Steam turbine and Hydraulic machine

A steam nozzle is a passage of varying cross-section that converts the heat (enthalpy) energy of steam into kinetic energy by expansion from a high to a low pressure. Its main use is in steam turbines (to produce jets), and also in steam injectors, ejectors, and jet condensers.

Types

Also, nozzles may be classified by cross-section (circular, rectangular) and by arrangement (single, ring).

Exit velocity (steady flow energy equation)

For adiabatic flow with no work, negligible change in height, the SFEE between inlet (1) and exit (2) is

$$h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}$$

Neglecting inlet velocity ($V_1\approx0$):

$$V_2=\sqrt{2(h_1-h_2)}\ \ \text{m/s},\ \ h\text{ in J/kg}$$

(or $V_2=44.72\sqrt{h_1-h_2}$ with $h$ in kJ/kg). For an isentropic expansion with $pv^n=\text{constant}$ ($n=1.3$ for superheated, 1.135 for dry saturated steam, Zeuner's law), $h_1-h_2=\dfrac{n}{n-1}(p_1v_1-p_2v_2)$, so

$$V_2=\sqrt{\frac{2n}{n-1}\,p_1v_1\left[1-\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}\right]}$$

Mass flow and maximum discharge

Mass flow per unit exit area, using $v_2=v_1(p_1/p_2)^{1/n}$:

$$\frac{m}{A}=\frac{V_2}{v_2}=\sqrt{\frac{2n}{n-1}\frac{p_1}{v_1}\left[\left(\frac{p_2}{p_1}\right)^{\frac{2}{n}}-\left(\frac{p_2}{p_1}\right)^{\frac{n+1}{n}}\right]}$$

Let $r=p_2/p_1$. The flow per unit area is a maximum when $d(m/A)/dr=0$, i.e. when the bracket is a maximum:

$$\frac{2}{n}r^{\frac{2}{n}-1}-\frac{n+1}{n}r^{\frac{1}{n}}=0\ \Rightarrow\ r=\left(\frac{2}{n+1}\right)^{\frac{n}{n-1}}$$

This is the critical pressure ratio $p_c/p_1$, reached at the throat. At it, the velocity at the throat equals the local velocity of sound, and the nozzle is "choked": lowering the back pressure further cannot increase the mass flow through a convergent nozzle.

Throat velocity: $V_t=\sqrt{\dfrac{2n}{n+1}p_1v_1}$.

Use of the result: if the back pressure is above $p_c$, a simple convergent nozzle is used, and the exit pressure equals the back pressure. If the back pressure is below $p_c$, a convergent-divergent nozzle is required to expand the steam fully to the back pressure, with a supersonic jet in the divergent part.

Given: $p_1=10$ bar $=10^6$ Pa, $v_1=0.2327$ m³/kg, $n=1.3$, $p_2=1.5$ bar, $m=1.5$ kg/s.

$p_1v_1=232.7$ kJ/kg.

(a) Throat (critical) pressure

$$\frac{p_t}{p_1}=\left(\frac{2}{n+1}\right)^{\frac{n}{n-1}}=\left(\frac{2}{2.3}\right)^{4.333}=0.5457$$ $$p_t=0.5457\times10=5.46\ \text{bar}$$

Since $p_2/p_1=0.150<0.5457$, the exit pressure is below the critical pressure, so a convergent-divergent nozzle is needed.

(b) Throat velocity and area

$$V_t=\sqrt{\frac{2n}{n+1}p_1v_1}=\sqrt{\frac{2\times1.3}{2.3}\times10^6\times0.2327}=512.9\ \text{m/s}$$

Specific volume at throat:

$$v_t=v_1\left(\frac{p_1}{p_t}\right)^{1/n}=0.2327\times\left(\frac{1}{0.5457}\right)^{1/1.3}=0.3708\ \text{m}^3/\text{kg}$$ $$A_t=\frac{m\,v_t}{V_t}=\frac{1.5\times0.3708}{512.9}=10.84\ \text{cm}^2\quad\Big(d_t=37.2\ \text{mm}\Big)$$

(c) Exit conditions

$$V_2=\sqrt{\frac{2n}{n-1}p_1v_1\left[1-\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}\right]} =\sqrt{8.667\times10^6\times0.2327\times\left[1-0.6455\right]}=845.6\ \text{m/s}$$ $$v_2=v_1\left(\frac{p_1}{p_2}\right)^{1/n}=0.2327\times\left(\frac{10}{1.5}\right)^{1/1.3}=1.0013\ \text{m}^3/\text{kg}$$ $$A_2=\frac{m\,v_2}{V_2}=\frac{1.5\times1.0013}{845.6}=17.76\ \text{cm}^2\quad\Big(d_2=47.6\ \text{mm}\Big)$$

Area ratio $A_2/A_t=1.64$.

Answer: (a) $p_t=5.46$ bar; (b) $V_t=512.9$ m/s, $A_t=10.84$ cm²; (c) $V_2=845.6$ m/s, $A_2=17.76$ cm², $d_t=37.2$ mm, $d_2=47.6$ mm.

Given: $V_1=900$ m/s, $\alpha=20^\circ$, $u=350$ m/s, blades symmetrical ($\theta=\phi$), $k=V_{r2}/V_{r1}=0.9$, $m=2$ kg/s.

Inlet triangle

$$V_{w1}=V_1\cos20^\circ=845.7\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=307.8\ \text{m/s}$$ $$V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{495.7^2+307.8^2}=583.5\ \text{m/s}$$

(a) Blade angles

$$\tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{307.8}{495.7}\ \Rightarrow\ \theta=31.8^\circ,\qquad \phi=\theta=31.8^\circ$$

Outlet triangle

 Inlet                    Outlet
      V1  Vf1                  V2    Vf2
        \ |                      \   |
   20deg \|  Vr1            Vr2   \  |
 --------*---------> u    <-------*------- u
  |<-Vw1->|                |Vw2|

$$V_{r2}=0.9\times583.5=525.2\ \text{m/s},\quad V_{f2}=V_{r2}\sin\phi=277.0\ \text{m/s}$$ $$V_{w2}=V_{r2}\cos\phi-u=446.2-350=96.2\ \text{m/s (opposite to blade motion)}$$

The exit whirl is opposite to the blade motion, so it adds to the inlet whirl:

$$V_{w1}+V_{w2}=845.7+96.2=941.9\ \text{m/s}$$

(b) Force and power

$$F_t=m(V_{w1}+V_{w2})=2\times941.9=1884\ \text{N}$$ $$P=F_tu=1884\times350=659\ \text{kW}$$

(c) Axial thrust

$$F_a=m(V_{f1}-V_{f2})=2\times(307.8-277.0)=61.6\ \text{N}$$

(d) Blade efficiency

$$\eta_b=\frac{\text{work done}}{\text{kinetic energy supplied}}=\frac{2u(V_{w1}+V_{w2})}{V_1^2}=\frac{2\times350\times941.9}{900^2}=81.4\%$$

(Check: $\frac12mV_1^2=810$ kW and $P/\text{KE}=81.4\%$.)

Blade speed for maximum efficiency

For a single-stage impulse turbine with equiangular blades, $\eta_b$ is a maximum at

$$u_{opt}=\frac{V_1\cos\alpha}{2}=\frac{900\times\cos20^\circ}{2}=423\ \text{m/s}$$

and the maximum efficiency (with no friction) is $\cos^2\alpha=88.3\%$. The given 350 m/s is below this value, so the efficiency is lower than the maximum.

Answer: (a) $\theta=\phi=31.8^\circ$; (b) $F_t=1884$ N, $P=659$ kW; (c) $F_a=61.6$ N; (d) $\eta_b=81.4\%$; $u_{opt}=423$ m/s.

Need for compounding

In a simple impulse turbine (De Laval), the whole pressure drop occurs in one nozzle, so the steam leaves at a very high velocity (1000 m/s or more). For best efficiency the blade speed must be about half the steam velocity component, so the rotor speed is very high (20000 to 30000 rpm), which gives severe centrifugal stresses, large vibration and bearing problems, and the exit kinetic energy loss is large. The speed must also be reduced by gearing.

Compounding is the method of dividing the total pressure drop or velocity drop into several stages, so that the steam velocity and the blade speed in each stage is moderate. It gives lower rotor speed and better efficiency.

Types

1. Pressure compounding (Rateau turbine)

The turbine has several stages in series. Each stage has a nozzle (fixed row) and a row of moving blades. The total pressure drop is divided among the nozzles; in each stage the steam is expanded in the nozzle, and its velocity is absorbed fully in the moving blades.

 P  \
     \___                     Pressure falls in nozzles
         \___                 only; constant over blades
  V   /\  /\  /\             Velocity rises in nozzle,
     /  \/  \/  \            falls in blades (saw tooth)
  [N1][B1][N2][B2][N3][B3]

Used in medium capacity turbines; each stage is separated by a diaphragm with nozzles.

2. Velocity compounding (Curtis turbine)

The whole pressure drop occurs in one nozzle. The high-velocity steam then passes through several rows of moving blades, with rows of fixed (guide) blades between them to redirect steam to the next row. The velocity falls in each moving row.

 P  \____________________      Pressure drops in nozzle only
                                 and stays constant after
 V  /\                          Velocity rises in nozzle
   /  \  /\  /\                 and falls in the moving rows,
  /    \/  \/  \                 slightly in fixed rows
  [N] [M1][F][M2]

Gives a large work from one stage, but efficiency is lower than pressure compounding because friction in the blades is high. Often used as the first (governing) stage of a large turbine.

3. Pressure-velocity compounding

Combines both. The total pressure drop is divided into several stages, and in each stage a velocity compounded Curtis wheel (two or three rows of moving blades) is used. It gives a shorter turbine than pure pressure compounding for a given pressure drop.

Comparison

In an impulse turbine the steam expands completely in fixed nozzles and strikes the moving blades as a high-velocity jet; the pressure is the same at inlet and exit of moving blades. In a reaction turbine the steam expands in both the fixed blades (which act as nozzles) and the moving blades, so the pressure drops in the rotor also; the moving blades work by the reaction of the accelerated steam.

Classification of steam turbines

1. By action of steam: impulse, reaction, impulse-reaction (combined; usually impulse at the high-pressure end and reaction at low pressure).
2. By steam flow direction: axial, radial (Ljungstrom).
3. By exhaust conditions: condensing, non-condensing (back pressure), pass-out (extraction).
4. By number of stages: single or multistage.
5. By steam pressure at inlet: low, medium, high, supercritical.
6. By arrangement: single casing, compound (tandem or cross compound), and single or double flow.

Applications

• Thermal and nuclear power stations: large reaction/impulse-reaction condensing turbines driving alternators.
• Combined heat and power (industries, sugar mills, paper mills): back-pressure or pass-out turbines, which supply process steam.
• Marine propulsion and boiler feed pumps (with reduction gearing).
• Small impulse turbines for driving pumps, blowers and compressors.

Given: $D=1.0$ m, $N=3000$ rpm, $\alpha_1=\phi=20^\circ$, $V_1=250$ m/s, $m=10$ kg/s. For a Parsons turbine the moving blades are the mirror image of the fixed blades, so the velocity triangles are symmetrical: $V_{r2}=V_1$, $V_{r1}=V_2$ and the exit angle of moving blades $\phi=\alpha_1=20^\circ$.

Blade speed

$$u=\frac{\pi DN}{60}=\frac{\pi\times1.0\times3000}{60}=157.1\ \text{m/s}$$

Inlet triangle (leaving fixed blades, entering moving blades)

$$V_{w1}=V_1\cos20^\circ=234.9\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=85.5\ \text{m/s}$$ $$V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{77.8^2+85.5^2}=115.6\ \text{m/s}$$

(a) Inlet angle of moving blades

$$\tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{85.5}{77.8}\ \Rightarrow\ \theta=47.7^\circ$$

Outlet triangle

 Inlet (moving blade)          Outlet
      V1   Vf                      V2   Vf
        \  |  Vr1                   \   |   Vr2 = V1
   20 deg\ |/                         \  |  /
   -------*----> u                ----*-----> u  (Vr2 backward)

$V_{r2}=V_1=250$ m/s at $\phi=20^\circ$. Exit whirl (backward):

$$V_{w2}=V_{r2}\cos20^\circ-u=234.9-157.1=77.8\ \text{m/s}$$

Exit velocity of steam: $V_2=\sqrt{V_{w2}^2+V_f^2}=115.6$ m/s, (equal to $V_{r1}$, as expected).

(b) Power

Work per kg of steam:

$$W=u\,(V_{w1}+V_{w2})=157.1\times312.8=49.13\ \text{kJ/kg}$$ $$P=mW=10\times49.13=491\ \text{kW}$$

(c) Blade efficiency

Energy supplied $=$ work done $+$ kinetic energy carried away by steam leaving the stage $\left(\frac12V_2^2\right)$:

$$\frac12V_2^2=\frac{115.6^2}{2}=6.69\ \text{kJ/kg}$$ $$\eta_b=\frac{W}{W+\frac12V_2^2}=\frac{49.13}{49.13+6.69}=88.0\%$$

The kinetic energy leaving this stage is not wasted when the next stage follows; it is used again by the next fixed blades.

Answer: (a) $\theta=47.7^\circ$; (b) $P=491$ kW (49.13 kJ/kg); (c) $\eta_b=88.0\%$.