work out the rest of your problem?

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    \$\begingroup\$ Tobias, please un-accept my answer for a day or two. That way you will encourage others to answer. This will give you other ways to look at the problem and you might get a better answer than mine. (Some parts of the world are asleep right now.) \$\endgroup\$ – Transistor Oct 25 '17 at 23:04
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1 kWh is 1 kW (power) for 1 hour, 1 hour is 3600 s, so 1 kWh is 3.6 MJ, but you don't need this as you are not asked to calculate the efficiency (Which is clearly poor).

The question is mostly a nasty excersize in units conversion, or in noticing the shortcuts. For example you need 14.5 MJ per kWh of electrical output, so at 300 MW electrical, you need 300,000 kW * 14.5 MJ/kWh of heat energy for 1 hour of operation, thus you can figure how many kg of coal or m^3 of gas will be burned in one hour, hence the prices. (Mind the units coal and gas are priced in).

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