Isentropic efficiency definition for turbine (expansion), compressor (compression) and nozzle (acceleration) processes. An example of refrigeration compression.
Thanks for this video. I'm finding a Sat. Vapor Entropy of 0.9354 (kJ/kg)K instead of 0.9447 like you show here. Yours is the value my table has for P=0.8 Bar instead of 1.2 Bar.
You need to find the vapor quality by using the isentropic "assumption" @ P=1MPa, and then again use the quality on the enthalpy to derive to h2s. The values are pretty logical, given that the pressure increases drastically (approx 10x from inlet to outlet), hence there would be more Liquid water than vapor ==> enthalpy closer to Liquid. And thanks to Bing Guo, very Nice lectures.
@@khan120495 it's superheated bruh, no need to find the quality(quality is 1)... since you know that S1=S2s , you must go to superheated table at 1mpa ,and interpolate within the known intervals of the entropy(s) and your unknown enthalphy(h) value, you'll find h2s to be 281.21
@@Wt520lyn I assume you are referring to the isentropic efficiency of a compressor in a vapor compression refrigeration cycle. The compressor's isentropic efficiency does affect the COP of the cycle. However, the temperatures of the hot reservoir and the cold reservoir also affect the COP. Begin with a "perfect" compressor that has 100% isentropic efficiency, and see what the COP of a cycle is; then study the case where compressor isentropic efficiency is less than 100% (the cycle will be different than first case even if you keep the lower and upper pressure limits constant). Your professor or a good tutor should be able to help you do this comparison with an example problem.
Would it make sense that the steam table you used had values that were almost 2.0% different from mine? I am getting entirely different values for the values that you search up.
Different tables may use different reference states, that is the state with zero enthalpy may be defined differently in different tables. What matters is the CHANGE of enthalpy. The enthalpy values of intermediate steps may be dependent on the tables you use for reasons given above, but your final results should be the same except for rounding errors.
you're a life saver, legend!
Thanks for this video.
I'm finding a Sat. Vapor Entropy of 0.9354 (kJ/kg)K instead of 0.9447 like you show here. Yours is the value my table has for P=0.8 Bar instead of 1.2 Bar.
wow it is very best explanition thinkyou all
This is my favorite video, thank you
Subscripts "a" and "s" stand for "actual" and "isentropic", respectively.
Sir what is the difference between isentropic efficiency and second law efficiency
@16:10 P2=1Mpa but how and where did you find h2s to be 281.21? I looked on the table at 1Mpa and the value for h is not there!
You need to find the vapor quality by using the isentropic "assumption" @ P=1MPa, and then again use the quality on the enthalpy to derive to h2s. The values are pretty logical, given that the pressure increases drastically (approx 10x from inlet to outlet), hence there would be more Liquid water than vapor ==> enthalpy closer to Liquid. And thanks to Bing Guo, very Nice lectures.
@@khan120495 it's superheated bruh, no need to find the quality(quality is 1)... since you know that S1=S2s , you must go to superheated table at 1mpa ,and interpolate within the known intervals of the entropy(s) and your unknown enthalphy(h) value, you'll find h2s to be 281.21
how to gat the enthalpy of h2s from table
Could u please put the videos in the right sequence?
is higher isentropic efficiency better for a compressor in a system?
Generally yes. The higher the isentropic efficiency of a compressor, the less work it consumes to achieve the same compression outcome.
@@bing9u0 but what if my isentropic efficiency is high, but my coefficient of performance is low?
@@Wt520lyn I assume you are referring to the isentropic efficiency of a compressor in a vapor compression refrigeration cycle. The compressor's isentropic efficiency does affect the COP of the cycle. However, the temperatures of the hot reservoir and the cold reservoir also affect the COP. Begin with a "perfect" compressor that has 100% isentropic efficiency, and see what the COP of a cycle is; then study the case where compressor isentropic efficiency is less than 100% (the cycle will be different than first case even if you keep the lower and upper pressure limits constant). Your professor or a good tutor should be able to help you do this comparison with an example problem.
@@bing9u0 thank you
Would it make sense that the steam table you used had values that were almost 2.0% different from mine? I am getting entirely different values for the values that you search up.
Different tables may use different reference states, that is the state with zero enthalpy may be defined differently in different tables. What matters is the CHANGE of enthalpy. The enthalpy values of intermediate steps may be dependent on the tables you use for reasons given above, but your final results should be the same except for rounding errors.
However, it is also possible that there was an error in my table lookup, according to a comment below. I need to take a look when I get a chance.