I would have never been able to understand so many topics, if it was not for your videos. Each time here on RUclips, I type the topic with your name, and always am happy to know that you have made a video on it. Thankyou so very much! Keep posting videos.
There is an error at min 1:34 when you said "adiabatic process and so temperature decreases"; in fact it increases by adiabatic compression from point 3 to point 4 to become T subscript H.
My physics book states that carnot's theorem states that "no cyclical heat engine has a greater efficiency than a reversible engine operating between the same two temperatures"... What does this mean?
means efficiency depends upon the difference of temperature it does not depend upon same temperature for better and maximum efficiency a difference of temperature is required
but if we take the sign convesion then we can write U= Q+w ( here teacher take heat is given to system is negative and workdone by the system is also negative but we can also take the oposit sign convesion)
I haven't watch the video, but I have been reading the physics chapters thouroughly so I hope this answers your question appropriately: you're given the fact that |Qc| = nRTcln(Vc/Vd) right? (I hope this [and more] is relevent.) Well when "the heat is discharged into the cold reservoir." (University Physics 423) I'm assuming you must make the assumption that |Qh|=0. Now "the net work done BY the gas[(system)] on the piston [(surrounding)] is equal to the heat input:W=|Qh|-|Qc|... This work is represented by the area enclosed within the cycle abcd...." (University Physics 423). If |Qh|=0 then W=> -|Qc|. -|Qc|=nRTcln(Vd/Vc) so +|Qc|=-nRTcln(Vd/Vc)=nRTcln[(Vd/Vc)^(-1)]=nRTcln[(Vc/Vd)^(+1)]... Now this is is important to know -- by-the-way |Qh|=nRThln(Vb/Va) --- because it helps to reinforce the idea of thermal equilibrium. Khan from Khan Academy has a video explaining how Vb/Va=Vc/Vd. I Hope that helped you and anyone who read this far! :)
yes, i wonder why others does not point this out. sample of exercises in Principles of Physics also uses u= q+w, which i wonder if its just a simple mistake, or is it actually not?
The reason why it is okay for him to write it as U = Q + W is because he explains that "W" is the work being done BY the system, which by default is a negative value. Therefore, making it U = Q - W anyway after inputing any work value.
I would have never been able to understand so many topics, if it was not for your videos. Each time here on RUclips, I type the topic with your name, and always am happy to know that you have made a video on it. Thankyou so very much! Keep posting videos.
Excellent and well-coordinated. I hope some professors might learn from this clear exposition of such a complex idea. Bravo.
great set of lectures
There is an error at min 1:34 when you said "adiabatic process and so temperature decreases"; in fact it increases by adiabatic compression from point 3 to point 4 to become T subscript H.
Q=U+W , process 1to 2 is isothermal hence temperature is constant hence internal energy is zero , U=0 from the above Q=W this is right
Thank you for being so succinct, unlike Khan academy this is easy to follow. thank you sir!
Thank you so much for explaining this! It's so clear
plyedness you're very welcome!
AK LECTURES I LOVE YOU!!!!!
My physics book states that carnot's theorem states that "no cyclical heat engine has a greater efficiency than a reversible engine operating between the same two temperatures"... What does this mean?
means efficiency depends upon the difference of temperature it does not depend upon same temperature for better and maximum efficiency a difference of temperature is required
Thanks Andrey Sir, Can you please discussion about how to improve the efficiency of Rankine cycle.
'eudjfjjd
superb explanation .thank you
Thanks! Glad to hear it!
you are welcome
but if we take the sign convesion then we can write U= Q+w ( here teacher take heat is given to system is negative and workdone by the system is also negative but we can also take the oposit sign convesion)
you are amazing. thank you so much.
thank !!!!!!....we hope u always help me ...
doctor thank you , i have a question about process c : work is poistive and Q is negative so Q = - nRTLln(v3/v4) is this right or not thank you ...
u're patly right but Q in this case goes line.....Q= -nrtL ln(v4/v3)
when we inverse the volumes fraction of logarithm the sign inverses
I haven't watch the video, but I have been reading the physics chapters thouroughly so I hope this answers your question appropriately: you're given the fact that |Qc| = nRTcln(Vc/Vd) right? (I hope this [and more] is relevent.) Well when "the heat is discharged into the cold reservoir." (University Physics 423) I'm assuming you must make the assumption that |Qh|=0. Now "the net work done BY the gas[(system)] on the piston [(surrounding)] is equal to the heat input:W=|Qh|-|Qc|... This work is represented by the area enclosed within the cycle abcd...." (University Physics 423). If |Qh|=0 then W=> -|Qc|. -|Qc|=nRTcln(Vd/Vc) so +|Qc|=-nRTcln(Vd/Vc)=nRTcln[(Vd/Vc)^(-1)]=nRTcln[(Vc/Vd)^(+1)]... Now this is is important to know -- by-the-way |Qh|=nRThln(Vb/Va) --- because it helps to reinforce the idea of thermal equilibrium. Khan from Khan Academy has a video explaining how Vb/Va=Vc/Vd. I Hope that helped you and anyone who read this far! :)
Sorry for the crossed out equations. "RUclips" does this when characters are embedded in hypens (-).
sorry.
I think Ql/QH = Tl/TH is reasonable not only in carnot cycle but also in other engines.
because Q=cnT
what is the difference between the carrot and Otto cycle
Eunice Adeyemi carrot?????!!!
What is your favourite movie?
PIEASE PERFECT
Thanks man! :-)
U=Q-W but u wrote this wrong equation U=Q+W
yes, i wonder why others does not point this out. sample of exercises in Principles of Physics also uses u= q+w, which i wonder if its just a simple mistake, or is it actually not?
The reason why it is okay for him to write it as U = Q + W is because he explains that "W" is the work being done BY the system, which by default is a negative value. Therefore, making it U = Q - W anyway after inputing any work value.