At that time i was attending your lecter ...i did understand about centrifugal compressor......now today i am working on centeifugal compressor ....that some point is missing than idid search on youtube ....now i am waching your video ...this video is very helpful for me
The only person who really understands the physical behavior!!!... the other people just learn from memory how to draw the velocity triangles (which do not help understand well the working principle) CONGRATULATIONS!!!
lecture is good. but there are few mistakes. such as it is not neccesrly 50% pressure rise in rotar and 50% in stator but fraction of pressure rise occurs in stator and rest in rotar. and another mistake is that pressure rise in rotar in not only because of increase in passage since there is external work involved. pressure rises to counter centrifugal force is also involved.
Arvind, I would like o clear your doubt 1) In the lecture I have said that its a 'common practice' to raise the pressure 50% in impeller and 50% in diffuser it means It is not necessary of course. 2) The static pressure rise of a moving gas takes place only due to 'diffusing action' that is by transforming K.E into Enthalpy. Suppose a pipe in which some fluid is flowing and there is turn in pipe . When fluid will pass that turn then; does at the exit of turn the static pressure of fluid will rise? the answer is no. At the moment the fluid was taking turn then it would experience centrifugal force on corresponding surface of pipe. When fluid will pass away the turn then pressure would be same is it was before entering the turn. But imagine that the pipe haves a turn together with diverging cross sectional area . In such case at the exit of turn it would experience the hike in pressure due to diffusing action. Workdone by impeller - rise in KE - Diffussing action - Rise in pressure
Amit Mandal thnx I got 1st point that was my mistake. but i couldn't get second point. yes there is pressure rise due to increase in area due to decrese in relative velocity. but what abut influence of change in cercrumfrencial velocity. U1 to u2. that will increase fluid pressure due to counter centrifugal action. in pipe there is no cercrumfrencial velocity involved so what u said is right but not in this case.
ashok ashok, P=F/A is true only when Force is restricted by perpendicular Area, that means the Pressure there rests on the blocking area. But when we talk about amount of mass of air flowing through an open Area, Pressure is not restricted therefore F = mass flow rate times velocity. In that case, according to Continuity Equation, the lesser Area embodied in that formula of mass flow rate (Mass Flow Rate = Density X Area X Velocity), the resulting Pressure must also be lesser due to conversion of Pressure formula as P = (0.5 X Density X Velocity squared) + given Static Pressure (due to atmospheric pressure). This is the Incompressible Bernoulli principle for convergent - divergent flow of air in stream tubes such as the one discussed in centrifugal compressors. If the velocity of air reaches more than 0.30 Mach, the equation becomes Compressible and Isentropic Flow equations must then be used instead. Do not confuse the formula P = F/A because it is not applicable on cross sectional areas that are open for fluid flow.
Thank you for this video. Can you provide me with an explanation showing what amendments can be done to change the centrifugal pump to a centrifugal compressor? OR can we use the same pump for compressing gas?
In which application water is being compressed in compressor... Can u tell me plz. Bcz as much I know compressors are only used to compress fluid having very small density like air and gases while for heavy liquid like water pump is used..
Sorry, please at what velocity does air strike the nozzle. Let's rage for instance in a turbocharger of an IC engine. At what force does the air enter the compressor please?
+Onyedika okpala- ideally the air enters the nozzle with 0 velocity.however practically its mass flow rate / (density at entry ×area at inlet of nozzle), the force at inlet = pressure of air at inlet × area.
Hi sir Is there any more explanation for the conversation of centrifugal action to the pressure rise.? Any equation or mathematical explanation ? Can you please provide .?
Sir you are telling in this video (cross sectional area decreases the pressure decreases it is not correct sir example if you take pressure is equal to force per area. So the pressure is inversely proportional to area so the area decreases in that place the pressure increases.
P=F/A is true only when Force is restricted by perpendicular Area, that means the Pressure there rests on the blocking area. But when we talk about amount of mass of air flowing through an open Area, Pressure is not restricted therefore F = mass flow rate times velocity. In that case, according to Continuity Equation, the lesser Area embodied in that formula of mass flow rate (Mass Flow Rate = Density X Area X Velocity), the resulting Pressure must also be lesser due to conversion of Pressure formula as P = (0.5 X Density X Velocity squared) + given Static Pressure (due to atmospheric pressure). This is the Incompressible Bernoulli principle for convergent - divergent flow of air in stream tubes such as the one discussed in centrifugal compressors. If the velocity of air reaches more than 0.30 Mach, the equation becomes Compressible and Isentropic Flow equations must then be used instead. Do not confuse the formula P = F/A because it is not applicable on cross sectional areas that are open for fluid flow.
there is big contradiction in your statement 1.velocity is directly proportional to area so as area increases so the velocity and ultimately pressure decreases (nozzle is used to increase the pressure not velocity) please reply on this as soon as possible,,,,emergency
sir,asking about 1.converging nozzle which has bigger area of intake than outlet 2.we use nozzle to increase pressure or velocity?? .morrow im having exam please clear it soon
sumeet more nozzles are generally used to gain kinetic energy by increasing fluid velocity thus pressure is converted into velocity inside a nozzle. Diffuser is opposite to that of nozzle.
At that time i was attending your lecter ...i did understand about centrifugal compressor......now today i am working on centeifugal compressor ....that some point is missing than idid search on youtube ....now i am waching your video ...this video is very helpful for me
Very incredible explanation 👍👍👍 One of the most simplest and detail explanation on Centrifugal compressor ever on youtube.. keep it up sir
Excellent Explanation bro...With proper schematic diagram 👍 thnks a lot
NICE explanation with simple examples.
Thanks a lot for this. Very well explained. Crisp and to the point.
Bahut bahut bdhiya. Well explained ❤🙏
Very well explained Amit! Thanks!
Thank you brother very useful information
Very nice explanation. I understood it very well.
Good explanation....i understood easily the basic principles.... Thank you sir..
Great lecture
Excellent presentation.
thanks very helpful.. I viewed just 30mins before exam n I understood..
u understood huh
very good study hard
The only person who really understands the physical behavior!!!... the other people just learn from memory how to draw the velocity triangles (which do not help understand well the working principle)
CONGRATULATIONS!!!
What an excellent explanation 👍👍👍full respect broo
thank you sir for making understand the working of centrifugal compressor..
Thank you. This was sooo helpful 😇😇
Thank you for this informative video.
Excellent!!!! Thank you
Very well explained
Thank you so much sir ❤️
Great video man.
GREAT EFFORT!
Such a helpful video 🙂
Thank you for presenting a very clear explanation 🙏
Super explanation sir
Thanks dear👍🏻👍🏻
Good work . Neat explanation. Thank you
Nicely done
very nicely explained, thank you
Thank You
You could have added the significance of each type of vanes as well. A nice informative video. 👍
Thanks brother
good basic info
Awesome!!!
thank you.
THANKS YOU ARE AWSOME!
Superb explanation
Very resourceful...
Thank you SiR....
thanks man
lecture is good. but there are few mistakes. such as it is not neccesrly 50% pressure rise in rotar and 50% in stator but fraction of pressure rise occurs in stator and rest in rotar. and another mistake is that pressure rise in rotar in not only because of increase in passage since there is external work involved. pressure rises to counter centrifugal force is also involved.
Arvind, I would like o clear your doubt
1) In the lecture I have said that its a 'common practice' to raise the pressure 50% in impeller and 50% in diffuser it means It is not necessary of course.
2) The static pressure rise of a moving gas takes place only due to 'diffusing action' that is by transforming K.E into Enthalpy. Suppose a pipe in which some fluid is flowing and there is turn in pipe . When fluid will pass that turn then; does at the exit of turn the static pressure of fluid will rise? the answer is no. At the moment the fluid was taking turn then it would experience centrifugal force on corresponding surface of pipe. When fluid will pass away the turn then pressure would be same is it was before entering the turn. But imagine that the pipe haves a turn together with diverging cross sectional area . In such case at the exit of turn it would experience the hike in pressure due to diffusing action.
Workdone by impeller - rise in KE - Diffussing action - Rise in pressure
Amit Mandal thnx I got 1st point that was my mistake. but i couldn't get second point. yes there is pressure rise due to increase in area due to decrese in relative velocity. but what abut influence of change in cercrumfrencial velocity. U1 to u2. that will increase fluid pressure due to counter centrifugal action. in pipe there is no cercrumfrencial velocity involved so what u said is right but not in this case.
Thank you so much..love it..!!
Excellent sir
ashok ashok, P=F/A is true only when Force is restricted by perpendicular Area, that means the Pressure there rests on the blocking area. But when we talk about amount of mass of air flowing through an open Area, Pressure is not restricted therefore F = mass flow rate times velocity. In that case, according to Continuity Equation, the lesser Area embodied in that formula of mass flow rate (Mass Flow Rate = Density X Area X Velocity), the resulting Pressure must also be lesser due to conversion of Pressure formula as P = (0.5 X Density X Velocity squared) + given Static Pressure (due to atmospheric pressure). This is the Incompressible Bernoulli principle for convergent - divergent flow of air in stream tubes such as the one discussed in centrifugal compressors. If the velocity of air reaches more than 0.30 Mach, the equation becomes Compressible and Isentropic Flow equations must then be used instead. Do not confuse the formula P = F/A because it is not applicable on cross sectional areas that are open for fluid flow.
P= F/A is one and only way to define pressure...
Thank you for this video. Can you provide me with an explanation showing what amendments can be done to change the centrifugal pump to a centrifugal compressor? OR can we use the same pump for compressing gas?
Using pump as compressor, low pressure rise
Accha smjaya sir aapne
Thanks sir
nice thank you
In which application water is being compressed in compressor... Can u tell me plz. Bcz as much I know compressors are only used to compress fluid having very small density like air and gases while for heavy liquid like water pump is used..
Liquids are incompressible
I know liquid are incompressible thats why pumps are used but you have mentioned water is being compressed in compressor.
How to water or liquid compressed ......only gas and vapour and air can compressed ...liquid can not compressed ...
that was so help full. i appreciate it , can you introduce me a good resource about compressors and turbines ?
+hossein hossein i have read many sources and the way i understood, i explained.
Sorry, please at what velocity does air strike the nozzle. Let's rage for instance in a turbocharger of an IC engine. At what force does the air enter the compressor please?
+Onyedika okpala- ideally the air enters the nozzle with 0 velocity.however practically its mass flow rate / (density at entry ×area at inlet of nozzle), the force at inlet = pressure of air at inlet × area.
Sir, the rotor also causes the increase in pressure, because of the reduce in velocity and increase in pressure, AM i correct?
Hi sir
Is there any more explanation for the conversation of centrifugal action to the pressure rise.? Any equation or mathematical explanation ? Can you please provide .?
You can go through NPTEL lectures.
What is the static head and dynamic head why we need this
Study stagnation pressure concept.
where is diffuser description?
what is the reason for different vane shapes.
Kourosh Khavari to have a variable and desired pressure
Sir you are telling in this video (cross sectional area decreases the pressure decreases it is not correct sir example if you take pressure is equal to force per area. So the pressure is inversely proportional to area so the area decreases in that place the pressure increases.
P=F/A is true only when Force is restricted by perpendicular Area, that means the Pressure there rests on the blocking area. But when we talk about amount of mass of air flowing through an open Area, Pressure is not restricted therefore F = mass flow rate times velocity. In that case, according to Continuity Equation, the lesser Area embodied in that formula of mass flow rate (Mass Flow Rate = Density X Area X Velocity), the resulting Pressure must also be lesser due to conversion of Pressure formula as P = (0.5 X Density X Velocity squared) + given Static Pressure (due to atmospheric pressure). This is the Incompressible Bernoulli principle for convergent - divergent flow of air in stream tubes such as the one discussed in centrifugal compressors. If the velocity of air reaches more than 0.30 Mach, the equation becomes Compressible and Isentropic Flow equations must then be used instead. Do not confuse the formula P = F/A because it is not applicable on cross sectional areas that are open for fluid flow.
helfull
there is big contradiction in your statement
1.velocity is directly proportional to area so as area increases so the velocity and ultimately pressure decreases (nozzle is used to increase the pressure not velocity)
please reply on this as soon as possible,,,,emergency
+sumeet more are you asking about nozzle part of compressor?
sir,asking about 1.converging nozzle which has bigger area of intake than outlet
2.we use nozzle to increase pressure or velocity??
.morrow im having exam please clear it soon
sumeet more nozzles are generally used to gain kinetic energy by increasing fluid velocity thus pressure is converted into velocity inside a nozzle. Diffuser is opposite to that of nozzle.
sumeet more when area increases presure decreases and velocity increases....
nozzle is used to form a jet with high velocity
what is inducer
+Roopinder Singh the first point of impeller vane from where the air interacts with it.
Hindi me explain kijia taki or gahray se samjme aajaye
है हिंदी में भी, सेम लेक्चर। find in same channel
Hindi me btaya kijiye....
Same lecture hindi me bhi hai....
Chuperrr
Thank you so much sir ❤️