Thermodynamics - Entropy 7.2 Increase in Entropy Principle

[@jacalynmorgan7008]

I think I'll use these videos (and your notes) to prepare for my PhD qualifying exam

[@DenisPinsonnault]

QL could be found using QH/TH = QL/TL ---> QL = QH X TL/TH because it is a reversible heat pump.

[@fozansharfulhaque1973]

u mean QL/QH=TL/TH

[@quratulainwarraich344]

Will you please make video lectures for chpater 8 and 9? Thank you

[@Michael-jv2gf]

Amazing playlist!

[@duakhan5882]

We were wondering if you could also please upload lectures for chapters 8 and 9?

[@engineeringdeciphered]

I’m sorry, I’ve never taught 8 and 9 - that would take some work, I don’t think I could get that done this semester. Maybe next.

[@duakhan5882]

@@engineeringdeciphered Oh that's alright! Your videos have helped alot, that is.

[@averyblasdale178]

@@engineeringdeciphered Hi! Wondering if you were able to do Ch 8+9? Thank you in advance!

[@sunmichoi6888]

Thank you so much >_

[@zaynmoaaz8504]

[@sourish_mitra]

Respected Sir We have to consider here the consideration of an isolated system: some establish the principle of increase of entropy from consideration of heat added and rejected by two heat reservoirs at different temperatures: how Q=0 idea and T1/T2 remains constant idea lead to same destination? Please help?

[@WritersDigest-b8f]

In your imagination, If u put your home on top of the condensing unit and the evaporator is left sitting in the background, now u have a heat pump heating your home. And we need heat pump in winters. The backyard becomes source reservoir and the home becomes the sink reservoir. When I was replacing my air conditioning units, I looked for an AC unit with heat pump but did not go for it because it’s more expensive, I settled for cooling only because the furnace takes care of heating in winters. The backyard becomes source reservoir and the home becomes the sink reservoir.

[@WritersDigest-b8f]

Entropy of the universe is increasing s gen? or increase in total entropy from heat transfer plus s gen? Or s gen is just another mathematical expression for total sum of entropy changes from all heat transfers in a closed system from source(s) to sink(s)?

[@yigitcan824]

Professor sorry to bother you again but I have a confusion here about 0:40 'Carnot devices' are internally reversible but there's only two isothermal stages, which are _isothermal expansion_ and _isothermal compression_ . So I thought carnot cycle is not totally isothermal.Am I right? Or I'm confusing it?

[@kylecatman7738]

Did you say isothermal is reversible, 17:32 mark. Is that Internally reversible??

[@aidanvine1633]

Im confused about question 7.2 question 2 as it uses q/t to solve the entropy change of the source and sink, are they not irreversible, why are we using an equation that is meant for reversible isothermal processes, not irreversible isothermal processes?

[@mina1718]

the process itself is reversible and since we're working on a reversible heat pump the equations come within as for a reversible process also it's pointed out at the beginning of the question.

[@Caroline-ot8jn]

I have a question about 7.3, in the problem statement it says : "Heat pump produces heat at a rate of 300 KW" shouldn't this be QL and not QH, since we take QH from a low temperature reservoir in heat pumps?

[@engineeringdeciphered]

Heat pumps take QL from a low temp reservoir, and pumps out QH into the higher temp area. The “H” in QH and TH is always associated with the higher temp area.

[@bountyhunter4985]

thanks a lot

[@MisterBinx]

Good video but I think Work in at 24:27 should have a dot over it to note that it's in rate form.

[@josephmcmahon7470]

Anything which is usually in units of Joules and is a rate form is a type of Work. It's just convenient to use Qdot for rate of heat transfer compared to just Work so you keep track of what is going on. And the Work the heat pump does will be some sort of transformation of energy. So, there is no need for Wdot when talking about Work itself.

[@yolo3004]

in the 3rd problem you could just use Qh/Qc=Th/Tc and get ΔS=0

[@kk2937]

I FRIGGIN LOVE U I WISH I CAN HUG UUUUUUU

[@sourish_mitra]

Respected Sir Do isolated systems have surroundings?what type of interactions do occur between them?please help

[@burakemreevrenos8385]

I don't know do you still need explanation but I Will try to give it. Isolated systems have surrouindings too but they do not interact with surroundings. Think a bottle of water which is in a thermos. If we take the system as water, the thermos is system boundary and the room is surrounding. Although there is surrounding, the isolated system does not interact with it