I graduated from NIT Calicut in 2016 (mech). Now am preparing the mechanical syllabus for my upcoming UPSC interview. I have not seen a teacher like you even at NIT. You are the best teacher to teach mechanical. Thanks a lot, sir for enlightening us with your amazing knowledge.
In usual ... we say carnot cycle's 2nd & 4th process are adiabatic ,which implies there is friction? Frictionless condition is mandatory or not for a carnot cycle. Why do " no friction" condition leads to constant entropy ?
It was 2 yrs ago.. Now i'm grown.. thankyou anyways. NOTE: We should name this process as "reversible adiabatic" process . Not by the name " isentropic". Actually it is isentropic also... but we couldn't name it as isentropic.. why? Because, every isentropic processes are not necessary to be adiabatic. [ ie, even if heat transaction is presented ( non adiabatic), the net entropy change ( due to entropy tranfer and entropy generation) leads to 0 in magnitude ( isentropic) That means isentropic doesn't imply that adiabatic. ] But for a carnot engine/Heat pump/ refrigerator... processes really adiabatic with no entropy generation... ie, reversible adiabatic process" So in this video, title of the 2nd and 4 processes really needs to be edited into " rev adiabatic" instead of isentropic" Its common old interview question also asked in gate exam also
Dude .. thats wrong... pls understand.. Adiabatic can be frictionless/ or with friction... If frictionless +adiabatic = reversible adiabatic If friction +adiabatic = irreversible adiabatic or say just adiabatic Then you may have a doubt that how can be a process both adiabatic and friction, where friction creats heat.. right? Understand , adiabatic means no heat transfer (gain/loss), doesn't mean that no heat generation ( where this generated heat forms into entropy) So even if there is friction , adiabatic condition is possible. But it should be irreversible adiabatic.. even if there is no exchange - gain/loss of heat from the surrounding, entropy may be generated due to internal irreversibities. Just consider a cylinder where a gas expands adiabatically but the piston provides friction, thus create heat generation (but not heat transfer) forms new extra entropy. So its an adiabatic process which is non isentropic. [ where, carnot cycle' s adiabatic process => frictionless ( reversible adiabatic) I don't know who you are, but may be helpful 👇 🐺Every reversible adiabatic = isentropic , no friction 🦊 isentropic process may / may not be adiabatic * isentropic, but non-adiabatic = should be irreversible 🐱 every adiabatic may /not be frictionless 🐶 reasons for change in entropy: irreversibities / heat transfer/mass transfer
In the case of isothermal compression, from one side the low temp source absorbs heat from air and from other side as volume decreases from V3-V4 the temp has to decrease according to charle's law (V/T= const).so in overall scenario the temp has to decrease right
During isothermal compression process the heat is getting rejected to the low temperature body but the piston moves inward and volume decreases. Due to this volume reduction, pressure rise will occur. So even though the heat is rejecting during this process the pressure rise will compensate that temperature drop and hence temperature remains constant..
I graduated from NIT Calicut in 2016 (mech). Now am preparing the mechanical syllabus for my upcoming UPSC interview. I have not seen a teacher like you even at NIT. You are the best teacher to teach mechanical. Thanks a lot, sir for enlightening us with your amazing knowledge.
😳
7:15 Not because of inertia (piston is assumed massless), it's due to pressure difference b/w air inside the piston and outside
Great
Right ✅
Sir super..... Plz continue all module
Siree isentropic compression temperature t4 to t1(decrease) akukayalle cheyyanate.. Enikke oru doubt tonni
Thank you sir. Great💯class
very useful classss
Super class... Thanks sir😊
Excellent
sir y u stopped with module 3 ... these are really best vdeos on ytube for bme pls do the rest modules also
We have only 3 modules in KTU syllabus for BME. Other 3 modules are BCE..
Thank you sir ,Great class
Super class
Sir e PPT koode evidelum upload cheyyuvo?
Thank you so much sir.waiting for your next videos
All videos were uploaded already.. please see the playlist.. all three modules uploaded..
In usual ... we say carnot cycle's 2nd & 4th process are adiabatic ,which implies there is friction?
Frictionless condition is mandatory or not for a carnot cycle.
Why do " no friction" condition leads to constant entropy ?
The process should be frictionless..
if friction is present process will not be reversible.
It was 2 yrs ago..
Now i'm grown.. thankyou anyways.
NOTE: We should name this process as "reversible adiabatic" process . Not by the name " isentropic".
Actually it is isentropic also... but we couldn't name it as isentropic.. why?
Because, every isentropic processes are not necessary to be adiabatic.
[ ie, even if heat transaction is presented ( non adiabatic), the net entropy change ( due to entropy tranfer and entropy generation) leads to 0 in magnitude ( isentropic)
That means isentropic doesn't imply that adiabatic. ]
But for a carnot engine/Heat pump/ refrigerator... processes really adiabatic with no entropy generation... ie, reversible adiabatic process"
So in this video, title of the 2nd and 4 processes really needs to be edited into
" rev adiabatic" instead of isentropic"
Its common old interview question also asked in gate exam also
if friction is present process cannot be adiabatic.
Dude .. thats wrong... pls understand..
Adiabatic can be frictionless/ or with friction...
If frictionless +adiabatic = reversible adiabatic
If friction +adiabatic = irreversible adiabatic or say just adiabatic
Then you may have a doubt that how can be a process both adiabatic and friction, where friction creats heat.. right?
Understand , adiabatic means no heat transfer (gain/loss), doesn't mean that no heat generation ( where this generated heat forms into entropy)
So even if there is friction , adiabatic condition is possible. But it should be irreversible adiabatic..
even if there is no exchange - gain/loss of heat from the surrounding, entropy may be generated due to internal irreversibities.
Just consider a cylinder where a gas expands adiabatically but the piston provides friction, thus create heat generation (but not heat transfer) forms new extra entropy. So its an adiabatic process which is non isentropic.
[ where, carnot cycle' s adiabatic process => frictionless ( reversible adiabatic)
I don't know who you are, but may be helpful 👇
🐺Every reversible adiabatic = isentropic , no friction
🦊 isentropic process may / may not be adiabatic
* isentropic, but non-adiabatic = should be irreversible
🐱 every adiabatic may /not be frictionless
🐶 reasons for change in entropy: irreversibities / heat transfer/mass transfer
Diesel cycle, petrol engine, diesel engine, two and four stroke engines and air cooling systems inte class upload cheyyumo?
Already uploaded.. please see the playlist
Sir slides ayitt kittumo
Sir examn isobaric process ,adiabatic process ,isochronic process, isothermal process derive chyan chothiko??
Illa...
very helpful sir
sir isothermalil temp change avillalo so t1 t2 t3 avre alle varu
Thank you so much sir 👍😍
Sir, two four stroke engine explain cheyyo
Yes...
@@VISHNUSHAJIKANNETH sir enna upload cheyya?
Kurach busy anu.. Tomorrow ic engine introduction video upload cheyyam
Sir ith 2019 scheme Alle plz rply
Yes it is
Sir 2023 new engineering studentsin ee cls kanda mathiyavumo
Yes
In the case of isothermal compression,
from one side the low temp source absorbs heat from air
and from other side as volume decreases from V3-V4 the temp has to decrease according to charle's law (V/T= const).so in overall scenario the temp has to decrease right
so the process cant be called as isothermal. pls do correct me if I'm wrong
No..It's an isothermal process.. so temperature has to remain constant throughout the isothermal compression process..
During isothermal compression process the heat is getting rejected to the low temperature body but the piston moves inward and volume decreases. Due to this volume reduction, pressure rise will occur. So even though the heat is rejecting during this process the pressure rise will compensate that temperature drop and hence temperature remains constant..
carnot cycle is a practcally impossible cycle.
Sir ppt onn upload cheyyavuoo
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Please upload all the modules in basics on mech engineering...
Sure.. 👍🏿👍🏿
Sir poli
OTTO CYCLE DIESEL CYCLE DUAL CYCLE CLASSUM VENAM SIR...PLZ..
Sure, this weekend upload cheyyam... 👍🏿
Sir next videoil ithile problems kudi idumo
Sure.. Problems will be discussed after completing cycles.. 👍🏿👍🏿
Sir bme important topics ethokaya module wise parayamo
Thanks sir ❤️
sir.e slides koodi share cheyyamo...
Sir, module 1 note onn share cheyyoo??please
❤
Sir, ee notes onn share cheyamo
Sir please give derivatons of 1st module notes as pdf please please
Excellent