- 122 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Energy Conversion' - quemby-warner

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Specific energy

The specific energy of a hydro power plant is the quantity of potential and kinetic energy which 1 kilogram of the water delivers when passing through the plant from an upper to a lower reservoir.

The expression of the specific energy is Nm/kg or J/kg and is designated as m2/s2.

Gross Specific Hydraulic Energy

In a hydro power plant, the difference between the level of the upper reservoir zres and the level of the tail water ztw is defined as the gross head:

Hgr = zres - ztw [m]

The corresponding gross specific hydraulic energy:

Gross Power

where:

Pgr is the gross power of the plant [W]

is the density of the water [kg/m3]

Q is the discharge [m3/s]

Impulse turbines(Partial turbines)

The hydraulic energy of the impulse turbines are completely converted to kinetic energy before transformation in the runner

Reaction turbines(Full turbines)

In the reaction turbines two effects cause the energy transfer from the flow to mechanical energy on the turbine shaft.

Firstly it follows from a drop in pressure from inlet to outlet of the runner. This is denoted thereaction partof the energy conversion.

Secondly changes in the directions of the velocity vectors of the flow through the canals between the runner blades transfer impulse forces. This is denoted the impulse part of the energy conversion.

c2y

c2x

c1y

dl

c1x

x

Reaction forces in a curved channelNewton’s 2.law in the x-direction:

where:

A is the area [m2]

c is the velocity [m/s]

is the density of the water [kg/m3]

Q is the discharge [m3/s]

Reaction forces in a curved channel

y

Newton’s 2.law in the

x-direction:

c2y

Rx

Fx

c2x

c1y

dl

c1x

x

Fx is the force that acts on the fluid particle from the wall. Rx is the reaction force that acts on the wall from the fluid:

Rx = -Fx

Ry

c2y

R

Rx

c2x

c1y

dl

c1x

x

Reaction forces in a curved channelIntegrate the forces in the x-direction:

Integrate the forces in the y-direction:

Using vectors give:

Let the channel rotate around the point ”o”. What is the torque ?

Let us define the u-direction as the normal of the radius

(or tangent to the circle)

Torque = force · arm:

r1

c1

cu1

cu2

c2

r2

a1

w

o

a2

r torque ?1

c1

cu1

cu2

c2

r2

a1

o

a2

Euler’s turbine equationPower : P = T•w [W]

Angular velocity: w[rad/s]

Peripheral velocity: u = w•r [m/s]

Euler’s turbine equation torque ?

Inlet and outlet velocity diagram for reaction turbines torque ?

Guidevanes

u1

a1

b1

c1

v1

u2

Runner vanes

b2

c2

v2

V torque ?1=C1- U

V2

Example1 torque ?Francis turbine

Head: 150 m

Q: 2 m3/s

Speed: 1000 rpm

D1: 0,7 m

D2: 0,3 m

B1: 0,1 m

h: 0,96

Find all the information to

draw inlet and outlet velocity

triangles

D1

D2

B1

Example1 torque ?Inlet velocity triangle

cu1

u1

u1 = 36,7 m/s

cu1 = 33,4 m/s

cm1 = 9,1 m/s

b1

cm1

w1

c1

Example1 torque ?Outlet velocity triangle

u2

We assume: cu2 = 0

and we choose: cm2 = 1,1· cm1

b2

c2

v2

Exercise1 torque ?Francis turbine

Head: 543 m

Q: 71,5 m3/s

Speed: 333 rpm

D1: 4,3 m

D2: 2,35 m

B1: 0,35 m

h: 0,96

cm2 : 1,1· cm1

Find all the information to draw inlet and outlet velocity triangles

D1

D2

B1

Download Presentation

Connecting to Server..