Question 11.6: The objective of this example is to numerically verify the t......

Referring to the inlet velocity diagram in Figure 11.25, we get

W_{\theta\ 1}=U_{1}=209.1\,\mathrm{m/s}

T_{t\ r\ 1}=T_{t\ 1}+{\frac{\left(W_{\theta\ 1}{}^{2}-V_{\theta\ 1}{}^{2}\right)}{2c_{p}}}=T_{t\ 1}+{\frac{W_{\theta\ 1}{}^{2}}{2c_{p}}}=496.8\,\mathrm{K}

W_{c\ r\ 1}=\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ r\ 1}}=407.8\,\mathrm{m/s}

W_{1}=W_{c r\ 1}=407.8\,\mathrm{m/s}

V_{1}=V_{z1}={\sqrt{W_{1}{}^{2}-U_{1}{}^{2}}}=350.2\,{\mathrm{m/s}}

V_{c r\ 1}={\sqrt{\left({\frac{2\gamma}{\gamma+1}}\right)R T_{t\ 1}}}=405.1\,{\mathrm{m/s}}

M_{c r\ 1}={\frac{V_{1}}{V_{c r\ 1}}}=0.88

\rho_{1}=\frac{p_{t\ 1}}{R T_{t\ 1}}\biggl[1-\biggl(\frac{\gamma-1}{\gamma+1}\biggr)M_{c r\ 1}{}^{2}\biggr]^{\frac{1}{\gamma-1}}=0.976\,\mathrm{kg/m}^{3}

\dot{m}=\rho_{1}V_{z1}(2\pi r_{1}h_{1})=5.14\,\mathrm{kg/s}

Referring to the exit velocity diagram in Fig. 11.25, we can calculate the following exit variables:

V_{c r\ 2}=\sqrt{\left(\frac{2\gamma}{\gamma+1}\right)R T_{t\ 2}}=478.2\,{\mathrm{m/s}}

V_{2}=V_{c r\ 2}=478.2\ {\mathrm{m/s}}

(V_{\theta\ 2})_{i d.}=U_{2}=457.4\,\mathrm{m/s}

(V_{\theta\ 2})_{a c t.}=\sigma_{s}(V_{\theta\ 2})_{i d.}=423.6\,\mathrm{m/s}

V_{r\ 2}={\sqrt{V_{2}{}^{2}-V_{\theta\ 2}{}^{2}}}=222.0\,\mathrm{m/s}

PARABOLIC INTERPOLATION OF V_{r} AND V_{θ}:
Let us define the nondimensional radial coordinate \bar{r} as follows:

{\bar{r}}={\frac{r-r_{1}}{r_{2}-r_{1}}}

Using the impeller-exit magnitudes of V_{r} and V_{θ} as boundary conditions, we get the following two parabolic relationships:

V_{r}=222.0\ \bar{r}^{2}

V_{\theta}=423.6\ \bar{r}^{2}

Referring back to Example 7 in Chapter 9, and noting that the rotational speed ω is opposite to the θ direction, we can express the combined centripetal and Coriolis acceleration components as

(a)_{n e t}=[(\omega^{2}r)-2(\omega W_{\theta})]e_{r}+[2\omega W_{r}]e_{\theta}

Noting that W_{θ} = V_{θ} − ω_{r} , we can now calculate the net acceleration components at the following five radial locations:

(a_{r})_{\bar{r}=0.2}=3.14\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}\quad\mathrm{and}\quad(a_{\theta})_{\bar{r}=0.2}=0.08\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}

(a_{r})_{\bar{r}=0.4}=3.38\times10^{6}\,\mathrm{m/s^{2}}\quad\mathrm{and}\quad(a_{\theta})_{\bar{r}=0.4}=0.31\times10^{6}\,\mathrm{m/s^{2}}

(a_{r})_{\bar{r}=0.6}=3.30\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}\quad\mathrm{and}\quad(a_{\theta})_{\bar{r}=0.6}=0.70\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}

(a_{r})_{\bar{r}=0.8}=2.95\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}\quad\mathrm{and}\quad(a_{\theta})_{\bar{r}=0.8}=1.24\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}

(a_{r})_{\bar{r}=1.0}=2.29\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}\quad\mathrm{and}\quad(a_{\theta})_{\bar{r}=1.0}=1.94\times10^{6}\,\mathrm{m}/\mathrm{s}^{2}

A plot of these results is shown in Figure 11.26, where the following conclusions can be drawn:

• The radial-acceleration component continues to decline as the impeller exit station is approached. This is due to the decline in the Coriolis-acceleration radial component. As a result, a fluid particle in this region will continually lose radial momentum.
• Over the same exit subregion, the tangential component of Coriolis acceleration continuously grows, causing the tangential shift of a fluid particle that is sketched and labeled “particle trajectory” in Figure 11.26.
• Combination of the two flow behavioral characteristics (above) gives rise to the slip phenomenon, which was qualitatively discussed earlier in this chapter.