Water Turbine

The Pelton wheel is a tangential-flow impulse turbine for high heads (above about 250 m) and small discharge. All the available head is converted into kinetic energy in the nozzle, and the jet at atmospheric pressure strikes buckets on the wheel rim.

        Penstock
           |
        [Spear valve]---> nozzle
              \  jet
               \             _
   Casing       \     ()  ()   ) buckets
                 ===>(  WHEEL  )
                      ()  ()
                    ______________
                    Tailrace (open)
   Deflector plate hinged at nozzle exit

Main components

1. Penstock: brings water from the reservoir to the nozzle.
2. Nozzle with spear (needle) valve: the nozzle forms a high-velocity jet. A conical spear moved axially by the governor changes the annular exit area, hence the discharge, while the jet velocity stays almost the same.
3. Runner and buckets: a disc carries double-hemispherical (twin cup) buckets on its rim. A splitter ridge in the middle divides the jet into two halves that leave from the sides. Each bucket has a notch (cut-out) so that the jet enters it freely and the following bucket does not cut off the jet too early. Buckets are of cast iron, bronze or stainless steel.
4. Casing: an airtight cover that prevents splashing and guides the water down to the tailrace. It does not run full of water.
5. Brake nozzle: a small jet applied to the back of the buckets to stop the wheel quickly after shutdown.
6. Deflector (jet deflector): when the load drops suddenly, it deflects the jet away from the buckets without closing the spear quickly, which avoids water hammer in the penstock.

Working

1. Water of net head $H$ flows through the penstock to the nozzle; the pressure head turns into jet velocity $V_1 = C_v\sqrt{2gH}$, $C_v\approx0.97$ to $0.99$.
2. The jet strikes the splitter at the centre of the bucket and is divided into two streams, which turn through about $165^\circ$ (not $180^\circ$, so that the water leaving a bucket does not hit the next bucket).
3. The change of momentum of the jet produces a tangential force on the buckets; the wheel rotates and drives the generator shaft.
4. Water leaves with low velocity and falls into the tailrace. The wheel is placed above the tailrace level, so a draft tube is not used.

Maximum efficiency occurs when the bucket speed is about 0.45 to 0.46 of the jet speed ($u\approx\tfrac12 V_1$ for a frictionless 180° bucket). Overall efficiency is 85 to 90 %.

Governing is by moving the spear (change of discharge) and, for sudden load rejection, the deflector.

Given: $H_g=450$ m, $L=2000$ m, $D=0.6$ m, $f=0.016$, $Q=0.5$ m³/s, $C_v=0.98$, $\phi=0.46$, $D_w=1.4$ m, deflection $165^\circ$, $k=0.95$ ($V_{r2}=kV_{r1}$), $\eta_m=0.97$.

(a) Net head

Velocity in penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=\dfrac{0.5}{0.2827}=1.768$ m/s.

$$h_f=\frac{fLV_p^2}{2gD}=\frac{0.016\times2000\times1.768^2}{2\times9.81\times0.6}=8.50\ \text{m}$$ $$H = H_g - h_f = 450 - 8.50 = 441.50\ \text{m}$$

(b) Jet diameter

$$V_1 = C_v\sqrt{2gH} = 0.98\times\sqrt{2\times9.81\times441.50} = 0.98\times93.07 = 91.21\ \text{m/s}$$ $$a=\frac{Q}{V_1}=5.482\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=83.5\ \text{mm}$$

(c) Speed of wheel

$$u=\phi\sqrt{2gH}=0.46\times93.07=42.81\ \text{m/s},\qquad N=\frac{60u}{\pi D_w}=\frac{60\times42.81}{\pi\times1.4}=584\ \text{rpm}$$

(d) Power developed and hydraulic efficiency

Velocity triangles: $V_{w1}=V_1=91.21$ m/s; $V_{r1}=V_1-u=48.40$ m/s; $V_{r2}=0.95V_{r1}=45.98$ m/s.

The bucket turns the jet by $165^\circ$, so the relative exit velocity makes $180-165=15^\circ$ with the direction opposite to the bucket motion.

$$V_{w2}=u-V_{r2}\cos15^\circ=42.81-44.41=-1.60\ \text{m/s (backwards)}$$ $$V_{w1}+|V_{w2}|=91.21+1.60=92.81\ \text{m/s}$$ $$\begin{aligned} P&=\rho Q\,u\,(V_{w1}+|V_{w2}|)=1000\times0.5\times42.81\times92.81=1986.7\ \text{kW}\\ \text{Water power}&=\rho g Q H=2165.6\ \text{kW}\\ \eta_h&=\frac{1986.7}{2165.6}=91.7\% \end{aligned}$$

(e) Shaft power

$$P_s=\eta_m P=0.97\times1986.7=1927.1\ \text{kW}$$

Answer: (a) 441.50 m; (b) 83.5 mm; (c) 584 rpm; (d) 1986.7 kW, $\eta_h=91.7\%$; (e) 1927.1 kW.

Given: $P=6000$ kW, $H=350$ m, $N=500$ rpm, $C_v=0.98$, $\phi=0.46$, $\eta_o=0.86$, number of jets $n=2$.

Discharge

$$\eta_o=\frac{P}{\rho g Q H}\ \Rightarrow\ Q=\frac{6\times10^{6}}{1000\times9.81\times350\times0.86}=2.032\ \text{m}^3/\text{s}$$

Jet velocity and bucket speed

$\sqrt{2gH}=\sqrt{2\times9.81\times350}=82.87$ m/s.

$$V_1=C_v\sqrt{2gH}=81.21\ \text{m/s},\qquad u=\phi\sqrt{2gH}=38.12\ \text{m/s}$$

Wheel (pitch circle) diameter

$$D=\frac{60u}{\pi N}=\frac{60\times38.12}{\pi\times500}=1.456\ \text{m}$$

Jet diameter

Discharge per jet $=1.016$ m³/s.

$$a=\frac{Q_j}{V_1}=12.511\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=126.2\ \text{mm}$$

Jet ratio $m=D/d=11.5$, which lies in the normal range of 10 to 14. Good.

Number of buckets

$$Z=15+\frac{D}{2d}=15+\frac{11.5}{2}=20.8\ \Rightarrow\ Z=21\ \text{(rounded up) per wheel}$$

Specific speed check

$$N_s=\frac{N\sqrt{P}}{H^{5/4}}=\frac{500\sqrt{6000}}{350^{1.25}}=25.6\ \text{for the whole wheel}$$

Based on the power per jet (3000 kW), $N_s=18.1$, which is within the Pelton range of about 10 to 35 (metric units, kW). The design is suitable.

Answer: $Q=2.032$ m³/s; wheel diameter $=1.456$ m; jet diameter $=126.2$ mm (two jets); buckets $=21$; $N_s=18.1$ per jet.

A Francis turbine is an inward radial (mixed flow in modern designs) reaction turbine used for medium heads (about 40 to 600 m) and medium discharge. Part of the head is converted to velocity in the guide vanes and the rest in the runner, so the pressure falls from inlet to outlet of the runner and the runner is full of water.

          Penstock
              |
       +------v-------+   Spiral casing
       |   ___ ___    |   (volute)
       |  /  \ /  \   |
       | | R  U  N | <-- guide vanes
       |  \__/ \__/   |      (wicket gates)
       +------+-------+
              | Shaft up
              | runner
          ====v====
         Draft tube (diverging)
              \     /
               \___/  Tailrace

Main parts

1. Spiral (scroll) casing: its cross-section decreases along the flow so that water enters the guide vanes with uniform velocity all around.
2. Stay ring and stay vanes: carry the load of the casing and guide water to the guide vanes.
3. Guide (wicket) vanes: adjustable vanes that give water the correct angle at runner inlet and regulate discharge by turning about their pivots (operated by the governor).
4. Runner: a wheel with 16 to 24 curved blades fixed between a crown and a band. Water enters radially inward and leaves axially.
5. Draft tube: an expanding pipe connecting the runner exit to the tailrace.

Working

Water from the penstock enters the casing, passes the guide vanes with high whirl velocity, and moves inward through the runner blades. The change of angular momentum, together with the drop of pressure in the runner (reaction), exerts torque on the shaft. The water leaves nearly axially with little whirl.

Function of draft tube

• Allows the runner to be placed above the tailwater level, for easy inspection, without losing head.
• Converts the kinetic energy at the runner exit into pressure energy by gradually increasing the area, thus lowering the pressure at the runner exit below atmospheric and increasing the effective head across the runner.
• Discharges water to the tailrace below the tailwater level, so the whole head between the runner and tailrace is used.

Efficiency of a Francis turbine is about 90 to 94 % at the design point.

Given: $H=45$ m, $N=400$ rpm, $D_1=1.0$ m, $D_2=0.5$ m, $B_1=0.12$ m, $\eta_h=0.93$, $V_{w2}=0$, $V_{f1}=V_{f2}=7$ m/s.

Blade speeds

$$u_1=\frac{\pi D_1N}{60}=\frac{\pi\times1.0\times400}{60}=20.94\ \text{m/s},\qquad u_2=\frac{\pi\times0.5\times400}{60}=10.47\ \text{m/s}$$

Inlet whirl velocity

For $V_{w2}=0$: $\eta_h=\dfrac{V_{w1}u_1}{gH}$

$$V_{w1}=\frac{0.93\times9.81\times45}{20.94}=19.60\ \text{m/s}$$

(a) Guide vane angle

$$\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{7}{19.60}\ \Rightarrow\ \alpha=19.7^\circ$$

(b) Runner vane angles

Since $V_{w1}<u_1$, the whirl component of the relative velocity ($V_{w1}-u_1$) is opposite to $u_1$, so the blade angle $\theta$ (measured from the direction of $u_1$) is obtuse, greater than 90 degrees.

$$\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{7}{1.34}\ \Rightarrow\ 180^\circ-\theta=79.1^\circ,\ \ \theta=100.9^\circ$$

At outlet ($V_{w2}=0$):

$$\tan\phi=\frac{V_{f2}}{u_2}=\frac{7}{10.47}\ \Rightarrow\ \phi=33.8^\circ$$

(c) Discharge

$$Q=\pi D_1B_1V_{f1}=\pi\times1.0\times0.12\times7=2.639\ \text{m}^3/\text{s}$$

(d) Width at outlet

Continuity: $\pi D_1B_1V_{f1}=\pi D_2B_2V_{f2}$, with $V_{f1}=V_{f2}$:

$$B_2=\frac{D_1B_1}{D_2}=\frac{1.0\times0.12}{0.5}=0.24\ \text{m}=240\ \text{mm}$$

(e) Power developed by runner

$$P=\rho Q\,V_{w1}u_1=1000\times2.639\times19.60\times20.94=1083.4\ \text{kW}$$

Check: water power $\rho gQH=1165.0$ kW, and $1083.4/1165.0=93.0\%=\eta_h$ (satisfied).

Answer: (a) $\alpha=19.7^\circ$; (b) $\theta=100.9^\circ$, $\phi=33.8^\circ$; (c) $2.639$ m³/s; (d) 0.24 m; (e) 1083.4 kW.

The Kaplan turbine is an axial-flow reaction turbine used for low heads (up to about 60 m) and very large discharge. Its runner looks like a ship's propeller, with blades that can be rotated about their own axes.

        Spiral casing
        +---------------+
        |  guide vanes  |  wicket gates
        |  ===========  |
        |      ||       |   water turns 90 deg
        |      \/       |   and flows axially
        |   [HUB]       |
        |  / | | | \    |  <- adjustable blades
        +--------+------+
                 |
            Draft tube

Construction

• Scroll casing and guide vanes like a Francis turbine; water enters radially and is turned through 90 degrees by the vanes before reaching the runner (a space without vanes, between guide vanes and runner, helps the whirl to become free-vortex).
• Runner: a hub (boss) carrying 3 to 8 blades. The hub is streamlined and is about 0.35 to 0.6 of runner diameter.
• Blade turning mechanism: inside the hub, a servo-motor operated by the governor turns the blades.
• Draft tube: usually elbow type; very important for recovering the exit kinetic energy.

Working

Water flows axially through the runner blades. The guide vane opening and the blade angle are adjusted together by the governor with the load, so the water always enters the blades without shock. The pressure drops across the runner, and the axial change of whirl momentum gives torque. Exit is axial and without whirl at design point.

Advantages

• High efficiency (about 90 %) over a wide range of load, since both guide vane and blade angles adjust. The efficiency curve is flat.
• High specific speed, so a small turbine and generator, and a high rotational speed, suit low head.
• Suitable for sites with variable head and discharge.

Kaplan versus propeller turbine

Given: $P=9000$ kW, $H=20$ m, $\eta_o=0.88$, $\eta_h=0.93$, $D_o=4.0$ m, $D_b=1.8$ m, $N=120$ rpm, $V_{w2}=0$, $u_1=u_2$ (axial flow).

Discharge

$$Q=\frac{P}{\rho g H\eta_o}=\frac{9\times10^{6}}{1000\times9.81\times20\times0.88}=52.13\ \text{m}^3/\text{s}$$

Flow velocity

Annular flow area:

$$A=\frac{\pi}{4}(D_o^2-D_b^2)=\frac{\pi}{4}(4.0^2-1.8^2)=10.022\ \text{m}^2$$ $$V_{f1}=V_{f2}=\frac{Q}{A}=\frac{52.13}{10.022}=5.201\ \text{m/s}$$

Blade speed at mean diameter

$D_m=\dfrac{4.0+1.8}{2}=2.90$ m

$$u_1=u_2=\frac{\pi D_mN}{60}=\frac{\pi\times2.90\times120}{60}=18.22\ \text{m/s}$$

Whirl velocity at inlet

$$\eta_h=\frac{V_{w1}u_1}{gH}\ \Rightarrow\ V_{w1}=\frac{0.93\times9.81\times20}{18.22}=10.014\ \text{m/s}$$

Angles

Guide vane angle:

$$\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{5.201}{10.014}\ \Rightarrow\ \alpha=27.4^\circ$$

Runner blade angle at inlet. Here $V_{w1}<u_1$, so the blade angle is obtuse:

$$\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{5.201}{8.21}\ \Rightarrow\ 180^\circ-\theta=32.4^\circ,\ \ \theta=147.6^\circ$$

Runner blade angle at outlet:

$$\tan\phi=\frac{V_{f2}}{u_2}=\frac{5.201}{18.22}\ \Rightarrow\ \phi=15.9^\circ$$

(Angles of blades measured from the direction of blade motion. Because $u$ changes from hub to tip, the blade is twisted; the angles at the other radii are found in the same way with the local $u$.)

Answer: $Q=52.13$ m³/s; $V_f=5.201$ m/s; guide vane angle $27.4^\circ$; blade angle $147.6^\circ$ at inlet and $15.9^\circ$ at outlet.

Definition: The specific speed $N_s$ of a turbine is the speed of a geometrically similar turbine that develops 1 kW of power under a head of 1 m.

Derivation

For geometrically similar turbines, using the dimensionless groups with $D$ = diameter, $N$ = speed, $H$ = head:

• Head coefficient: $\dfrac{gH}{N^2D^2}=\text{constant}\ \Rightarrow\ D\propto\dfrac{\sqrt H}{N}$
• Power coefficient: $\dfrac{P}{\rho N^3D^5}=\text{constant}\ \Rightarrow\ P\propto N^3D^5$

Substituting $D$:

$$P\propto N^3\left(\frac{\sqrt H}{N}\right)^5=N^{-2}H^{5/2}$$

so

$$P\propto\frac{H^{5/2}}{N^2}\ \Rightarrow\ N\propto\frac{\sqrt P}{H^{5/4}}$$

Writing the constant of proportionality as $N_s$ (the value when $P=1$ kW and $H=1$ m):

$$\boxed{N_s=\frac{N\sqrt{P}}{H^{5/4}}}$$

with $N$ in rpm, $P$ in kW (shaft power) and $H$ in m (net head). (If $P$ is in horse power, as in some old books, the numbers differ.)

For a Pelton wheel with more than one jet, $P$ is the power per jet when comparing a single-jet equivalent.

Typical ranges (metric units: rpm, kW, m)

Significance

• It is a type number for the turbine: it identifies the shape of the runner (low Ns: impulse or narrow radial; high Ns: axial).
• Helps to select the type of turbine for given head, power and speed.
• Allows comparison and performance prediction of different turbines, since two similar turbines have the same Ns.
• Gives the speed at which the turbine will run: higher Ns means a higher speed, smaller machine and cheaper generator for the same power.

Given (prototype): $N=140$ rpm, $H=30$ m, $P=15000$ kW, $\eta_o=0.9$. Model: $D_m/D=0.2$, $H_m=6$ m.

(a) Specific speed of prototype

$$N_s=\frac{N\sqrt P}{H^{5/4}}=\frac{140\times\sqrt{15000}}{30^{1.25}}=\frac{140\times122.47}{70.21}=244.2$$

(medium-speed Francis turbine range.)

(b) Discharge of prototype

$$Q=\frac{P}{\rho gH\eta_o}=\frac{15\times10^{6}}{1000\times9.81\times30\times0.9}=56.63\ \text{m}^3/\text{s}$$

(c) Model

For geometrically similar turbines (same $\psi=gH/N^2D^2$, $\phi=Q/ND^3$, same efficiency):

$$\frac{N_m}{N}=\sqrt{\frac{H_m}{H}}\cdot\frac{D}{D_m}\ \Rightarrow\ N_m=140\times\sqrt{\frac{6}{30}}\times\frac{1}{0.2}=140\times0.4472\times5=313\ \text{rpm}$$ $$\frac{P_m}{P}=\left(\frac{H_m}{H}\right)^{3/2}\left(\frac{D_m}{D}\right)^2\ \Rightarrow\ P_m=15000\times0.00358=53.7\ \text{kW}$$ $$\frac{Q_m}{Q}=\left(\frac{D_m}{D}\right)^2\sqrt{\frac{H_m}{H}}\ \Rightarrow\ Q_m=56.63\times0.01789=1.013\ \text{m}^3/\text{s}$$

Check: $N_{s,\text{model}}=\dfrac{313\sqrt{53.7}}{6^{1.25}}=244.2$, the same as the prototype ($244.2$), as required for similarity.

Answer: (a) $N_s=244.2$; (b) $Q=56.63$ m³/s; (c) model speed $313$ rpm, power $53.7$ kW, discharge $1.013$ m³/s.

Governing is the automatic regulation of the turbine speed to keep it constant (so that the generator frequency stays constant) when the electrical load varies. When load falls, the speed rises, and the governor reduces the water flow; when load rises, it increases the flow.

Principle

Output power $=\rho g Q H\eta$. With the head fixed, power is changed by changing the discharge $Q$:

• Pelton turbine: by moving the spear (needle) in the nozzle, and by the jet deflector for sudden load drops.
• Francis and Kaplan turbines: by changing the opening of the guide vanes (wicket gates); in Kaplan, runner blade angle is also adjusted.

Oil-pressure relay governor

 Speed        Pilot valve     Servomotor (relay cylinder)
 sensor  ->   +--+          +--------------------+
 (fly-ball   -|  |<-oil in  |  piston ->  rod ---+--> spear /
 or electric)  +--+         +--------------------+    gate ring
   ^             |  |                                     |
   |             v  v                                     |
   +---- feedback linkage (restoring) <-------------------+
   Oil sump <- gear pump -> oil under pressure

1. A fly-ball (or electronic) speed sensor driven from the turbine shaft senses the speed.
2. If the load decreases, the speed increases; the balls move outward and raise the sleeve, which moves the pilot (control) valve.
3. The pilot valve admits high-pressure oil (from a pump driven by the turbine) to one side of the servomotor piston; oil on the other side goes to the sump.
4. The piston moves and, through the linkage, moves the spear forward (Pelton) or closes the guide vanes (Francis, Kaplan), reducing discharge and bringing the speed back.
5. A feedback (restoring) mechanism returns the pilot valve to the neutral position as soon as the correct opening is reached, which prevents overshoot and hunting.
6. For a sudden large load rejection on a Pelton, the jet deflector acts first to divert the jet, and the spear closes slowly, so water hammer in the penstock is avoided. For reaction turbines, a surge tank and slow closing of gates do the same.

The governor handles a power far above that which a direct mechanical linkage could, because oil pressure supplies the force.

Qualities of a good governor

• Stability: returns the speed to normal without hunting (continuous oscillations).
• Sensitivity: responds to small speed changes.
• Quick response (promptness): small time lag in changing the gate opening, but not so fast that water hammer is caused.
• Small speed regulation (droop): the change of speed between no load and full load should be small, typically 3 to 5 %.
• Isochronism: ability to give constant speed at all loads (ideal; approached by the use of the restoring mechanism).
• Reliability, simple maintenance and the ability to be used for parallel operation of generators.

The cross-flow turbine (Banki-Michell or Ossberger turbine) is a partial-admission impulse-reaction turbine with a drum-shaped runner, used in small hydro plants.

        water in
          |  nozzle with
          v  guide vane
        +-----+
       /   ___ \
      |  /     \ |   <- 1st pass: outside to inside
      | |  ( )  ||
      |  \_____/ |   <- 2nd pass: inside to outside
       \_________/
           |
        tailrace

Construction

• A cylindrical runner made of two end discs joined by 20 to 30 curved blades (flat strips bent to arc form, about 30 mm to 50 mm apart on rim).
• A rectangular nozzle of the width of the runner, with an adjustable guide vane (flow regulator), directs water at about 16 degrees to the tangent.
• The casing leads water to the tailrace. Head range is about 2 to 100 m.

Working

Water from the nozzle strikes the blades near the rim and flows inward through the runner (first pass, about 70 to 80 % of the power), crosses the empty space inside the drum and passes through the blades outward a second time (about 20 to 30 %). It thus crosses the runner twice, hence the name. The pressure at entry and exit is nearly atmospheric.

Advantages and applications

• Simple, cheap, easy to build locally and repair; suitable for micro and mini hydro (up to a few MW).
• Flat efficiency curve (about 80 % over a wide load range), by splitting the runner into sections of different width (1/3 and 2/3 as in Ossberger design).
• Self-cleaning, since the leaves and sand fall off in the second pass.
• Used for low to medium heads and widely used in rural electrification schemes.
• Disadvantage: peak efficiency (about 80 to 87 %) is lower than Pelton or Francis turbines.

In an impulse turbine the available head is fully converted into kinetic energy before the water reaches the runner; in a reaction turbine only part of the head is converted into kinetic energy and the rest remains as pressure energy, which is converted to work inside the runner.

Impulse and reaction turbines

Pelton, Francis and Kaplan

A draft tube is a gradually expanding tube fitted at the exit of the runner of a reaction turbine; its lower end is submerged in the tailwater.

Functions

1. The runner can be placed above the tailwater level for easy maintenance without losing the head between the runner and the tailrace.
2. It converts the large kinetic energy at the runner exit into pressure energy (by the diverging area), so the water is released to the tailrace at low velocity.
3. It creates a pressure below atmospheric at the runner exit. This increases the pressure drop across the runner and hence the effective head.

Head gain (theory)

Let $H_s$ be the height of the runner exit (section 1) above tailwater, $V_1$ the velocity at runner exit, $V_2$ the velocity at the draft tube outlet, $h_f$ the loss in the tube. Applying Bernoulli's equation between section 1 and the tailwater surface (section 2, at atmospheric pressure):

$$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+H_s=\frac{p_{a}}{\rho g}+\frac{V_2^2}{2g}+h_f$$ $$\frac{p_1}{\rho g}=\frac{p_a}{\rho g}-H_s-\left(\frac{V_1^2-V_2^2}{2g}-h_f\right)$$

So the pressure at runner exit is below atmospheric by $H_s$ plus the recovered kinetic energy $(V_1^2-V_2^2)/2g - h_f$. The efficiency of the draft tube is

$$\eta_d=\frac{(V_1^2-V_2^2)/2g-h_f}{V_1^2/2g}$$

Also $H_s$ must be limited (about 5 to 6 m) so that the pressure at the runner exit does not fall to the vapour pressure, which would cause cavitation.

Types

 Conical      Simple elbow    Moody spreading   Elbow with
 (straight)                   (bell mouth)      varying section
   \  /          |  |           \      /          |   \
    \/           |  |_____       \    /           |    \_____
  (vertical)     \________       \__/ (hollow      \________
                 horizontal      center, flared)