Cavitation

Contents:

Occurence of cavitation in turbomachinery with liquids: movie Examples of Cavitation

Cavitation effects in turbomachinery performance

Net Positive Suction Head (NPSH)

Suction specific velocity

Thoma’s Coefficient

Exercises

Cavitation

Cavitation: evaporation, followed by

condensation (almost instantaneously)

How to detect cavitation:

Change in performance curves

Visual observation of bubble formation

Noise and vibrations

Cavitation effects:

Noise, vibration

Material erosion

Performance reduction (efficiency, etc.)

Occurrence of Cavitation in pumps

In the figure below p1 is less than the atmosferic

pressure. If p1 < pvap (T) Evaporation

However minimum pressure occurs inside the pump

pmin < pvap (T) Evaporation cavitation

Pumps

Pump

Head loses in suction

pipe

Occurrence of Cavitation in pumps

vappppp 1min

to avoid cavitation

Minimum

Pressure

Compression face

Suction face

Occurrence of Cavitation in pumps

To avoid cavition:

vapppp 1ppp vap 1

g

V

g

p

g

p

g

V

g

p vap

22

2

1

2

11

Piping System (Hs) Pump (NPSH)

po

asps Zeg

V

g

p

g

p

2

2

101

g

pZe

g

p vap

asps

0

Hs > Hsi

Pump

Occurrence of Cavitation in pumps

Suction Head (𝐻𝑠):

it only depends on piping system g

p

g

V

g

pH

vap

s

2

2

11

NPSH (also 𝐻𝑠𝑖 critical value of 𝐻𝑠):

it only depends of the pump g

V

g

pHNPSH si

2

2

1

Curves of Hsi are given by the pump supplier:

Hsi = F(Q,N,D,,, geometric parameters)

Occurrence of Cavitation in pumps

NPSH - Hsi

(it only depends on the

pump)

g

V

g

pH si

2

2

1

Plots of H, and Hsi for the pump

original diameter (260 mm) and

after turning (“torneamento”)

Occurrence of Cavitation in turbines

Now, head loses in the diffuser (part of the supply) are included in the NPSH definition

No cavitation if

Hs > Hsi

Hydraulic Turbines

Applying Bernoulli’s equation:

Occurrence of Cavitation

NPSH:

it only depends of the machine g

V

g

pH si

2

2

1

Hsi = F(Q,N,D,,, geometrical parameters)

Applying dimensional analysis:

2

3,ND

ND

Qf

H

H sii

negligible Re

influence

Critical Thoma´s

Coefficient

No cavitation if: > i

Thoma´s Coefficient = Hs/H only depends on piping system

Occurrence of Cavitation

Suction specific velocity:

43

sgH

QNS

Critical suction specific velocity

343ND

QF

gH

QNS

si

i

No cavitation if S < Si

We can use Si or i

Occurrence of Cavitation

Critical suction specific velocity:

3ND

Qf

H

H sii

Turbomachines of same geometrical family have

equal (Si)max and (i )max

Critical Thoma´s Coefficient:

343ND

QF

gH

QNS

si

i

Occurrence of Cavitation

Reference values:

Pumps: 2,5 < (Si)max = F() < 3,5 (Si)max = 3

Turbines: 3,5 < (Si)max = F() < 5,2 (Si)max = 4

Occurrence of

Cavitation

Pump

Turbine

Problems 5 and 8 Problems 5 and 8 – The pump, with the attached performance curves and a

D=265 mm rotor, delivers a flow of 260 m3/h when installed in a given piping

system (reservoirs open to the atmosphere and with 12 m head difference).

It is intended to decrease the flow rate to 180 m3/h; 4 possible solutions for this

are considered:

a) Partially close discharge pipe valve;

b) Adjustment of the rotation speed;

c) Turning the rotor diameter (use given performance curves for rotor diameters

less than 265mm);

d) Installing a re-circulating circuit by connecting the pump outlet to its inlet in

such a way that some pump flow recirculates and the flow in the main duct is

the desired flow rate of 180 m3/h.

Assuming fully turbulent flow at the pipe, find the pump power consumption and

the maximum height above the upstream reservoir free-surface for each of the

four possible control processes used to deliver 180m3/h to the downstream

reservoir. Assume that atmospheric pressure applies at the free-surface of the two

reservoirs and Zasp=0,5 m (for 180 m3/h), pvap=2,45 kPa and patm=101,3 kPa)

Problems 5 and 8 –piping system characteristic

0 17

pump

Piping system

characteristics

𝐻 = 12 + 𝑘𝑄2

𝑘 = 7.4 × 10−5𝑚 𝑚3 ℎ 2

Problems 5 and 8 a) – Valve adjustment

1

pump

Power: 𝑃 = 𝜌𝑔𝑄𝐻 𝜂 = 13646 𝑊

Maximum height:

𝑒𝑠 ≤ 9.59 − 𝐻𝑠𝑖 = 7.79𝑚 1.8

Problems 5 and 8 b) – Rotational speed

adjustment

2

3 Operation point (2):

𝑄 = 180𝑚3 ℎ

𝐻 = 14,4 𝑚

Efficiency of point 2?

𝜂 = 80%

𝐻 = 4.44 × 10−4𝑄2

Power: 𝑃 = 𝜌𝑔𝑄𝐻 𝜂 = 8820 𝑊

Maximum height:

𝑒𝑠 ≤ 9.59 − 𝐻𝑠𝑖 = 8.12𝑚

𝐻𝑠𝑖 = 1.47 𝑚

Computed from same

Thoma’s Coefficients between

points 2 and 3

Problems 5 and 8 c)– Rotor diameter adjustment

2

Operation point (2):

𝑄 = 180𝑚3 ℎ

𝐻 = 14,4 𝑚

Efficiency of point 2?

𝜂 = 75%

Power: 𝑃 = 𝜌𝑔𝑄𝐻 𝜂 = 9408 𝑊

Maximum height:

𝑒𝑠 ≤ 9.59 − 𝐻𝑠𝑖 = 8.12𝑚

𝐻𝑠𝑖 = 1.8 𝑚

Problems 5 and 8 d) – Recirculation circuit

4

recirculation

Operation point (4)?

𝑄 = 280𝑚3 ℎ 𝐻 = 14,4 𝑚

𝜂 = 68%

Power: 𝑃 = 𝜌𝑔𝑄𝐻 𝜂 = 16142 𝑊

Maximum height:

𝑒𝑠 ≤ 9.59 − 𝐻𝑠𝑖 = 8.12𝑚

𝐻𝑠𝑖 = 3.5 𝑚