 A reversible heat pump is to be used to heat a house and maintain it at 20 degrees Celsius in the winter. On a day when the average outdoor temperature remains about 2 degrees Celsius, the house is estimated to lose heat at a rate of 82,000 kJ per hour. If the heat pump consumes 8 kW of power while operating, determine first how long the heat pump ran that day, then the total heating costs, assuming an average price of electricity of 8.5 cents per kWh, and the heating costs for that same day if we were using electric resistance heating and heating the entire house. So we have a refrigeration cycle operating as a heating device, or heat pump. So the heat pump is pushing heat from the cool outside air, which is 2 degrees Celsius, to the warm inside air, which is 22 degrees Celsius, which is 20 degrees Celsius. I was told that the house is naturally losing heat to its atmosphere at a rate of 82,000 kJ per hour. By the way, that is a terrible rate of heat rejection. I mean, a gigantic house with no insulation is going to reject less heat than that. For it to reject that much heat, I mean we basically have to have a hole in the ceiling and the floor for good measure. That is a lot of heat loss. But the goal here is to compare and contrast the heat pump operation to the electric resistance heater, and to make that contrast as stark as possible. We are considering having to heat a lot. And that logic works like this. If we are putting in as much energy as the house is losing, the temperature of the house will remain constant. If we are adding more heat than the house is losing, it's going to increase in temperature over time. And if we are losing more heat than we're adding to the house, it's going to be decreasing in temperature over time. So the fact that it loses heat at a rate of 82,000 kJ per hour implies to us that we have to heat it at an average rate of 82,000 kJ per hour in order to hold the temperature of the house at 20 degrees Celsius. Furthermore, we know that it is a reversible heat pump, which implies to us that the coefficient of performance of this heat pump is the maximum that could occur. Because it is reversible, i.e. perfect, we have a Carnot heat pump heating our house. Array for us. We can relate that heat pump to the temperatures driving the heat transfer on either side and write COPHP max for this specific case because it is reversible as 1 over 1 minus TL over TH. So that's going to be 1 over 1 minus 2 plus 273.15 over 20 plus 273.15. So our coefficient of performance for this perfect heat pump is 1 over 1 minus 2 plus 273.15 divided by 20 plus 273.15. And we have a coefficient of performance of 16.286. That is a really, really good heat pump. Terrible house, terrible insulation, really good heat pump. We were also told that the heat pump consumes 8 kW of power when it's operating. So this, writing it out like this, doesn't work because when it's running it has to consume 8 kW. We can't turn this heat pump up or down. What we're turning up or down is the holding temperature of our house. The way that your thermostat works with a system like this is it turns the heating on or off. In order to average 82,000 kJ per hour, we're figuring out how much the heat pump has to run on that day. So cycling on and off to maintain 82,000 kJ per hour is going to give us how long the heat pump ran that day. So you could think of that in a couple of different ways. You could figure out how much heat is rejected over the course of a day by taking 82,000 kJ per hour multiplied by 24 hours. And then you could use the relation between Q out and network in to figure out how much work is going to be consumed. And then you could use 8 kJ per second. That's the rate at which work is applied to figure out how long the heating system ran. You could also approach it by figuring out how long the heat pump would run in a single hour if it was averaging 82,000 kJ of heat rejection and then use that amount of time multiplied by 24 to figure out how long it ran in a day. By the way, we are assuming that the temperature of the outside air is constant the whole day. Either case will produce the same result. For our simplification here, let's call it a 24-hour period, in which case the magnitude of heat lost by the house is going to be 82,000 kJ per hour multiplied by 24 hours. So calculator if you would be so kind. 82,000, that's 81,000 calculator. Nope, that's 820,000. There you go. Multiply by 24. We get 100, excuse me, 1,968,000 kJ of heat lost over the day. And then, because COPHP can represent Q dot out over rate of network in and or the magnitude of heat transfer out to the magnitude of network in, we can say the network in is just going to be that quantity of heat lost divided by our COP of our system. So that's 1,968,000 kJ divided by the COP that we calculated earlier, which is 16.2861. And we figure out that the magnitude of electrical work consumed by the heat pump is 120,839. And then when the pump is running, it is consuming work at a rate of 8 kW, and that represents the magnitude of work per unit time. So we were to take 120,839 kJ and divide by 8 kW. We would produce an amount of time that it would take to consume that much electricity. So 120,000 divided by 8 kJ per second. KJ will cancel and I'll be left with a quantity in seconds. 15,104.9 seconds. So we can make that a little bit readable by dividing by 3,600. And I get 4.196 hours. So the heat pump will run 4.196 hours over the course of the day. And that will average the make-up heat required if the heat lost by the house is 82,000 kJ per hour. Part B asks us to calculate the total heating costs if we're paying for electricity at a rate of 8.5 cents per kWh. So every hour that we consume power, we are paying 8.5 cents per kW of electricity drawn. So if we are using 8 kW for 4.12 hours, we can figure out the electrical magnitude that we're paying for. So 4.1958 hours multiplied by 8 kW gives us 33.5664 kWh. And for each of those, we are paying the electrical company 8.5 cents. Which means that we are paying 285 cents or $2.85 over the course of the day. Lastly, I wanted us to compare that to what it would cost to heat the house if we were using electric resistance heating. So what we're assuming here is that we are converting electricity to heat in the electric resistance heaters at a rate of 1 to 1. So for every 1 unit of electricity we put through the heater, we get out 1 unit of heat. That sounds great when you say it has an efficiency of 100%. But note that we're comparing that against a system that has an efficiency of 1,630%. So 100% is actually relatively small in this context. Which means we're going to have to pay more money because we're not getting that return on our investment of electricity. So there, we're going to have to figure out how much electricity over the course of the day is going to make up for 1,968,000 kJ of heat. Well, if we're converting at a rate of 1 to 1, that means we're going to have to consume 1,968,000 kJ of electricity. And the electrical company is billing us in kilowatt hours, which is a unit of energy. So we might as well convert that while we're here. 1,968,000 kJ. And I know that a kilowatt is a kJ per second. And I know that there are 3,600 seconds in one hour. And I know that a kilowatt hour is a kilowatt times an hour. So if I take 1,968,000 and divide it by 3,600, I will yield an answer to how much heat, or excuse me, how much work are we being billed for in kilowatt hours. And we get a fraction, thank you calculator. We get 546.67 kilowatt hours, which at a rate of 8.5 cents per kilowatt hour means we are paying 4,646 cents or $46.47. That's quite a difference, right? So let's reflect on this problem for a minute. We have a lot of different ways that we can analyze this problem. We have lots of different methods that will get us to the same answers. I mean, for example, in our heating pump, we are getting a return on investment of 16.29, which means that we're paying 16.29 times less to run that heating system. So we could have taken 2.85 multiplied by 16.289 and yielded the answer for part C. Also note, we could have analyzed a single 24-hour period, or we could have analyzed a 24-hour period, or any time in between. We could also try to think of it as a fraction of time spent. We could figure out the fraction of time that the heat pump needs to run in order to output 82,000 kJ per hour by figuring out how much heat it's rejecting by multiplying 16.286 times 8 kilowatts, converting that to kJ per hour, and then taking 82,000 divided by that quantity. We would figure out a proportion of the day in percentage at that point, and we could multiply that by 24 hours to figure out how many hours the heat pump ran that day. There are a lot of ways to approach this. Also note, the equations we quote, built, unquote for this problem are not really equations that came from anywhere. We are relating the quantities we know based on how this system works. This sort of analysis is what makes thermo hard for a lot of new students. It's not just an algorithm, you're not just applying the same equations. You are reasoning through a situation using your understanding of the fundamentals. The easiest way to get good at this is practice, building up some experience. The last thing I'll point out while we're here is the fact that a heat pump is going to have a coefficient of performance of greater than one, which means that it's going to be better than electric resistance heating. But we are doing the electric resistance heater a disservice by doing a direct comparison. Because while, yes, the electric resistance heater is far less efficient, you gain a little bit of return on that investment by being able to only run it in the rooms that you're actually occupying. So if you had a big mansion and you spent all day in your bedroom watching Netflix, you could just heat the room and not the rest of the house. So it's not exactly an apples to apples comparison. Also, the heat pump also takes more initial cost to install, so it really becomes a return on an investment over time. The advantages and disadvantages of paying more for a year for a less expensive system to install. In almost all cases, the heat pump is worth it in the long run compared to an electric resistance heater. Unless, of course, you're in a situation where you're not the one paying the electricity bill. If you have tenants, electric resistance heaters are easily the way to go. You can offload the heating onto them. You don't have to pay as much upfront and also it's a much easier maintenance. So pros and cons depending on your perspective. We could also compare this to say a natural gas furnace by looking at the combustion efficiency of the furnace, figuring out how much fuel would have to be consumed using the heating value of the fuel and the efficiency to figure out how much fuel has to be burned to output 1,968,000 kilojoules. We could use the cost of natural gas to figure out how much it would cost to operate that system. And again, we have a different type of insulation which costs a different amount of money. We have a different lifespan. So we compare and contrast the long term benefits and disadvantages of each of the systems.