Пікірлер
@jimleflar3166
@jimleflar3166 6 сағат бұрын
I have a typo in the calculation of the volume of the saturated liquid. Vf should equal 0.0234 m3 and not 0.00234 m3. The final solution is correct.
@mustachcastach697
@mustachcastach697 11 күн бұрын
thank you so much
@Donald1991
@Donald1991 2 ай бұрын
How to find the critical temperature
@jimleflar3166
@jimleflar3166 2 ай бұрын
The critical temperature of a any substance is the highest temperature listed in the saturation tables for that substance. For instance, the last entrance in the water saturation tables is 705.4 degrees Fahrenheit, which is its critical temperature. You can also find the critical temperature of substances in many other property tables, or you can just use any search engine on the internet to find this property of a substance.
@nabilatanzim
@nabilatanzim 8 ай бұрын
Thank you
@nasralselwi967
@nasralselwi967 10 ай бұрын
Great job sir ❤❤❤❤
@user_.b
@user_.b Жыл бұрын
Thank you so much for this
@jamaicamaysuficiencia6667
@jamaicamaysuficiencia6667 Жыл бұрын
hiii. at example b, why did you assume at 50 celcius? cant we directly use the given u and locate it at the superheated table at p 10 bar and then interpolate using u and s values?
@jimleflar3166
@jimleflar3166 Жыл бұрын
Hi Jamaica May, Thank you for your question. You are correct in your statement about how to solve this problem. I also solved it as you suggest but, perhaps, did not explain what I was doing sufficiently. We know that two properties define a state and this water’s given properties are a pressure of 10 bar and an internal energy of 3124.4 kJ/kg. We need to find a state in the superheat tables with these properties, so begin by looking in the sub-table of 10 bar. We are looking down the column for internal energy looking for a value of 3124.4. We would not normally find this exact value in the table but would find it bracketed by two rows and then have to interpolate a value of entropy. However, in this case, the value of 3124.4 is found exactly in the row whose first column is 500C. So, I did not assume that the temperature was 500C; I found that at the given properties of pressure and internal energy, the temperature actually was 500C. I was simply using the temperature as a marker for the row where our entropy value could be found, which is 7.7622 KJ/Kg*K Perhaps you are using a table with slightly different property values (as different thermodynamic tables do offer slightly different property values) and, in that case, you would have to interpolate the entropy value directly from the internal energy values in your table, just as you suggest. You have a good grasp of this problem. Thank you for your question. Regards, Jim
@jamaicamaysuficiencia6667
@jamaicamaysuficiencia6667 Жыл бұрын
@@jimleflar3166 we have the same reference. shapiro is my textbook too. and thankk you so much for taking time to replyy to this and actually ding what i thought.. god job on your videos! it actually helped me on my finals last january 18. i was ramming and have no time to digest the book, so your playlist on chapter 5-6 is what i just watched! best regards!
@eadoinburns3527
@eadoinburns3527 2 жыл бұрын
awesome dam I needed this fr !!!!!!!