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Nozzles and Steam Turbines 73

NOZZLES AND STEAM TURBINESNOZZLES AND STEAM TURBINESNOZZLES AND STEAM TURBINESNOZZLES AND STEAM TURBINES 2222

3.1 NOZZLES

A nozzle is a mechanical device or orifice designed to accelerate afluid to a high velocity by means of a drop in pressure of the fluid.

A nozzle is often a pipe or tube of varying cross sectional area

and it can be used to direct or modify the flow of a fluid (liquid or

gas). Nozzles are frequently used to control the rate of flow, speed,

direction, mass, shape, and/or the pressure of the stream thatemerges from them.

3.2 Types of nozzles

In Figure 3.1 is shown two of the common nozzle shapes.

i) Convergent-divergent nozzle (Figure 3.1 a): It will beseen that from entry area the nozzle converges down to a

minimum area. The nozzle has a tapered cross-section

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reducing to the minimum size called throat and then it has adivergent tapered length upto the maximum size called exit.

ii) Convergent nozzle (Figure 3.1. b): The cross section ofthis nozzle is converges down from entry area to a

minimum area which is the exit.

(a) (b)

Figure 3.1 (a) convergent-divergent nozzle, (b) convergentnozzle

3.3 Flow analysis through nozzlesFigure 3.2 shows a nozzle with the following conditions:

Entry

conditions

Exit

conditions

Area, m2

A1 A2Velocity, m/s V

1V

2Pressure, N/ m2 P1 P2Specific volume, m

3/kg ,1 ,2

Temperature,oC T1 T2

Specific enthalpy, kJ/kg h1 H2

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Figure 3.2 Nozzle with entry and exit conditions

Apply the steady- flow energy equation for a nozzle, neglecting:

change in potential energy and work done, W = 0 (no work is done

in a nozzle), Q = 0 (there is a little time for heat exchange between

the substance and its surrounding).Then, steady flow energy equation for unit mass flow.

Wgz2

Qgz2

2

2221

211 ++++++++====++++++++++++

VhVh (3.1)

21

22

22

2hh

VV====

The entry velocity to a nozzle is small compared with the exitvelocity. Neglecting V1, so:

21

22

2hh

V=

(((( )))){{{{ }}}}21 h-hV 22 ==== (3.2)

For a gas,

(((( )))) (((( ))))2121p21 TTR1-

T-TCh-h ========$

$

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Since1-

RCp

$

$====

Therefore,

(((( ))))

(((( ))))

====

====

========

====

====

====

====

$

$

$

$$

%$

$

%$

$

%

%%%

%

%%

$

$

%%%$

$

$

$

1

1

2

1

211

1

2

1

1

211

1

2

1

1

22211

11

2211

2211

112

P

P

P

P-1P1-

2

P

P

P

P-1P

1-

2

P

P,PPBut

P

P-1P

1-

2

RTPsinceP-P1-

2

T-TR1-

2V

====

$

%$

$11

1

211

P

P-1P

1-

2

====

$

$

%$

$

1-

1

2112

PP-1P

1-2V

velocity,Exit

(3.3)

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Apply continuity equation at any section of nozzle,

m 3 = A V

Where, m = mass flow, kg/sec

3 = specific volume, m3/kg

A = cross-sectional area, m2

V = velocity, m/sec.

Then, %

V

A

m====

which is the mass flow/ unit area at any section.

3.4 Maximum discharge through nozzleThe convergent-divergent nozzle may be suitably designed for a

maximum discharge.In this condition, the cross-sectional area must be a minimum

which will be the throat of the nozzle.

Let, Pt = throat pressure

3t = specific volume at the throat, m3/kg

Vt = velocity at the throat, m/sec.

Then,

V

A

mBut,

P

P-1P

1-

2V

t

t

t

1-

1

t

11t%

%$

$$

$

====

====

Therefore,

====

$

$

%$$

%

1-

1

t11

tt PP-1P

1-21

Am

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P

Por,PPTherefore,

C.PVgasaofexansionadiabaticFor

1

t

11t21

$

$

t$

t

$

%%%%

========

====

Then,

====

$

1-$

1

t1$

t

11

t PP-%P

1-$$2

P

P%

1Am 111

P

P-

P

PP

1-

21

P

P-1

P

PP

1-

21

P

P-1P

1-

2

P

P1

1

1

t

2

1

t11

1

1-

1

t

2

1

t11

1

1-

1

t11

1

1

t

1

====

====

====

++++

$

$$

$

$$

$

$$

%$

$

%

%$

$

%

%$

$

%

t

11

1

1

t

2

1

t

t

AmmaximumforThen

equation.thisinconstantsareandP,

maximumbemustP

P-

P

Pthen,

A

mmaximumFor

%$

$

$$

++++

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P

P1

P

P2

0P

P1-

P

P2

0P

P-

P

P

dx

d

1

1

t

2

1

t

1

1

t

2

1

t

1

1

t

2

1

t

$$$

$$

$

$

$$

$

$

$

$

$

$

++++====

====

++++

====

++++

1

2

P

P

P

P

2

1

t

1

1

t

++++

====

$

$

$

$

1

2

P

P1-

1

t

++++====

$

$

$

ratiopressurecriticalthe1

2

P

P 1-

1

t====

++++

====$

$

$(3.4)

presurecriticalP1

2P 1

1-

t ====

++++

====

$

$

$

For air, the average value of = = 1.4

Therefore,

P0.528P11.4

2P 11t ==== ++++====

14.14.1

The variation of mass flow rate with pressure ratio is shown

graphically in Figure 3.3.

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Figure 3.3 Mass flow of steam vsP2/P1

The Critical temperature

etemperaturcriticalT1

2T

ratioetemperaturcritical

1

2

1

2

P

P

T

T

1t

1

1-1

1

t

1

t

====

++++

====

====

++++

====

++++

====

====

$

$$

$

$

$

$

$

$

==== 1-

T

TRT

1-

2

t

1t

$

$

Now, from equation (3.2)

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(((( )))){{{{ }}}} (((( )))){{{{ }}}}

(((( ))))

====

========

t1

t1pt1t

T-TR1-

2

T-TC2h-h2V

$

$

{{{{ }}}} soundofvelocitytheshowsThis,RT

1-2

1RT

1-2

t

t

$

$

$

$

====

++++====

(3.5)

Sonic velocity is said to be Mach 1 (Ma = 1). From criticalpressure, if P2 < Pt , then the nozzle will be convergent-divergent.From entry to the throat, the velocity will be subsonic (Ma1).

If P2 > Pt , then the nozzle will be convergent only. The velocity

will be subsonic from inlet to exit (Ma

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=1

1-$$

tP

1$

2P

++++

==== , n= 1.135

= 0.577P1n= 1.035 + 0.1 x for wet steam with dryness fraction x

In Figure 3.4 the steam is expand from initial state A at P1 to exit

pressure P2. Process A-B represents the isentropic frictionlessexpansion. Process A-C is the reduced heat drop due to

friction. Actually the energy lost in friction is transferred into heatwhich tends to dry or superheat the steam. The exit state will be

represented by point C not by point B. The dryness friction by

point C is higher than that at B. Thus, the actual state line ofthe steam through the nozzle is A-C. In practice it is possible to

achieve a heat drop of only about 0.9 to 0.95 of the theoretical

value.

Figure 3.4 Expansion in nozzle.

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Nozzle efficiency or efficiency of expansion

In ideal case, flow through nozzle is isentropic. But in actual case,

friction exists and affects in following ways:1. reduces the enthalpy drop2. reduces the final velocity of steam3. increases the final dryness fraction4. increases specific volume of the fluid5. decreases the mass flow rateNozzle efficiency =

h-h

h-h

dropheatisentropic

dropheatactual

BA

CA==== (3.6)

The efficiency of nozzle is depends upon many parameters , i.e:material it is made of, smoothness, size and shape, angle of nozzle

divergence, nature of fluid flowing and its state, fluid velocity,

turbulence in nozzle flow.

Velocity coefficient of nozzle Kn: is the ratio of the actual exitvelocity to isentropic velocity obtained for the same pressure drop.

yeffciciencnozzle

h-h

h-V

velocityisentropic

velocityactual

BA

CAC

====

============

A

AB

n

h

VK

3.6 Supersaturated flow in NozzlesAs steam expands in the nozzle, its pressure and temperature

drop, and it is expected that the steam start condensing when it

strikes the saturation line. But this is not always the case. Owing to

the high velocities, the residence time of the steam in the nozzle is

small, and there may not sufficient time for the necessary heat

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transfer and the formation of liquid droplets. Consequently, thecondensation of steam is delayed for a little while. This

phenomenon is known as supersaturation, and the steam that exists

in the wet region without containing any liquid is known as

supersaturated steam.The locus of points where condensation will take place regardless

of the initial temperature and pressure at the nozzle entrance is

called the Wilson line. The Wilson line lies between 4 and 5percent moisture curves in the saturation region on the h-s diagram

for steam, and is often approximated by the 4 percent moistureline. The supersaturation phenomenon is shown on the h-s chart in

Figure 3.5.

Figure 3.5 Supersaturated expansion of steam

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Example 3.1

A convergent-divergent nozzle is required to pass 1.8 kg of steam

per second. At inlet the steam pressure, temperature and the speed

are 7 bar, 200oC and 75 m/sec. respectively. Expansion is stable

throughout to exit pressure of 1.1 bar. There is no loss by friction

in the converging section of the nozzle, but loss by friction

between throat and outlet is equivalent to 71 kJ/kg of steam.

Calculate, assuming throat pressure of 4 bar :i) The required area of throat;ii) Te required area of outlet;iii) The overall efficiency of the nozzle.

Solution

Figure below illustrate the expansion process of this problem

(i) From steam tables at P1 = 7 bar and t1 = 200 oC ,h1 = 2845 kJ/kg,1 = 0.3 m3/kgs1 = 6.886 kJ/kg

From Mollier chart h1 h2s = 73.7 kJ/kg

t2 = 160oC

,2 = 0.481 m3/kg

[Also, kg%% /m0.481473

433

4

70.3

T

T

p

p 3

1

2

2

112 ============ ]

Apply steady flow energy equation between 1 and 2,Hence,

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(((( ))))[[[[ ]]]]

(((( ))))

m0.002213391.2

0.4811.8

V

3mATherefore,

m/sec.391.2

73.71000275

h-h2VVor2

Vh

2

Vh

2

2

22

2

s21212

22

2

21

1

====

========

====

++++====

++++====++++====++++

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(ii) From the Mollier charth1h3s = 216 kJ/kg

x3 = 0.9633g3 = 1.549 m

3/kg (from steam tables at P3=1.1 bar)

Apply steady flow energy equation between 1 and 3s sections ofnozzle,

(((( ))))

cm49.38

m0.004938543.7

1.5490.9631.8

V

xmAhence,

m/sec.543.7V

2

75-V1071-216

2

V-Vlosses-h-h

losses2

Vh2

Vh

2

2

3

33

3

3

2233

21

23

3s1

23

3s

21

1

====

====

========

====

====

====

++++++++====++++

g%

(iii)

67.1%0.671

216

71-216

dropheatIsentropic

dropheatActualefficiencyOverall

========

========

Example 3.2

A group of convergent-divergent nozzles are supplied with steam

at 20 bar and 325oC. Supersaturated expansion according to the

law PV1.3=constant occurs in the nozzle down to an exit pressureof 3.6 bar, steam is supplied at the rate of 7.5 kg/sec. Determine:

a)The required throat and exit areas;b)The degree of undercooling at exit.

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Solution

The given data is clearly shown in below Figure.

(a) From steam table, ,1 = 0.132 m3/kgHence,

bar10.92200.546P13.1

2

P1

2P

1

1-3.13.1

1

1-nn

t

========

++++

====

++++

====

$

Throat area

m/sec.546

0210.92-10.1321020

1-.311.32

P

P-1P

1-n

2nV

3.11-.31

5

n1-n

1

t11t

====

====

==== %

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/kgm0.2192.10

200.132

P

P 33.11n

1

t

11t ====

====

==== %%

areathroatmm2890

m0.00289545

0.217.5

V

mA

2

2

t

tt

========

====

========%

Exit area

m/sec.86502

3.6-10.1321020

1-.31

1.32

P

P-1P

1-n

2nV

3.11-.31

5

n1-n

1

2112

====

====

==== %

/kgm0.4956.3

200.132

P

P 33.11n

1

2

112 ====

====

==== %%

areaexitmm4280

m0.00428866

495.07.5

V

mA

2

2

2

22

========

====

========%

(b)

C129K40220

3.6273)(325

P

PTT o

3.113.1n1n

1

212 ========

++++====

====

At 3.6 bar from steam table, saturation temperature, ts = 139.9oC

Therefore, degree of undercooling at exit = 139.9 129 = 10.90C

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