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# Energy Conversion - PowerPoint PPT Presentation

Energy Conversion. 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.

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### Energy Conversion

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.

Hgr

zres

ztw

Reference line

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:

where:

Pgr is the gross power of the plant [W]

 is the density of the water [kg/m3]

Q is the discharge [m3/s]

h1

abs

c1

ztw

z1

Reference line

Impulse turbines(Partial turbines)

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

Impulse turbines(Partial turbines)

Pelton

Turgo

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.

Reaction turbines(Full turbines)

Francis

Kaplan

Bulb

c2y

c2x

c1y

dl

c1x

x

Reaction forces in a curved channel

Newton’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]

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 channel

Integrate the forces in the x-direction:

Integrate the forces in the y-direction:

Using vectors give:

Reaction forces in a curved channelForce-Momentum Equation

R1

y

c2

R

R2

c1

x

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 equation

Power : P = T•w [W]

Angular velocity: w[rad/s]

Peripheral velocity: u = w•r [m/s]

Euler’s turbine equation torque ?

Output power from the runner

Available hydraulic power

Velocity triangle torque ?

cu

v

cm

c

Francis turbine torque ?

cu1

u1

b1

cm1

v1

c1

u2

b2

c2

D1

D2

v2

v torque ?1

c1

u1

v2

c2

u2

w

v torque ?1

c1

u1

w

v2

c2

u2

w

Guidevanes

Runnerblades

u torque ?2

c2

v2

v1

c1

u1

c

w

Guidevanes

u1

a1

b1

c1

v1

u2

Runner vanes

b2

c2

v2

V torque ?1=C1- U

V2

Example1 torque ?Francis turbine

D1

D2

B1

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

u1

D1

v1

c1

Example1 torque ?Inlet velocity triangle

cu1

u1

b1

cm1

D1

w1

c1

B1

Example1 torque ?Inlet velocity triangle

cu1

u1

We assume cu2 = 0

b1

cm1

w1

c1

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

Exercise 2 torque ?Francis turbine

Speed: 666 rpm

D1: 1,0 m

h: 0,96

c1: 40 m/s

a1: 40o

Find: H

b1