Aula 24 - Cavitation

8/17/2019 Aula 24 - Cavitation

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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


2/19

Cavitation

Cavitation: evaporation, followed bycondensation (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.)


3/19

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


4/19


vap p p p p 1min

to avoid cavitation

Minimum

Pressure

Compression face

Suction face


5/19


To avoid cavition:

vap p p p 1 p p p vap 1

g

V

g

p

g

p

g

V

g

p vap

22

21

211

Piping System (H s ) Pump ( NPSH )

p o

asp s Z e g

V

g

p

g

p

2

2

101

g

p Z e

g p vapasp s

0

H s > H si

Pump


6/19


Suction Head ():

it only depends on piping system g

p

g

V

g

p H vap

s

2

211

NPSH (also critical value of ):it only depends of the pump

g

V

g

p

H NPSH si 2

2

1

Curves of H si

are given by the pump supplier:

H si = F(Q,N,D, , , geometric parameters)


7/19


NPSH - H si

(it only depends on the

pump) g

V

g

p H si

2

2

1

Plots of H , and H si for the pumporiginal diameter (260 mm) and

after turning (“torneamento”)


8/19

Occurrence of Cavitation in turbines

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

No cavitation if

H s > H si

Hydraulic Turbines

Applying Bernoulli’s equation:


9/19

Occurrence of Cavitation

NPSH:

it only depends of the machine g V

g

p H si

2

2

1

H si = F(Q,N,D, , , geometrical parameters)

Applying dimensional analysis:

2

3 , ND

ND

Q f

H

H sii

negligible Re

influence

Critical Thoma´s

Coefficient

No cavitation if: > i

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


10/19


Suction specific velocity: 43 s gH

Q N S

Critical suction specific velocity

343 ND

Q F gH

Q N S si

i

No cavitation if S < S i

We can use S i or i


11/19


Critical suction specific velocity:

3 ND

Q

f H

H sii

Turbomachines of same geometrical family have

equal (S i)max and ( i )max

Critical Thoma´s Coefficient:

343 ND

Q F

gH

Q N S

si

i


12/19


Reference values: Pumps: 2,5 < (S i)max = F () < 3,5 (S i)max = 3

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


13/19

Occurrence of

Cavitation

PumpTurbine


14/19

Problems 5 and 8Problems 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 diametersless 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)


15/19

Problems 5 and 8 – piping system characteristic

017

pump

Piping system

characteristics = 12 +

= 7.4 × 10− ℎ


16/19

Problems 5 and 8 a) – Valve adjustment

1

pump

Power: =

= 13646

Maximum height:

≤ 9.59 = 7.79 1.8


17/19

Problems 5 and 8 b) – Rotational speed

adjustment

2

3Operation point (2):

= 180 ℎ

= 14,4

Efficiency of point 2? = 80%

= 4.44 × 10−

Power: =

= 8820

Maximum height: ≤ 9.59 = 8.12

= 1.47

Computed from same

Thoma’s Coefficients between

points 2 and 3


18/19

Problems 5 and 8 c) – Rotor diameter adjustment

2

Operation point (2):

= 180 ℎ

= 14,4

Efficiency of point 2? = 75%

Power: =

= 9408

Maximum height: ≤ 9.59 = 8.12

= 1.8


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Problems 5 and 8 d) – Recirculation circuit

4

recirculation

Operation point (4)?

= 280 ℎ = 14,4

= 68%

Power: = = 16142

Maximum height:

≤ 9.59 = 8.12

= 3.5

Aula 24 - Cavitation

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