Question 14.2: A single-stage axial flow gas turbine has the following data......

The gas properties are

C_{\text{P}}=1148 \text{ J/kg}  ·  \text{K}, \quad R=287 \text{ J/kg}  ·  \text{K}, \quad \gamma=\frac{4}{3}.

1. The blade loading is

\psi=\frac{C_{\text{P}}\Delta T_0}{U^2} =\frac{1148 \times 144}{298^2} \\ \psi=1.8615

2.  T_{02}=T_{01}=1100 \text{ K}

T_{03}=T_{01}-\Delta T_0=1100-144=956 \text{ K} \\ \frac{P_{03}}{P_{01}} =\left[1-\frac{\Delta T_0}{\eta_{\text{t}} T_{01}} \right] ^{\gamma/(\gamma-1)}=\left[1-\frac{144}{0.9 \times 1100} \right] ^4 =0.533

The pressure ratio in the turbine is

\frac{P_{01}}{P_{03}} =1.875, \quad \text{or} \quad P_{03}=1.813 \text{ bar}

3. Since the nozzle is chocked,

(i) C_2=\sqrt{\gamma RT_2}

(ii) \frac{T_{02}}{T_2} =\frac{\gamma+1}{2}=1.165

The static temperature at the nozzle outlet is then T_2 = 944.2 K.
The absolute speed of the gases leaving the choked nozzle is C_2=\sqrt{\gamma RT_2} = 600.3 m/s.
Since the nozzle losses are expressed by the relation

T_2-T^\prime_2=\lambda_{\text{N}}\frac{C_2^2}{2C_{\text{P}}} =\frac{0.05 \times 600^2}{2 \times 1148} =7.839 \text{ K} \\ T^\prime_2= 936.4 \text{ K}

The pressure ratio is

\frac{P_{01}}{P_2} =\left(\frac{T_{01}}{T^\prime_2} \right) ^{\gamma/(\gamma-1)}=\left(\frac{1100}{936.4} \right) ^4=1.904 \\ P_2=1.785 \text{ bar} \\ C_{\text{a}2}=U \phi=298 \times 0.95=283 \text{ m/s} \\ \cos \alpha_2 =\frac{C_{\text{a}2}}{C_2} =\frac{283}{600} =0.4716 \\ \alpha_2=62^\circ

Since

\frac{U}{C_{\text{a}}} =\tan \alpha_2- \tan \beta_2=\frac{1}{\phi} \\ \tan \beta_2 = \tan \alpha_2-\frac{1}{\phi} =0.828 \\ \beta_2=39.6^\circ

The specified work W_{\text{s}} is expressed as

W_{\text{s}} =C_{\text{P}} \Delta T_0=U( C_{\text{u}2}C_{\text{u}3})=UC_{\text{a}}(\tan \alpha_2+\tan \alpha_3) \\ \tan \alpha_3=\frac{C_{\text{P}}\Delta T_0 }{UC_{\text{a}}} – \tan \alpha_2 =\frac{1148 \times 144}{298 \times 283} -1.8807=0.0793 \\ \alpha_3=4.54^\circ \\ \frac{U}{C_{\text{a}}}=\tan \beta_3 -\tan \alpha_3 \\ \tan \beta_3=\frac{1}{\phi} +\tan \alpha_3=1.0526+0.0794=1.132 \\ \beta_3=48.54^\circ