Hey very quick question. So I understand that if we have a air standard analysis we cannot assume constant specific heat. However, if we have a cold air analysis, we can assume specific heat. Does this mean that I cannot use "k" with my isentropic equations (at 9:35) since it is the ratio of specific heat and I can't assume specific heat?
thanks for the video first, I know this might sound stupid to some, but i am dying to get a satisfying answer, at 11:08 you used the Cp version to extract the equation, why not Cv ?. Also, what i took in Otto cycle, process isentropic---> q=0 and work = -Cv delta T, why Cv not Cp here.
+Seven4 Seven There are two general equations to calculate the change in entropy for an ideal gas assuming constant specific heats. Both equations are valid. There are also three equations relating T&p, T&v, or p&v, and are only valid for ideal gas undergoing an isentropic process. The vid concerns the Otto cycle, so the students are already familiar with these equations. At around 11:08, students are reminded how to move from s2-s1 =cp*ln(t2/t1)-R*ln(p2/p1) to the equation: T2/T1 = (p2/p1)^((k-1)/k) . Same can be said for moving from s2-s1 = cv*ln(t2/t1)+R*ln(v2/v1) to t2/t1 = (v1/v2)^(k-1). For your second question, start from 1st Law for closed system undergoing a process. u2-u1 = q1-2 - w1-2. When q=0, w1-2 = u2-u1, not h2-h1. For ideal gas, u2=u1=cv*(t2-t1).
You are a great professor!! You are great, Professor! Thanks!
Randall, thank you very much for the superb explanation! :D
Hey very quick question. So I understand that if we have a air standard analysis we cannot assume constant specific heat. However, if we have a cold air analysis, we can assume specific heat. Does this mean that I cannot use "k" with my isentropic equations (at 9:35) since it is the ratio of specific heat and I can't assume specific heat?
Or to make it more clear, I can't use the equations at 9:35 since I can't assume specific heat
Amazing video thank you
If i added turbocharger what it will be for the analysis?
thanks for the video first,
I know this might sound stupid to some, but i am dying to get a satisfying answer, at 11:08 you used the Cp version to extract the equation, why not Cv ?.
Also, what i took in Otto cycle, process isentropic---> q=0 and work = -Cv delta T, why Cv not Cp here.
+Seven4 Seven There are two general equations to calculate the change in entropy for an ideal gas assuming constant specific heats. Both equations are valid. There are also three equations relating T&p, T&v, or p&v, and are only valid for ideal gas undergoing an isentropic process. The vid concerns the Otto cycle, so the students are already familiar with these equations. At around 11:08, students are reminded how to move from s2-s1 =cp*ln(t2/t1)-R*ln(p2/p1) to the equation: T2/T1 = (p2/p1)^((k-1)/k) . Same can be said for moving from s2-s1 = cv*ln(t2/t1)+R*ln(v2/v1) to t2/t1 = (v1/v2)^(k-1).
For your second question, start from 1st Law for closed system undergoing a process. u2-u1 = q1-2 - w1-2. When q=0, w1-2 = u2-u1, not h2-h1. For ideal gas, u2=u1=cv*(t2-t1).
+Randall Manteufel thanks for fast reply.
I get now, but one more thing, when do we start with h2-h1 then ?
open systems use "h", also closed system will result in "h" for the case of positive pressure heating, as in the Diesel cycle
+Randall Manteufel ah i see, thanks very very much