 A thin, double-pipe, counter-flow heat exchanger is used to cool oil from 150 degrees Celsius to 40 degrees Celsius at a rate of 2 kilograms per second. This cooling is done by water, which enters at 22 degrees Celsius at a rate of 1.5 kilograms per second. Determine the rate of heat transfer within the heat exchanger and the exit temperature of the water. So we recognize that we have a heat exchanger. We are trying to pull heat out of the oil, and that heat has to go somewhere, it is going into the water. We recognize from our list of study fluid devices that a heat exchanger is a device which transfers heat from one fluid to another. Heat there is kind of in quotation marks what we are really doing is pulling energy out of one fluid and pushing it into another. This heat exchanger is pulling heat out of oil, pushing it into water, but that energy is the same. If we figure out how much the energy of the oil has to decrease in order to yield an exit temperature of 40 degrees Celsius, we can apply that same amount of energy to the inlet of the water, and that would allow us to figure out how much the temperature of the water will increase as a result of absorbing that much energy. That logic of applying one energy to the other is going to require that we assume the heat exchanger itself is adiabatic, which is kind of a strange assumption when you think about it. We are assuming there is no heat transfer in the device that is trying to transfer heat. What we are referring to here is that there is no heat transfer between the hot oil on the outside of this heat exchanger and the air around it. In reality there is likely to be a little bit of heat transfer, but compared to the amount of heat transfer going into the water, that amount is likely to be very small. The water is being driven over the heat exchanger, which means that there is more of it moving, which means a higher convection coefficient, so it is likely to pull more heat out. There is a thin walled heat exchanger, which means that there is likely to be much less resistance between the oil and the water. Also the water itself has a higher heat capacity than the air around the oil. There are just many opportunities for the water to pull more heat out of the oil than the air to pull heat out of the oil. As a result of that, we are assuming that there is no heat transfer between the oil and the air. Another way to think about this is what we are figuring out is essentially the worst case scenario. We are saying if all of the heat which was pulled out of the oil went into the water, how much temperature increase results from that. That essentially gives us a worst case scenario for the hottest water that could occur if that amount of heat was pulled out of the oil. The presence of any other losses to the air or to other opportunities for energy to be driven out of the system would decrease the temperature of the water. Or you could think of it like if we ensure that the water exits at that temperature, that means that we are going to be pulling that much heat out of the oil. If we are trying to cool the oil, then any additional heat loss is just going to lower the temperature of the oil at the outlet. You could think of it like if we figure out how much heat transfer we have to pull out in order to yield a temperature drop of the oil of 110 degrees Celsius, and we pull that amount of heat out, then we are guaranteeing that the outlet of the oil is going to be the highest it could be as 40 degrees Celsius, and it's likely temperatures to be lower than that. Regardless though, we are going to be treating this as two separate changes in enthalpy. We have the oil on the outside of the tube, which is losing energy, to water on the inside of the tube. To make things a little bit more convenient here, I'm going to refer to the inlet and outlet of the oil and water as state points. For consistency here, I will call the oil letters. So the inlet of the oil will be A, the outlet of the oil will be B, and I will call the water numbers. So the inlet of the water will be one, the outlet of the water will be two. So if we can figure out how much the energy will decrease from A to B, that will give us our first answer, and if we figure out how much that energy would affect the temperature between one and two, that will give us our second answer, the temperature at state two. To set up our analysis, we're going to be involving a change in enthalpy, which means that we're going to have to determine a change in enthalpy from A to B and one to two. To do that, we have a couple of different options. The best option would be to look up the enthalpies in property tables. To do that, we would need to know, first of all, property tables for oil, which we don't have. We do have property tables for water, but we don't have two independent intensive properties. We only know the temperature of the water. It's 22 degrees Celsius at the inlet, and presumably something higher than that at the outlet. In order to perform a property lookup, we would have to determine something else about the water, likely as a result of an assumption for the case of this problem. We would have to assume, for example, that it's a saturated liquid and figure out what its enthalpy is there, or we would have to assume a pressure and then use the pressure and temperature to determine an enthalpy, but even that is going to be looking up a property that lies between the saturated liquid properties and the compressed liquid properties, which is going to mean interpolating very broadly between two sets of property tables, which is going to incur its own error, and that error is incurred after a pretty big assumption, too. So it's going to be most convenient for us to just assume that the specific heat capacity of the water is constant. That is a reasonable assumption to make, first of all, because it's one of the better options we have for the oil anyway. I mean, if we don't have property tables for the oil, the only way that we can figure out a change in enthalpy is by assuming that the heat capacity of the oil is constant, and looking that up, I guess a better thing to do would be to determine how the specific heat capacity of the oil changes as a function of temperature and perform that integral, but we don't have that value for the oil. All we have is a single specific heat capacity, so that's the best we can do. And since we are already incurring that error with the oil, we might as well apply it to the water as well. There's no point in us applying additional precision to the water if we've already incurred that error. You could also think about the fact that the water is unlikely to be any other phase than liquid at the inlet and the outlet. If it doesn't incur enough heat to increase its temperature beyond the boiling point at whatever pressure it's at, then it's not going to have changed phase, which means that across the liquid region, we are dealing with a relatively small change in temperature, which means that the error induced by assuming constant specific heats is also pretty small. That's a long way of saying it's pretty reasonable for us to assume constant specific heats for both the oil and the water. If we write out some assumptions about the problem that we're setting up, let's start with that. Specific heat capacities of the oil and water are constant. Furthermore, I will assume that they're constant at 300 Kelvin. Making that assumption is going to allow us to determine which of the specific heat capacities to use in the book, and that additional parameter here at the end of that assumption is based on the fact that the specific heat capacity I have for oil is only available at 300 Kelvin. The better thing to do here would probably be to figure out what the temperature is at the outlet of the water, if we use a value of specific heat capacity at 300 Kelvin, and then use that new T2 to figure out an average temperature of the water, and use that average temperature to look up a new specific heat capacity, and repeat that guess and check process until we converge on a better answer. But there's no point in us being extra precise about a problem that we have already induced error in our assumptions anyway. This model is of the problem, and as a result of it being a model, it is going to be a certain degree of inaccurate. Next, I will assume steady state analysis. I will assume that there are no opportunities for work, and that the heat exchanged outside of the heat exchanger is zero. That is, the heat exchanged from the outside shell from the oil to any surrounding air is zero. I can add to my list of assumptions here that any changes in kinetic and potential energy are likely to be very small relative to the magnitude of the energy change of the enthalpy of the oil and water themselves, and this should be good enough to get us set up with a mass and energy balance. Note that when I set up my energy balance, I have a couple of different ways to approach this. I could establish a control volume around the entire kit and caboodle. Doing that would allow me to neglect heat transfer altogether, and my energy balance would simplify down to the sum in of m.h is equal to the sum out of m.h. I would have two inlets, one and a, and two outlets, a, b and two. That is useful if what I'm looking for is the temperature at the outlet of the water, but it doesn't allow me to figure out the amount of heat exchanged, because that heat exchanged, the heat going from the oil to the water, I could call that qHx, heat exchanged by the heat exchanger, qHx here does not appear in my energy balance, because it's not crossing the boundary. So to figure out qHx, I would have to establish an energy balance where in qHx cross the boundary. To do that, I could either set up an energy balance on the oil, at which point qHx would be crossing the boundary in the outward direction, or an energy balance on the water. If I did that, then qHx would be crossing the boundary on the inward direction. So I essentially have three options for defining my control volume. Each of them have different advantages and disadvantages. Since the problem is asking me for both the temperature of the water at the outlet, and the amount of heat exchanged within the heat exchanger, I'm going to have to use two of them. So since I already have to use two of them, I mean I think it only makes sense for us to talk through all three. Good character building if nothing else. So I'm going to use this set of assumptions to set up a mass balance and an energy balance for all three variations. One where my energy balance is around the whole thing, my control volume is around the whole thing. One where the control volume is around just the oil, and one where the control volume is around just the water. So I will add some big vertical lines to split my page into three sections, and then I will make my best attempt at art. So let's go with method one, kit and caboodle. K-A-B-O-O-D-L-E question mark. Alexa, how do you spell caboodle? Caboodle is spelled C-A-B-O-O-D-L-E. C. And I was wondering, did you have to try more than once to get my attention? You know I didn't. Thanks for your feedback. We learned a thing today. Caboodle is spelled with a C. Anyway, in the kit and caboodle variation, I'm drawing a control volume around the whole thing. I have a water pipe, I have an oil pipe, an inlet and outlet for the water. States one and two respectively. And then an inlet and outlet for the oil. States A and B respectively. Beautiful drawing. And my control volume is around all three. I guess all four. Ingresses and egresses. So the heat exchange within the heat exchanger, QHX, doesn't cross my boundary so it doesn't appear in my energy balance. For the mass balance, I'm going to start with the change in mass of my control volume. Is equal to mass in minus mass out. I have no opportunities for mass to change within the control volume. So when I divide everything by DT, DMDT disappears because of our assumption of steady state. Then m dot in divided by DT becomes m dot in. And m out divided by DT becomes m dot out. M dot in is the sum of all the mass flow rates entering my system, which is going to be m dot one plus m dot a. And m dot out is going to be the sum of all the mass flow rates exiting my system, which is going to be m dot two and m dot b. So in this relationship, I'm just saying that the mass flow rate of the water at the inlet plus the mass flow rate of the oil at the inlet, it has to equal the mass flow rate of the water at the outlet plus the mass flow rate of the oil at the outlet. This isn't particularly helpful, but you know for character building we can always draw it out. Slightly more useful is the energy balance. On the kit and caboodle, the energy balance begins the same way. I have the energy change of the control volume is equal to the energy entering the control volume minus the energy exiting the control volume. Because I'm dealing with steady state analysis, it's probably more useful to use the rate form of the energy balance, which means I'm going to divide everything by DT. If I divide everything by DT, I end up with zero on the left and e dot in minus e dot out on the right. Therefore, e dot in is going to equal e dot out. e dot in could be power input, could be heat transfer in, could be the sum in of m dot theta. e dot out could be power output plus the rate of heat transfer out plus the sum out of m dot theta. I'm neglecting work and because I've drawn my control volume around the entire kit and caboodle, I have no heat transfer as well. Furthermore, within my theta terms, I have enthalpy plus kinetic energy plus potential energy. By neglecting any changes in kinetic or potential energy, whatever the kinetic energy is on the left hand side of the equation, it's going to be the same as it is on the right hand side of the equation, so it disappears. So I'm left with the sum in of m dot h is equal to the sum out of m dot h. That would be m dot one h one plus m dot a h a is equal to m dot two h two plus m dot b h b. So the combination of the mass flow rate and the enthalpy at the two inlets has to equal the combination of the mass flow rate and the enthalpy at the outlets. For convenience here, I can rearrange this and group the oil and the water together to do that, I probably want to write a positive change. Therefore, I recognize that the water is increasing in energy, therefore two minus one would yield a positive quantity. So I'll bring the waters to the right hand side of the equation and the oil is going to be decreasing in energy, so I'm going to want to write a positive quantity as a minus b, so I'll bring the b over to the left hand side of the equation. Therefore, I have m dot oil, a and b are both the mass flow rate of the oil times h a minus h b is equal to m dot water times h two minus h one. I've assumed constant specific heats, which means that I can make a substitution here. For delta h, I'm going to use a heat capacity times a temperature difference. For liquids, it doesn't matter. It's not cp, it's not cv, it's just c. So I'm going to write that as mass of oil times heat capacity of oil times ta minus tb is equal to the mass flow rate of water times the heat capacity of the water times t two minus t one. In this setup, I have the ability to look up the heat capacity of the oil and the heat capacity of the water. I know t one, I know ta, I know tb, I know both mass flow rates. I know everything I need to calculate t two. So t two would be m dot oil times heat capacity of the oil divided by m dot water times heat capacity of the water times ta minus tb plus t one. I will leave that aside for the moment and let's consider how this mass and energy balance differs when we start looking at one fluid at a time. So let's do oil next. In analyzing the oil, I have one inlet and one outlet and it is exchanging heat in the outward direction. Fuzzy lines, then I'm going to zoom in, brace yourselves. I have heat transfer crossing my boundaries in the outward direction and I've decided to call that qhx. That's the heat exchanged within the heat exchanger. So the control volume is a hollow cylinder, a tube, and that control volume has one inlet, one outlet, and a single heat transfer out. My mass balance and energy balance are going to start the same. So I am going to copy and paste those values to make that a little bit faster. And I recognize that I only have one inlet, it is a one outlet, it is b. So m dot a is equal to m dot b. That is the conclusion of my mass balance on this control line. I can add that I'm just going to call this m dot oil, this quantity and recognize that it was the value given. I believe it was two kilograms per second. I could scroll up and look at it but I don't want to nauseate the viewer any more than I have to. In the energy balance, I recognize that there are still no opportunities for work or changes in kinetic or potential energy but this time I have to accommodate a heat transfer out. So let's definitely save some effort here by just deleting that line. And that Q out is actually Q H X. So when I write my sum in of m dot theta, what I'm writing is m dot a H A because again kinetic and potential energy are both neglected is equal to Q H X as a dot plus m dot b times H B. I want to make this the substitution allowed by the assumption of constant specific heats which means that I have to write this as a delta H so I will solve for Q dot H X and recognize that that's equal to m dot oil because it could be B or a times H A minus H B. So the amount of heat exchange could be calculated by figuring out how much the energy of the oil decreases. If we're assuming that that energy decreases only able to go into the water then that decrease in energy is the energy exchanged within the heat exchanger. Interesting the mass flow rate times the heat capacity of the oil times the temperature change of the oil is the heat exchanged. I wonder if that'll be a pattern. Just for consistency sake here I will write m dot oil times heat capacity of oil times T A minus T B definitely a B. So two mass and energy balances down one more to go. This time just on the water. Let's get to it shall we. Water are we waiting for? On the water I have a tube with one inlet stay one one outlet stay two and a heat transfer input Q H X. My control volume has two opportunities for energy to cross the boundary in the outward direction m dot one H one and Q H X. And one opportunity for energy to cross the boundary in the outward direction m dot two H two. And like before I will accelerate this process a little bit by copying over the start of my mass and energy balance beautiful. And then I recognize that m dot in is just m dot one and m dot out is just m dot two. So m dot one is equal to m dot two and I can refer to that mass flow rate which is the same as m dot water. In the energy balance the Q H X term is still relevant but this time it appears as an inlet. So I'll move Q dot H X over to the other side. I will waste a ton of time trying to avoid rising it and starting over neglect Q dot out. So Q dot in plus m dot one H one is equal to m dot two H two Q dot H X plus m dot one H one is equal to m dot two H two. Since I want to make the assumption of constant specific heats and plug in that substitution I want to write this as a delta H so I will rearrange and write this as Q dot H X is equal to m dot two H two minus m dot one H one m dot water m dot H two O times H two minus H one which would be m dot H two O times C water times T two minus T one. Interestingly that is the right hand side of this equation. So both of these sides of this equation give me Q dot H X. I could also have gotten to that by recognizing that the heat leaving the oil is going into the water so this quantity has to be the same as this quantity therefore I could write this is equal to this and that would have yielded the same result as the mass and energy balance on the whole kit and caboodle. Whichever way I decide to approach this though I'm going to have to look up the heat capacity of the oil and the water in order to be able to finish the problem and for that we will jump into our property tables. The most relevant property table here is likely going to be table A 19 so if I jump down to table A 19 I see that I have specific heat capacity values for a variety of liquids including oil the best value I have is unused engine oil so we're going to be using those properties as the oil in this problem even though the specific oil type isn't specified and then because we have 300 kelvin as the property for the oil we're going to use 300 kelvin for the water as well. So our two specific heat capacities are going to be 1.909 and 4.179 respectively and those quantities are in kilojoules per kilogram kelvin so in my work here I will add that maybe to the upper left hand corner we'll say from table A 19 the heat capacity of oil at 300 kelvin is 1.909 kilojoules per kilogram kelvin and the heat capacity of water at 300 kelvin is going to be 4.179 kilojoules per kilogram kelvin with those two properties looked up all that's left to do is actually tabulate a number for q dot hx I could approach that as the result of the energy balance on the oil or the result of the energy balance on the water however note that I don't actually have enough information to finish the energy balance in the water because I don't know t2 yet therefore my only real option is using the result of the energy balance on the oil so for that I need to plug in the mass flow rate of the oil which was two kilograms per second question mark slowly scrolling two kilograms per second multiplied by 1.909 kilojoules per kilogram kelvin multiplied by the temperature difference of the oil which would be 150 minus 40 degrees celsius kilograms will cancel kilograms degrees celsius will cancel kelvin note this kelvin and this degree celsius cancel because a temperature change in kelvin is the same as a temperature change in degrees celsius we could take 150 plus 273.15 minus 40 plus 273.15 but the result of that subtraction is still going to be 110 that leaves me with kilojoules per second which is a kilowatt so two times 1.909 times 110 will be my answer to the first part of the problem and for that I will pull up my calculator two times no two times 1.909 times the quantity 150 minus 40 we get 419.98 419.98 kilowatts and that is q.hx halfway through the problem we have one of the answers whoo for the exit temperature of the water I could take that number and plug that into this equation over here and solve for t2 or I could use this equation they should produce equal answers at the end the advantage of using the one on the left is the fact that I'm not incurring any error from a miscalculation over here the downside of plugging this into this equation is that if I made an error here that error is going to be compounded so since I have the opportunity to do either I'm going to use this equation on the left so t2 is equal to the mass flow rate of the oil which is going to be two kilograms per second divided by the mass flow rate of the water which is one and a half kilograms per second kilograms cancels kilograms seconds cancel seconds so I have a unitless proportion multiplied by the specific heat capacity of the oil which was 1.909 kilojoules per kilogram Kelvin divided by the specific heat capacity of the water which is 4.179 kilojoules per kilogram Kelvin the kilojoules cancel the kilojoules the kilograms cancel the kilograms I guess actually this kilogram cancels this kilogram and this kilogram cancels this kilogram because they're technically different kilograms but regardless all four kilograms disappear then Kelvin cancels Kelvin leaving me with a unitless proportion so I'm taking a unitless proportion I'm multiplying it by a change in temperature and I'm adding that change in temperature to a temperature which will yield a temperature and I'm going to need a little bit more room so I'm going to slide this a little bit to the left cha cha real slow multiplied by ta minus tv ta was 150 minus tv was 40 and I'm adding to that 22 degrees Celsius so I'm figuring out a change in temperature in Celsius I'm adding it to a temperature in Celsius which means that my result will be in Celsius because I'm dealing only with temperature differences here I don't really have to worry about the distinction between Celsius and Kelvin best practice is generally to use Kelvin anytime that you're using a math in your calculation I mean an absolute temperature will not lead you wrong in these sorts of things but because it's all delta t's I can be a little bit lazy and leave it in Celsius and recognize that the result is still going to be in Celsius so I will have a result in degrees Celsius so I'm going to take 2 times 1.909 divided by 1.5 times 4.179 I'm taking that result and I'm multiplying it by the quantity 150 minus 40 and I'm adding that to 22 and what I get is 89 degrees Celsius so since the pressure of the water is likely to be relatively close to atmospheric pressure or higher the temperature at state two is not high enough to draw concern if that temperature had been something like say 120 degrees Celsius there is the possibility that we would have had a phase change occurring which would mess up our calculation remember of course that you cannot assume constant specific heats across a phase change so we would have had to back up our analysis with some lookups of HF at the beginning of the process and figure out what the enthalpy would be at the end and use an approximated pressure to try to figure out what the phase might be but if we had a complicated enough setup to warrant that we would probably need more information about the water so since we don't have to worry about it we might as well not worry about it the amount of heat exchange was 420 kilowatts the temperature of the outlet was 90 degrees Celsius or so