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In the book of Kondepudi & Prigogine, Modern Theormodynamics, at page 87, in question 2.18, it is asked that

$$ \begin{array}{l}{\mathrm{O}_{2} \text { is flowing into a nozzle with a velocity } v_{\mathrm{i}}=50.0 \mathrm{m} \mathrm{s}^{-1} \text { at } T=300.0 \mathrm{K} \text { . The temperature of the gasflowing out of the nozzle is } 270.0 \mathrm{K} \text { (a) Assume the ideal gas law for the flowing gas and calculate }} \\ {\text { the velocity of the gas flowing out of the nozzle. (b) If the inlet diameter is } 5.0 \mathrm{cm}, \text { what is the outlet diameter? }}\end{array} $$

For the part $a-)$, I thought since there is no energy loss or gain is mentioned, the total energy of the substance must be preserve. Moreover, since the velocity is constant when the gas goes in/out, we can take a section of the nozzle, and consider the energy of the gas in that section. This implies that, for the gas whose volume is $\pi (2,5cm)^2 dx = dV$ has a total energy $$E = U + KE = 2.5 * NRT + 0.5 N(32g)v^2.$$

By energy conservation, this implies $$2,5 NR(300K) + 0.5 N (32g) (50m/2)^2 = 2,5 NR(270K) + 0.5 N(32g)v_f^2,$$ hence we can determine the velocity of the gas when it goes out of the nozzle; I've found this quantity as ~$203m/s$.

However, I have almost no idea why we can obtain the radius of the nozzle (part b-)). I mean, knowing the internal energy allows us to calculate $pV$, and from that, if we know the pressure, we can determine the radius from the volume, but I don't see any way to calculate $p$ without knowing $V$.

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You should be using the open system (control volume) version of the 1st law to solve the first part. For this system, it reduces $$\dot{m}\Delta h+\dot{m}\Delta \left(\frac{v^2}{2}\right)=0$$where h is the enthalpy per unit mass and $\dot{m}$ is the mass flow rate. From this, it follows that $$\Delta h+\Delta \left(\frac{v^2}{2}\right)=0$$For an ideal gas, the change in enthalpy per unit mass is given by $$\Delta h=\frac{C_p}{M}\Delta T$$where Cp is the molar heat capacity at constant pressure and M is the molar mass. So, in your equations, the 2.5 should be a 3.5.

In part b, if the nozzle is operating adiabatically and reversibly, you also have $$C_pdT=\frac{RT}{P}dP$$From this, you can get the pressure ratio.

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  • $\begingroup$ Thanks for your answer @Chet ; however, I'm little confused: why are you using enthalpy instead of energy in your first equation ? $\endgroup$
    – Our
    Commented Sep 1, 2019 at 16:46
  • $\begingroup$ Also, for the part b, how did you obtained that last equality ? $\endgroup$
    – Our
    Commented Sep 1, 2019 at 16:47
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    $\begingroup$ I'm using enthalpy in the first equation because it is called for in the open system (flow) version of the first law of thermodynamics. The work in pushing gas into and out of the section of nozzle has been lumped into this term in the conventional derivation of the open system version of the 1st law. Didn't they cover this in your course? In part b, we use $dh=C_pdT=Tds+vdP$. For an adiabatic reversible nozzle, Tds = 0, and, from the ideal gas law, v=RT/P. $\endgroup$
    – Chet Miller
    Commented Sep 1, 2019 at 19:27
  • $\begingroup$ I see, thanks for the response @Chet. By the way, I didn't take a course on theormodnamics, nor do I plan to; I'm selft-studying the subject. $\endgroup$
    – Our
    Commented Sep 1, 2019 at 19:33
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    $\begingroup$ OK. Most good thermo textbooks contain a section on the open system version of the 1st law. I particularly like Fundamentals of Engineering Thermodynamics by Moran et al. $\endgroup$
    – Chet Miller
    Commented Sep 1, 2019 at 19:40

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