SOLUTION 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 temperature. 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,
\frac{d A}{A}=\frac{d P}{\rho}\left(\frac{1}{V^{2}}-\frac{d \rho}{d P}\right) (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
From Eq. 12–4,
\frac{d P}{\rho}+V d V=0 (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 \cong 164 m / s\end{aligned}
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{3.00 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,
c=\sqrt{k R T} (Eq. 12-11)
Ma =\frac{V}{c} (Eq. 12-12)
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
The results for the other pressure steps are summarized in Table 12–1 and are plotted in Fig. 12–7.
TABLE 12–1 |
Variation of fluid properties in flow direction in the duct described in Example 12–2 for m⋅ = 3 kg/s = constant |
P, kPa |
T, K |
V, m/s |
𝜌, kg/m3 |
c, m/s |
A, cm2 |
Ma |
1400 |
473 |
0 |
15.7 |
339.4 |
∞ |
0 |
1200 |
457 |
164.5 |
13.9 |
333.6 |
13.1 |
0.493 |
1000 |
439 |
240.7 |
12.1 |
326.9 |
10.3 |
0.736 |
800 |
417 |
306.6 |
10.1 |
318.8 |
9.64 |
0.962 |
767* |
413 |
317.2 |
9.82 |
317.2 |
9.63 |
1.000 |
600 |
391 |
371.4 |
8.12 |
308.7 |
10.0 |
1.203 |
400 |
357 |
441.9 |
5.93 |
295.0 |
11.5 |
1.498 |
200 |
306 |
530.9 |
3.46 |
272.9 |
16.3 |
1.946 |
* 767 kPa is the critical pressure where the local Mach number is unity. |
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.