Introduction of Turbomachime and Dynamic Action of Fluid

A turbomachine is a device in which energy is transferred continuously between a rotating element (rotor or runner) and a flowing fluid, through the dynamic action of the fluid on the blades. The change in the fluid's angular momentum gives the torque on the shaft.

Classification

1. By direction of energy transfer

• Power-producing (turbines): fluid gives energy to the rotor. Examples: water turbine, steam turbine, gas turbine, wind turbine.
• Power-absorbing (pumps, fans, blowers, compressors): the rotor gives energy to the fluid.

2. By the fluid handled

• Hydraulic (incompressible fluid, mainly water): water turbines, centrifugal pumps.
• Thermal (compressible fluid): steam and gas turbines, compressors, fans.

3. By the action of the fluid in the rotor

• Impulse type: the whole pressure drop occurs in the nozzle; the rotor works at constant (atmospheric) pressure. Example: Pelton wheel, De Laval steam turbine.
• Reaction type: pressure drops also in the rotor, so the rotor is full of fluid. Example: Francis and Kaplan turbines, Parsons turbine.

4. By the direction of flow through the rotor

  Axial            Radial           Tangential
  ||| -> |||       -> O <-          jet
                                      \  (  )
                                       -> (Pelton)

5. By the number of stages: single-stage and multistage (e.g. multistage boiler feed pump).

6. By shaft position: horizontal, vertical, inclined.

Both transfer energy between a machine and a fluid, but the mechanism is different. In a turbomachine the energy exchange is due to the dynamic action of the fluid on rotating blades, whereas in a positive displacement machine a fixed volume of fluid is trapped, moved and pushed out by a moving boundary.

In a turbomachine the delivered head depends on speed and discharge, so closing the delivery valve gives a limited head. A positive displacement pump will build up pressure until it fails, so it needs a relief valve.

Water power is one of the oldest sources of mechanical energy. The development went from simple wheels to the high-efficiency turbines used today.

Water wheels

• Horizontal (Norse or Greek) wheel: earliest form, with a vertical shaft; the stream strikes paddles directly. Low efficiency (below 15%), but no gearing needed to drive millstones.
• Vertical wheels (Roman, about 1st century BC): a horizontal shaft driven through gears.
  • Undershot wheel: water flows under the wheel, striking the lower blades. Used on low heads; efficiency about 20 to 30%.
  • Breast wheel: water enters at about the shaft level. Efficiency about 40 to 60%.
  • Overshot wheel: water falls into buckets at the top; works by weight of water. Efficiency up to 60 to 70%.
• Smeaton (1750s) tested wheels with models and showed that overshot wheels were better than undershot, starting scientific design.

Water turbines

Later improvements include the Banki (cross-flow) turbine for small plants, draft tubes, automatic governors and better materials. Modern turbines reach efficiencies of 90 to 95%, much higher than the early wheels.

Momentum principle

Newton's second law: the rate of change of linear momentum of a body is equal to the resultant force acting on it, and takes place in the direction of that force.

$$\vec F = \frac{d(m\vec V)}{dt}$$

If the force $F$ acts for a short time $dt$, then $F\,dt = d(mV)$. The left side is the impulse and the right side is the change of momentum. This is the impulse-momentum equation.

Control volume form (steady flow)

Consider a fixed control volume with one inlet (section 1) and one outlet (section 2), through which fluid of density $\rho$ flows steadily. The mass flow rate is $m = \rho Q = \rho A_1 V_1 = \rho A_2 V_2$.

In time $dt$ the mass $dm = \rho Q\,dt$ enters with velocity $V_1$ and the same mass leaves with velocity $V_2$. Momentum change of the fluid in the given direction $x$:

$$dm\,(V_{2x} - V_{1x}) = \rho Q\,dt\,(V_{2x}-V_{1x})$$

Dividing by $dt$, the net external force on the fluid in the control volume (pressure forces, body force and force by the solid surface) is

$$\sum F_x = \rho Q\,(V_{2x} - V_{1x}), \qquad \sum F_y = \rho Q\,(V_{2y} - V_{1y})$$

Force of jet on a stationary body

For a free jet, the pressure is atmospheric everywhere on the control surface, so pressure forces cancel and the body force is negligible. The only force on the fluid is the force $R$ exerted by the body on the fluid. By Newton's third law the force of the jet on the body is equal and opposite:

$$F_x = \rho Q\,(V_{1x} - V_{2x})$$

Hence the force on a body equals the rate at which the fluid loses momentum in that direction. Direction of the force is found by the sign of $(V_{1x}-V_{2x})$.

Special results for a jet of area $a$ and velocity $V$:

• Flat plate normal to jet (fluid leaves along plate, so $V_{2x}=0$): $F = \rho a V^2$.
• Symmetrical curved vane turning the jet through angle $\theta$ (measured from the jet direction): $F = \rho a V^2(1-\cos\theta)$.

These expressions are the basis for the analysis of Pelton wheel buckets and other impulse machines.

Given: $d = 0.05$ m, $V = 25$ m/s, $u = 8$ m/s, $\rho = 1000$ kg/m³.

Jet area: $a = \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}$ m².

Kinetic energy of jet per second (input power): $\frac12\rho a V^3 = 15340$ W.

(a) Stationary plate

The jet loses all its momentum normal to the plate.

$$F = \rho a V^2 = 1000\times 1.9635\times10^{-3}\times 25^2 = 1227.2\ \text{N}$$

(b) Single plate moving at 8 m/s

Only the relative velocity $(V-u)$ acts, and the mass hitting the plate per second is $\rho a (V-u)$ (the plate runs away from the jet, so some water is not caught).

$$\begin{aligned} F &= \rho a (V-u)^2 = 1000\times1.9635\times10^{-3}\times 17^2 = 567.5\ \text{N}\\ W &= F\,u = 567.5\times 8 = 4540\ \text{W}\\ \eta &= \frac{F u}{\tfrac12\rho a V^3} = \frac{4540}{15340} = 29.6\% \end{aligned}$$

(c) Series of plates on a wheel

With many plates, a plate is always in front of the jet, so the whole discharge $\rho a V$ strikes the plates, but each kilogram loses only the relative velocity $(V-u)$.

$$\begin{aligned} F &= \rho a V (V-u) = 1000\times1.9635\times10^{-3}\times25\times 17 = 834.5\ \text{N}\\ W &= F u = 6676\ \text{W}\\ \eta &= \frac{2u(V-u)}{V^2} = \frac{6676}{15340} = 43.5\% \end{aligned}$$

The maximum efficiency of a series of flat plates is 50 %, at $u = V/2$.

Answer: (a) 1227.2 N; (b) 567.5 N, 4540 W, 29.6 %; (c) 834.5 N, 6676 W, 43.5 %.

Given: $d=0.06$ m, $V=30$ m/s, deflection $\theta=160^\circ$, so the exit angle with the original jet direction is $180^\circ-20^\circ$ and the exit velocity component along the jet is $-V\cos20^\circ$ (opposite to the jet).

Jet area $a=\frac{\pi}{4}(0.06)^2=2.827\times10^{-3}$ m².

(a) Fixed vane

The vane is symmetrical and the jet splits into two equal halves, which leave with velocity $V$ (no friction, no change in speed). Change of velocity along the jet per kg: $V - (-V\cos20^\circ) = V(1+\cos20^\circ)$.

$$\begin{aligned} F &= \rho a V^2 (1+\cos 20^\circ)\\ &= 1000\times2.827\times10^{-3}\times 30^2\times(1+0.9397)\\ &= 4936\ \text{N} \end{aligned}$$

(b) Vane moving at 10 m/s in the jet direction

A single vane moving away receives mass flow $\rho a (V-u)$ per second, and the relative speed $(V-u)=20$ m/s is unchanged along the vane (no friction). Relative to the vane the jet is turned in the same way, so

$$\begin{aligned} F &= \rho a (V-u)^2 (1+\cos20^\circ)\\ &= 1000\times2.827\times10^{-3}\times 20^2\times(1+0.9397)\\ &= 2194\ \text{N}\\ P &= F\,u = 2194\times10 = 21937\ \text{W} = 21.94\ \text{kW} \end{aligned}$$

Answer: (a) 4936 N (fixed vane); (b) 2194 N and 21.94 kW (moving vane).

Euler's equation is the basic equation of all turbomachines. It follows from the angular momentum principle: the torque on the rotor equals the rate of change of angular momentum of the fluid passing through it.

Velocity triangles

At inlet (1) and outlet (2): $V$ = absolute velocity, $u$ = blade (peripheral) velocity, $V_r$ = relative velocity, $V_w$ = whirl component (along $u$), $V_f$ = flow component (radial for a radial machine), $\alpha$ = angle of $V$ with $u$, $\beta$ = angle of $V_r$ with $u$ (blade angle at inlet is $\theta$, at outlet $\phi$).

  Triangle (same form at inlet 1 and outlet 2)

              apex
             /|\
        V   / | \  Vr
           /  |  \
          /   |Vf \
         / a  |  b \
        *-----+-----*------> blade motion
        |<-Vw->|
        |<----- u ----->|

  V  = u + Vr (vector sum),  Vw = V cos a,  Vf = V sin a

Derivation

Let a mass flow $m=\rho Q$ pass through the runner per second. Inlet radius $r_1$, outlet radius $r_2$.

Angular momentum per second at inlet: $m V_{w1} r_1$. At outlet: $m V_{w2} r_2$.

Torque on the runner:

$$T = \rho Q\,(V_{w1} r_1 - V_{w2} r_2)$$

Angular velocity $\omega = u_1/r_1 = u_2/r_2$, so the power given to the runner is

$$P = T\omega = \rho Q\,(V_{w1} u_1 - V_{w2} u_2)$$

Work done per second per unit weight of water (weight flow $= \rho g Q$) is the Euler head:

$$\boxed{H_e = \frac{1}{g}\,(V_{w1} u_1 - V_{w2} u_2)}$$

Notes

• If the water leaves without whirl ($V_{w2}=0$, radial discharge), $H_e = V_{w1}u_1/g$.
• For an axial machine $u_1=u_2=u$, so $H_e = u(V_{w1}-V_{w2})/g$.
• $V_{w2}$ is taken negative if it is opposite to the blade motion.
• For a pump the sign is reversed: $H_e = (V_{w2}u_2 - V_{w1}u_1)/g$.
• Hydraulic efficiency of a turbine: $\eta_h = H_e/H$, where $H$ is the net head.

Given: $d=50$ mm, $V=V_1=40$ m/s, $u=15$ m/s (same at inlet and outlet), outlet vane angle $\phi=20^\circ$, no friction so $V_{r2}=V_{r1}$.

Jet area $a=\frac{\pi}{4}(0.05)^2=1.9635\times10^{-3}$ m². Because the vanes form a series on a wheel, the entire jet flow is used: $m=\rho a V=78.54$ kg/s.

Inlet triangle

The jet enters tangentially, so $V_{w1}=V=40$ m/s and $V_{r1}=V-u=25$ m/s.

Outlet triangle

 Outlet (vane moves right, water leaves backwards)

   Vr2 = 25 m/s
        \
         \ 20 deg
   <------*-------------> u = 15 m/s
   Vw2 = u - Vr2 cos20  (negative = backwards)

$V_{r2}=V_{r1}=25$ m/s. Component of $V_{r2}$ opposite to motion: $V_{r2}\cos20^\circ=23.49$ m/s.

$$V_{w2} = u - V_{r2}\cos\phi = 15 - 23.49 = -8.49\ \text{m/s}$$

The negative sign means the exit whirl is opposite to the direction of the vane. The change in whirl velocity is $V_{w1} - V_{w2} = 40 + 8.49 = 48.49$ m/s.

Force, power, efficiency

$$\begin{aligned} F &= \rho a V\,(V_{w1}-V_{w2}) = 78.54\times48.49 = 3809\ \text{N}\\ P &= F u = 3809\times15 = 57129\ \text{W} = 57.13\ \text{kW}\\ \text{Jet power} &= \tfrac12\rho a V^3 = 62.83\ \text{kW}\\ \eta &= \frac{P}{\text{jet power}} = \frac{57.13}{62.83} = 90.9\% \end{aligned}$$

Check with the formula $\eta=\dfrac{2u(V-u)(1+\cos\phi)}{V^2}=\dfrac{2\times15\times25\times(1+0.9397)}{1600}=90.9\%$, which agrees.

The remaining energy is lost as the kinetic energy of the water leaving the vanes, with $V_2=\sqrt{V_{w2}^2+V_{f2}^2}=12.05$ m/s.

Answer: Force $=3809$ N, power $=57.13$ kW, efficiency $=90.9\%$.