 A steam power plant with a power output of 150 megawatts consumes coal at a rate of 60 tons per hour. If the heating value of the coal is 30,000 kJ per kilogram, determine the overall efficiency of this plant. Then, if the flame temperature of the combusting coal is about 500 degrees Celsius, and the heat in the condenser is rejected to ambient air at 22 degrees Celsius, determine the theoretical maximum thermal efficiency of the plant in those conditions. The first thing I'm going to do is recognize that I have a heat engine. I know that I have a heat engine because, first of all, it's in the example problem statement, but also because I'm consuming heat transfer in order to produce work. I am tapping into the natural transfer of heat from the high temperature burning coal to the low temperature air to produce power. As a result, I can represent it with my heat engine diagram. I've heat driven from the high temperature to the low temperature, which for this problem I know are 600 degrees Celsius and 22 degrees Celsius respectively. Excuse me, 500 degrees Celsius and 22 degrees Celsius. And remember also that I can only plug those in when I'm evaluating the maximum case analysis. I know that the plant has a power output of 150 megawatts, which is going to represent the net work output. It is getting its heat by burning coal. We're consuming coal at a rate of 60 tons per hour, which is a mass flow rate. And we were told the heating value of the coal is 30,000 kilojoules per kilogram. So every kilogram of coal that we burn is producing 30,000 kilojoules of energy and we are burning coal at a rate of 60 tons per hour. We can multiply those two quantities together to come up with an expression for how much heat is emitted by the burning coal per unit time. That will give us Q dot in. Once I have Q dot in, I will be using that to calculate the thermal efficiency, which means I'm going to divide my network out of 150 megawatts by the number we calculate. And because I want a unitless proportion, it will probably be easiest if I calculate Q dot in in megawatts. In order to make that conversion, I have to make an assumption about the tons. There are a couple of different types of tons that we consider. We have the imperial ton, which is 2,000 pounds. We have the metric ton, which is 1,000 kilograms. We also have a ton of refrigeration, which is 211 kilojoules of refrigeration per minute. We can rule that one out immediately, but we have to deduce whether this is an imperial ton or a metric ton based on the context. The fact that every other unit in the problem is a metric unit implies to us that we should treat this as a metric mass flow rate. Therefore, I'm going to call that ton the metric ton. And the metric ton, which I can spell as T-O-N-N-E, is 1,000 kilograms. Then I recognize one hour is 60 minutes. And one minute is 60 seconds. That will give me kilojoules per second. And I want megawatts, which is a megajoule per second. And one megajoule is 1,000 kilojoules. So for that, I will use my calculator, which we have avoided using as long as possible. And calculator, we need you to take 60 times 30,000 times 1,000 divided by 60 times 60 times 1,000. Calculator says 500 megawatts. Thank you, calculator. Now we take our net power output, which we were told is 150 megawatts. And we divide by our shiny new q.in, which is 500 megawatts. And we get a thermal efficiency of 30%. Just to verify, I will let the calculator second my calculation. Look at that, I get three tenths. Thank you, calculator. How about a decimal? 0.3. So assuming that all of the coal produces 30,000 kilojoules per kilogram of heat transfer and all of that heat transfer is being acquired for the purposes of this power production, the thermal efficiency of this facility is 30%. Well, that begs the next question. Is that a good thermal efficiency or not? To evaluate that, we should calculate the best thermal efficiency that could be achieved under these operating conditions. Doing that will give us a ruler to compare against. If we determine that the theoretical maximum thermal efficiency is 32%, then we can say that this is a great power plant. If the theoretical maximum is 60% or 70%, then we can say that there are a lot of opportunities for improvement. If we calculate that the theoretical maximum is 28%, then we know that we cannot have the operation of this facility. It cannot consume coal at this rate and produce that much power. Doing so would violate the second law of thermodynamics. So using my equation from earlier, to evaluate the best case scenario, we're going to be treating this as a Carnot heat engine. And when we do that, we are going to be taking 1-TL over Th. My low temperature reservoir is operated at 22 degrees. My high temperature reservoir is operated at 500 degrees. And remember, anytime I do math, I'm going to need absolute temperature. So let's see if my calculator can work this one out. I have low expectations. Calculator says 61.8%. So we determine that operating between these two temperatures, we theoretically could, if everything were perfect and there were no losses, have a thermal efficiency of 61.825%. By the way, while we're here, knowing that the rate of heat transfer in is 500 megawatts and the rate of net power output is 150 megawatts, how much heat is rejected? If you said 350, you're right. We are bringing in 500 megawatts of heat. We are rejecting 150 of that as net power output, which means that the remaining 350 must leave as heat transfer output.