Velocity Vector Diagram

A turbine blade is an aerofoil-shaped body that turns and accelerates (or only turns) the flowing fluid. Cascades of such blades form the nozzle (stator) ring and the rotor ring of a stage.

     leading edge        camber line
         ______---------___
        /  ___-----------__ \__
       (  (_________________)  > trailing edge
        \_____________________/
        |<------- chord c ------>|

  cascade:   |<- pitch s ->|
          \     \     \          blade
           \     \     \         row
        ---> flow, angle measured from axial

Terms

• Chord ($c$): straight line joining the leading and trailing edges.
• Camber line: mean line midway between the pressure (concave) and suction (convex) surfaces. More camber means more turning.
• Pitch ($s$): distance between corresponding points of adjacent blades in the row.
• Stagger angle ($\xi$): angle between the chord and the axial direction.
• Blade inlet and outlet angles: angles of the camber line (tangent) at the leading and trailing edges. Flow angles $\alpha$, $\beta$ are the actual fluid directions; the difference is the incidence or deviation.
• Solidity ($\sigma = c/s$): ratio of chord to pitch; it fixes how well the blades guide the flow. Turbines use $\sigma \approx 1$ to 2.
• Aspect ratio: blade height to chord.
• Turning angle ($\theta$): $\beta_1 + \beta_2$ for the rotor, the total deflection of the flow.

Impulse and reaction profiles

Reaction blades give a smooth acceleration of the flow in the passage, so profile loss is lower, but leakage over the tips is higher because of the pressure difference across the rotor.

Derivation

Consider a rotor with fluid entering at radius $r_1$ with absolute velocity $C_1$ and leaving at $r_2$ with $C_2$. The whirl (tangential) components are $C_{w1}$ and $C_{w2}$. For a steady mass flow $\dot m$, the moment of momentum equation gives the torque on the rotor:

$$T = \dot m (r_1 C_{w1} - r_2 C_{w2})$$

Power $= T\omega$ and the blade speed is $U = \omega r$:

$$P = \dot m (U_1 C_{w1} - U_2 C_{w2})$$

Work per unit mass (Euler equation for a turbine):

$$\boxed{w = U_1 C_{w1} - U_2 C_{w2}}$$

For an axial machine $U_1 = U_2 = U$, so $w = U(C_{w1} - C_{w2})$. For a compressor or pump the sign is reversed: $w = U_2 C_{w2} - U_1 C_{w1}$.

Three energy components

From the velocity triangle, the cosine rule gives $U C_w = \tfrac12 (C^2 + U^2 - V_r^2)$, where $V_r$ is the relative velocity. Substituting at inlet and outlet:

$$w = \frac{C_1^2 - C_2^2}{2} + \frac{U_1^2 - U_2^2}{2} + \frac{V_{r2}^2 - V_{r1}^2}{2}$$

  inlet triangle           outlet triangle
       C1  Vr1                 C2  Vr2
        \  /|                   \  /|
         \/ | Ca1                \/ | Ca2
   ------U1--                ----U2--

1. $\dfrac{C_1^2 - C_2^2}{2}$ - change in absolute kinetic energy (impulse effect).
2. $\dfrac{U_1^2 - U_2^2}{2}$ - centrifugal effect, work from change of radius (zero in axial machines).
3. $\dfrac{V_{r2}^2 - V_{r1}^2}{2}$ - change of relative kinetic energy, the acceleration inside the rotor passage (reaction effect).

For an axial impulse turbine, terms 2 and 3 are (nearly) zero, and the work comes only from the first term. The equation holds whatever the losses, since it depends only on the velocity triangles.

Velocity diagrams

Steam leaves the nozzle at $C_1$ at angle $\alpha_1$ to the blade direction. Relative velocity at inlet $V_{r1}$ (angle $\beta_1$); at outlet $V_{r2}$ (angle $\beta_2$); absolute outlet velocity $C_2$.

 Inlet                      Outlet
      C1                       Vr2  C2
     /|\                        \   /|
    / | \ Ca1                    \ / | Ca2
   /a1| b1\Vr1                    \b2|
  +---U---+----                 ---U--+--
  |<--Cw1 ----->|            |<--Vw2-->|

Work done

Take the blade speed $U$ as positive. Whirl of the relative velocity at inlet: $V_{w1} = C_1\cos\alpha_1 - U$. At outlet the relative velocity has whirl $V_{w2} = V_{r2}\cos\beta_2$ in the direction opposite to $U$. With blade friction, $V_{r2} = k V_{r1}$ and $V_{r1}\cos\beta_1 = V_{w1}$.

Change of whirl velocity $= V_{w1} + V_{w2}$, so work per kg:

$$w = U (V_{w1} + V_{w2}) = U (C_1\cos\alpha_1 - U)\left(1 + k\frac{\cos\beta_2}{\cos\beta_1}\right)$$

Diagram efficiency

$$\eta_d = \frac{w}{C_1^2/2} = \frac{2U (C_1\cos\alpha_1 - U)(1 + kc)}{C_1^2}, \quad c = \frac{\cos\beta_2}{\cos\beta_1}$$

With blade speed ratio $\rho = U/C_1$:

$$\eta_d = 2\rho(\cos\alpha_1 - \rho)(1 + kc)$$

Condition for maximum

$$\frac{d\eta_d}{d\rho} = 2(1 + kc)(\cos\alpha_1 - 2\rho) = 0 \;\Rightarrow\; \rho = \frac{\cos\alpha_1}{2}$$

that is, $U = \tfrac12 C_1\cos\alpha_1 = \tfrac12 C_{w1}$. The second derivative is negative, so this is a maximum.

Maximum efficiency

$$\eta_{d,max} = \frac{\cos^2\alpha_1}{2}(1 + kc)$$

For a symmetrical blade ($\beta_1 = \beta_2$, so $c = 1$) with no friction ($k = 1$):

$$\boxed{\eta_{d,max} = \cos^2\alpha_1}$$

For $\alpha_1 = 20^\circ$ this is $0.883$. With friction ($k < 1$) the value is lower. This low optimum speed ($U \approx 0.47C_1$) is why a single-stage impulse turbine runs at very high rotor speed and why velocity compounding or pressure compounding is used.

Data: $\alpha_1 = 20^\circ$, $C_1 = 600$ m/s, $U = 250$ m/s, $k = 0.9$, $\dot m = 10$ kg/s. Symmetrical blades: $\beta_1 = \beta_2$.

(a) Inlet velocity triangle and blade angles

$$\begin{aligned} C_{w1} &= C_1\cos 20^\circ = 563.8\ \text{m/s}, \quad C_{a1} = C_1\sin 20^\circ = 205.2\ \text{m/s} \\ V_{w1} &= C_{w1} - U = 313.8\ \text{m/s} \\ V_{r1} &= \sqrt{V_{w1}^2 + C_{a1}^2} = 375.0\ \text{m/s}, \quad \tan\beta_1 = \frac{C_{a1}}{V_{w1}} \Rightarrow \beta_1 = 33.2^\circ \end{aligned}$$

Blade inlet angle = blade outlet angle = 33.2 degrees.

(b) Outlet velocity triangle

$$\begin{aligned} V_{r2} &= k V_{r1} = 0.9 \times 375.0 = 337.5\ \text{m/s} \\ V_{w2} &= V_{r2}\cos\beta_2 = 282.4\ \text{m/s (backward)}, \quad C_{a2} = V_{r2}\sin\beta_2 = 184.7\ \text{m/s} \\ C_{w2} &= V_{w2} - U = 32.4\ \text{m/s (backward, opposite to } U\text{)} \end{aligned}$$ $$C_2 = \sqrt{C_{w2}^2 + C_{a2}^2} = 187.5\ \text{m/s}$$

The absolute exit velocity is 187.5 m/s at an angle $\alpha_2 = \tan^{-1}(C_{a2}/|C_{w2}|) = 80.0^\circ$ to the blade direction.

(c) Power

$$w = U(V_{w1} + V_{w2}) = 250 \times (313.8 + 282.4) = 149.1\ \text{kJ/kg}$$ $$P = \dot m w = 10 \times 149.1 = 1491\ \text{kW}$$

(d) Diagram efficiency

$$\eta_d = \frac{w}{C_1^2/2} = \frac{149.1}{180.0} = 0.828$$

(e) Axial thrust

$$F_a = \dot m (C_{a1} - C_{a2}) = 10 \times (205.2 - 184.7) = 205\ \text{N}$$

Answer: blade angles $33.2^\circ$; exit velocity 187.5 m/s at $80.0^\circ$; power 1491 kW; $\eta_d$ = 82.8%; axial thrust 205 N.

Degree of reaction

The degree of reaction $R$ is the ratio of the enthalpy drop in the moving blades to the enthalpy drop in the whole stage:

$$R = \frac{h_2 - h_3}{h_1 - h_3} = \frac{\Delta h_{rotor}}{\Delta h_{stage}}$$

$R = 0$ for impulse; $R = 0.5$ for a Parsons turbine, where fixed and moving blades are identical and mirror images of each other.

Velocity diagrams (50% reaction)

Because fixed and moving blades are identical: $\alpha_1 = \beta_2$ and $\beta_1 = \alpha_2$, so $V_{r2} = C_1$ and $C_2 = V_{r1}$.

 Inlet                   Outlet
     C1                     Vr2=C1   C2=Vr1
    /|\                       \      /|
   / | \ Vr1                   \    / | Ca
  /a1|b1\                       \b2/  |
 +---U---+                      +--U--+

Work done per kg

$$C_{w1} = C_1\cos\alpha_1, \quad V_{w2} = V_{r2}\cos\beta_2 = C_1\cos\alpha_1$$ $$w = U(C_{w1} + V_{w2} - U) = U(2C_1\cos\alpha_1 - U)$$

Blade efficiency

Energy supplied = enthalpy drop in the fixed blades ($C_1^2/2$, entry velocity neglected) + enthalpy drop in the moving blades $(V_{r2}^2 - V_{r1}^2)/2$. This equals the work done plus the exit kinetic energy, $w + C_2^2/2$. With $C_2^2 = V_{r1}^2 = C_1^2 - 2UC_1\cos\alpha_1 + U^2$:

$$\text{Supply} = w + \frac{C_2^2}{2} = \frac{C_1^2}{2} + UC_1\cos\alpha_1 - \frac{U^2}{2}$$ $$\eta_b = \frac{2U(2C_1\cos\alpha_1 - U)}{C_1^2 + 2UC_1\cos\alpha_1 - U^2} = \frac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2}, \quad \rho = \frac{U}{C_1}$$

Condition for maximum

Putting $d\eta_b/d\rho = 0$ gives $\rho = \cos\alpha_1$, i.e. $U = C_1\cos\alpha_1$.

$$\boxed{\eta_{b,max} = \frac{2\cos^2\alpha_1}{1 + \cos^2\alpha_1}}$$

For $\alpha_1 = 20^\circ$, $\eta_{b,max} = 2(0.883)/(1.883) = 0.938$. This is higher than the single-stage impulse maximum $\cos^2\alpha_1 = 0.883$, and the optimum blade speed is about twice as high, so reaction stages are used in multistage turbines at moderate speed.

Data: $\alpha_1 = \beta_2 = 20^\circ$, $U = 200$ m/s, $\rho = U/C_1 = 0.7$, $\dot m = 12$ kg/s. For a Parsons stage: $V_{r2} = C_1$, $C_2 = V_{r1}$, $\beta_1 = \alpha_2$.

(a) Inlet velocity triangle

$$\begin{aligned} C_1 &= U/0.7 = 285.7\ \text{m/s} \\ C_{w1} &= C_1\cos 20^\circ = 268.5\ \text{m/s}, \quad C_{a} = C_1\sin 20^\circ = 97.7\ \text{m/s} \\ V_{w1} &= C_{w1} - U = 68.5\ \text{m/s} \\ \tan\beta_1 &= C_a/V_{w1} \Rightarrow \beta_1 = 55.0^\circ \end{aligned}$$ $$V_{r1} = \sqrt{68.5^2 + 97.7^2} = 119.3\ \text{m/s} = C_2$$

(b) Work and power

$$w = U(2C_1\cos\alpha_1 - U) = 200 \times (2 \times 268.5 - 200) = 67.4\ \text{kJ/kg}$$ $$P = 12 \times 67.4 = 809\ \text{kW}$$

(c) Energy supplied (entry velocity neglected)

$$\frac{C_2^2}{2} = \frac{119.3^2}{2} = 7.12\ \text{kJ/kg}$$ $$\Delta h = w + \frac{C_2^2}{2} = 67.4 + 7.12 = 74.5\ \text{kJ/kg}$$

(d) Blade efficiency

$$\eta_b = \frac{w}{\Delta h} = \frac{67.4}{74.5} = 0.904$$

Check with the formula $\eta_b = \dfrac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2} = \dfrac{1.4 \times 1.1794}{1.8256} = 0.904$.

Answer: $\beta_1 = 55.0^\circ$; $w = 67.4$ kJ/kg; $P = 809$ kW; $\Delta h = 74.5$ kJ/kg; $\eta_b = 90.4\%$.

Data: $U = 340$ m/s, $C_a = 250$ m/s, $\alpha_2 = 65^\circ$ (nozzle exit), $\alpha_3 = 0$, $T_{01} = 1200$ K, $\eta_{tt} = 0.88$. Stations: 1 nozzle inlet, 2 nozzle exit/rotor inlet, 3 rotor exit.

 Rotor inlet                 Rotor exit
      C2                         Vr3   C3=Ca
     /|\ Vr2                      \    |
    / | \                          \b3 |
   /65|b2\                          \  |
  +---U---+                       ---U-+

(a) Blade angles

$$C_{w2} = C_a\tan 65^\circ = 250 \times 2.1445 = 536.1\ \text{m/s}$$ $$\tan\beta_2 = \frac{C_{w2} - U}{C_a} = \frac{196.1}{250} \Rightarrow \beta_2 = 38.1^\circ$$

At exit $C_{w3} = 0$, so the relative whirl is $U$:

$$\tan\beta_3 = \frac{U}{C_a} = \frac{340}{250} \Rightarrow \beta_3 = 53.7^\circ$$

(b) Work and temperature drop

$$w = U(C_{w2} + C_{w3}) = 340 \times 536.1 = 182.3\ \text{kJ/kg}$$ $$\Delta T_0 = \frac{w}{c_p} = \frac{182.3}{1.148} = 158.8\ \text{K}$$

(c) Degree of reaction

$$R = \frac{C_a}{2U}(\tan\beta_3 - \tan\beta_2) = 1 - \frac{C_{w2}}{2U} = 1 - \frac{536.1}{680} = 0.212$$

(d) Pressure ratio

From $\eta_{tt} = \dfrac{\Delta T_0}{T_{01}(1 - (p_{03}/p_{01})^{(\gamma-1)/\gamma})}$:

$$\frac{p_{01}}{p_{03}} = \left[1 - \frac{\Delta T_0}{\eta_{tt}T_{01}}\right]^{-\gamma/(\gamma-1)} = \left[1 - \frac{158.8}{0.88 \times 1200}\right]^{-4.003} = 1.92$$

(e) Power

$$P = \dot m w = 25 \times 182.3 = 4557\ \text{kW} = 4.56\ \text{MW}$$

Stage loading $w/U^2 = 1.58$ and flow coefficient $C_a/U = 0.74$.

Answer: $\beta_2 = 38.1^\circ$, $\beta_3 = 53.7^\circ$; $w = 182.3$ kJ/kg, $\Delta T_0 = 158.8$ K; $R = 0.212$; $p_{01}/p_{03} = 1.92$; $P = 4.56$ MW.