TURBOMACHINE

Definition- All devices in which energy is transferred either to, or from, a continuously flowing fluid by the dynamic action of one or more moving blade rows.

Turbo - means which spins

Open turbo machines – that influences an indeterminate quantity of fluid Such as propellers, wind mill and un-shrouded fans.

Enclosed turbo machines- which influences finite quantity of fluid passing through a casing.

Classification 1. Power transfer (a) Those absorbing power (fans, compresses, pumps).

(b) Those producing power (hydraulic, steam & gas turbines). 2. Nature of flow path- (a) when through flow over rotor is wholly or mainly parallel to the

Axis of rotation – AXIAL FLOW TURBOMACHINES.

(b) When through flow is wholly or mainly in a plan perpendicular to the Rotation axis – RADIAL FLOW TURBOMACHINES.

(c) When through flow has both axial and radial direction – MIXED FLOW TURBOMACHINES.

3. Pressure changes are absent (IMPULSE) or present (REACTION) in the flow Through the rotor.

BASIC PHYSICAL LAWS

The continuity equation.The First Law of Thermodynamics.Newton’s Second Law of Motion.The Second Law of Thermodynamics.

The Equation of Continuity dm = ρCndA

m = ρ1Cn1A1 = ρ2Cn2A2 = ρCnA

C

Flow direction

Cn

dA

The first Law of Thermodynamics Through a cycle ∫ (dQ-dW) = 0 During Change of state from 1 to 2, there is change in the

property i.e. internal energy

E2-E1 = ∫12 (dQ-dW).

For on infinitesimal change of state

dE = dQ – dW

The steady flow energy equation

m

m

w

Q1

2

Control volume

Q- W = m [ ( h2- h1 ) + ½ (C22-C1

2) + g (z2-z1)] Enthalpy kinetic Energy Potential Energy

Apart from hydraulic machines the Contribution of potential Energy change is negligible.

Q- W = m [ ( h2- h1 ) + ½ (C2

2-C12)]-------------------------------------(1)

Stagnation enthalpy ho = h + ½ C2

Q-W = m ( h02-h01) Most turbomachinery flow processes are (nearly) adiabatic [Q = 0]. For work Producing machines (turbines). W = Wt = m (h01-h02) For work absorbing machines (Compressors ). -W = WC = m (h02-h01)

Newton’s Second Law of motion 1).for a system ∑ Fx = d/dt { M Cx } = [(M/∆t). ∆(Cx)] ( For control volume).

= m {CX2-CX1} [m mass/sec]▬

2). Euler’s equation of motion 1/ρ dp + cdc + gdz = 03). Bernoulli’s equation ∫1

2 1/ρ dp + ½ (C22-C1

2) + g (Z2- Z1) = 0

For incompressible fluid ρ is constant 1/ρ (p2-p1) + ½ (C2

2-C12) + g (Z2- Z1) = 0

1/ρ (p02-p01) + g (Z2- Z1) = 0

Where po = stagnation pressure = p + ½ ρc2

If change of gravitational potential is negligible 1/ρ (p02-p01) = 0

For incompressible fluid p02=p01 = p0

4 ) MOMENT OF MOMENTUM ------- Rate of change of moment of momentum on system is equal to the moment of the force acting.

OR 2 Vector sum of moment of all external forces acting on the system is equal to

rate of change of angular momentum {or moment of momentum}. Cr T = d/dt (r MCn ) = M d/dt ( r Cn ) C = (M/ ∆ t) . ∆ ( r Cn) [ for control volume] Cn =m (r2 Cn2-r1Cn1) r Work done /sec O 1 W = T ω = m(u2 Cn2-u1Cn1) Where u = blade speed =ωr This is referred as Euler’s turbo-machine equation.

The second law thermodynamics:---

In order to specify the degree of imperfection of actual processes in Turbo machines, an ideal process is required for comparison. The second law of thermodynamic defines an ideal process by introducing the concept of entropy.

The inequality of Clausius states that. ∫dQ/T ≤ 0 Where dQ is an amount of heat transferred to the system at an absolute

temp T. For reversible processes in the cycle ∫ dQR/T= 0 . The property called entropy, for a finite change of state is then defined as S2-S1 = ∫1

2 dQR/T= 0

For an incremental change state

dS = mds = dQR/T; dQR /m =Tds For steady one dimensional flow through a control volume in which the fluid experiences a change of state from condition l at entry to 2 at exit ∫1

2 dQ/T ≤ S2-S1

≤ m ( s2-s1 )

If the process is adiabatic dQ = 0 so s2 >s1

If the process is reversible adiabatic s2 =s1  Thus for a flow which is adiabatic, the ideal process will be one in which the entropy remain unchanged during the process.

From First Law of thermodynamicsdE = dQ –dW i.e. dQ = dE + dW

For reversible process for unit mass, dQR /m =Tds = (dE = dW )/m =du + pdv

With h = u +pv ; dh = du +vdp = Tds + vdpOr Tds = dh – dp/ρ = dh –dp/ρ 

Stagnation properties

• Stagnation (or Total) enthalpy

• Stagnation temperature

• Stagnation Pressure

Stagnation (or Total) enthalpy

Stagnation enthalpy h0 is the enthalpy which a gas stream of enthalpy ( h ) and velocity (C) would possess when brought to rest adiabatically (Q = 0) and without Work transfer (W = 0).

Energy equation (1) then reduces to (h0 - h) + ½ (0-C2) = 0And thus ho is defined by h0 = h+C2/2

Stagnation temperature When the fluid is a perfect gas CpT can be substituted for h, and the

corresponding concept of Stagnation (or total) temperature To is defined as h0 = h+C2/2

To = T+C2/2Cp C2/2Cp is called dynamic temp.

T-Static temp. [For atmospheric air, Cp = 1.005KJ/kg k flowing at 100m/s

To – T = 1002 / (2*1.005*103) = 50K ]From energy equation with no heat & work transfer. Q – W = 0 = m [ h02-h01] (h02 = h01 & T02 = T01)h0 & T0 Remains Constant . It is easier to measure Stagnation temperature of high velocity stream than

the static temp.

Stagnation Pressure:-

When the gas is slowed down and the temperature rises there is a simultaneous rise in pressure. The stagnation (or total) pressure p0 is defined in a similar way to T0,but with the added restriction That the gas is imagined to be brought to rest not only adiabatically but also reversibly i.e. Isentropically.

The stagnation pressure is thus defined by p0/p = (T0/T) γ/γ-1

Stagnation pressure, unlike stagnation temp, is constant in a stream flowing without heat or work transfer only . If friction is absent; the drop is Stagnation pressure can be used as a measure of the fluid friction.

p0 is not same as usual pitot pressure p0*, defined for incompressible flow by p0* = p +ρC2/2

From earlier equation p0/p = (T0/T) γ/γ-1 = [(T+C2/2Cp)/T] γ/(γ-1) Substitute [Cp= γR/(γ-1)]

= [1+C2(γ-1)/(2γRT)] γ/(γ-1) with [ a2= γRT] = [1+C2(γ-1)/(2 a2 )] γ/(γ-1) = (1+ M2 (γ-1)/2) γ/(γ-1)

= 1+γ/2 M2 +γ/8 M4 +γ(2-γ)/48M6+ _ _ _ _ _ _ _ _ _ _ _ _ (p0/p) -1 = ( p0-p)/p = γ/2 .M2 +γ/8 .M4 +γ(2-γ)/48.M6+_ _ _ _ _ _

( p0-p)/[p(γM2/2)] = 1+ M2/4+ (2-γ)/24M4 +_ _ _ _ _ _ _ _ _ _ _ _

But p(γM2/2) = ρRT (γ/2) C2/ γRT = ½(ρC2) Therefore pressure coefficient is( p0-p)/(½ ρC2) = 1+ M2/4+ (2-γ)/24M4 +---------------------------

=1+ M2/4+ 1/24M4 + ----------------------- [for γ= 1.4]

M 0.1 0.2 0.3 0.4 0.5 1.0

Percentage Deviation

0.3 1.0 2.3 4.1 6.4 27.5

For incompressible flow

( p*o -p)/(½ ρC2) = 1

From this equation the percentage deviation of the pressure coefficient ( po -p)/(½ ρC2) from incompressible flow value [ ( po -p)/(½ ρC2) = 1] with Mach no. can be calculated.

Max energy

Difference possible

Energy transferred to

rotor

Mechanical energy Available

at the shaft

Definition of Efficiency

Efficiency of turbines Fluid Energy ▬converted to mechanical energy of rotation .

Adiabatic or Mechanical Hydraulic efficiency efficiency

Efficiency of turbines

Enthalpy entropy diagrampo1

p1

p02

p2

01

1

02s

2s

C12 /2

C2s2 /2

C22 /2

Turbine expansion process

h

ss1 s2

02

2

WaWi

For turbines there are several ways of expressing adiabatic efficiency depending upon whether the exit Kinetic energy is usefully employed or is wasted. The examples of kinetic energy not wasted are

(a) Last stage of an aircraft gas turbine where it contributes to jet propulsive thrust.

(b) The exit kinetic energy from one stage of a multistage turbine is used in the next stage.

In these two cases the turbine and stage adiabatic efficiency is known as the total-to-total efficiency.

ηtt = (h0l-h02)/(h01-h02S)

If The difference between the inlet and outlet kinetic energy is small i.e.

½ C12 ≈ ½ C2

2 ≈ ½ C22S

ηtt = (hl-h2 )/ (h1-h2S ) When the exhaust kinetic energy is not usefully employed, the

relevant adiabatic efficiency is the. Total-to-static efficiency ηts = (h0l-h02 )/(h01-h2S) for small difference of inlet and outlet kinetic energies ηts = (hl-h2 )/(h1-h2s + ½ C1

2 ).A situation where the outlet kinetic energy is wasted, is a turbine

exhausting directly to the Surroundings.

Efficiency of compressorsIn compressors the compressed air carry its kinetic energy along with and the only meaningful efficiency is the total - to - total efficiency which is ηC = (h02S-h01 ) /(h02-h01)

For small difference of kinetic energy at

inlet and outlet

ηC = (h2S-h1) / (h2-h1) = Cp (T2S-T1)/Cp (T2-T1)

= (T2S-T1)/ (T2-T1)

02

2

po2

p2

p01

p1

02s

2s

01

1C12 /2

C2s2 /2

C22/2

h

sCompression process

Small stage or polytropic efficiency ( ηP ) compression process ηP = dhs /dh dhFor a reversible process Tds = dh - vdpfor one isentropic process Tds = o = dhS –vdpOr dhS = vdp Therefore ηP = vdp/(CpdT) =(RT/p)dp/(CpdT) [ Substituting v = RT/P] dT/ T = [(γ-1)/γηP ]( dp/ p) [Substituting CP = γR/γ-1] Integrating T2/T1 = (p2/p1) (γ-1)/η

pγ --------------- (A)

For a compressor adiabatic efficiency is (For equal velocity at inlet & outlet) ηc = (h2S-h1 ) /(h2- h1 ) = (T2S-T1 ) /(T2-T1 )

h

s

dhs

p+dp

p

For ideal compression process 0.9 T2S/T1 = (P2/P1) (γ-1)/γ

which can also be obtained from (A) by 0.8 substituting ηP =1 Therefore ηC

ηC = (T2S-T1 ) /(T2-T1 ) = [T1 (T2S/T1-1)] / [T1 (T2/T1-1)] 0.7 = [(P2/P1) (γ-1)/γ -1] /[ (P2/P1) (γ-1)/γη

P - 1]

0.6

This value for different values of ( P2/P1 ) is as shown

The actual irreversible adiabatic expansion process can be considered as equivalent to a polytropic Process (hence the term polytropic efficiency) with index n (pvn=const)

Pr. Ratio P2/P1

ηP =0.8

ηP =0.7

ηP =0.9

Therefore T2/T1 = (P2/P1) (γ-1)/(η

Pγ)

= (p2/p1) (n-1)/n

Equating the indices (γ-1)/γηP = (n-1)/n ηP = [ (γ-1)/γ] * [ n/(n-1)] n = γηP / [1- γ(1- ηP)] Turbine expansion process – The equivalent relation will be T2/T1 =(P2/P1) η

P(γ-1)/γ

n = γ / [γ-(γ-1) ηP]

ηt =[1-(P2/P1) ηP

(γ-1)/γ] / [1- (p2/p1) γ-1/γ] [ DERIVE THESE RELATION ]

Nozzle efficiency

In a large number of turbo machine components the flow process can be regarded as a purely nozzle flow in which the fluid receives an acceleration as a result of drop in Pressure.

From steady flow energyequation with no shaft work h02-h01= 0

so h01=h02

h1+ ½ C12=h2+ ½ C2

2

h1-h2 = ½ (C22-C2

1) (A)

For equivalent reversible Adiabatic process h01 = h02S

h1-h2S= ½(C2S2-C1

1) (B)

the nozzle efficiency can be define as ηN = ( h1-h2)/(h1-h2S) ( Substituting from (A) & (B) )

= [½(C22-C1

2)] / [½(C2S2-C1

2)

= (C22-C1

2) / (C2S2-C1

2)

p1

p2

1

2s 2

C22 /2

C12 /2

C1s2/2

s

h

01 02

For an isentropic change Tds = dh - vdp=0 . If the fluid is treated as incompressible, the change of state from l to 2s ∫1

2Sdh =1/ρ∫12S dp

h1-h2S =1/ρ (P1-P2)

h2-h2S = (h1-h2S)-(h1-h2)

= 1/ρ(P1-P2)-1/2(C22-C1

2)

=1/ρ(P01-P02)

ηN = (h1-h2)/(h1-h2S )

=[ (h1-h2S)-(h2-h2S)] / (h1-h2S)

=1- (h2-h2S) / (h1-h2S)

=1- ( p01-p02) / (p1-p2 )

Diffuser efficiency For steady adiabatic flow in a stationary passage h01= h02 ;

i.e. h1+ ½ C12=h2+ ½ C2

2

h2-h1= ½(C12-C2

2)

h2s-h1= ½ (C12-C2s

2)ηD = ( h2s-h1)/(h2-h1) = [ (C1

2-C2s2)] / (C1

2-C22)

For isentropic in compressible flow h2s-h1 =1/ρ(p2-p1) ηD = 2(p2-p1) /ρ(C1

2-C22)

01 02 p2

C22 /2C1

2 /2s

1

22s

p1

h

s

h2-h2s = (h2-h1) - (h2s-h1) = ½(C1

2-C22) - 1/ρ(P2-P1)

=1/ρ(p01-p02) ηD = (h2s-h1) / [ (h2s-h1)- (h2s-h2) ] = 1 / [1- (h2s-h2) / (h2s-h1) ] = 1/ [1+ (p01-p02) / (p2-p1) ]For an incompressible flow energy equation (Bernoulli’s equation)

written as p1/ρ + ½ C1

2 = p2/ρ + ½ C22 + ∆p0 /ρ

Where the loss in total pressure ∆p0 = p01-p02

Compressible Flow

Sonic velocity a = √(dp/dρ) = √(γP/ρ) =√(γRT)

[For isentropic process ] Mach no M = C/a

Isentropic Flow through varying area passage Continuity Eq. ρAC= Const ; Which gives d ρ / ρ +dA/A +dC/C = 0 (1)

Euler’s equation; dp/ρ +C dC = 0 or (dp/d ρ).(d ρ / ρ) +C dC = 0 (2) Or a2 .(d ρ / ρ) +C dC = 0

Substituting d ρ / ρ from (1) ; - a2 .(dA/A +dC/C ) +C dC =0 dA/A =C dC/ a2 -dC/C = (dC/C).(C2/a2 -1) dA/A =(dC/C) (M2-1) = - (dp/ρC2) [M2-1] [Euler’s eq. dC = - dp/ρC ]

For accelerating flow or flow through nozzle (dC = + ve) For M < 1; dA is – ve or area must decrease For M > 1; dA is + ve or area must increase For decelerating flow or flow through diffuser (dc = -ve)

For M < 1; dA is + ve or area must increase For M>1; dA is - ve or area must decrease

dA = o for M =1 indicating minimum area or throat area .

Converging-diverging nozzle

Inlet pressure

Critical pressure

M<1

M<1

M<1

M>1

Subsonic diffusion

Normal shock

Normal shock

pr

Length

M=1

• From steady state energy equation• • h01 =h 02

• h1+ ½ C1 2 =h2+ ½ C2

2

• Putting C = M.a =M√(γRT)• CpT1+ ½ M1

2γRT1=CpT2+ ½ M22γRT2

• Putting Cp= γR/γ-1 and rearranging• T2/T1=(1+ ½(γ-1)M1

2)/(1+ ½(γ-1)M22)

• For isentropic flow• P2 /P1= (T2/T1)γ/γ-1=[ (1+ ½(γ-1)M1

2)/(1+ ½ (γ-1)M22) ] γ/γ-1

• Stagnation properties• h0=h + C2/2• T0=T+C2/2Cp• T0/T= 1+ C2/2CpT= (M2 .γRT)/[2γR/(γ-1) = 1+[½(γ-1)]M2

• • p 0/p =( T0/T)γ/(γ-1) =(1+{½(γ-1)}M2 ) γ/(γ-1)

• • ρo /ρ = (P0/P)1/γ = (1+{½(γ-1)}M2 )1/(γ-1)

• • Mass flow rate per unit area • m =ρAC

• m/A =ρC =(P/RT)C = (p0/p) .( po/√To).√ (T0/T) .{C/√(γRT)} .√(γ/R) • Substituting for (p0/p) and (T0/T) from above and simplifying

• m/A =√(γ/R) .po/√To . M . (1+ ½ (γ-1)M2 ) –(γ+1)/2(γ-1)

• Mass flow rate per unit area is maximum When• d (m/A)/dM =0 • This condition gives 1-M2 =0.• Or M =1.• Which is the condition of max mass flow at the throat. This condition is

called CHOCKED FLOW.• • The properties for this condition are denoted by (*), • (P*,ρ*,T*,A*) and are terms as a critical properties.• For this condition; ( M=1)• m/A* ==√(γ/R) *Po/√To * (2/(γ+1) )(γ+1)/2(γ-1)

• • T0/T* = (γ+1)/2• P0/P* = ((γ+1)/2) γ/(γ-1)

• ρ 0/ρ =((γ+1)/2) 1/(γ-1) • A/A* =1/M [(2+(γ-1)M2)/γ+1] (γ+1)/2(γ-1)

• • Where A is the area at a section and A* is the area at the throat

• GAS TABLE • The valve of M*(value at a point/Critical value

at throat ) =(C/A*) , A/A*, T0/T, P0/P & ρ 0/ρ computed for an ideal gas having γ =1.4 for various value of mach no. M from above equations. These are given in the gas table for isentropic flow.

Flow over an aerofoil

M<1M<1

M>1

Shock wave

p p+Δp

Low speed High speed

AB ----Loss of head due to shock waveC-------Peak loss due to separation because of shock wave

A

B

C

• PLAIN NORMAL SHOCK WAVE• • When shock waves occur normal to the axis of the flow, they are

discontinuities which occupy a finite but very short length of duct for this reason they can be treated as adiabatic frictionless processes in a duct of constant cross sectional area .In general, three conservation laws are satisfied without satisfying reversibility.

• Flow in a duct Flow at entrance to nozzle

21

STEADY STATE ENERGY EQATION h02-h01 = 0

Cp(T2-T1) +1/2(C22-C1

2) = 0………………….(4)

From eq.….(3) p2+ ρ2C2

2 =p1 + ρ1C11

p2(1+( ρ2C22 )/p2) =P1 (1+ ρ1C1

2 / p1)

p2/p1 = (1+ ρ1C12 γ/γp1) /(1+ ρ2C2

2 γ/γ p2)

= [1+γC12/a1 2]/ [1+γC2

2 /a22 ]

• = (1+γM12)/(1+γM2

2)………………………………..(5)

From eqn ….(4) , Energy equation Since T02 = T01

T2/T1 = (T2/T02) * (T01/T1)

= (1+(γ-1)/2M12)/(1+(γ-1)/2M2

2)

= (2+(γ-1) M12)/(2+(γ-1) M2

2)………………………………..(6)

But M2/M1 =(C2/a2)/(C1/a1)

= C2/C1 * √(γRT1)/√(γRT2)

= ρ1/ρ2 *√(T1/T2) ( From eqn ..2)

= (p1/RT1)/(p2/RT2) *√(T1/T2)

= p1/p2 * √(T2/T1)

Substituting from (5) and (6) M2/M1 = (1+γM2

2)/(1+γM12) *√[ (2+(γ-1) M1

2)/(2+(γ-1) M22)]

Simplifying M2

2 = [ (γ-1) M12)+2] /[2γM1

2-(γ-1) ] ………………(7)

Substituting for M2 in eqn ..(5)& eqn ..(6)Across the shock wave P2/P1 = ( 2γM1

2-(γ-1) )/ (γ+1)………………………………………………(8) T2/T1 = [ 2γM1

2-(γ-1) ] * [ (2+(γ-1) M12)/(γ-1) M1

2)]………….(9) P02/P1 = P02/P2 * P2/P1

= (1+(γ-1)/2M2

2)γ/( γ-1) * ( 2γM12-(γ-1) )/ (γ+1)

= [1+ (γ-1)/2 *( (γ-1) M1

2)+2) /(2γM12-(γ-1) ) ] γ/( γ-1) *[ ( 2γM1

2-(γ-1) )/ (γ+1)] = [(γ+1) M1

2)] γ/( γ-1) / [( 2γM12-(γ-1) )/ (γ+1)] 1/( γ-1) ……..(10)

• Evaluation of equation (7) shows that for M1 > 1 M2 < 1;

• while for M1 < 1, M2 > 1• But calculation of change of entropy• • ∆s/Cv = γ ln T2/T1 - (γ-1) ln P2/P1

• Will give for M1 > 1 , +ve value of ∆s where as for M2< 1, -ve valve of ∆s which is impossible therefore normal shock wave can only arise in a supersonic flow.

• • For different value of M1 and for γ = 1.4 . The value of M2 ,

P2/P1 , T2/T1 , P02/P1 ρ 2/ρ1, P02/P01 computed from equation (7 ), (8),(9) etc. are given in gas tables for normal shock wave .

EXAMPLEA stream of air flows in a duct of 100 mm dia at a rate of 1kg/s. The stagnation temp. is 370C. At one section of the duct the static pressure is 40 kPa . Calculate the mach no. , Velocity, and stagnation pressure at this section.

SOLUTIONGivenT0 = 37+273 = 310k, P = 40 kpa, γ = 1.4. Mass flow rate per unit area ism/a = ρv = p/RT .√(γRT) .M = √(γ/R) . PM/√T0 .√T0/T = √(γ/R) . PM/√T0 . (1+(γ-1)/2M2)1/2

1/[(π/4)(0.1)2] = √(1.4/(0.287.1000).40 kN/m2 . M/√310 .[1+0.2 M2]1/2

127.39 = √(1.4/[(0.287 . 1000) .310] . 40 .103 M . [1+0.2 M2]1/2

M . [1+0.2 M2]1/2 = 0.803 M . [1+0.2 M2] = 0.645 ; M4 + 5M2 – 3.225 = 0 M2 = -5± √(25+12.9)/2 = (-5+6.16)/2 = 0.58 M = 0.76 Ans.

• T0/T1 = 1+(γ-1)/2 M2 = 1+0.2 . (0.76)2 = 1.116• • T = 277.78 K. Ans.• • C = √γRT • = √(1.4 . 0.287 . 1000 . 277.78)• = 334.08 m/s Ans.• • V = CM = 334.08 . 0.76• = 253.9 m/s Ans.• P0/P =( T0/T ) γ/( γ-1) = (1.116)1.4/0.4 = 1.468• • P0 = 40 . 1.468 = 58 .72 kpa. Ans.•

EXAMPLE- 2

A conical air diffuser has an intake area of 0.11 m2 , an exit area of 0.44 m2 . Air enter the diffuser with a static pressure of 0.18 Mpa , static temp. of 370C , and velocity of 267 m/s.

Calculate a. Mass flow rate of air through the diffuser. b. The mech no. , Static Temp. , and static pressure of air leaving the

diffuser.

2

2

1

1

T1 =370C

P1 = 0.18MPa V1 = 267m/sA1 = 0.11 m2

A2 = 0.44m2

M2 = ?T2 = ?P2 = ?

V2

• Sol.

• m = ρAV • = p/(RT1) * A1 * V1• = (0.18 *106 * 0.11 * 267)/(287*310)• = 59.42 kg/s Ans.• C = √γRT • = √(1.4 * 287 *310) • = 352 m/s Ans.• • M1 = V1/C1• = 267/352• = 0.76. Ans.• From gas table for the isentropic flow of air (γ=1.4).• When M1 = 0.76;• A1/A* = 1.0570 , P1/P01 = 0.68207, T1/T01 = 0.89644 • A2/A1 = 0.44/0.11 = 4 = A2/A* / A1/A* •

A2/A* = 4 * 1.0570 = 4.228

• Again from isentropic flow table when A2/A* = 4.228• • M2 = 0.135,

• P2/P02 = 0.987 , T2/T02 = 0.996• • P2/P1 = (P2/P02 )/ (P1/P01 ) = 0.987/0.682 = 1.447

• P2 = 1.447 *0.18• = 0.26 Mpa. Ans.• • T2/T1 = (T2/T02 )/ (T1/T01)

• = 0.996/0.89644• = 1.111• T2 = 1.111* 310• = 344K Ans

• EXAMPLE 3.• A convergent – divergent nozzle has a throat area 500 mm2. Air

enters the nozzle with stagnation Temp. of 360K and stagnation pressure of 1 Mpa. Determine the Max. flow rate of air that nozzle can pass, and the static pressure, static temp, mach no. , and velocity at the exit from the nozzle if

• (a)The divergent section acts as a nozzle. And • (b) the divergent section acts as a diffuser.

1

1

*

*

2

2

T01 = 360KP01 = 1MPa

A* = 500mm2

A2 = 1000mm2

• Sol • A2/A* = 1000/500 = 2• • From isentropic flow rate , when A2/A* = 2 There are two value of the mach no.,• One for supersonic flow when the divergent section acts as a nozzle , and the

other for subsonic flow when divergent section acts as a diffuser, which are • M2 =2.197&M2 =0.308.• • (a)When M2 =2.197 Ans.• • P2/P0 = 0.0939, T2/T0 = 0.5089 [P0= P01 , T0=T0 1]

• P2 = 0.0939 *1000 • = 93.9 Kpa. Ans.

T2 = 0.5089 * 360 = 183.2 K Ans.C2 = √γRT2 = 20.045 * √183.2 = 271.2 m/s.V2 = 271.2*2.197 = 596 m/s. Ans.

Mass flow rate m = ρ*v*A = ρ2v2A 2 = ρ1v1A1

For throat M* =1, from gas table (γ=1.4) for air P0*/P0 = 0.528 & T*/ T0 = 0.833

ρ* = p0*/(RT*) = (0.528 * 106)/(287 * 0.833 *360) = 6013 kg/m3.

T* = 360 *0.833 = 300KV* = √γRT* = 347.2 m/s. m = 500 * 10-6 * 6.13 * 347.2 = 1.065 Kg/s. Ans.

(b) M2 =0.308P2/P0 = 0.936 , T2/T0 = 0.9812P2 = 0.936 *1 = 0.936 Mpa.T2 = 0.9812 *360 = 35302 K.C2 = √γRT2 = 376.8 m/s.V2 = 376.8 *0.308 = 116 m/s.m = 1.065Kg/s Ans.

Example 4 A convergent – divergent nozzle operate at off design condition while

conducting air from a high pressure tank to a large pressure tank to a large container .A normal shock occurs in the divergent part of the nozzle at a section where the cross sectional area is 18.75cm2. if the stagnation pressure and stagnation temp. at the inlet of the nozzle are 0.21 Mpa and 360C resp. and the throat area is 12.50cm2 and the exit area is 25 cm2 estimate the exit mach no. , exit pressure , loss of stagnation pressure and entropy increase during the flow between two tanks

1

1 A* = 12.5 cm2

*

*

2

A2 = 25cm2

2

MX My

• Sol • • At shock section • Ax/A* = 18.75/12.5 = 1.5• Up to the shock, the flow is isotropic, from isentropic flow table, when• A/Ax = 1.5 ;

• Mx = 1.86 Px/P0 x = 0.159

• Px = 0.159*0.21*1000 = 33.4 Kpa• From the gas table on normal shocks • When Mx = 1.86; My = 0.604

• Py/P x = 3.87

• P0y/P x = 4.95, P0y/P0 x = 0.786

• Py = 3.87 * 33.4• = 129.3 Kpa• P0y = 4.95*33.4• = 165.3 Kpa

• From the shock section to the exit of the nozzle, the flow is again isentropic.• My = 0.604 from the isentropic table

• Ay/A* =1.183

• A2/A* = (A2/Ay) *( Ay/A* ) • = 25/18.75 * 1.183• 1.582• A2/A* = 1.582 , from isentropic flow tables M2 = 0.402 Ans.

• P0/P 0y = 0.895 .

• P2 = 0.895 * 165.3• = 147.94 Kpa.• Loss in stagnation pressure occurs only across the shock • • P0x - P0y = 210-165.3 = 44.7 Kpa• • Entrpy increase Sy – Sx = R ln P0x/P 0y

• = 0.287 ln 210/165.3• = 0.00686KJ/Kg K Ans.