Introduction to Chemical Engineering Thermodynamics - Energy and Entropy Balance in Open System Example 4 (Chapter 6, Problem 4)

Here we will work through our fourth example energy and entropy balance in an open system, in this case applied to a turbine. The end of the video is longer where we first compute the lost work as the difference between the actual and reversible work (although I had a sign error the first time), and then reconcile with the expression provided in the text involving the use of a bath. This is from Chapter 6: Balances in Open Systems. A copy of my work is provided below: https://drive.google.com/file/d/1uoT2Og_xNLFQ8ejI_wEm7FZ990CNm08g/view?usp=sharing https://drive.google.com/file/d/1GcqPOnV3VhH1VwnwzXU6QIODHgNBmahK/view?usp=sharing Please know that the initial reasoning to compute the lost work is reasonable and correct. The issue is just in reconciling with the expression provided in the textbook, as the value to use for the temperature of the bath is not clear. After the video, solving for T0 we find that both expression are consistent when we have a bath temperature of T0=274.73 oC. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Subscribe: https://www.youtube.com/c/AndrewPaluch?sub_confirmation=1