Intermediate Thermodynamics Questions & Answers  
Question by Student 201427145
I would like to ask about applying polytropic process when I get the heat transferred to air in Assignment Q#6-(b). $\delta$Q, dV, dT, dP is not Zero. So I assumed it's a Polytropic Process that various processes were mixed. Given that $T_1$=294K, $V_1=0.0283m^2$, $V_2=0.038m^3$(It coule be induced through 6-(a)), $P_1=103kN/m^2, P_2=206kN/m^2, n_{air}$=1.2mole(From Ideal gas law), $C_{v.air}$=717J/kgK
Definition of Polytropic Process:
$P_1V_1^n=P_2V_2^n$---n=-2.352(It's negative)
Ideal Gas Law-$PV=n\bar{R}T$:
1st Law of Thermodynamics:
d(me)+PdV=$\delta$Q ($\delta$W=0 because it's frictionless)
Recall, $PV^n=C(Polytropic), u=c_vT$(Calorically Perfect gas) and Integrate on both side:
It's 1.5% error with answer even though Polytropic exponent is negative. In the wikipedia, negative exponents reflect a process where work and heat flow simultaneously in or out of the system. For this reason, Can not I apply the polytropic process?
It's a good effort but your solution and the answer are off. You can not assume $n$ is constant in this process for the air. As the heat is added, $n$ will vary. You need to rather set the pressure of the argon equal to the one of the air. If you assume a polytropic process for the air, I'll give you A0 at the most. Also, please use \$\$ instead of \$ for long mathematical expressions — this will make your post easier to read.
Question by Student 201900041
Hello professor, I would like to ask you a question about assignement #4. In the question #6, it is asked to find the temperature at the exit of the duct and then the heat transfer. So I started with the conservation of energy equation but because it not equal to 0 (because of the heat) I suppose I have to find the temperature seperatly and then substitute it to equation in order to find the heat transfer. But I guess it can't be as simple as P = pRT. Could you give me a hint of the assumption that can be make ? Also, during the lecture, we only studied examples where the velocity of fluid was assumed as low, so it simplified a lot the equation (uniform property among the volume). But which speed do you consider as low and what is limit for which you can not do this assumption?
You should take into consideration kinetic energy (flow speed) when it's possible to do so. Thus, if there is enough information to determine velocity of the gas, then take into account kinetic energy. Otherwise, there is no choice but to neglect it.
Question by Student 201900041
Professor, for the question #6 Assignement 4, can we make the Steady State assumption ? So the system is independant regarding to the time? Thanks for your previous quick answer.
As much as possible try to set things up so that the system is at steady-state — this will simplify things. Thus, when solving a problem choose the reference frame of the control volume such that the properties don't vary in time anywhere within that volume.
Question by Student 201327106
Professor I have solved all problems needed to be solved my own. But when it comes to problems you did not give answer, I am not sure what I did is totally correct.
So can you give me answer of #4, #5 of Assign. 1, #7 of Assign. 2, and #4, #5 of Assign. 3?
Thank you.
These problems are easier than those I gave answers to. You should be able to get the right answer on your own. There are no answers when solving new engineering problems in industry: it's good to start practicing for these situations now.
Question by Student 201327106
Firstly, for non-ideal gas e.g.2 with the generated compressibility charts, you used below equations. $$P_{c}= \rho _{c}RT_{c}, \, {v_{c}}'=\frac{RT_{c}}{P_{c}}$$ But at table A-1, there is $Z_{c}$. Why didn't you use below equations? $$P_{c}= Z_{c}\rho _{c}RT_{c},\,{v_{c}}'=\frac{Z_{c}RT_{c}}{P_{c}}$$ $$\\$$ Secondly, why is there prime mark with ${v_{c}}'$? $$\\$$ Thank you and have a nice holiday!
But $v_c \neq v_c^\prime$: $$ v_c = \frac{Z_c R T_c}{P_c} $$ and $$ v_c^\prime \equiv \frac{R T_c}{P_c} $$
Question by Student 201427145
Professor, I have thought about Thermally Equilibrium that you had mentioned at last class. We can assume initially Diaphram X(closed system) divided in two Volume by membrane and each has Gas A, B. And then membrane has eliminated and Gas A and B mixed perfectly. Diaphram X, gas A, B have energy of $$E_{X}, E_{A}, E_{B}$$ each other. First, $$Define:\Omega(E)\equiv$$[Number of Microstate according to Energy]. If system has Energy(E), what we can see is macrostate but it is one of a lot of microstate case of probability distribution. Macrostate can be easily occured at E which makes Maximum number of Microstate. $$ \therefore\frac{d\Omega(E_{X})}{dE}=0 $$ Here Maximum Number of Microstate occurs at Energy of X, A and B. For the convenient, Let's focus on Energy of A. $$ \therefore\frac{d\Omega(E_{X})}{dE_{A}}=0 $$ Note: Closed system- $$ E_{X}=E_{A}+E_{B} \Rightarrow \Omega(E_{X})=\Omega(E_{A}+E_{B}) $$ State A and B is independent- $$ \Omega(E_{A}+E_{B})=\Omega(E_{A})\Omega(E_{B})\Rightarrow \Omega(E_{X})=\Omega(E_{A})\Omega(E_{B}) $$ $$\therefore \frac{d\Omega(E_{x})}{dE_{A}}=\frac{d\Omega(E_{A})\Omega(E_{B})}{dE_{A}}=\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{A}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{B}}\frac{dE_{B}}{dE_{A}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=0 $$ $$ Note: E_{X}=E_{A}+E_{B}=const\Rightarrow dE_{A}+dE_{B}=0 \Rightarrow\frac{dE_{B}}{dE_{A}}=-1 $$ $$\therefore -\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{B}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=0 $$ $$ \frac{1}{\Omega(E_{A})}\frac{d\Omega(E_{A})}{dE_{A}}=\frac{1}{\Omega(E_{B})}\frac{d\Omega(E_{B})}{dE_{B}} $$ $$Note: \frac{dln{\Omega(E)}}{d\Omega(E)}=\frac{1}{\Omega(E)} \Rightarrow\frac{d\Omega(E)}{\Omega(E)}=dln{\Omega(E)} $$ $$ \therefore\frac{dln{\Omega(E_{A})}}{d\Omega(E_{A})}=\frac{dln{\Omega(E_{B})}}{d\Omega(E_{B})} $$ $$Define: \frac{1}{k_{B}T}\equiv\frac{dln{\Omega(E)}}{d\Omega(E)} $$ where $$ k_{B}$$ is Boltzmann constant $$ \therefore\frac{dln{\Omega(E_{A})}}{d\Omega(E_{A})}=\frac{dln{\Omega(E_{B})}}{d\Omega(E_{B})}=\frac{1}{k_{B}T} $$ $$\therefore T_{A}=T_{B}=T-Answer $$
This is a good effort. It's a bit unclear whether $$ \therefore\frac{d\Omega(E_{X})}{dE_{A}}=0 $$ is correct. But it's on the right track.
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