Thermal And Hydraulic Machine Part2

Name: Thermal And Hydraulic Machine Part2
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Author: Em4nuel3

Revisionato il 04/05/2026

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Appunti utili per la preparazione dell'esame di Thermal and Hydraulic Machines del professor Claudio Don Giovanni al Politecnico di Torino, Prima Facoltà di Ingegneria, corso di laurea in …

Esame Thermal and Hydraulic Machines

Facoltà Ingegneria i

Dal corso del Prof. Don Giovanni Claudio

Università Politecnico di Torino

A.A. 2013-2014

143 pagine

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Axial pump

Axial pumps are used to supply water when there is a little head but a high mass flow rate is desired.

Velocity triangle

Axial because the cross-sectional area is constant and the angle is the same.

_{Li =}¹_{γ g h_m}

The head is few meters, so the deflection (β) is very low (~10°).

They present a very big diameter and few blades (4 ÷ 6), allowing a large passage between them. This type of machine is sensitive when we change the mass flow rate, as it introduces an incidence.

At the exit of rotor

Axial because the cross-sectional area is constant and the angle is the same.

L_i = ¹⁄ g h_m

The head is few meters, so the deflection (Δβ) is very low (~ 10°).

They present a very big diameter and few blades (4 ÷ 6), allowing a large passage between them. This type of machine is sensitive when we change the mass flow rate, as it introduces an incidence.

In design condition

In off-design conditions, remember to decrease G by decreasing C₁ because ρ = const and A = const. To avoid incidence, an axial pump has a tool that can change the setting angle of the rotor blade. This is possible because there are only a few blades, achieved by using a fluid power system. This axial pump is called a Kaplan pump, and adjustments are made using a fluid power machine.

Lesson 25/11/2013

Last lesson we saw axial pump velocity triangle, Kaplan because we have incidence, we have losses, so we can adjust the rotating blade, in this way change the velocity triangle also in off-design. Remember:

_Ns = _^ω√Q⁄_{(g_Hm)^3/4}

Q = πD_e·c₁

L_i = ¹⁄_γ g H_m = μ (C_u2 - C_u1) = μ C_u1

So we can write Ns ∝ √ e/D tgδ₂

Where

K = ^M_m⁄_{√2gH_m} (peripheral coefficient: 1.4 ÷ 2)
e/D = 0.25 ÷ 0.56
δ₂ = 60° ÷ 75°
Approximately N_S ≈ 3 ÷ 9

We had also seen

N_C = ^{n [rpm] √P_obs [kW]} H_m^5/4 [m]

Number evaluated with reference to have maximum efficiency number used in Europe (N_C ≈ 3.65), where:

P_obs

P_obs = ¹ ψ_m ̅G ⋅ ̅Li = ¹ Q ρLi m

With reference

N_S ≈ 3 ÷ 9 => N_C ≈ 600 ÷ 1800 rpm

Easy to pass from N_S to ̅H_c

N_C = 1.96N_S

Referencing

N_S ∝ √^f tg d₂

If N_S ≈ 3, the shape of the pump is ^D

If N_S ≈ 9, the shape is generally the same but changes in proportion e/d, or angles, and so on.

With reference to cavitation

We have to verify NPSH ≥ NPSH_min = h_o

Plant ^p₁-p_v/_r + ^c_1²/_2g

Characteristics of pump

λ^w'₁²/_2g + ^c'₁²/_2g

We have also defined the max head to avoid cavitation in the pump:

∆z < ^p_amb-p_v/_r - y_a->1 - h_o

h_o = χ ^w'₁²/_2g + ^c'₁²/_2g

But h_o can be written as a ratio of velocity:

h_o = ^u'₁²/_2g [λ (^w'₁/_u'₁)² + (^c'₁/_u'₁)²] are constant under similitude, so h_o ∝ u₁² => ^h_o/_{H_u n²}

We can define the Thomas parameter σ_v

Anteprima

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