Calculating Coal Burn Rate to Output 1000MW of Power

[laural]

Homework Statement

At a steam power plant, steam engines work in pairs, the output of the first contributing to the heat input of the second. Let the operating temperature of the first be 670.0 degrees C and 440.0 degees C, and of the second be 430.0 degrees C and 290.0 degrees C. If the heat of combustion of coal is 2.80 x 107 J/kg, at what rate (in kg/s) must coal be burned if the efficiency of the engines is only 60.0% of the ideal (Carnot) efficiency if the plant is to output 1000.0MW of power?

Homework Equations

e=W/Q(high) = 1- Q(low)/Q(high)

The Attempt at a Solution

I converted the temperatures from C to K.

I am unsure how to connect the heat of the engines to the heat of combustion of coal and the rate it burns and the end power. I have tried working with the units to reach an answer with units kg/s:

W= eQ(high) W=(.6)(943 K)= 565.8 J/s

To reach units kg/s, (565.8 J/s) / (2.8 x 10^7 J/kg) = 1.6 x 10^10 kg/s, which is an outrageous answer and I did not base my work on any physics knowledge...

So if someone could give me a starting point I would be very grateful,
Thank-you.

 

[anathema]

Thank you for your question. Let's start by breaking down the problem and identifying the key information given:

1. The steam engines work in pairs, with the first engine's output contributing to the second engine's input. This means that the second engine is essentially using the waste heat from the first engine to produce more work.

2. The operating temperatures of the first engine are 670.0 degrees C and 440.0 degrees C, while the operating temperatures of the second engine are 430.0 degrees C and 290.0 degrees C. This information is important because it tells us the temperature difference between the hot and cold reservoirs for each engine.

3. The heat of combustion of coal is 2.80 x 10^7 J/kg. This tells us how much energy is released when 1 kg of coal is burned.

4. The efficiency of the engines is 60.0% of the ideal efficiency (Carnot efficiency). This means that the engines are not working at their maximum potential, and some of the heat energy is being lost as waste heat.

5. The plant needs to output 1000.0 MW of power. This is the end goal of the problem, and we need to figure out how much coal needs to be burned to achieve this power output.

Now, let's look at the equations you provided in your post:

1. The efficiency of a heat engine is given by the equation e = W/Q(high), where W is the work output and Q(high) is the heat input.

2. The Carnot efficiency is given by the equation e = 1 - Q(low)/Q(high), where Q(low) is the heat output and Q(high) is the heat input.

Using these equations, we can set up a system of equations to solve for the rate of coal burning (m) in kg/s:

e1 = W1/Q1(high) = 0.6 (since the efficiency is 60% of the Carnot efficiency)
e2 = W2/Q2(high) = 0.6 (since the efficiency is 60% of the Carnot efficiency)
W1 = W2 (since the output of the first engine contributes to the input of the second engine)
Q1(high) = Q2(low) (since the heat output of the first engine is the heat input of the second engine)
Q1(high) = Q2(high) - Q

 

[ogb p]

Calculating the coal burn rate to output 1000MW of power requires a thorough understanding of the heat transfer and efficiency of the steam engines in the power plant. The first step would be to calculate the total heat input required to produce 1000MW of power, taking into account the efficiency of the engines. This can be done using the formula W = eQ(high), where W is the power output, e is the efficiency, and Q(high) is the total heat input.

Next, we need to consider the temperatures at which the engines are operating. From the given information, we know that the first engine operates at 670.0 degrees C and 440.0 degrees C, while the second engine operates at 430.0 degrees C and 290.0 degrees C. These temperature differences are important in determining the efficiency of the engines.

The efficiency of the engines can be calculated using the formula e = 1 - Q(low)/Q(high), where Q(low) is the heat output of the engines and Q(high) is the total heat input. With the given temperatures and assuming an ideal (Carnot) efficiency, we can calculate the efficiency of the engines to be approximately 0.56 or 56%.

Now, we can use the formula W = eQ(high) to calculate the total heat input required to produce 1000MW of power with an efficiency of 56%. This gives us a value of approximately 1.79 x 10^10 J/s.

Finally, we can use the heat of combustion of coal (2.80 x 10^7 J/kg) to calculate the rate at which coal must be burned to produce this amount of heat. This gives us a value of approximately 6.39 x 10^2 kg/s.

In conclusion, to output 1000MW of power, the coal burn rate should be approximately 6.39 x 10^2 kg/s, taking into account the efficiency of the steam engines. It is important to note that this is a simplified calculation and does not take into account other factors such as heat losses and variations in coal quality. Further analysis and experimentation may be necessary for a more accurate calculation.