Correction: No need to use specific volume method, just linear interpolate using temperature (Max_A-Min_B)/(Max_A-X_(given_z ) )=(Max_(A_2 ) -Min_(B_2 ))/(Y_(unknown_w )-Min_(B_2 ) ) (5-0.01)/(5-3)=(0.8725 -0.6117)/(P_2-0.6117) P_2=0.768 kPa
Thanks for watching, as a former struggling mechanical engineering student who eventually graduated. I understand the struggle. The playlist and series has evolved initially it was aimed at students who knew what they were doing and wanted to revise. As more questions came in from viewers the videos became step by step tutorials
Here is my (incorrect) thought process for this problem: The mass is constant, so the relation (P1 V1)/ T1 = (P2 V2) / T2 can be used. Since we aren't given the initial temperature, we refer to the tables and find that this is a superheated vapor, and has initial T1 = 250 C Solving for P2 gives 35.2 kPa. Why is this answer wrong? Is it because of the ideal-gas assumption? T_R = 0.42. Since ideal-gas behavior is best assumed for gasses with high temperature and low pressure, does it follow that T_R = 0.42 leads to the assumption that this gas does not behave as an ideal gas?
For this question, the approach would be find the specific volume in the evacuated (empty) part of the chamber v=V/m=(1.1989 m^3)/(1 kg)=1.1989 m^3/kg v_2=(3) v_1=3(1.1989 m^3/kg)=3.5967 m^3/kg for the empty chamber For Final pressure check table, you will notice you need to linear interpolate because 3 degrees is between those temperature values in your table. (Max_A-Min_B)/(Max_A-X_(given_z ) )=(Max_(A_2 ) -Min_(B_2 ))/(Y_(unknown_w )-Min_(B_2 ) ) (5-0.01)/(5-3)=(0.8725 -0.6117)/(P_2-0.6117) P_2=0.768 kPa
Correction: No need to use specific volume method, just linear interpolate using temperature
(Max_A-Min_B)/(Max_A-X_(given_z ) )=(Max_(A_2 ) -Min_(B_2 ))/(Y_(unknown_w )-Min_(B_2 ) )
(5-0.01)/(5-3)=(0.8725 -0.6117)/(P_2-0.6117)
P_2=0.768 kPa
if i had these videos the first time i took the class, i would be graduated by now. great work thank you.
Thanks for watching, as a former struggling mechanical engineering student who eventually graduated. I understand the struggle. The playlist and series has evolved initially it was aimed at students who knew what they were doing and wanted to revise. As more questions came in from viewers the videos became step by step tutorials
Here is my (incorrect) thought process for this problem:
The mass is constant, so the relation (P1 V1)/ T1 = (P2 V2) / T2 can be used.
Since we aren't given the initial temperature, we refer to the tables and find that this is a superheated vapor, and has initial T1 = 250 C
Solving for P2 gives 35.2 kPa. Why is this answer wrong?
Is it because of the ideal-gas assumption? T_R = 0.42. Since ideal-gas behavior is best assumed for gasses with high temperature and low pressure, does it follow that T_R = 0.42 leads to the assumption that this gas does not behave as an ideal gas?
For this question, the approach would be find the specific volume in the evacuated (empty) part of the chamber
v=V/m=(1.1989 m^3)/(1 kg)=1.1989 m^3/kg
v_2=(3) v_1=3(1.1989 m^3/kg)=3.5967 m^3/kg for the empty chamber
For Final pressure check table, you will notice you need to linear interpolate because 3 degrees is between those temperature values in your table.
(Max_A-Min_B)/(Max_A-X_(given_z ) )=(Max_(A_2 ) -Min_(B_2 ))/(Y_(unknown_w )-Min_(B_2 ) )
(5-0.01)/(5-3)=(0.8725 -0.6117)/(P_2-0.6117)
P_2=0.768 kPa
where is the pinned comment?
Good question, not sure why it was removed maybe I was it editing it and forgot to post it. I'll repost it
@@SuemonKwok could you copy paste a link here?
@@catsc4978 pinned comment is up for 3-37
awesome
Thanks for watching