LAB - RADIAL FLOW REACTION TURBINE

Objectives:

To measure volume flow rate, input head, hydraulic input power, torque, brake power & turbine efficiency when the brake force is increase

Theoretical Background:

Radial flow

When a fluid flows radially inwards or outwards from a centre, between two parallel planes as in Fig. 6.21, the streamlines will be radial straight lines and the streamtubes will be in forms of sectors. The area of flow will therefore increase as the radius increases, causing the velocity to decrease. Since the flow pattern is symmetrical, the total energy per unit weight H will be the same for all streamlines and for all points along each streamlines if we assume that there is no loss of energy.
[pic]

Figure 6.21
If v is the radial velocity and p is the pressure at any radius r,

H = p/ρg + v²/2g = constant …………………………(6.31)

Applying the continuity of flow equation flow and assuming that the density of the fluid remains constant, as would be the case for the fluid,

Volume rate of flow, Q = area x velocity = 2πrb x v

where b is the distance between the planes. Thus,

v = Q/2πrb

and substituting in equation………(6.31)

p/ρg + Q²/ 8π²r²b² = H p = ρg [ H – ( Q²/ 8π²b² ) x ( 1/r² )] …………………..(6.32)

If the pressure p at any radius r is plotted in Fig. 6.21(c), the curve will be parabolic and is sometimes referred to as Barlow’s curve.

If the flow discharges to the atmosphere at the periphery, the pressure at any points between the plates will be below atmospheric; there will be a force tending to bring the two plates together and so shut off flow. This phenomenon can be observed in a case of a disc valve. Radial flow under the disc will cause the disc to be drawn down onto the valve seating. This will cause the flow to stop, the pressure between the plates will return to atmospheric and the static pressure of the fluid on the upstream side of the disc will push it off its seating again. The disc will tend to vibrate on the seating and the flow will be to intermittent.

Flow in a curved path. Pressure gradient and change of total energy across the streamlines.

Velocity is a vector quantity with both magnitude and direction. When a fluid flows in a curved path, the velocity of the fluid along any streamline will undergo a change due to its change of direction, irrespective of any alteration in magnitude which may also occur. Considering the streamtube (shown in fig. 6.22),

[pic]

fig. 6.22:

as the fluid flows round the curved there will be a rate of change of velocity, that is to say an acceleration, towards the centre of curvature of the streamtube. The consequent rate of change of momentum of the fluid must be due, in accordance with Newton’s second law, to a force acting radially across the streamlines resulting from the difference of pressure between the sides BC and AD of the of the streamtube element.

In Fig. 4.2, suppose that the control volume ABCD subtends an angle δθ at the centre of curvature O, has length δs in the direction of flow and thickness b perpendicular to the diagram. For the streamline AD, let r be the radius of curvature, p the pressure and v the velocity of the fluid. For the streamline BC, the radius will be r + δr, the pressure p + δp and the velocity v + δv, where δp is the change of pressure in a radial direction.

From the velocity diagram,

Change of velocity in radial direction, δv = v δθ

Or since δθ = δs/r,

Radial change of velocity = v δs/r between AB and CD

Mass per unit time flowing = mass density x area x velocity through streamtube = ρ x ( b x δs ) x v

Change of momentum per unit = mass per unit time x radial change of time in radial direction velocity = ρbδrv² δs/r ………………………..…6.33

this rate of change of momentum is produced by the force due to the pressure difference between faces BC and AD of the control volume:

force = [(p + δp) – p ] bδs

equating equations (6.33) and (6.34), according to Newton’s second law,

δpbδs = ρbδrv²δs/r δp/δr = ρv²/r

for an incompressible fluid, ρ will be constant and equation (6.35) can be expressed in the of the pressure head h. since p = ρgh, we have δp = ρgδh.
Substituting in equation (6.35),

pg δh/δr = ρv²/r, δh/δr = v²/gr,

or, in the limit as δr tends to zero,

rate of change of pressure head in radial direction

= dh/dr = v²/gr ……………………………………(6.36)

to produce the curve flow shown in Fig. 6.22, we have seen that there must be a change of pressure head in a radial direction. However, since the velocity v along the streamline AD is different from the velocity v + δv along BC, there will also be a change in the velocity head from one streamline to another:

rate of change of velocity head radially:

= [(v + δv)² - v²] / 2gδr = v/g x δv/δr, neglecting products of small quantities, = v/g x δv/δr, as δr tends to zero. ……………….(6.37)

in streamlines are in a horizontal plane, so that changes changes in potential head do not occur, the change of total head H – i.e. the total energy per unit weight – in a radial direction, δH/δr, is given by,

δH/δr = change of pressure head + change of velocity head.

Substituting from equations (6.36) and (6.37), in the limit,

Change of total energy with radius, dH/dr = v²/gr + (v/g) x (dv/dr)

dH/dr = v²/gr + (v/g) x (dv/dr) ……………………(6.38)

the term (v/r + dv/dr) is also known as the vorticity of the fluid.
In obtaining equation 6.38, it has been assumed that the streamlines are horizontal, but this equation also applies to cases where the streamlines are inclined to the horizontal, since the fluid in the control volume is in effect weightless, being supported vertically by the surrounding fluid.

If the streamlines are straight lines, r = œ and dv/dr = 0. From equation (6.38) for a stream of fluid in which the velocity is uniform across the cross-section, and neglecting friction we have dH/dr = 0 and the total energy per unit weight H is constant for all points on all streamlines. This applies whether the streamlines are parallel or inclined, as the case of radial flow.

Procedures:

1) The model of the RADIAL FLOW REACTION TURBINE which is connected to the computer system is used to run the experiment. Lab assistant showed how to use the machine.
2) The machine is switched on; knob is used to adjust the brake force until we got the required steady value as shown on the monitor.
3) When the value of brake force is steady, the value of volume flow rate, input head, hydraulic input power, torque, brake power and turbine efficiency measured by computer will be taken.
4) The brake force is increased by adjusting the knob.
5) The procedures of 3 to 4 are repeated until we got the table of graph.
6) We can get the required value calculated by the computer automatically.

Sample of calculation:
Comparison tables:

Given: g = 9.81 m/s d = 0.09 m Cd = 0.63 r = 0.024m ρw = 998.2

Volume flow rate, Q v = Cd π d 2 √( 2 ρw d Po ) = (0.63) (π) (0.09)2 √(2(998.2)(34.5) 4πw 4 π (2 π (154.37)

= 0.000345 m3 s

Impact head, H1 = P1 = 243089 = 24.824m ρw g (998.2)(9.81)

Hydraulic Input Power, Ph = ρw g Qv H1 = (998.2) (9.81) (0.000345)(24.824) = 80.947 watt

Torque, T = Fb r = 0.44 x 0.024 = 0.01 Nm

Brake power,Pb = 2 π n = 2 π (154.37) (0.01) = 9.699 watt

Turbine efficiencies, Et(%) = Pb x 100 Ph = 9.699 x 100 80.947

= 11.98%

Discussion:

From our observation of the experiment result, it is found that the increasing of the break forces, Fb the rotational speed of turbine, Hz is also increase .The increasing of rotational speed of turbine caused by the increasing of the input head, Hi, slowly. The turbine efficiency, Et, is increase gradually. The hydraulic input Power, Ph , decrease gradually. The torque, T, of turbine is increase slowly. However the brake power, Pb , increase gradually and volume flow rate, Qv decrease slowly. All the data we get are from the computer. But from the theoretical calculation, we found small different value compared with computer data. We assume the error is come from sensor of equipments. As we know before, this equipment has some problem.

Conclusion:

From the experiment we can conclude that the rotational speed of turbine, Hz and the increase proportionally with brake power, Pb ,efficiency,Et, input head, Hi, and torque, T, but decreased proportionally with volume flow rate, Qv and hydraulic input power,Ph .The theoretical data has a small different value with the experimental value caused equipment error.

Reference:

1) Fluid Mechanics (4th edition), John F. Douglas, Janusz M. Gasiorek, John A. Swaffield, Prentice Hall 2001.

2) Fluid Mechanics (4th edition), Frank M. White, McGraw Hill International Edition.