Nozzle

NOZZLE

THERMPDYNAMICS-II

Abdul Ahad Noohani (MUCET KHAIRPUR)

Mehran University of Engineering and Technology SZAB Campus Khairpur Mir’s

LECTURE NO: 01

Nozzle

Nozzle is duct of smoothly varying cross-sectional area in which a steadily flowing fluid can be made to accelerate by a pressure drop along the duct.

Applications:

steam and gas turbinesJet engines Rocket motorsFlow measurement

When a fluid is decelerated in a duct causing a rise in pressure along the stream, then the duct is called a diffuser.

Two applications in practice in which a diffuser is used are :

The centrifugal compressor Ramjet

Diffuser

One dimensional flow In one dimensional flow it is assumed that the fluid velocity, and the fluid properties, change only in the direction of the flow.

This means that the fluid velocity is assumed to remain constant at a mean value across the cross section of the duct.

Nozzle Shape

Applying Steady flow energy equation:

…………………………….. (1)

If the area at section X - X is A, and the specific volume is v, then , using equation:

Substituting the valve of C in above equation,

Area per unit mass flow

In order to find the way in which the area of the duct varies,

At any section X-X We need to evaluate :

specific volume v enthalpy h

For evaluating specific volume and enthalpy we must know the process undergone between section 1 and X-X.

For an ideal frictionless case:

since the flow is adiabatic and reversible, the process undergone is an isentropic process

For isentropic process :

vγ = c /p

S1 = (entropy at any section X-X) = s & pv γ = c

Enthalpy is taken from table with respect to the given Pressure and temperature.

h @ p, T

vγ = c /p

Enthalpy is taken from table with respect to the given Pressure and temperature.

(h @ p, T)

LECTURE NO : 02

Cross-sectional area, velocity, specific volume variations with pressure through a nozzle

When v increases more rapidly than C, then the area decreases; When v increases more rapidly than C, than the area increases A nozzle in which area varies as in above fig; is called convergent divergent nozzle.

Critical pressure ratio:

Ratio of pressure at the section where sonic velocity is attained to the inlet pressure of a nozzle is called the critical pressure ratio.

Consider convergent divergent nozzle,

Inlet conditions of Nozzle be: (P1 ; h1 ; C1) Conditions at any other section X-X: (P ; h ; C)

C1 << C therefore neglecting C1

Then,

For perfect Gas:

From perfect gas equation

….1 …..2

For an isentropic process for a perfect gas:

substituting

In above equation:

A / m’

Substituting the values in eq: 2

A / m’ …3

To find the value of pressure ratio for which the area is minimum we differentiate the eq:3 w.r.t x

For carbon dioxide at 10 bar the convergent nozzle requires back pressure of 5.457 bar for sonic flow at exit

Critical temperature ratio:

Since,

LECTURE NO : 03

20

:Problem:01

DATA:

P1 = 8.6 barT1 = 190 ‘cm’ = 4,5kg/sP = 1.03 bar

Calculate:

Throat and exit cross-sectional areas.

21

(I) We know that area of the throat can be found as:

From this eq: we can find Tc:

we can find Pc by using

…………(1)

22

Substituting the values in equation (1)

23

(II) We know that area at the exit can be found as:

LECTURE NO : 04

Convergent nozzle with back pressure variation

Consider a convergent nozzle expanding into a space, the pressure of which can be varied, while the inlet pressure remains fixed.

When, Pb = P1 : then no fluid can flow through the nozzle

As Pb is reduced : mass flow through the nozzle increases, since the enthalpy drops and hence the velocity increases.

When the Pb = Pc: no further reduction in back pressure can affect the mass flow and velocity at the exit is sonic.

MAXIMUM MASS FLOW

Maximum mass flow through a convergent nozzle is obtained when the pressure ratio across the nozzle is the critical pressure ratio.

Also for a convergent – divergent nozzle , with sonic velocity at the throat, the cross-sectional area of the throat fixes the mass flow through the nozzle for fixed inlet conditions.

When a nozzle operates with the maximum mass flow it is said to be choked.

FOR EXAMPLE : When air at 10 bar expands in a nozzle, the critical pressure can be shown to be 5.283 bar. When the Pb = 4 bar then nozzle is chocked and is passing the maximum mass flow. If the Pb is reduced to 1 bar, the mass flow through the nozzle is unchanged

CHOCKING

PROBLEM NO: 02

………….. eq no: 1

First we find the critical pressure to find whether nozzle is chocked.

For this we need specific heat ratio of Helium

m’ for Helium = 4

………….. eq no: 2

Putting values in equation no: 01

The actual back pressure is 3.6 bar, hence in this case the fluid does not reach the critical conditions and the nozzle is not chocked

Substituting values in equation: 02

For exit temperature we use following expression

Velocity at the exit of Nozzle:

Now substituting values in equation no: 1

LECTURE NO : 05

Nozzle effeciency is defined as:

“ The ratio of the actual enthalpy drop to the isentropic enthalpy drop between the same pressures. “

Due to friction between the fluid and the walls of the nozzle and friction within the fluid itself, the expansion process is irreversible although still approximately adiabatic.

In nozzle design it is usual to base all the calculations on isentropic flow and then to make an allowance for friction by using a co efficient or an effeciency.

Nozzle Effeciency

Nozzle expansion processes for a vapour (a) and a perfect gas (b) on a T-s diagram

Typical expansion between P1 and P2 in a nozzle are shown on T-s diagram in fig. For a vapour and for a perfect gas respectively.

The line 1-2s on each diagram represents the ideal isentropic expansion, and the line 1-2 represents the actual irreversible adiabatic expansion.

For a perfect gas this eq: reduces to

If The actual velocity at exit from the nozzle is C2 and Velocity at exit when the flow is isentropic is C2s . Then, using the steady –flow energy equation in each case we have

Substituting these vales in eq: 1

When inlet velocity C1 is negligibly small then

Velocity Co-efficient:

It is the ratio of actual exit velocity to the velocity when the flow is isentropic b/w the same pressures.

Velocity Co-efficient is square root of the nozzle efficiency when the inlet velocity is assumed to be negligible.

“ Ratio of the actual mass flow through the nozzle m’, to the mass flow which would be passed if the flow were isentropic, m’s. “

Co-efficient of discharge :

Angle of divergence:

The included angle of divergent duct is usually kept below about 20*.

If the angle of divergence of a convergent – divergent nozzle is made too large, then breakaway of the fluid from the duct is liable to occur, with consequent increased friction losses.

Problem : 3

(i) Area at the throat:

Since,Co-efficient of discharge = 0.99m’ = 18 kg/s

Substituting values we get

(ii) Area at the Exit:

LECTURE NO : 06

Steam Nozzle Properties of steam can be obtained from tables or from h-s chart.

But in order to find the critical pressure ratio, and hence the critical velocity and the maximum mass flow rate, approximate formulae may be used.

We assume that steam follows an isentropic law = Pvk

(k = isentropic index for steam) For steam k = 1.135, dry saturated k = 1.3, for superheated

k = 1.135, dry saturated

k = 1.3, for superheated

The temperature at the throat i.e. the critical temperature, can be found from steam tables at the value of Pc and Sc = S1.

The critical velocity can be found as before from equation:

Where, hc is read from tables or the h-s chart at Pc and Sc

Problem: 04

LECTURE NO : 07

Supersaturation or Supersaturated expansion

The type of expansion in which expansion of vapour after the saturated vapour line continues as if it did not exist, and then at a certain point it condenses suddenly and irreversibly.

The state of vapour during this type of expansion is called metastable state.

Generally, Condensation with in the vapour begins to form when the saturated vapour line is reached.

Below the saturated vapour line (in wet region) Dryness fraction becomes smaller.

Superheated steam expanding into the wet region on (a) T-S and (b) h – S diagrams

In case of Nozzle: Condensation is so quick that condensation does not occur at point A.Then it suddenly condenses irreversibly at certain point.That point may be outside or within the nozzle.

Supersaturated expansion of steam on (a) T-s and (b) h-as diagram

Assuming the isentropic flow,

Line 1-2 represents expansion with equilibrium

Line 1-R represents supersaturated expansion

Line 1-R intersects the pressure line P2 produced from superheat region.

It can be seen that the temperature of the super- saturated vapour at P2 is tr, which is less than the saturation temperature t2, corresponding to P2.

Supersaturated expansion of steam on (a) T-s and (b) h-s diagram

The vapour after Supersaturation is called super cooled vapour.

Degree of Supercooling or super saturation = t2 –tr

Sometimes It is also written as:

“The ratio of the actual pressure P2 to the saturation pressure corresponding to tr.”

Degree of Supercooling or super saturation

Enthalpy drop comparison

It can be seen from figure that the enthalpy drop in supersaturated flow (h1 – h2) is less than the enthalpy drop under equilibrium conditions.

Since the velocity at exit C2 is given by equation

It follows that exit velocity for supersaturated flow is less than that for an equilibrium flow.

Nevertheless the difference in enthalpy drop is small, and since the square root of the enthalpy drop is used in equation than the effect on the exit velocity is small

P-V diagram for The equilibrium expansion and the supersaturates expansion

It can be seen from fig that the specific volume at exit with supersaturated flow vr is less than the specific volume at exit v2.

Mass flow rate for supersaturated flow

It has been pointed out that C2 and Cr are very nearly equal

therefore since Vr < V2, it folllows that the mass flow with

supersaturates flow is greater than the mass flow with equilibrium flow.

Problem: 05

h – s Diagram for the problem

(i) Exit Area when the flow is in equilibrium throughout:

h1

h2

Solution of pb no : 05 on P-h diagram