Thermal And Hydraulic Machine Part2

Axial Pump

Used to supply water when we have little head but we want high mass flow rate

Velocity Triangle

Axial because cross section area is constant and the angle is the same

L_i = 1/γ g h_m

The head is few meter so the deflection (Δβ) is very low (~10°)

Presents very big diameter and few blades (4 ÷ 6) so the passage between them is big. So this type of machine are sensible when we change mass flow rate, because when we do it we will introduce an incidence.

IN DESIGN CONDITION

In off-design

Remember G = ρ A C1

To decrease G we have to decrease C1 because ρ = const, A = const

To avoid incidence axial pump has a tool able to change the angle of rotor blade it is possible because we have only few blades by using a fluid power system.

This axial pump is called

KAPLAN PUMP.

Adjust by using fluid power machine

IN THIS WAY WE WRITE

COMBINING THESE 2

EMPIRICALLY WE FIND:

COMBINING with the formula before

SO IN GENERAL WE DON’T KNOW, BUT USING THIS WE CAN FIND

TO EVALUATE THE VELOCITY AT THE EXIT OF NOZZLE (C₁)

WE CAN WRITE WELL

H⁰ = H₁ + γ_nozzle = H⁰_1D

A⁰ = z₁ + _p1/^γ + _C²1D/^2g + _C²2/^2g = H⁰₁

=> H⁰ = H⁰₁ + _C²1D/^2g - _C²/^2g

H⁰ CAN BE WRITTEN AS

H⁰ = z₂ + _p2/^γ + _C'1D'²/^2γ + _C'2-C'2'²/^2γ

=> H⁰ = H⁰₂ + _C²1D/^2g - _C²/^2g

SO WE OBTAIN

H_u = H⁰ - H⁰₂ = _C²1D/^2g - _C²/^2g

φ = _C1/^C_1,1D

=> H_u = _C²/^{2g φ2}

REARRANGING

C₁ = φ √ 2g H_u + C²

FOR PELTON

H_u >> _C²/^2g

=> C₁ = φ √ 2g H_u

WHERE

Q = i A₁ C₁     i = NUMBER OF NOZZLE

WE CAN CHANGE THE CROSS SECTION AREA USING DOUBLE NEEDLE

[Illustration of double needle mechanism]

H_t = H_t^o = H_t^o + γ z_pen + γ z_nozzle = CONST

^{z₁ +P₁/_σ +c₁2/_2g => C₁ IS CONSTANT}

P₁ = P_amb

WE HAVE ALSO THAT IN REAL APPLICATION THE TURBINE IS CONNECTED WITH ELECTRICAL GENERATOR

[Diagram showing connection]

IT HAS f = 50Hz

THE RELATIONSHIP BETWEEN n AND f DEPEND ON THE NUMBER OF POLE OF GENERATOR

[Generator illustration]

P = 1 (N. OF POLES) n = f

[Generator illustration]

P = 2 n = ^f/₂

So we must stop the machine. To do this

The only way to stop is to close G, but the head is about some hundred of meter, therefore we have a pipe of Km and we don't stop immediately the flow. We need time

t ≈ 4; 5 s

τ = _2L / _a speed of sound

• where

a ≈ 1000 m/s

L = 1 Km

τ ≈ 2 s

To obtain this we can use a deflector in front of own needle

Now we have to think of the diffuser

_da2

C_m2

_da3

From this we can understand that we have an equal flux

Flow Continuity Equation

C_a2 * dA₂ = C_a3 dA₃

Applying Angular Momentum Equation

∮ r₂ C_m2 = ∮ r₃ C_m3

d_A2/A3

Therefore

• From the first

C_a2 = C_a3 * A₂/A₃ = C_a2 * (V₂/V₃)²

• From the second

C_m3 = C_m2 * V₂/V₃

So the kinetic energy at discharge of diffuser

C₃²/2 = C_a3² + C_m3²/2 = 1/2 * [ C_a2² (V₂/V₃)⁴ + C_m2² (V₂/V₃)⁴ ]

=> C₃^2/2 = 1/2 C_a2² [ (V₂/V₃)⁴ + C_m2^{2/Ca22 ] = 1/2 C_a22 [ A₂/A₃ + ctn2 α₂ ]}

Having N_s we can compute

Low Speed Francis Turbine (LSF)

• α₁ is smaller
• d₁ = 0.04
• β₁ = 60°
• N_s = 0.3
• n_c = 60 rpm

e₁ = 0.55 / D₁ = 130°

• ω_c = ^n_c√P_m / _(Hm)^3/4

High Speed Francis Turbine (HSF)

• α₁ is bigger
• d₁ = 40°
• N_s = 2.3
• n_c = 1450 rpm

In the middle we have the so called Normal Speed Francis Turbine (NSF)

It is interesting to draw the velocity triangle for these 3 kinds of speed.

Low Speed Francis Turbine

• α₁ (small)
• ∆β = very high deflection
• Height of the blade is important in comparison to the diameter

High Speed

• d₁ ≈ 40°
• ∆β = very small deflection

Lezione 5/12/2013

Kaplan Turbine

Also called propeller turbine - The inlet is like Francis

As in Francis we can adjust the setting angle of down nozzle

Usually we have small number of blade (3≅5)

In propeller turbine in the past the rotor blade was fixed, but we solve this problem with Kaplan in which we can adjust rotor blade to improve the performance in off-design

We want to analyze the shape, but first of all we have to see:

• H₀ = const. (The energy doesn't change by changing elevation)
• H_i = const. (Also at the exit of distributor
• The angular momentum at the inlet in distributor doesn't change by changing the elevation

r₀ C_m0 = const

r_1' C_m1' = const

So we start to study from 1' to 1 writing:

• Continuity equation
• Angular momentum conserv. equation
• Mechan. energy conserv. equation

We define own control volume:

Adjustment

Useful power is

P_m = μ_t ρ Q H_m         μ_m = ω_y

Changing Q we change H_m and change also a little bit ψ_y because it is:

ψ_y = L_i / gH_m

Analyzing propeller, the velocity triangle

Design

But we have adjust the mass flow rate, so change the axial component

We have to assume (Hypothesis)

μ is the same

We have incidence at the inlet of rotor, but attention we have a limit because it could provides a stall

For the exit we have to assume P'₂ ≅ P₂