 Helium is to be compressed from 120 kilopascals in 310 kelvin to 700 kilopascals in 430 kelvin. A heat loss of 20 kilojoules per kilogram occurs during the compression process. Neglecting kinetic energy changes determine the power input required for a mass flow rate of 90 kilograms per minute. Parsing that out, we recognize that our steady flow device here is a compressor, as evidence from the name, as well as the description of the process. Furthermore, we know that our working fluid is going to be helium. We don't have tables for helium, which means that we're going to have to treat helium as an ideal gas and assume constant specific heats. We are told an inlet temperature and pressure, as well as an outlet temperature and pressure. We're given a specific rate of heat rejection, so a little cue out. We are told a mass flow rate in kilograms per minute. Furthermore, we recognize that because this is a steady flow device with one inlet and one outlet, the mass flow rate entering has to equal the mass flow rate exiting, which means 90 kilograms per minute describes both the mass flow rate in and the mass flow rate out. Jumping into our steady flow device's table, we can see that for a compressor, the goal of the device is to increase the enthalpy of the fluid. That primarily occurs in the form of pressure, but the goal is to increase enthalpy. The source of energy for that increase in enthalpy is work input. So we are putting in work to get out a change in enthalpy. As a result of that, we would expect to see a work input, which is implied here as the thing that we're looking for, and a change in enthalpy from state 1 to state 2. I will start with a rough diagram, and for a compressor, we typically draw a shape with a converging cross-sectional area, and then an inlet and an outlet on the big and small sides. Typically, that is drawn as though the fluid is going from left to right, but that doesn't really matter. On this diagram, I can indicate that there will be a work input. I'm drawing that as a specific work. It doesn't really matter. I could draw that as a power input as well. I'm going to describe my inlet state point as state 1, and my outlet state point as state 2. I will indicate that my control volume here includes an inlet, state 1, and an outlet, state 2, as well as a work input in the form of shaft work and a heat rejection in the form of a specific hue out. Furthermore, while I'm here, I could indicate that helium is the working fluid, and I could parse my given information into an inlet and outlet set of state point properties. At state 1, we have a temperature of 310 Kelvin and 120 kilopascals, and for state 2, I have a temperature of 430 Kelvin and a pressure of 700 kilopascals, and I will add into that my mass flow rate at both state points, 90 kilograms per minute, and that my specific hue out is 20 kilojoules per kilogram. I'm looking for a power input, which implies that I'm going to be using an energy balance to relate how the enthalpy increase corresponds to the power input. To do that, I will write out my energy balance. My energy balance is on the control volume, and that control volume is an open system, so I'm going to have to account for energy crossing the boundary as heat and or work and or the energy associated with a moving mass. I'm also going to be treating this as being steady state, which means I'm neglecting the effects of time. So the compressor is assumed to be fully warmed up and is running steadily, so regardless of when you look at it, the temperature of the inlet and the outlet don't change. I will add to my list of assumptions that I'm treating it as an open system, even though I don't necessarily have to. And then I could think through what terms are going to be appearing in the energy balance. Since I'm using steady state analysis, I'm likely going to be using the rate form of the energy balance, which means that I'm probably going to end up with a rate of heat transfer, a rate of work, and then a change in energy of the fluid itself, which can include enthalpy and kinetic energy and potential energy. While I'm listing assumptions, it might be beneficial to me to assume that there's no heat transfer in, that is, that the heat transfer that's occurring is only in the outward direction, and that there is no work output, that is, that the only work occurring is in the inward direction. We weren't given enough information to know of any other sources of work or heat transfer, so it's reasonable to assume that none occur. Lastly, I will assume that any changes in kinetic or potential energy are negligibly small. I don't necessarily have to think through all these assumptions right now. If I wanted to, I could start my energy balance and then at each step of the process, eliminate some of the terms, but by thinking through it ahead of time, I can build a plan for my energy balance and think about what it is that I'm actually trying to accomplish with that setup. So back in my energy balance, the change in energy of the control volume is equal to the energy entering minus the energy exiting. I'm going to divide all three terms by dt, at which point I have the rate of change of the energy of the control volume is equal to the rate of energy entering minus the rate of energy exiting. The rate of change of energy of the control volume is zero because of the assumption of steady state, which means e.in is equal to e.out. Since it's an open system, energy can cross the boundary in the form of heat, work, or the energy associated with the moving mass. So I can write out q.in plus work.in plus the sum in of m.theta is equal to q.out plus work.out plus the sum out of m.theta. Next, I can neglect the terms that I've assumed are negligibly small. Since I have a rate of heat rejection, I'm neglecting q.in. That was assumption number three. Since I have a power input, I'm neglecting a power output. That was assumption number four. So I have power input plus the summation of the mass flow rate times the specific energy of the moving mass for all of the inputs, which of course there is only one. It is state one. So I can write that as power input plus m.1 theta one is equal to q.out plus the sum of the mass flow rate times the specific energy of the flowing mass for all of the outlets, which of course there is only one. It is state two. Then I can expand my theta term. Remember that theta could be specific enthalpy. It could be specific kinetic energy, and it could be specific potential energy. Sometimes neglecting changes in kinetic and potential energy, whatever the kinetic energy is at the inlet, it is going to be the same at the outlet. And kE in one and kE in two are going to be the same. It's important to note that I'm not saying that there is no kinetic energy. I'm saying that whatever the kinetic energy is, it isn't changing. Because it appears on both sides of the equation, it disappears. It's not that there is no kinetic energy. It's that there is no change in kinetic energy. Making the substitution I have m.1 times h1 specific kE1 plus specific pE1. Note that I could substitute in there 1 half times velocity squared plus gravity times height. But since I'm going to be cancelling it anyway, there's really no point. We'll move this over to keep my equal signs aligned. And then I will recognize that my mass flow rates are going to be the same. I know that because of the mass balance. To formalize that, I could write out a mass balance. The mass balance begins the same way. I have delta m is equal to m in minus m out. I divide all three terms by dt at which point I have dM dt which is zero because nothing can change with respect to time. Is equal to the sum of all the mass flow rates entering minus the sum of all the mass flow rates exiting. That means m.in has to equal m.out because there's only one inlet. It is state 1. And because there is only one outlet, it is state 2. m.1 and m.2 must be the same. I will abbreviate that as just m. And in fact, I know that quantity. It is 90 kg per minute. So over here I can group together my mass flow rates and factor out the mass flow rate itself. Since what I'm looking for is power input, I will rearrange to solve for power input and then I can neglect changes in kinetic and potential energy as a result of assumption number 5. Therefore, the power input is going to be q.out plus the mass flow rate times the change in enthalpy. So let's think about it. We are increasing the enthalpy, which means that we should have a positive change in this delta here. That represents how much power we have to add. And the presence of heat transfer means that we have to increase how much power we are adding. So losing some of the energy in the process means that we have to add more work in the process to reach the same endpoint. That makes sense. Let's see if we can really set up correctly. In my q.out term I recognize that what I'm actually doing is going to be plugging in mass flow rate times specific q.out which means that at this point I know mass flow rate I know specific q.out the only unknown is delta H. When I evaluate a delta H I have 3 options. Option 1 would be to determine H1 and H2 look them up in tables and subtract those results directly. I could approximate delta H by coming up with a line of best fit for its specific heat capacity and perform that integral. Or option 3 would be to assume that the specific heat capacity doesn't change. That is, whatever it is it is the same across this entire temperature change. Therefore it would come out of the integral. That comes from our definition of Cp which is del H del T. And in fact because this is an ideal gas internal energy and enthalpy are only functions of temperature which means that we only have to use dH dT. Furthermore we don't need this to be a constant pressure or constant volume process because it's an ideal gas when we evaluate delta H what we're using is Cp. If we were evaluating a delta U we would use Cv. H is to P U is to V. Cp is dH dT Cv is du dT Harry Potter ultraviolet In fact I should add helium is ideal to my list of assumptions at which point I can write dH is equal to Cp times dT and integrate both sides. On the left hand side I just have delta H. On the right hand side I have the integral of Cp as a function of temperature dT. If I make one more assumption here that the specific capacity of the helium is constant then that Cp term would come out of the integral and I would be left with Cp times T2 minus T1. So my energy balance is now mass flow rate which I can factor out of the heat transfer term and the change in enthalpy times specific Q out plus Cp of helium times T2 minus T1. While we're here let's talk a little bit about why we're assuming Cp is constant. If we had helium tables the best thing to do in this circumstance would be to look up H1 and H2. That would provide the best answer. However we don't happen to have a set of tables. In the appendices of our textbook the best thing we have are specific heat capacity values. If I jump into table A21 I can see that I have specific heat capacity values for a variety of gases and this is the best data I have for helium. Note this bottom row for monatomic gases includes helium, neon and argon. So we're going to be using Cp over R is equal to 2.5. That's the best thing we can do because it's the only thing that we can do and still evaluate a delta H. Therefore we're making the assumption of constant specific heats. We know that for an ideal gas across a relatively small temperature change this is only a temperature change of around 100 degrees celsius which isn't much for an ideal gas this should provide relatively accurate results. So it's an okay assumption even at that. So my Cp value for helium is going to come from table A21 and that is specifically the molar specific Cp value divided by the molar specific gas constant that's what's implied by this horizontal bar above the value is equal to alpha plus beta times temperature plus gamma times temperature squared plus delta times temperature cubed plus epsilon times temperature to the fourth power where alpha is 2.5 beta is 0 gamma is 0, delta is 0 and epsilon is 0. Therefore the molar specific Cp value divided by the molar specific gas constant is 2.5 Therefore the molar specific value for helium would be 2.5 times the molar gas constant. The molar gas constant or universal gas constant is 8.314 kilojoules per kilomole kelvin you can get that from the inside of our front cover of our textbook. Down here at the bottom we have a couple different values for the universal gas constant we have kilojoules per kilomole kelvin feet pound force per pound mole rankine and BTUs per pound mole rankine kilojoules per kilomole kelvin is going to be the most convenient here because I have a problem in metric units. So if I take 8.314 kilojoules per kilomole kelvin and multiply that by 2.5 I get a result for the molar specific heat capacity of helium for that I will use my calculator and I get 20.785 kilojoules per kilomole kelvin that is all well and good but I recognize that I want to yield a value in kilowatts in my energy balance calculation which means that I eventually need the kilograms in the mass flow rate to cancel so I am going to want kilojoules per kilogram kelvin instead of kilojoules per kilomole kelvin so I want to convert this value from a molar specific specific heat capacity to a mass specific specific heat capacity to do that I will divide by the molar mass so mass specific heat capacity is going to be molar specific heat capacity divided by molar mass which means I am going to take 20.785 and divide by the molar mass for helium for the molar mass for helium I will jump into table A1 which is about the closest thing I have to a periodic table on table A1 I see that the molar mass of helium is 4.003 kilograms per kilomole excuse me kilograms per kilomole at which point I am left with a result in kilojoules per kilogram kelvin so 20.785 divided by 4.003 that yields 5.192 kilojoules per kilogram kelvin and I will note this value came from table A1 just to try to make that clearer for students who aren't following along with the video so I am taking 90 kilograms per minute I am multiplying that by the quantity 20 kilojoules per kilogram plus 5.2 kilojoules per kilogram kelvin times a temperature change from state 2 to state 1 which would be 430 minus 310 for that let me clear up some space much more better and take 90 kilograms per minute and multiply by 20 kilojoules per kilogram times 5.192 5.192 kilojoules per kilogram kelvin multiplied by 430 minus 310 multiplying 5.2 kilojoules per kilogram kelvin by a temperature change in kelvin will yield a result in kilojoules per kilogram multiplying a quantity in kilojoules per kilogram by a quantity in kilograms per minute will yield a result in kilojoules per minute I want kilowatts which would be a kilojoule per second so I am going to divide by 60 out front here that means I will be left with kilojoules per second which is a kilowatt now let me jump back into my calculator and take 90 divided by 60 times the quantity plus 5.19236 times 430 minus 310 which yields 964.625 so it takes just under a megawatt to run this compressor and result in a temperature change of about 120 degrees celsius with a mass flow rate of 90 kilograms per minute that makes sense as 90 kilograms of helium every minute is quite a lot of helium remember that the density of a gas like this is relatively low so that means a lot of volume has to be processed by this compressor also note we didn't actually need the pressure all we actually cared about was a change in enthalpy and for an ideal gas specific internal energy and specific enthalpy are only functions of temperature as a result of that all that really mattered was the temperature we don't even really care about the pressures