Thermodynamics - Entropy 7.3 Pure Substances

[@mebawubeshet6729]

Foe anyone wondering why the specific volume stays the same , it is not because of the rigid tank but because of the rigid tank + mass stays same. rigid tank alone tells us only the volume stays the same not the specific volume

[@kylecatman7738]

Wrong, the spec. vol. stays the same as well, how else are you gonna find "X"???

[@SyedMuhammadAliAkbar-f7e]

its a closed system so mass remains the same hence if V remains the same then so does V/m

[@rutvikshah9922]

Thank you so much for this video..!! This was all I needed..!!

[@kylecatman7738]

#6 S = 3450 BTU/ Tsat @ 20 pisa which is 3450 BTU/688 R = 5.01 ish (This is quick way, if it was isothermal it'd be 3450 / 530 R but it's not so you have to divide by the sat temp at constant P of 20 psia)

[@enesberkkocak5621]

on question 5,i cant verify the vfg that you wrote as .18255 i find it as 0.19254 on the table.Can you tell where is my mistake?Where should i look on the property tables?

[@yigitcan824]

I guess his property table shows different values some times.

[@aminzamani7866]

thanks alot to you for helping us undrestanding thermodynamics better.i have question. I use van wylen property table and there is a huge difference between that and which you are using.i am somehow confused which one is better to use??

[@gabriellajasmine4165]

hello, would it be possible if u share the tables that you use ? thanks a lot 🙏🏻

[@engineeringdeciphered]

www.dropbox.com/s/tuqy5e8657ysoda/Property%20Tables%20-%20Appendices%201%20and%202.pdf?dl=0

[@gabriellajasmine4165]

@@engineeringdeciphered thats very helpful, thanks again !

[@aliozaltn5928]

how did you find the s2 and u2 in the last question you solved ?

[@engineeringdeciphered]

If you know the temp (40 C) and quality (0.45) then you can calculate everything else. Go to the temp table, find the line for 40C, and calculate s = sf + x(sfg). Same thing with u. Does that make sense?

[@burakemreevrenos8385]

In question 7, how do we know there's no entrophy generation? ΔS= ∫ δQ/T +σ(entrophy generation term). If T is constant, the integral becomes Q/T but we also have another unknown which is the entrophy generation term

[@engineeringdeciphered]

I think because it’s an isothermal heat transfer, there is no entropy generation.

[@burakemreevrenos8385]

@@engineeringdeciphered thanks sir

[@yigitcan824]

@@burakemreevrenos8385So it must be totally reversible?

[@anasthasyatalitha4209]

good day, i would like to ask for the units that are used for number 6. should we convert F, psia, BTU to C, MPa and kJ/kg? or are there any different tables that were used ? thank youu

[@engineeringdeciphered]

We have tables for English units: appendix 2 here: www.dropbox.com/s/tuqy5e8657ysoda/Property%20Tables%20-%20Appendices%201%20and%202.pdf?dl=0

[@anasthasyatalitha4209]

@@engineeringdeciphered thank youuu this is really what i need thank you so much

[@lifexsarai]

for #7... for a closed system, if it is isothermal, I thought that the internal energy goes to 0? Therefore, q=w. Is this wrong?

[@zaydbhikhu4129]

For an isothermal process, isn't change in internal energy zero?

[@engineeringdeciphered]

For gases, yes, usually internal energy is defined by temperature. But for liquids and especially liq-vapor mixtures, no - something can be isothermal but it is changing states and internal energy is definitely not constant.

[@boramerttigli3215]

Initially, thank you so much for your helpful videos. At the last part of last question, why did you used u instead of h? I am confused because the tank is not rigid.

[@engineeringdeciphered]

For problem 7: you’ve got to use u because it is not constant pressure. We use H when it is a constant pressure process, use u for any other process.

[@boramerttigli3215]

@@engineeringdeciphered Thank you for the answer but at the next video (Thermodynamics - Entropy 7.4 Isentropic processes) you used h in the 8.question(at 6.00 sec). And it is not a constant pressure process.

[@engineeringdeciphered]

@@boramerttigli3215 oh yes, definitely use h for steady flow devices and flowing fluid as well. I was just talking about closed systems. H combines u and pv so that takes care of both internal energy and flow work (for a flowing fluid) or internal energy and constant pressure boundary work (for closed, constant pressure systems).

[@boramerttigli3215]

Ok i see now. Thank you very much for your help. 😊

[@sokundavannchriv3967]

How did you interpolate to find entropy in question 6 since temperature is not given!

[@eniyajaber1029]

I want to know too!!!

[@mrtylmz2209]

It is assumed to be superheated. So, Table A-6E P=20 psia section is used for interpolation.

[@ibra9025]

on q7 how did u know it is u and not h?

[@brysontate553]

^

[@hihisaybye1073]

For problem 7: you’ve got to use u because it is not constant pressure. We use H when it is a constant pressure process, use u for any other process

[@kylecatman7738]

#5 0.07182 is completely wrong, Sf at 101.3 kPa = 0.8690... also S1 = 1.8625 not 1.0625.

[@engineeringdeciphered]

again, I'm not sure what R134 tables you are looking at. The tables from Cengel and Boles show what I've got in the video.

[@kylecatman7738]

# 7 LOLOL, You have some weird values for R-134a, but problem seven is solved just like you did it, find Q = S2 - S1 X Temp... then U2 - U1... ah you know the rest.... but those values you're throwing out there for this substance are suspect. LOL

[@engineeringdeciphered]

I am using the property tables from Cengel and Boles. The property tables from Moran and Shapiro look slightly different, but in the same ballpark. Honestly, this is just a problem (and solution) striaght from the Cengel and Boles textbook. I would double check your tables. engineering.wayne.edu/mechanical/pdfs/thermodynamic-_tables-updated.pdf

[@digitallightbulb]

No 7. S1 is wrong.