FIGURE P21.57 shows the cycle for a heat engine that uses a gas h...

Question

FIGURE P21.57 shows the cycle for a heat engine that uses a gas having r = 1.25. The initial temperature is T₁ = 300 K, and this engine operates at 20 cycles per second.a. What is the power output of the engine?

Accepted Answer

Hello, let's go through this practice problem. A prototype thermal engine operates on a closed cycle that includes one isobaric, one isoc chic and one isothermal process. As illustrated in the figure below the working substance inside the engine is one mole of air which has a molar heat capacity ratio of 1.40. The cycle starts with an initial temperature T sub A of 290 Kelvins. The engine runs at a rate of five cycles per second. Determine the output power of the thermal engine option. A 0.87 kilowatts B 4.33 kilowatts, C, 16.5 kilowatts and D 21.7 kilowatts. So this problem is kind enough to give us a lovely PV diagram showing the processes. So we start at point A where our temperature is 290 Kelvins, we have a pressure of 1.5 multiplied by 10 to the power of five pascals and we have a volume of 40 multiplied by 10 to the power of negative three cubic meters. So we're starting at point A and then after the isobaric process, that means constant pressure, we move to point P. Then after the iso choric process at constant volume, we move down to point C and then we move from C to point A through an isothermal process, which means that the temperature remains constant. And we can find some mathematical relations between the three points using the ideal gas law, which states that the pressure multiplied by volume is equal to the number of moles of the gas multiplied by the ideal gas constant multiplied by the gases. Temperature. We can use this equation to narrow down a few variables and relationships between the three different points. Let's start by talking about the isothermal process that goes from point C to point A because the process is isothermal. That means that the temperature at point C is the same as the temperature at point A, both of which are equal to 290 Kelvins. So what that tells us with respect to the ideal gas law is that if the temperature remains constant throughout that process, then that means the entire right side of the equation remains constant because R always has the same value and N the number of moles of the gas isn't changing in this problem. So if the right hand side of this equation is constant, then that means that the left hand side of the ideal gas law is also constant. So for the process from C to A, that means that the PV, the pressure multiplied by volume product must stay the same. So this tells us that the product of P sub A and V sub A is the same as the product of P sub C and V sub C. We can use this relationship to solve for the pressure at point C because that is not apparent from the graph. So algebraically dividing both sides of the equation by V sub C, the pressure at point C is equal to P sub A multiplied by V sub A divided by V sub C. To solving this equation. For P sub C, we take the pressure at point C which is 1.5 multiplied by 10 to the power of five pascals. That's the pressure at point A multiplied by the volume at point A which is 40 multiplied by 10 to the power of a negative three cubic meters all divided by the volume at point C which is 80 multiplied by 10 to the power of negative three cubic meters. Put that into a calculator and the pressure at point C is 0.75 multiplied by 10 to the power of five pascals. Now let's talk for a moment about the isobaric process from point A to point B where the pressure is constant. With this pro with this process, the pressure is constant. So the pressure at point A is equal to the pressure at point B. During this process, we can see that the volume doubles as it moves from 40 multiplied by 10 to the power of negative three cubic meters to 80 multiplied by 10 to the power of negative 30 cubic meters, negative 30 cubic meters. So this means that V sub B, the volume at point B is equal to double the volume at point A. Once again, we will refer to the combined Ideal gas law to find a relationship between these variables which will help us find the temperature at point B. So the combined Ideal gas law states that P sub A multiplied by V sub A divided by T sub A should be constant. So this is equal to P sub B multiplied by V sub B divided by T sub B solving this equation for T sub B by multiplying both sides of the equation by T sub B to bring it into the numerator multiplying both sides by T sub A and dividing both sides by P sub A V sub A to get the T sub B on its own. And we can find that T sub B is equal to P sub B multiplied by V sub B divided by P sub A multiplied by V sub A. And then all this is multiplied by T sub A. And as we've already established, P sub B is equal to P sub A. So the pressures cancel out and V sub B is equal to twice of V sub A. So this V sub B divided by V sub A can just be replaced with A two. So we can see that T sub B is just equal to two multiplied by T sub A. The temperature at point B is just double the temperature at point A and the temperature at point A is 290 Kelvins. So that means that the temperature at point B is twice that it is 580 Kelvins. So now that we've determined the variables for each of the three states. Now let's talk about how to figure out the work done by one cycle. The total work done during one cycle is going to be equal to the work. The sum of the work done by each of the three segments. So first, let's talk about the work done during the process from A to B. Now for an isobaric process where the pressure is constant work is equal to that pressure. So P sub B in this case multiplied by the change in volume. So P sub B multiplied by V sub B minus V sub A. So the work during this segment is just P sub B 1.5 multiplied by 10 to the power of negative five pascals multiplied by the change in volume. And the change in volume is from 40 to 80 multiplied by 10 to the power of negative 3 m. So that's a change of 80 minus 40 is 40. So that's 40 multiplied by 10 to the power of negative three cubic meters, put that into a calculator and we find the work to be about 6000 jewels. So that's the work done by the engine during the first part of the cycle. Now, we'll talk about the part we'll remove from point B to point C. Now, this is the point where the volume is kept constant. This is the isoc coric section and this is the easy part because generally, when we have volume being kept constant, no work is being done. So W sub two is equal to zero. Lastly for the final leg of the cycle from C back to a, it's an isothermal process. And the work done during an isothermal process is equal to the number of moles of the gas multiplied by the ideal gas constant multiplied by the temperature multiplied by the natural log of the final volume divided by the initial volume. So let's plug our values into this equation. Now, we have one mole of the gas. The ideal gas constant is 8.31 joules per mole. Kelvins, the temperature is 290 Kelvins at points C and A and multiplied by the natural log of the ratio of the volumes. So that's the natural log of 40 multiplied by 10 to the power of a negative three cubic meters divided by 80. And when we put this into a calculator, we find the work done here to be negative 1670 jewels. So now that we know the work done during each individual segment of the cycle's journey, now we can find the total work done during one cycle by adding them together. So W sub one plus W sub two plus W sub three or 6000 jewels plus zero jewels plus negative 1670 jewels which putting into a calculator gives us 4330 jewels. But the problem isn't asking for the work done during one cycle. It's asking for the output power and we're told that the engine runs at a rate of five cycles per second. So to find the power, we're going to have to take the work we just found. So that's 4330 jewels per cycle and multiply that by the cycle rate. So that is five cycles per second. And this gives us 21,700 jewels per second, which can be abbreviated as 21.7 kilowatts. And that is the output power of the engine which corresponds to option D. So option D is our answer and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

Struggling with Physics?

FIGURE P21.57 shows the cycle for a heat engine that uses a gas having r = 1.25. The initial temperature is T₁ = 300 K, and this engine operates at 20 cycles per second.a. What is the power output of the engine?

Watch next

Heat Engines & PV Diagrams practice set