Thermodynamics - Entropy 7.4 Isentropic processes

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Thermodynamics - isentropic processes
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Опубликовано:

11 июл 2018

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Комментарии : 29

@nicolefrancia7728 3 года назад

this is so helpful. Thanks a lot!

@diyaazaghloul756 3 года назад

Amazing video thank you

@kinburnfields6759 Год назад

The pressure of 0.15MPa @120C is not in my table A-6 for superheated water.. When I look at the T-s diagram (figure A-9) you can kinda approximate the values (120C @ 1.5bar).. They look to be what he has, but since on a chart its hard to get an exact value.

@idowuaina7674 3 года назад

Hello, please is there a comprehensive steam table that has temerature in the superheated range.. I am in dire need of one

@sunmichoi6888 2 года назад

Thank you so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

@zaynmoaaz8504 4 месяца назад

@laurenpadron74 3 года назад

Hi, for the second question (#9), how do we find the specific enthalpy and entropy values? I went to A-5 and found that at state one the temperature is much higher than the saturated temperature so it must be a superheated liquid. But when I go to table A-6, I don't know how to interpolate between 0.1MPa and 0.2MPa because they don't share the same temperature ranges (0.2MPa actually doesn't list 120 deg. Celcius at all).

@kk2937 2 года назад

just replying for anyone whos interested in the future ur probably done with it; find the enthalpy in the table for 0.1MPa using interpolation between temps 100 and 150, then use that value to interpolate again with the value of enthalpy from the 0.2MPa table where its temperature at saturation is almost 120 so its safe to use, (Make the temperature constant (120C) and interpolate using pressures and enthalpy) same thing for entropy

@hiteshadookarun8941 Год назад

@@kk2937 use the 120 almost at saturation did not get the answer

@robed3697 3 месяца назад

at 8:05, how to get h2? .I can't find the values of h2 on my tablele

@user-dh3fd8iq6m Год назад

Quick question, why is the first law written with ΔH sometimes and other times with ΔU? I saw this occurring in previous chapters but my book did not explain it at all; it only states that it is used for a constant pressure process. That being said, This process is NOT constant pressure... How I derived the formula for work: E_in - E_out = dE_sys / dt E_in - E_out = 0 E_in = E_out (m\dot)h_1 = (W\dot)_out + (m\dot)h_2 Work out rate = mass flow rate(h_1 - h_2) ∴ Work_out = h_1 - h_2

@engineeringdeciphered Год назад

I have a video about when to use U and when to use H. In general, for flowing fluids (inlets and outlets) use H because H includes both U and flow work. For closed, non-moving processes, use H when it is a constant pressure process because H includes both U and constant pressure boundary work. For other processes use U.

@fawazalmalki8239 Год назад

hello doc. can you please give quick explanation about how you found h1 at 11:23 ??? the 150 kPa isn't there in the tables

@Johnnyy832 4 месяца назад

You probably have to interpolate the values.

@mebawubeshet6729 Месяц назад

@@Johnnyy832 is it double interpolate

@blueandgreen8928 Год назад

I dİd not find last question h1 please sir answr my question I have exam tomorrow how could find h1 I look all tables dont find it is superheated ?

@MrOhMyGawwd 3 года назад

my superheated tables skip from 4mpa to 6mpa

@yourordinarynerdpersonal2775 Год назад

Can someone explain why on number 8, when we try and find h2, why do we interpolate between 250 c and 300 c ? Because we don't have any given temperatures except for 450 c but that's for state 1 and at 5 MPa or 5000 kPa. And for 1.4 MPa the phase is superheated vapor still, but we don't have any given temperatures except for the t-sat found at 1400 kPa or 1.4 MPa for state 2. So how do we know to use 250 c and 300 c ? EDIT: still need help, but the only connection I see that could make us interpolate between 250 c and 300 c is that in state 1 the t sat is 263.94. but even then that's for state one not state 2. So where is the connection to use those numbers ?

@Eggobaco Год назад

Hey i was having the same problem. But you need to use the fact that the process is isentropic. Such that s1=s2. You then find where the entrolpy lies on the tables for me it was superheated water 1.4MP between T250-300. Given that the s = 6.8210 it lays between the 2 values. You can use an equation to find the enthalpy given you know the upper limits, lower limits and the target value. in basic form it is Y = YL + [X-XL/XU-XL]*[YU-YL] where Y is the what you want, YL is the lower limit of what you want, YU is upper limit. XL is lower limit of what you know X is what you know and XU is the upper limit. so for this problem the equations comes out to be H2 = 2927.9 (Lower limit) + [6.8210 (what we know) - 6.7488 (lower limit) / 6.9553 (upper limit) - 6.7488] * [3040.9 ( Upper limit) - 2927.9 (lower limit) ] = 2967.41 kJ/kg. Hope this helps :)