Question 14.3: A single-stage axial flow turbine has a mean radius of 30 cm......

1. The given angles are \alpha_2=65^\circ,\beta_2=25^\circ, \beta_3=60^\circ.
From Equation 14.1c,

\frac{U}{C_{\text{a}}} =\tan \alpha_2-\tan \beta_2=\tan \beta_3-\tan \alpha_3 \quad \quad \quad (14.1 \text{c}) \\ \tan \alpha_3=\tan \beta_3-\tan \alpha_2+\tan \beta_2=0.0539

Then

\alpha_3=3.08^\circ

2. Since C_{\text{a}}=C_2 \cos \alpha_2 = 600 cos 65 = 253.57 m/s
From Equation 14.1c

U=C_{\text{a}}(\tan \alpha_2-\tan \beta_2)=253.57 \times (\tan 65-\tan 25)=425.5 \text{ m/s}

But since U = πND/60, then

N = 60 × 425.5/π × 0.6 = 13,54576 rpm

3. From Equation 14.9c, with \beta_2 = 25°, β = 60°, and

\Delta T_{0\text{s}}=\frac{UC_{\text{a}}(\tan \alpha_2+\tan \alpha_3)}{C_{\text{p}}} =\frac{UC_{\text{a}}(\tan \beta_2+\tan \beta_3)}{C_{\text{p}}} \quad \quad \quad (14.9\text{c}) \\ Cp=\gamma R/(\gamma -1)=1168.8 \text{ J/kg}\cdot \text{K} \\ \Delta T_{0\text{stage}}=\frac{425.5 \times 253.37}{1168.8} (\tan 25+\tan 60)=202.77 \text{ K}

From Equation 14.14a, the stage pressure ratio is

\pi_{\text{stage}}=\frac{P_{01}}{P_{03}} =\frac{1}{(1-(\Delta T_{0\text{s}}/\eta_{\text{s}}T_{01}))^{\gamma/(\gamma-1)}} =\frac{1}{(1-(202.77/(0.88 \times 1200)))^4} =2.346

4. The flow coefficient Φ = C_{\text{a}}/U = 0.6.
The blade-loading coefficient ψ is given by the relation

\Psi=\frac{C_{\text{p}}\Delta T_{0\text{stage}}}{U^2} =\frac{1168.8 \times 202.77}{(425.5)^2} =1.309

The degree of reaction (Equation 14.32a)

\Lambda =\frac{C_{\text{a}}}{2U} (\tan \beta_3-\tan \beta_2) \quad \quad \quad (14.32\text{a}) \\ \Lambda =\phi(\tan \beta_3-\tan \beta_2)/2=0.41

5. From Equation 14.1c, \tan \beta_3-\tan \beta_3=U/C_{\text{a}}

\tan \alpha_3=\tan 60-\frac{425.5}{253.57} =3.1^\circ \\ C_3=\frac{C_{\text{a}}}{\cos \alpha_3} =\frac{253.57}{\cos 3.1} =253.94 \text{ m/s}

Assuming that C_1=C_3 , the static conditions at the stage inlet are calculated as follows:

T_1=T_{01}-\frac{C_1^2}{2C_{\text{p}}} =1200-\frac{(253.94)^2}{2 \times 1168.8} =1172.4 \text{ K} \\ P_1=P_{01}\left(\frac{T_1}{T_{01}} \right) ^{\gamma/(\gamma-1)}=1900 \times 10^3\left(\frac{1172.4}{1200} \right) ^4=1731 \times 10^3 \text{ Pa} \\ \rho_1=\frac{P_1}{RT_1} =\frac{1731 \times 10^3}{290 \times 1172.4} =5.09 \text{ kg/m}^3

The annulus area at inlet is A_1=\pi D_{\text{m}}h = π × 0.6 × 0.06 = 0.1131 m² .
The mass flow rate is \dot{m}=\rho_1 C_{\text{a}}A_1 = 5.09 × 253.57 × 0.1131 = 145.97 kg/m³ .
6. Power = \dot{m} C_{\text{p}} \Delta T_{0\text{stage}} = 145.97 × 1.1688 × 202.77 = 34600 kW
Power = 34.6 MW.