 A 3-horsepower air compressor is attached to the top of a 100-gallon rigid tank and periodically needs to be filled with compressed air. Initially, the tank is completely discharged and has been allowed to sit long enough to reach ambient temperature. The compressor is able to maintain a sustained output of air at 35 mps, 42°C, and 700 kPa, regardless of the pressure in the tank. Determine A, the rate of mass flow into the tank. B, the rate of heat loss to the room. C, the total mass of air added to the tank if the compressor stops operating when the tank reaches 600 kPa in 42°C. And D, the amount of time it would take to do that. I'm going to be breaking this analysis into a steady flow device, an air compressor, and a transient analysis on the tank itself. I'll start by drawing a system diagram. So I have my steady flow device processing air from the ambient conditions into the room to the outlet condition of the compressor, which is also the inlet to the tank. I'm defining the inlet to the compressor as a state point, which I'm calling 1, and the outlet of the compressor as a state point I'm calling 2. Those are both steady state state points. On the tank itself, I'm considering a transient analysis. How long does it take for the tank to fill with compressed air until it reaches 600 kPa in 42°C? I have to establish two different state points in time. Both of them will be referring to the air in the tank, just at different points of time, one being the beginning, one being the end. In an effort to try to reduce confusion between calling them like state point 3 and 4, I will call them something else. Maybe A and B or alpha and beta. So the tank's analysis, which I can call control volume 2, is going to be an open transient analysis. The compressor analysis is going to be open and steady, and now I can begin to propagate information about my state points. I know my inlet conditions are the ambient conditions in the room, which are 22°C and atmospheric pressure. And at state 2, I know the outlet is 42°C and 700 kPa. Furthermore, I was told the velocity at the inlet and the outlet of the compressor and the velocity of the outlet is going to be 35 m per second. And if the compressor is pulling air from an approximately stagnant room, then I can say V1 is pretty close to zero. This is for the same reasons that it was in the analysis of our hairdryer. It's not that the velocity actually is zero, it's just that we're assuming all the change in kinetic energy comes from the compressor. For the air tank itself, it has been allowed to cool to room temperature, which is still 22°C. And if it's completely allowed to equalize with the room, it will be at one atmosphere. I think the problem describes it as having fully discharged. So discharged here means that it has reached equilibrium with the room, not that it was evacuated entirely. Then at the end of the process, what I'm calling beta, I was told the temperature and pressure are 42°C and 600 kPa this time. That gives me my two independent intensive properties at all four state points, from which I can fully define the states. I can use my outlet information to supply an inlet to the tank for my analysis. I'm probably going to end up with a mass and energy balance on both the tank and the compressor, control volume 1 and control volume 2. So I will start a list of assumptions. First of all, I can list that the compressor itself is operating steadily. Next, I can assume that the air is ideal, then I will neglect all changes in potential energy. I will neglect any work in the tank. I will establish an atmospheric pressure of, let's say, around one atmosphere. That should be enough information to get started. Okay, so I'm going to perform a mass balance and an energy balance on control volume 1. I will use the information I learned about the outlet of the compressor to figure out how long it will take for the tank to reach 600 kPa in 42°C. My mass balance on control volume 1 is the same as it's been for every other steady flow device. We have steady state operations, we divide everything by dt or derivate it by with respect to time. That leaves me with dm dt is equal to m dot in minus m dot out. I'm saying dm dt is zero because of steady state operation, which means that the entering mass also has to leave. i.e. m dot in equals m dot out. I have one inlet, it's state 1, I have one outlet, it's state 2. Therefore m dot 1 and m dot 2 are the same. For the energy balance, I begin in the same way. I divide everything by dt or derivate it with respect to time. At which point I have dE dt, dE dt here referring to the control volume is equal to e dot in minus e dot out. Therefore e dot in has to equal e dot out because it's an open system. Both e dot in and e dot out could be heat transfer or work for the energy associated with the moving mass. I'm neglecting workout. I have no changes in potential energy. I include my enthalpy change. I include my kinetic energy change, a work term. And then a heat transfer, if possible. Let's consider heat transfer for a moment. I have two independent intensive properties, which fully define states 1 and 2. However, I don't know a mass flow rate. Since I don't know the mass flow rate, I have to assume that I know everything else about the operation of the device in order to solve for a mass flow rate. If there was a way for me to calculate mass flow rate on its own, then I would do that. But since I don't, I cannot proceed unless I simplify further with an assumption. That's sort of the logic of the steady flow devices. We are trying to model behavior and that model is going to be inaccurate. And as long as we accept the error knowingly, then that's okay. We are establishing assumptions, building a model and solving the model. So here, since we don't have any more information, the only thing that we can do to proceed is to neglect heat transfer. If we had covered chapter 6, we could make an assumption about the compressor operating ideally, but even if we do that, we are neglecting heat transfer. Because heat transfer represents a type of loss for the compressor. So we are adding adiabatic to our list of assumptions so that we can proceed. We will be doing this frequently. When we don't know enough about the operation of a device, we will simplify it by assuming it operates ideally. That assumption is so frequently that it is in place unless we have enough information to dispute it. If we have enough information to say that this device doesn't operate perfectly, then we include it. But if we don't, we assume it's perfect. Then the power input is going to be equal to the mass flow rate times the change in enthalpy and change in kinetic energy. It comes as a result of power input plus m dot 1 times theta 1. Remember that theta could be enthalpy, could be specific kinetic energy, could be specific potential energy. We are neglecting changes in potential energy. Furthermore, we are assuming that the velocity at the inlet is relatively close to zero. Quote, very small unquote, which means that we have power input plus m dot 1 h1 is equal to m dot 2 times h2 plus velocity 2 squared over 2. We know velocity 2, we can figure out h, we know the power input, we can use this information to solve for the mass flow rate. Since the mass flow rates are the same, they can be factored out, solving for mass flow rate leaves power input divided by the specific work of the compressor, which is h2 minus h1 plus v2 squared over 2. All I need to be able to calculate a mass flow rate is the difference in h. Once again, we find ourselves considering a delta h. We have three options. Option one would be to look up h1 and h2. Option two would be to determine how the specific heat capacity of air changes as a function of temperature and perform an integral. Option three would be to assume that it doesn't change as a function of temperature and bring the specific heat capacity out of the integral, at which point we would just have delta h is equal to cp delta t. Generally speaking, since we have air and a relatively small change in temperature, it would be reasonable to assume constant specific heats. But just for fun and the opportunity to practice some more property table lookups, let's look up h1 and h2. At state 1, I have a temperature of 22 degrees Celsius. At state 2, I have a temperature of 42 degrees Celsius, which means that I'm going to jump over into my property tables. The properties of air are going to be table A22, where I can look up the enthalpy of air as a function of temperature. Remember that for ideal gases, internal energy and specific enthalpy are only functions of temperature. So h1 is determined at 22 degrees Celsius. 22 degrees Celsius would be about 295 Kelvin. Yup, 295. So I'm interpolating between 295 and 300. So I will use my calculator for that math so that you can read the interpolation a little bit more clearly. That'd be 295.15 minus 295 divided by 300 minus 295. And we're saying that that's equal to x minus the enthalpy value at 295, which is 295.17 divided by 300.19 minus 295.17. We get a value of 295.321. 295.321. If you watched the hairdryer video, I talk a little bit more about the inaccuracy of trying to approximate the h or other methods of determining delta h than looking up the property tables directly. But since we're here, we will continue to look up values this time at h2, wherein I have a temperature of 42 degrees Celsius. So 42 plus 273.15 is going to be 315.15. So I will do essentially the same interpolation except this time between 315 and 320. So I get an enthalpy value at state 2 of 315.42. And just to be clear, if this were an exam, you would be more than welcome to assume constant specific heats. And just for best practice here, let's sanity check our number. We should have a number between 315.27 and 320.29. The fact that we do is a good sign. Same is true of 295 and 300. We should have a number between 295.17 and 300.19, which we do. Now that we have the power input, the delta h and the velocity of state 2, I can compute a mass flow rate. So that would be my power input to the compressor, which is 3 horsepower. In fact, I can write that in my diagram. 3 horsepower divided by the enthalpy difference, which would be 315.421 minus 295.321 plus velocity at state 2, which is 35 meters per second. And then I recognize that I need to be able to add kilojoules per kilogram to my specific kinetic energy term. Right now it's in meters squared per second squared, which means I can't add them together. I'm going to convert meters squared per second squared to kilojoules per kilogram by starting with kilojoules and working backwards. Kilojoules of 1,000 joules. A joule is a Newton meter. A Newton is a kilogram meter per second squared. So kilojoules are going to be left. Joules cancels joules. Meter squared cancels meter squared. Second squared cancels second squared. I'm left with kilojoules per kilogram. So when I take 350.421, 315.421 minus 295.321, then take that result and add it to one half times 35 squared divided by 2,000, excuse me, divided by 1,000. I get 20.7125. So now my calculation is 3 horsepower divided by 20.7125 kilojoules per kilogram. And my goal is going to be to get a mass flow rate in kilograms per second. Yep, kilograms per second, which means that I need horsepower to cancel kilojoules and leave me with time. For that I have to convert horsepower to something else. By looking at my conversion sheet, I see that a horsepower can be written as 0.7457 kilowatts. I will use that. One horsepower is equal to 0.7457 kilowatts. And I know a kilowatt is a kilojoule per second. So kilojoules cancels kilojoules, kilowatts cancels kilowatts, horsepower cancels horsepower, leaving me with kilograms per second. So I can take 3 times 0.7457, divide that by 20.7125 times 1, and I get 0.108-ish, 0.108 kilograms per second. Next up, I'm asked to figure out the rate of heat loss to the room. I have concluded that my compressor must be operating ideally for the purposes of this analysis for me to be able to calculate a mass flow rate. Therefore, there's no heat rejected by the compressor. Now I can consider the energy balance on the tank and see if I have enough information to calculate a heat transfer from the tank. If there is no heat transfer on the tank, then I would have 0 plus 0 is equal to 0 for Part B. I will begin a mass and energy balance on the next page and just so that I don't have to reference it, I will copy this information over. So on the air tank, we have a mass flow rate entering the tank at a rate of 0.108-007 kilograms per second. The mass balance on control volume 2 is going to start with delta M of CB2 is equal to the mass entering minus the mass exiting. Because I have a transient analysis, it may not be helpful for me to derivate everything with respect to time. Instead, I should look at how I can include the mass flow rate as a mass in-term. Well, I recognize that the mass flow rate in is going to be the mass entering over a duration. Therefore, I can write mass entering as a magnitude as a rate of entering mass times duration. I recognize that I have no opportunities for mass to leave the tank. Therefore, the rate of change of the mass entering multiplied by the duration will give me how much the mass changes between state alpha and state beta. So I can write the mass for my energy balance begin with delta E is equal to E in minus E out. Because it's a transient process, delta E on the left could be delta U plus delta K E plus delta P E. Then because it's open, energy can cross the boundary as heat, work, the energy associated with the moving mass, and energy could exit in the form of heat, work and the energy associated with an exiting mass. Note that these are all magnitudes of energy. So I'm not writing m dot out theta. I'm not writing q dot out. I'm not writing work dot out. I'm just writing the magnitudes. Similarly, the delta U, the delta K E and the delta P E on the left-hand side of the equation are also magnitudes. Now in this setup, I have already neglected the work terms and inside of the theta entering, I'm going to have an enthalpy and a kinetic energy, but I'm assuming no changes in potential energy. Furthermore, by neglecting any mass exiting my system, I can get rid of this term entirely, and by neglecting any changes in potential energy, I can neglect that term. I have enough information to be able to figure out delta U by recognizing that I can figure out the specific internal energy change between states 1 and 2, and then I can multiply that by how much mass there is, and then I have enough information to be able to calculate the enthalpy and kinetic energy of any entering mass. I can figure out the amount of mass that enters as a result of knowing the difference in mass between the end condition and the beginning condition. Furthermore, I know that I can relate delta U of the control volume itself to the mass and internal energy because I know temperature and pressure at state alpha and state beta, I can determine those. I can pretty reasonably neglect changes in kinetic energy of the tank itself because presumably it's not moving, and even if it is moving, it's not moving as a result of what's happening to the tank. That leaves me with heat transfer, mass and energy entering as a result of the entering mass, and change in internal energy of the tank. I can make my analysis a little bit easier by simplifying my heat transfer into a single term. It's unlikely that I have a complicated enough setup to warrant multiple opportunities for heat transfer, therefore I'm just going to use one. I'm going to pick a heat transfer, and if I get a negative number, that means that I picked the wrong direction. You can also think of this as establishing a net heat transfer, either in the inward direction or the outward direction, and a negative heat transfer out would be the same as a positive heat transfer in. I'm going to semi-arbitrarily leave Q out and neglect Q in. I'm assuming that the tank is going to naturally try to heat up as it goes from a temperature of 22 degrees Celsius to a temperature of 42 degrees Celsius, and as a result of that increase in temperature there's going to be heat transfer out from the higher temperature to the temperature of the surroundings. Now I can write delta U of control volume 2 is equal to the mass entering, which from our mass balance we can write as the change in mass of the control volume times the theta term for the entering mass, which is going to be H in plus specific Ke1, excuse me, specific Ke in. I'm neglecting the potential energy within the theta, so all I have to do now is subtract Q out. I recognize that my properties of the entering mass are state 2, so I can write this as M in times H2 plus specific kinetic energy 2, and then instead of change in internal energy on the left, I can write that as U beta minus U alpha. Now I know H2, I know Ke2. I can figure out the mass entering by figuring out the mass at alpha and the mass at beta. I can write mass of beta times specific internal energy at beta minus mass of alpha times specific internal energy at alpha. I can look up U alpha, I can look up U beta, I can calculate the mass of alpha, and I can calculate the mass of beta. Once I know those masses, the mass entering will equal the change in mass of control volume 2, which is going to be mass of beta minus mass of alpha. Therefore, I will have everything I need to calculate Q out. All I need to do between then and now is look up U alpha, look up U beta, and use the ideal gas law to calculate the mass in the tank at alpha and the mass in the tank at beta. Easy peasy. First up, I'll look up U alpha and U beta. That's the internal energy of air at 22 degrees Celsius and 42 degrees Celsius. For that, I need to go back into my property tables. 22 degrees Celsius is going to be 295.15 again, which means that I'm interpolating between 295 and 300, except this time I'm interpolating for the internal energy. So my interpolation would go 295.15 minus 290 divided by 300 minus 295. Excuse me, that should be a 295, not a 290. And then I'm saying that that's equal to what I'm looking for minus the internal energy at 295, which was 210.49 divided by the internal energy at 300, 214.07 minus the internal energy at 295, which is 210.49. Solving that for X. And I get 210.597. Let's sanity check that number. I double checking that it's between 210 and 214. It is, which is a good sign. Next, I can repeat the process for 42 degrees. 42 degrees is going to be 315.15, which means that I'm doing the same interpolation as earlier, except this time for internal energy. I repeat the same process with internal energy at state 2. And I get an internal energy at state beta of 224.957. Next, I need to calculate the mass at state alpha and the mass at state beta. You might be tempted in this moment to try to look up a specific volume, but remember that we can't look up a specific volume for air. If we look at our property tables, we don't have a column for a specific volume. We have this column, which is kind of similar, but is not specific volume. So instead, we have to use our ideal gas laws. For that, I'm going to need a little bit more space. So I'm going to calculate the mass at state alpha. We remember that the ideal gas law says pressure times volume is equal to mass times specific gas constant times temperature. Then mass would be pressure times volume divided by a specific gas constant times temperature. Furthermore, remember that specific gas constant is universal gas constant divided by molar mass. The universal gas constant comes from the inside of the front cover of the textbook. The molar mass comes from table A1. You can look both of those up just for good practice. I see from my conversions table that my universal gas constant could be represented as 8.314 kilojoules per kilomole kelvin. From table A1, I see that the molar mass of air is 28.97 kilograms per kilomole. When I take 8.314 and divide by 28.97, the kilomoles cancel and I'm left with kilojoules per kilogram kelvin. 8.314 divided by 28.97 gives me a specific gas constant of air of 0.286987 or about 0.287. I know the volume of the tank is 100 gallons. I know the temperature and pressure at both alpha and beta, which means calculating the mass is just a matter of doing the math. So I start off with the pressure at alpha. I can see the pressure at alpha was one atmosphere. Then I'm multiplying by the volume of the tank that was given in the problem as being 100 gallons. Not one gallons, 100 gallons. Then I'm dividing by the specific gas constant for air which we just calculated as this mass. Under most circumstances, I usually round it, but in this case, just for fun. You can just copy and paste it down. You could also, of course, just leave the calculation symbolically and write universal gas constant in the denominator and the molar mass in the numerator. And then our temperature at state alpha was 22 degrees Celsius. I have to convert that to kelvin and let's see what units cancel so far. Kelvin cancels kelvin and that's it. Cool. First of all, I can recognize that if I'm looking for a mass, I'm probably going to be looking for a mass in kilograms, which means that I'm probably going to leave this term in the numerator. So converting between atmospheres, gallons, and kilojoules is really what I'm doing. As a general rule of thumb, I would encourage you to break everything down into its primary units and at that point everything will cancel if they're in the same unit system. So I can convert from gallons into cubic meters. For that, I go back to my conversion stable. See a gallon is 3.7854 times 10 to the negative third cubic meters. So one gallon is 3.7854 times 10 to the negative third cubic meters. And then I'm going to be inverting atmospheres into base symmetric units. One atmosphere can be represented as 14.696 pounds of force per square inch or 1.01325 bar. So one atmosphere is 1.01325 bar and then a bar is 10 to the fifth Newton's per square meter and then I can represent a kilojoule as being 1,000 joules, but for that I will need more space. I will scooch just a little bit more to the left and I can write a kilojoule as being 1,000 joules and a joule as a Newton times a meter. Newton's cancels Newton's joules cancels joules kilojoules cancels kilojoules and square meters and meters cancels cubic meters. Atmospheres cancels atmospheres and I'm left with kilograms in the numerator. So wake up calculator. We have work to do. 1 times 100 times 3.7854 times 10 to the negative third. Well done me. Remember that the negative sign is different from the minus sign on my calculator. 1 to the fifth then I divide by 0.286987 times 1222 plus 273.15 times 1,000 and I get a mass of 0.4528 kilograms. Let's write that down here. 0.4528 kilograms. Now I get to repeat the process for state beta. The only thing that will be different is that I will have a different pressure and a different temperature. So I will start by just copying and pasting everything that I did above. Copying and then pasting everything that I did above and then switching out atmospheres for 600 kilopascals and then before I forget I will go get rid of the conversion from atmospheres to bars and I will write instead 1 kilopascal is 1000 pascals and that a pascale is a Newton meter excuse me, a Newton per square meter. I'll do my best to make it look like I never deleted a thing and rewrote it. I'll cancel pascals, kilopascals, cancels, kilopascals. Then my temperature is different. This time instead of 22 degrees Celsius it's 42. So all I have to do is swap that 2 with a 4 and I am made in the shade. Yeah, kilopascals, pascals, pascals and Newton per square meter, kilogels, cancels, kilogels, joules, cancels, joules, Newton's, kilograms, we got kilograms. So I will take 600 multiplied by 100 multiplied by 3.7854 times 10 to the negative third. Remember the correct? minus sign times a thousand and then I divide that by 0.286987 times 42 plus 273.15 times 1000. And I get 2.511 kilograms. So while I'm here, now that I have the change in mass and I know that that's equal to my mass entering from the mass balance on control volume 2 I might as well just compute that. Why don't I write 0.4528 again? That's strange. That was of course 2.5112 and then I'm subtracting those two numbers and I get a mass entering, a magnitude of entering mass as 2.05839 kilograms hashtag unnecessary decimal places. And with that, I now have everything I need for the left hand side of my energy balance. I have the mass entering. I know H2 and specific kinetic energy to from my properties of state two from the previous page. Meaning the only thing left to do is calculate a Q out. So let's see, how do I make this as least confusing as possible? Just draw a big arrow. I'm sure that will help. So Q out is going to be when I solve that equation to the left. So it'd be M in times H2 plus kinetic energy 2. So M in times H2 plus V2 squared over 2 then I'm subtracting U beta minus U alpha which is mass of beta times specific internal energy beta minus mass alpha times specific internal energy alpha. So I'm taking minus mass of beta times specific internal energy beta and then I'm subtracting within the subtraction. So I'm writing that as a plus mass of alpha times why am I writing that as a viscosity? That's strange. Specific internal energy beta and specific internal energy alpha then plugging in what I know I have the mass entering as 2.05839 2, actually why don't I draw a big horizontal line before I get 2 carried away and that ladies and gentlemen is where I draw the line. 2.05839 kilograms multiplied by H2. For that I'm going to swap back to my previous page. H2 was 315.421 314.421 kilojoules per kilogram plus I should really write that both in the numerator since I'm adding them together plus one half times the velocity of state 2 which was 35 which was 35 meters per second I'm going to write that as 35 squared meter squared per second squared then I want to be able to add that to kilojoules per kilogram which means that I have to get this into kilojoules per kilogram which means I'm going to start with kilojoules and write that as a thousand joules and a joules a Newton times a meter and a Newton is a kilogram meter per second squared joules cancels joules cancels Newtons meters and meters cancels square meters second squared cancels second squared leaving me with kilojoules per kilogram and then I'm going to be subtracting the mass of beta mass of beta was 2.5112 probably just grab that from the calculator wake up calculator so let me get rid of my rest of my words on the line and write plus 2.0 nope 2.51121 kilograms then I'm multiplying that by specific internal energy of beta which was 224.957 224 24.957 957, right? yeah, 957 and then I'm going to be switching my positive to a negative because that was subtracted and then I'm adding my second term which was the mass of alpha times the specific internal energy at alpha mass of alpha was 0.4528 and the specific internal energy at alpha was 210.597 210.597 kilojoules per kilogram kilograms cancels kilograms leaving me with kilojoules kilograms cancels kilograms leaving me with kilojoules and on the first time kilograms and kilograms cancel kilograms leaving me with kilojoules so my Q out term is going to be 2.05839 times 315.421 plus 0.5 times 35 squared times 1 over 1000 minus 2.51121 times 224.957 plus 0.4528 times 210.597 so let me double check that I got all the parentheses in the right spot minus n plus that looks right to me I get 180.964 note that that's a positive quantity which means that our direction choice was correct so the amount of heat rejected is going to be 180.964 that is the amount of heat rejected from the tank again because we assume that the amount of heat rejected by the compressor is zero that means the total heat rejected to the room as a result of this sequence of processes is just 180.96 so next up the total mass of air added to the tank if the compressor stops operating on the tank which is 600 kPa in 42°C we figured that out already it is 2.05 I can write that as my part C I'll draw another arrow who doesn't love arrows running all over their example problems I think that makes things more clear and in the interest of making things more clear I'll make that look more like a parentheses less like another C so the amount of mass entering is going to be 2.05 2.05 let's just call it 2.06 kilograms then on to page number 3 next up the problem wants to know the time it would take for the tank to reach 600 kPa in 42°C well we know from our mass balance that 2.05839 kg of mass had to be added and we know that mass is being added to the tank at a rate of 0.108 kg per second I can combine my mass flow rate and my magnitude of mass added to figure out the duration because m dot is just mass, magnitude of mass over duration that means duration is going to be magnitude of mass added divided by the mass flow rate so I'm going to take 2.05 839 kg divided by the rate at which mass was added which was again 0.108 kg per second 0.08 kg per second kilograms cancels kilograms meaning if I take 2.05839 divided by 0.108 I will end up with an answer in seconds 2.05839 divided by I wonder if I have that mass flow rate up here somewhere hey look at that I do so I get 19.06 seconds 19.06 seconds so the compressor kicks on runs for just under 20 seconds and then kicks back off I think that I can fit that on the bottom of this page it would be easier for people who want to print this cut and paste there we go and that's the air compressor problem oh wait it's not part b asks for the rate of heat rejection not the magnitude of heat rejection so for that I have to come back after I know a duration so the magnitude of heat rejected divided by the duration will give me the rate of heat rejected excuse me the magnitude of heat rejected divided by the duration will give me the rate of heat rejected on average because remember as the temperature increases the rate of heat rejection increases so we're calculating an average so I'm taking 180.964 964 kilojoules and I'm dividing by 19.06 seconds a kilojoule per second is a kilowatt therefore I will get an answer in kilowatts see if I can make this more clear so rate of heat rejected is 180.964 divided by 19.06 so 180.964 divided by 19.06 and I get 9.495 let's call that 9.5 kilowatts and that's the air compressor problem