At a steam power plant, steam engines work in pairs, the heat ...

Question

At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750°C and 440°C, and of the second 415°C and 240°C. If the heat of combustion of coal is 2.8 x 10⁷ J/kg, 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 ideal (Carnot) efficiency.

Accepted Answer

Welcome back. Everyone in this problem. Imagine a thermal power plant relying on two steam engines working in tandem. The heat released by the first engine closely matches the heat absorbed by the second engine. One operates between temperatures of 805 100 °C. While engine two operates between 483 100 °C. The plant burns natural gas to fuel this operation with a 5.6 multiplied by 10 to the seventh joules per kg combustion heat. Given the plant's goal of producing 1200 megawatts of electricity. And considering that each engine operates at around 70% efficiency compared to its ideal carno efficiency. How much natural gas must be consumed per hour to achieve this power output? A says 234,000 kg per hour. B 324,000 kg per hour, C 432,000 kg per hour and d 430,000 kg per hour. Now, there's a lot of information that we have here, but ultimately, remember we're trying to figure out how much natural gas must be consumed per hour to achieve this power output. No, for this power output. OK. For this power output, then essentially we're trying to find the rate of natural gas consumption needed to produce a certain amount of heat. And we know that the rate of natural gas consumption, OK? M OK. Where we put that, that represented the rate of the mass as is gonna be equal to the amount of heat that's being absorbed. OK? Divided by the combustion heat. OK. Divided by the combustion heat. Now, so far in this scenario, we know what the combustion heat is. Our problem told us that it's 5.6 multiplied by 10 to the seventh joules per kilogram. So if we can figure out the amount of heat that's being produced, then we can find the rate of natural gas consumption. So let's start there. Now, before we figure out what that rate is, we have to ensure we understand the efficiency of our system. Ok? And so far our problem gives us our temperatures that both engines operate between. And it also tell us, it also tells us that each engine operates at around 70% efficiency compared to its ideal car. No efficiency. So already this problem, we know that the actual efficiency, OK. Actual efficiency is going to be 70% of the car. No Carno efficiency. Ok. In other words, it's going to be 0.7 eta Now the question is, what do we know about our Carno efficiency? Well, recall. Ok. The, the Carno efficiency is equal to one minus the temperature when it's cold. Or in other words, the low, the lowest, the lower temperature or lowest temperature divided by the temperature when it's hot. In other words, the highest temperature, so we can find the efficiency for both engines because no, that tells us then that the actual efficiency OK. ETA is gonna be equal to 0.7 multiplied by the Carno efficiency. That's one minus TC divided by th. So we can figure out the actual efficiency by converting our temperatures to Kelvin and then solving for ETA. So in this case, OK, then starting with engine one, its actual efficiency is going to be equal to 0.7 multiplied by one minus the temperature when it's cold. For engine one, we know it's 500 °C, converting that to Kelvin by adding 273. That would be 773 Kelvin. And we are dividing that by its highest temperature of 800 °C or 1073 Kelvin. And when we do that, we get engines one, engine one's efficiency to be 0.1957. Now what about engine two? Well, for engine two, that's going to be 0.7 multiplied by one minus its lowest temperature 300 °C or 573 Kelvin divided by its highest temperature of 483 sorry 480 °C or 753 Kelvin. And again, when we go ahead and solve here, then we get the efficiency for engine two to be 0.1673. So now we have the actual efficiency of both engines. Next, we need to figure out the power output of each engine. Now, what do we know about the poor output? Well, again, we know that the power output can be expressed in terms of the heat absorbed and efficiency. In other words, the power output is going to be equal to the efficiency multiplied by the heat that's absorbed. Qh which eventually what we, what is, what we're trying to find to figure out a rate of natural gas consumption. OK. Now, here we know that for our engine, the power output, since the engines are operating in tandem, the power output or the total power output, P one P two is 1200 megawatts. OK. Then if we use this formula to figure out what P one, sorry to figure out what P one and P two are in terms of the efficiency and the heat absorbed, then we can eventually solve for the heat that's being absorbed. OK. So now applying this formula, then that means ETA one, Q, one K plus eta two Q two is going to be equal to P one plus P two. OK. And again, we know what that value is. So I'll rewrite 1200 megawatts uh when we get to the end of our problem. But now let's just keep it in variables. Now, here we have the engine absorbed by, sorry, the heat absorbed by engine one and the heat absorbed by engine two. Can we find a way to express both in the same variable? Well, again, we can use what we know. Let's try to relate the power output to the heat input. We know, let me put a star here, we know that the absorbed heat by the second engine Q two is equal to the heat released by the first engine QC one, the heat released by the first engine is equal to the heat absorbed by the first engine multiplied by one minus the efficiency or the actual efficiency of the first engine. So in other words, qh two is equal to qh one multiplied by one minus ETA one. Now we can go ahead and substitute that into our term here because now we can rewrite it then as E A one, Qh one plus ETA two multiplied by our expression qh one multiplied by one minus ETA one. OK. And all of that is no equal to the total power output. Notice that both terms have qh one in common. So we can factor Qh one out and that's going to leave us with ETA one plus ETA two multiplied by one minus ETA one on our left hand side of the equation. No, if we're solving, then for qh one, we can divide both sides by the expression on the left hand side, that's multiplying qh one to get it to be equal to the total or power output divided by ETA one plus ETA two multiplied by one minus ETA one. OK. And no, since we know all of these values, we can solve for the heat absorbed by engine one because here that means we know our total power output is 1200 megawatts. The efficiency of our first engine is 0.1957 for the second engine is 0.1673. And now we're multiplying that by one minus 0.1957. Now, when we go ahead and solve here, OK. Then oops, that's definitely doesn't look like a, a fraction line. Let me fix it up a bit. Now when we go ahead and solve, OK, then we get the heat absorbed by the first engine to be equal to 3633 multiplied by 10 to the sixth watts. Know that we have a value for the heat that's absorbed. We can substitute it into our formula for the rate of natural gas consumption. OK? Because no, even though the heat absorbed by the first engine, it stands for the heat that's being absorbed. Qh OK. So calculating the rate of natural gas consumption, this means then that the rate OK to produce Q amount of heat qh amount of heat, assuming that combustion heat of natural gas. As given in the problem statement, it's gonna be equal to 3633 multiplied by 10 to the six watts divided by 5.6 multiplied by 10 to the seventh joules per kilogram. When we go ahead and divide, then we get our rate to be equal to 64.875 kg per second to complete our problem. Let's convert this to kilograms per hour like the rest of the units in our answers. Ok. So converting to kilograms per hour, we can take, ok, let me just write it here. We can take our rate 64.875 kg per second and multiply it by 3600 seconds for every hour, which gives us a rate of 233,550 kg per hour or to three significant figures like the rest of our answers 234,000 kg per hour. So to achieve the desired output of 1200 megawatts from the thermal power plant, 234,000 kg per hour of natural gas must be consumed. If you look back at our answer choices, that means a is the correct answer. Thanks for watching everyone. I hope this video helped.

Key Concepts

Carnot Efficiency

Thermal Efficiency

Power Output and Fuel Consumption

Watch next