Page:EB1911 - Volume 14.djvu/114

Let V_i, V_o be the velocities of the wheel at the inlet and
outlet surfaces,
    v_i, v_o the velocities of the water,
    u_i, u_o the velocities of flow,
    v_ri, v_ro the relative velocities,
    h_i, h_o the pressures, measured in feet of water,
    r_i, r_o the radii of the wheel,
      α the angular velocity of the wheel.

At any point in the path of a portion of water, at radius r, the velocity v of the water may be resolved into a component V = αr equal to the velocity at that point of the wheel, and a relative component v_r. Hence the motion of the water may be considered to consist of two parts:—(a) a motion identical with that in a forced vortex of constant angular velocity α; (b) a flow along curves parallel to the wheel vane curves. Taking the latter first, and using Bernoulli’s theorem, the change of pressure due to flow through the wheel passages is given by the equation

h′_i + v_ri² / 2g = h′_o + v_ro² / 2g;
h′_i − h′_o = (v_ro² − v_ri²) / 2g.

The variation of pressure due to rotation in a forced vortex is

h″_i − h″_o = (V_i² − V_o²) / 2g.

Consequently the whole difference of pressure at the inlet and outlet surfaces of the wheel is

h_i − h_o = h′_i + h″_i − h′_o − h″_o
= (V_i² − V_o²) / 2g + (v_ro² − v_ri²) / 2g.

(17)

Case 1. Axial Flow Turbines.—V_i = V_o; and the first term on the right, in equation 17, disappears. Adding, however, the work of gravity due to a fall of d ft. in passing through the wheel,

h_i − h_o = (v_ro² − v_ri²) / 2g − d.

(17a)

Case 2. Outward Flow Turbines.—The inlet radius is less than the outlet radius, and (V_i² − V_o²)/2g is negative. The centrifugal head diminishes the pressure at the inlet surface, and increases the velocity with which the water enters the wheel. This somewhat increases the frictional loss of head. Further, if the wheel varies in velocity from variations in the useful work done, the quantity (V_i² − V_o²)/2g increases when the turbine speed increases, and vice versa. Consequently the flow into the turbine increases when the speed increases, and diminishes when the speed diminishes, and this again augments the variation of speed. The action of the centrifugal head in an outward flow turbine is therefore prejudicial to steadiness of motion. For this reason r_o : r_i is made small, generally about 5 : 4. Even then a governor is sometimes required to regulate the speed of the turbine.

Case 3. Inward Flow Turbines.—The inlet radius is greater than the outlet radius, and the centrifugal head diminishes the velocity of flow into the turbine. This tends to diminish the frictional losses, but it has a more important influence in securing steadiness of motion. Any increase of speed diminishes the flow into the turbine, and vice versa. Hence the variation of speed is less than the variation of resistance overcome. In the so-called centre vent wheels in America, the ratio r_i : r_o is about 5 : 4, and then the influence of the centrifugal head is not very important. Professor James Thomson first pointed out the advantage of a much greater difference of radii. By making r_i : r_o = 2 : 1, the centrifugal head balances about half the head in the supply chamber. Then the velocity through the guide-blades does not exceed the velocity due to half the fall, and the action of the centrifugal head in securing steadiness of speed is considerable.

Since the total head producing flow through the turbine is H − ɧ, of this h_i − h_o is expended in overcoming the pressure in the wheel, the velocity of flow into the wheel is

v_i = c_v √ {2g (H − ɧ − (V_i² − V_o² / 2g + (v_ro² − v_ri²) / 2g) ],

(18)

where c_v may be taken 0.96.

From (14a),

v_ro = V_o √ (1 + u_o² / V_o²).

It will be shown immediately that

v_ri = u_i cosec θ;

or, as this is only a small term, and θ is on the average 90°, we may take, for the present purpose, v_ri = u_i nearly.

Inserting these values, and remembering that for an axial flow turbine V_i = V_o, ɧ = 0, and the fall d in the wheel is to be added,

vi = cv ${\sqrt {\Big \{}}$ 2g ${\Big (}$ H −	Vi2	${\Big (}$ 1 +	uo2	${\Big )}$ +	ui2	− d ${\Big )}{\Big \}}$.
	2g		Vo2		2g

For an outward flow turbine,

vi = cv ${\sqrt {\Big [}}$ 2g { H − ɧ −	Vi2	( 1 +	uo2	${\Big )}$ +	ui2	${\Big \}}{\Big ]}$.
	2g		Vi2		2g

For an inward flow turbine,

vi = cv ${\sqrt {\Big [}}$ 2g ${\Big \{}$ H −	Vi2	${\Big (}$ 1 +	uo2	${\Big )}$ +	ui2	${\Big \}}{\Big ]}$.
	2g		Vi2		2g

§ 194. Angle which the Guide-Blades make with the Circumference of the Wheel.—At the moment the water enters the wheel, the radial component of the velocity is u_i, and the velocity is v_i. Hence, if γ is the angle between the guide-blades and a tangent to the wheel

γ = sin⁻¹ (u_i/v_i).

Fig. 196.

This angle can, if necessary, be corrected to allow for the thickness of the guide-blades.

§ 195. Condition determining the Angle of the Vanes at the Inlet Surface of the Wheel.—The single condition necessary to be satisfied at the inlet surface of the wheel is that the water should enter the wheel without shock. This condition is satisfied if the direction of relative motion of the water and wheel is parallel to the first element of the wheel vanes.

Let A (fig. 196) be a point on the inlet surface of the wheel, and let v_i represent in magnitude and direction the velocity of the water entering the wheel, and V_i the velocity of the wheel. Completing the parallelogram, v_ri is the direction of relative motion. Hence the angle between v_ri and V_i is the angle θ which the vanes should make with the inlet surface of the wheel.

§ 196. Example of the Method of designing a Turbine. Professor James Thomson’s Inward Flow Turbine.—

Let H = the available fall after deducting loss of head in pipes and channels from the gross fall;
   Q = the supply of water in cubic feet per second; and
  η = the efficiency of the turbine.

The work done per second is ηGQH, and the horse-power of the turbine is h.p. = ηGQH/550. If η is taken at 0.75, an allowance will be made for the frictional losses in the turbine, the leakage and the friction of the turbine shaft. Then h.p. = 0.085QH.

The velocity of flow through the turbine (uncorrected for the space occupied by the vanes and guide-blades) may be taken

u_i = u_i = 0.125 √2gH,

in which case about 1/64th of the energy of the fall is carried away by the water discharged.

The areas of the outlet and inlet surface of the wheel are then

2πr_od_o = 2πr_id_i = Q / 0.125 √ (2gH).

If we take r_o, so that the axial velocity of discharge from the central orifices of the wheel is equal to u_o, we get

r_o = 0.3984 √ (Q/√H),
d_o = r_o.

If, to obtain considerable steadying action of the centrifugal head, r_i = 2r_o, then d_i = 1/2d_o.

Speed of the Wheel.—Let V_i = 0.66 √2gH, or the speed due to half the fall nearly. Then the number of rotations of the turbine per second is

N = V_i / 2πr_i = 1.0579 √ (H √ H/Q);

also

V_o = V_ir_o / r_i = 0.33 √2gH.

Angle of Vanes with Outlet Surface.

Tan φ = u_o / V_o = 0.125 / 0.33 = .3788;
φ = 21º nearly.

If this value is revised for the vane thickness it will ordinarily become about 25º.

Velocity with which the Water enters the Wheel.—The head producing the velocity is

 H − (V_i² / 2g) (1 + u_o² / V_i²) + u_i² / 2g
= H {1 − .4356 (1 + 0.0358) + .0156}
= 0.5646H.

Then the velocity is

V_i = .96 √2g (.5646H) = 0.721 √2gH.

Angle of Guide-Blades.

Sin γ = u_i / v_i = 0.125 / 0.721 = 0.173;

γ = 10° nearly.;

Tangential Velocity of Water entering Wheel.

w_i = v_i cos γ = 0.7101 √2gH.

Angle of Vanes at Inlet Surface.

Cot θ = (w_i − V_i) / u_i = (.7101 − .66) / .125 = .4008;
θ = 68° nearly.

Hydraulic Efficiency of Wheel.

η = w_iV_i / gH = .7101 × .66 × 2
 = 0.9373.

This, however, neglects the friction of wheel covers and leakage. The efficiency from experiment has been found to be 0.75 to 0.80.

Impulse and Partial Admission Turbines.

§ 197. The principal defect of most turbines with complete admission is the imperfection of the arrangements for working with less than the normal supply. With many forms of reaction turbine the efficiency is considerably reduced when the regulating