Question 5.5: Figure 5.15 shows a schematic of an adiabatic single-stage t......

Part a:

V_{1}=M_{c}r_{1}\,V_{c}r_{1}=529.0\,\mathrm{m/s}

U_{m}=\omega r_{m}=356.0\,\mathrm{m/s}

V_{\theta\ 1}=V_{1}\sin\alpha_{1}=505.9\,\mathrm{m/s}

V_{z}=V_{1}\cos\alpha_{1}=154.7\,\mathrm{m/s}

W_{\theta\ 1}=V_{\theta\ 1}-U_{m}=149.9\,\mathrm{m/s}

W_{1}=\sqrt{W_{\theta\ 1}{}^{2}+V_{z}{}^{2}}=215.4\,\mathrm{m/s}

T_{t\ r\ 1}=\,T_{t\ r\ 2}=\,T_{t\ r}=\,T_{t\ 1}-\left(\frac{V_{1}{}^{2}-W_{1}{}^{2}}{2c_{p}}\right)=1169.1\ K (axial stage)

p_{t\ r\ 1}=p_{t\ 1}\biggl(\frac{T_{t\ r\ 1}}{T_{t\ 1}}\biggr)^{\frac{\gamma}{\gamma-1}}=8.31\,\,\mathrm{bars}

Counting the total relative pressure loss, we have

p_{t\ r\ 2}=(1-0.062)p_{t\ r\ 1}=7.79\,\mathrm{bars}

W_{c r_{2}}=\sqrt{\frac{2\gamma}{(\gamma+1)R}T_{t\ r\ 2}}=618.9\,\mathrm{m/s}=W_{2} (rotor is choked)

\beta_{2}=\cos^{-1}\!\left(\frac{V_{z}}{W_{2}}\right)=-75.5^{\circ}

The negative sign (above) was inserted in view of Figure 5.15. Note that any traditionally designed turbine stage will always give rise to a negative rotor-exit relative flow angle.
Now, applying the continuity equation at the rotor-exit station in the rotating frame of reference, using relative flow properties, we have

\frac{\dot m\sqrt{T_{t\ r\ 2}}}{p_{t\ r\ 2}[\pi\left(r_{t\ {2}}{}^{2}-r_{h_{2}}{}^{2}\right)\cos\beta_{2}]}=\sqrt{\frac{2\gamma}{(\gamma+1)R}}\bigg(\frac{W_{2}}{W_{c r\ 2}}\bigg)\bigg[1-\frac{\gamma-1}{\gamma+1}\bigg(\frac{W_{2}}{W_{c r\ 2}}\bigg)^{2}\bigg]^{\frac{1}{\gamma-1}}

Note that we could, at least theoretically, apply the continuity equation, at the same station, in the stationary frame of reference (using absolute flow properties). However, we first would have had to translate the rotor-choking statement in terms of absolute flow properties (i.e., V_{2}, α_{2}, etc.).
Substituting in the chosen version of the continuity equation (above), we get

{\dot{m}}=4.27\,{\mathrm{kg/s}}

The “corrected” temperature θ_{0} and pressure  δ_{0} for the turbine stage can be calculated as

{\theta_{0}=T_{t\ 0}/T_{S T P}=4.41}

{\delta_{0}=p_{t\ 0}/p_{S T P}=11.6}

Now, the corrected mass-flow rate (\dot m_{C}) and corrected speed (N_{C}) can be calculated:

{\dot{m}}_{C}={\frac{\dot m{\sqrt{\theta_{0}}}}{\delta_{0}}}=0.773\,\mathrm{kg/s}

N_{C}={\frac{N}{{\sqrt{\theta_{0}}}}}\approx20,000\,\mathrm{rpm}

Using the provided turbine map, we get

{\frac{p_{t\ 0}}{p_{t\ 2}}}=2.38

Note that this is on the high side of the total-to-total pressure ratio for an axial-flow turbine. Therefore, and despite the moderate-to-high mass-flow rate, the specific speed  N_{s} (to be computed later) may be too low for an axial-flow turbine.
Let us now calculate the rotor-exit thermophysical properties:

W_{\theta\ 2}=W_{2}\sin\beta_{2}=-599.2\,\mathrm{m/s}

V_{\theta\ 2}=W_{\theta\ 2}+U_{m}=-243.2\,\mathrm{m/s}

Applying the Euler/energy-conservation relationship, we have

c_{p}(T_{t\ 1}-T_{t\ 2})=w_{s}=U_{m}(V_{\theta\ 1}-V_{\theta\ 2})

which, upon substitution, yields

T_{t\ 2}=1039.5\ \mathrm{K}

We are now in a position to calculate the stage total-to-total efficiency as follows:

\eta_{T}=\frac{1-(T_{t\ 2}/T_{t\ 0})}{1-(p_{t\ 2}/p_{t\ 0})^{\frac{\gamma-1}{\gamma}}}=93.7%

Part b: In order to calculate the specific speed, we first have to calculate the static magnitude of the rotor-exit density, as follows:

V_{2}={\sqrt{V_{\theta\ 2}{}^{2}+V_{z}{}^{2}}}=288.2

V_{c r_{2}}={\sqrt{\left({\frac{2\gamma}{\gamma+1}}\right)R T_{t\ 2}}}=583.6\,{\mathrm{m/s}}

M_{c r_{2}}=\frac{V_{2}}{V_{c r_{2}}}=0.494

\rho_{2}=\left(\frac{p_{t\ 2}}{R T_{t\ 2}}\right)\left(1-\frac{\gamma-1}{\gamma+1}M_{c r_{2}}{}^{2}\right)^{\frac{1}{\gamma-1}}=1.467\ \mathrm{kg/m}^{3}

Finally, we can calculate the stage specific speed (N_{s}) by direct substitution in expression (5.28) as follows:

N_{s}={\frac{N{\sqrt{\frac{\dot m}{\rho e x}}}}{(\Delta h_{t\ i d})^{\frac{3}{4}}}}   (5.28)

N_{s}=0.609\ \mathrm{radians}

Referring to Figure 5.9, we see that this N_{s} magnitude qualifies the stage to be of the radial type.