How I Approach Understanding Thermodynamics

[@mohamedjessar5770]

This is the best explanation ever, Hats off

[@davidcora2751]

Thanks for enlightenment!! A good video to enlighten potential engineers. Thanks for the knowledge 💕❤️

[@dmcafx]

I have been engineer for many years and I did not know this information. Thanks a lot!!! understood many things which I didn’t know.

[@ChemEngWeekly]

Thanks for sharing Pat! I can definitely say I understand more about thermodynamics now!

[@KhalidQureshi-z5q]

Wonderful! You have changed the way I look at thermodynamics.

[@carultch]

I never thought I'd see the value in the P-h diagrams, until I saw your videos. I'm so used to seeing cycles on T-s and P-V diagrams, that when I opened the P-h diagrams in my software that they looked so foreign to me. Seeing the heat exchanger processes take place on simple straight lines, makes it so much easier to draw than the isobars on T-s diagrams. The one disadvantage is that you no longer have area under the plot, corresponding to heat or work, like you do on T-s and P-V diagrams.

[@GuessWhoBC]

One quick clmment: measuring air temperature at the outlet of a vacuum cleaner, do we take into consideration the heat generated by the electrical motor? Unless, of course, that particular vacuum cleaner has separate casing for its motor and correspondingly separate cooling outlet.

[@DaveMody]

Pat, your vacuum cleaner is probably using the air flowing through it to cool the motor, so while you're right, that's not very good vacuum cleaner to illustrate the problem. I do like how you take items in your everyday environment and explain principles with them. Keep that up :)

[@ProcesswithPat]

You are totally right. Thanks for pointing it out! I certainly need to keep a running list of mistakes I’ve made that should be corrected or clarified somehow.

[@piotrektabo4750]

Absolutely amazing explanation skills!

[@mikefochtman7164]

Nice explanation and discussion. I'd just one little 'nitpick'. Early at 1:30 you show C-O-C for CO2. While CO2 is a linear molecule (unlike water), it should be O-C-O :) After all, it has two oxygens and one carbon.

[@xse345]

Nice explanation there. Just one clarification, at 22:01 you says at high enough pressure, It takes no energy to boil water, or i.e. its easier to vaporize with increasing pressure ? That doesn't seem right. Correct me if I am wrong.

[@ProcesswithPat]

Correct! Check it out in a steam table and prove it to yourself... Hear are a few heats of vapourisation at different pressures: 1 bar (atmospheric): 2258 kJ/kg 100 bar: 1317 kJ/kg 200 bar: 584 kJ/kg 221.2 bar: 0 kJ/kg (these are according to my steam table, I see a slight difference if I use some online versions, but they are small). It seems unintuitive because you think higher pressure steam, higher temperature more energy. Higher pressure steam simply contains energy that is available at a higher temperature. Even though 1 bar steam may have a higher latent heat, it is only available at 100 °C. Processes that utilise steam for heating need to be heated more than this.

[@xse345]

@@ProcesswithPat oh... it seemed unintuitive to me because, lower pressure --> lower boiling point --> less temp required --> less energy needed. Still confuses me.

[@axeld53]

@@xse345 A lower boiling point only tells you that the heat source you need to make that liquid boil doesn't have to be too high. However, it doesn't tell how how much energy you must give that liquid in order to make it boil. In the Pressure-Enthalpy diagram (P-h), you can see that the boiling point of the water is the line that becomes a horizontal in the two-phase region, meaning it is constant at constant pressure. You can see that " lower pressure --> lower boiling point --> less temp required" is indeed correct. But the amount of energy needed to make the liquid boil off if the difference of enthalpy between the vapor (right side of the two-phase region) and the liquid (left side of the two-phase region), which corresponds to the width of the two phase region. As you can see, as pressure goes up, the width of the two-phase region decreases, therefore the amount of energy required to completely boil the liquid decreases. As you approach the critical point, the width approaches zero, and when you reach the critical point, there is no difference between the enthalpy of the liquid and that of the vapor, hence the amount of energy needed to boil water at this point is zero. Further from that point there is no distinction between the liquid and vapor phases and the whole system becomes what is called a supercritical fluid. There are a lot of videos here on KZbin showing this transition happening.

[@AbhijithMR-j1n]

Bro it's not my question How to enjoy thermodynamics Like before starting thermodynamics i should know trigonometry/integration or differentiation etc Just like that how to approach this

[@fredkuijper6986]

Excellent video, much more educational than any other youtube videos I've seen so far. One note: would you mind removing that annoying background "music"? :)

[@rasmussverdrup7479]

I think that the institute of standards and technology has decided that if you are calling one end of an object the "business end" the other must be "the fun end."

[@ProcesswithPat]

By backup plan was the sucky and blowy ends. I definitely approve of the fun end!

[@Bigalinjapan]

2 years of studies in 28 minutes. One more proof some professors just suck...