Question 12.3: Gas Flow through a Converging–Diverging Duct Carbon dioxide ...

Carbon dioxide enters a varying cross-sectional area duct at specified conditions. The flow properties are to be determined along the duct.
Assumptions 1 Carbon dioxide is an ideal gas with constant specific heats at room temperature. 2 Flow through the duct is steady, one-dimensional, and isentropic.
Properties For simplicity we use c_{p}=0.846 kJ / kg \cdot K \text { and } k=1.289 throughout the calculations, which are the constant-pressure specific heat and specific heat ratio values of carbon dioxide at room temperatures. The gas constant of carbon dioxide is R = 0.1889 kJ/kg · K.
Analysis We note that the inlet temperature is nearly equal to the stagnation temperature since the inlet velocity is small. The flow is isentropic, and thus the stagnation temperature and pressure throughout the duct remain constant. Therefore,

T_{0} \cong T_{1}=200^{\circ} C =473 K

and

P_{0} \cong P_{1}=1400 kPa

To illustrate the solution procedure, we calculate the desired properties at the location where the pressure is 1200 kPa, the first location that corresponds to a pressure drop of 200 kPa.
From Eq. 12–5,

T=T_{0}\left(\frac{P}{P_{0}}\right)^{(k-1) / k}=(473 K )\left(\frac{1200 kPa }{1400 kPa }\right)^{(1.289-1) / 1.289}=457 K

\frac{P_{0}}{P}=\left(\frac{T_{0}}{T}\right)^{k /(k-1)}                       (12–5)

From Eq. 12–4,

\begin{aligned}V &=\sqrt{2 c_{p}\left(T_{0}-T\right)} \\&=\sqrt{2(0.846 kJ / kg \cdot K )(473 K -457 K )\left(\frac{1000 m ^{2} / s ^{3}}{1 kJ / kg }\right)} \\&=164.5 m / s\end{aligned}

T_{0}=T+\frac{V^{2}}{2 c_{p}}                       (12–4)

From the ideal-gas relation,

\rho=\frac{P}{R T}=\frac{1200 kPa }{\left(0.1889 kPa \cdot m ^{3} / kg \cdot K \right)(457 K )}=13.9 kg / m ^{3}

From the mass flow rate relation,

A=\frac{\dot{m}}{\rho V}=\frac{3 kg / s }{\left(13.9 kg / m ^{3}\right)(164.5 m / s )}=13.1 \times 10^{-4} m ^{2}=13.1 cm ^{2}

From Eqs. 12–11 and 12–12,

\begin{aligned}c &=\sqrt{k R T}=\sqrt{(1.289)(0.1889 kJ / kg \cdot K )(457 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)}=333.6 m / s \\Ma &=\frac{V}{c}=\frac{164.5 m / s }{333.6 m / s }=0.493\end{aligned}

c=\sqrt{k R T}                 (12–11)

Ma =\frac{V}{c}                   (12–12)

The results for the other pressure steps are summarized in Table 12–1 and are plotted in Fig. 12–13.
Discussion Note that as the pressure decreases, the temperature and speed of sound decrease while the fluid velocity and Mach number increase in the flow direction. The density decreases slowly at first and rapidly later as the fluid velocity increases.