Kiraben Raj Ravinddaran Based on watching his previous videos, I believe he places the negative on the outside to denote Heat Rejection as a negative process. Q out is still (u1-u4), because the first law states DU= Q-W. And the final process is going back to state 1, so final-initial is u1-u4. (No work done during an isovolumetric heat addition or rejection)
Sir, would you please help me with this: Air is compressed isothermally from 13 psia and 90 degree F to 80 psia in a reversible steady-flow device. Calculate the work required in Btu/lbm, for this compression. Thank you,
You have the start pressure and temp and its isothermal so the temp doesn't not change so you can your u's and apply it to your formula and then you're good to go
Can someone answer this. In finding V3, there are two ways that I know. One is in the process 2-3 where the pressure is constant. That gives T3/T2 = V3/V2. The other is that V3=V2+cVd. Now, the question is what is more accurate. I mean, what should I use? For example, you all had the values in both ways. Trust me, there's a discrepancy when use both. Not a small, but quite large discrepancy. Can someone explain. Thank you!
Im just starting thermo 2 and I think I'm going to be here more than in my lecture.
Same here bro.😂
I hope by now that you are successful in your field. Just started thermo 2 as well
@@ainerberedo2365 ha! Not exactly my field but close enough. And I was right. I only showed up to thermo 2 for the exams. That professor was terrible
@@sunnygcat13 can relate to that, some youtube vids are more helpful than the one’s discussed in class. hope your doing well and best regards!
thank you for awesome lectures i learned everything from you.
Great video. Gratitude.
thank you sir
There is a mistake, -q(Out) = u1-u4 so q(out)=u4-u1
Kiraben Raj Ravinddaran Based on watching his previous videos, I believe he places the negative on the outside to denote Heat Rejection as a negative process. Q out is still (u1-u4), because the first law states DU= Q-W. And the final process is going back to state 1, so final-initial is u1-u4. (No work done during an isovolumetric heat addition or rejection)
Thanks bro!
Sir,
would you please help me with this:
Air is compressed isothermally from 13 psia and 90 degree F to 80 psia in a reversible steady-flow device. Calculate the work required in Btu/lbm, for this compression.
Thank you,
You have the start pressure and temp and its isothermal so the temp doesn't not change so you can your u's and apply it to your formula and then you're good to go
@@Lance.2451 bruh, 2 years ago
@@lucalalsie5653 lmao pretty sure radu already solved that years ago
thank you
Can someone answer this.
In finding V3, there are two ways that I know. One is in the process 2-3 where the pressure is constant. That gives T3/T2 = V3/V2. The other is that V3=V2+cVd.
Now, the question is what is more accurate. I mean, what should I use?
For example, you all had the values in both ways. Trust me, there's a discrepancy when use both. Not a small, but quite large discrepancy.
Can someone explain. Thank you!
Why is process 2-3 isobaric ?
th-cam.com/video/qeGQgPQsJJI/w-d-xo.html that vid may help you visualize it better.