Recent content by TimoD

Yup I get that as well. Ok well since our method is sound I will note this in my assignment and see what happens. Thanks once again for all your help

Oh right I see, I misread p0 as pout. Oh I thought that the average may be above even though the instantaneous is always below pin. With a 200kPa exit pressure though, I get an entropy change of 797.7J/K. Maybe my integral value is off? For that I get quite a high 674.4J/K.

Thank you! Could you possibly just clarify for me one last time; in post #11 where you derive an expression for pin, I see it comes from the equation ##Cp(T)\frac{dT}{T}=R\frac{dp}{p}##, which comes from ##ds = Cp(T)\frac{dT}{T}-R\frac{dp}{p}## where change in entropy is zero. Why is this so...

Yes that's the one sorry. Chestermiller thank you so much for your step by step help through all of this as well as your patience through all my blunders. I really appreciate it! God bless you sir

Yes I think the values are close enough. Ok so if I use that parallelogram approximation to find the integral, then I can solve for pin? What about the iterative method that the question speaks of?

Great! I have the values in a table, and I think I see where the iterative part comes in. Now how to numerically integrate? No thank you!

Ok great I think I understand all that, I just didn't distribute the R into the brackets. Where to from here? What do we do with the sTin in the integral?

After simplifying I get $$dS=-dmΔs=R(\ln(p_{out}) - \ln(p_{0}) - \frac{s_{T_{in}}-s_{T_{0}}}{R})dm$$

Ah yes of course! My mistake. Ok so I have an expression for pin, and I also know that ##\Delta s = \int\frac{Q}{T} + S_{gen} = 100J/K## since the process is adiabatic. I'm still struggling though to find an expression for dm that I can use in our equation...

Yes exactly that. Sorry still getting used to the syntax for equations

In part 2.1 I used ##\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}##, and then substituted standard entropy values for the cp integral, following which I solved for sT2 (because everything else is known, and ##\Delta s= 0##) and then used tabulated data to interpolate and solve for T2.

Ok I think I see where this is going, but I can't seem to find a way to write Tin in terms of pin without using the ideal gas equation? As for the integral of cp with temperature dependence, we have an appendix of tabulated standard entropy values which we are encouraged to use due to their...

##m=\frac{p_{in}V}{RT_{in}}##

Well if enthalpy is strictly a function of temperature then it seems temperature is constant through the valve also. Is assuming this a bit of an approximation, since h = u + Pv, clearly pressure and volume have an influence too? In that case though, ##\Delta s=-R\ln \frac{p_{out}}{p_{in}}##...

Ok: I know how to find change in specific entropy using a Gibbs equation manipulation ( Δs = ∫cvdT/T + Rln(v2/v1) ). Then, using the first law for a single flow system, I get that hin = hout. I think it is correct to assume an adiabatic process here? So I guess change in enthalpy is zero. As...