A: Well, let's draw the saturation curve to sort of enrich this discussion. So if you were going to assume as you did that it was superheated, my question to you is, what would you assume? If it were to be superheated, for you to define a state, you need two pieces of information, right?

A: So, I copy and paste it from the reference handbook, what the equation that we're talking about is the efficiency of a compressor. And then I copy and paste it a little snip from the written solution as well so we can kind of look at those side by side. And I guess the challenge of this particular problem was trying to align the subscripts that the reference handbook uses with the, I think it was the Brayton cycle.

A: So I don't know that you'll find these formulas directly in the reference handbook. I think you'll find m. delta H, but probably not under a steam heat exchanger. I only was able to find explicitly by looking for turbines, which is a little different than what this problem is about, but I think it shouldn't be too much of a reach to use these on the basis that an energy flow rate or a rate of energy transfer heat transfer is a mass flow rate times a change in enthalpy.

A: So on one side we had a water that we're looking to heat cold water coming in we'll call that state one and then heated water coming out we'll call that state two and then on the other side we had the steam so saturated steam coming in call that state three and then we were told that we have saturated water coming out we'll call that state four and we were told that the saturated steam entering was at a pressure of 5 psi g and the question is how do you know that what's happening here on the steam side is a constant pressure process? So what's going on here we're obviously transferring heat from the steam to the water so the direction of heat is this Q is crossing the boundary in the heat exchanger and the water side if you think about the right side of this is undergoing sensible heating of liquid water. Liquid water comes in it's heated it's temperature changes it's temperature increases there's some delta t there there's no phase change it enters this liquid water it leaves this liquid water and that's the whole story it's at a higher temperature so you can think about mcp delta t or 500 gpm delta t however you want to analyze that that's not really what your question was about your question was about the steam side the steam side undergoes latent cooling latent cooling what is that that's phase change it comes in as steam and it leaves as liquid water and actually both of these processes are constant pressure because there's nothing that would change the pressure pressure there's no compression there's no turbine there's no other you know mechanical action in the system and interestingly the the the water being heated has a delta t but the steam is also constant temperature latent cooling occurs at constant temperature so you really have to think about that if if you haven't thought about that before the idea that heat can be removed through phase change at constant temperature and constant pressure and why I always come back to that kind of helps me just get my head around that is the vapor dome so on the vapor dome we typically would have this BTS temperature and entropy and if we draw an isobar a line of constant pressure and isobar line of constant pressure would look like this so let's suppose we had out in this well not suppose we do have out in this region super heated vapor and then as we cool we're doing sensible cooling of the super heat of vapor until it gets to this point and we call that saturated steam or saturated vapor that corresponds to what we're saying we have at state three entering the heat exchanger on the steam side then we undergo constant temperature constant pressure cooling latent cooling from three to four this is saturated liquid or saturated water at this point that's phase change and then if we continue to cool we're still at constant pressure but we're no longer at constant temperature now we're doing the sensible cooling of liquid water as we move into the subcooled region so what's actually changing as we go from super heated vapor to three we have sensible cooling of steam but no latent then as we go from three to four we have latent cooling of steam but no sensible because the temperature is not changing and then as we go past four continuing to cool we have the sensible cooling of water but no latent it's only changing temperature it's not changing phase anymore and this this line that I drew is constant p it's an isobar so just to put a really fine point on this as we go from three to four we have constant temperature constant pressure latent cooling what's changing well the quality is changing right the quality goes from one saturated vapor at state three to zero saturated liquid at state four what else changes while the entropy is reduced we know that because we're moving from right to left the horizontal axis is entropy and it's not shown here we'd have to draw this on hf on what we draw it on something else but the enthalpy is also decreasing and I think that might be the missing piece in the understanding because when you first think about this you're like hey wait a minute temperature is not changing how is this giving up heat well the enthalpy is changing the enthalpy is changing during phase change in fact it's changing dramatically we get more energy transfer as we change phase condensing steam until liquid water then we do cooling steam sensibly we don't get very much out of that but we get a lot going from three to four which is why sometimes in similar problems if you have a super heated steam condition you can almost ignore to some degree or at least maybe not ignore but round down the contribution of the sensible cooling saturated steam to point three and really just focus in on the phase change because that's where the magic happens follow up questions on that one and then this is trial just soaking that in so no questions here thank you okay