principle of thermodynamics and fluid flow

Table of Contents

Chapter 1: Principle of Thermodynamics and Fluid Flow........31.1 Mass Conservation Principle.............................3Figure 1.1.1 Turning of flow by steam nozzles.............4

1.2 First Law of Thermodynamics.............................5Figure 1.2.1 Flow through a diffuser......................7

1.3 Entropy.................................................81.3.1 Isentropic Relations................................81.3.2 Gas Tables.........................................10

1.4 Equations of State.....................................111.4.1 Ideal Gases........................................11Figure 1.4.1 Specific heat for air and their ratio.......12

1.5 Efficiency.............................................12Figure 1.5.1 Thermodynamic states used to define turbineefficiency...............................................13Figure 1.5.2 Thermodynamic states used to define compressorefficiency...............................................151.5.1 Thermodynamic Losses...............................20Figure 1.5.3 An infinitesimal irreversible process inturbine and compressor...................................211.5.2 Small Stage or Polytrophic Efficiency..............22Figure 1.5.4 Compression process by small stages.........22Figure 1.5.5 Incremental change of state in a compressionprocess..................................................24Figure 1.5.6 Isentropic efficiency, Polytropic efficiencyand pressure ratio for compressor........................25Figure 1.5.7 Turbine Isentropic efficiency...............26

Principle of Thermodynamics and Fluid Flow

1.6 Newton’s 2nd Law of Motion............................291.7 Summary and Key Points...............................30

Chapter 1: Principle of Thermodynamics and Fluid Flow

In this chapter we introduce the mass conservation principlefollowed by the basic physical laws of thermodynamics asapplicable to turbomachines. Two important equations of statefor ideal gases are suggested as isentropic relation ofentropy for a pure substance. We also aim to finding thevarious measures of efficiency, both isentropic and polytropicfor turbo-machinery flows and their relation to thermodynamiclosses. Newton’s second law of motion is proposed for a linearmotion.

1.1 Mass Conservation Principle

The mass flow rate (ṁ) for a uniform flow of a fluid isexpressed in terms of density (ρ), velocity (V), and the cross-sectional area of the flow channel (A) as:

ṁ = ρ Vn A

Specific volume (ν), also known as the reciprocal of density,is used in the above equation for analysis of flows.

Here,

subscript n is the direction normal to the flow area.

Vn A is the scalar product V . n = V cos Ɵ,


Where n is a unit normal vector on the surface A, and Ɵ is theangle between the normal and the direction of the velocityvector. Therefore, the scalar product can also be written inthe two alternative forms as:

V. n A = VA cos Ɵ = Vn A = V An

An is the area normal to the flow. Thus, for a uniform steadyflow through a control volume with one inlet and one exit, theprinciple of conservation of mass can be expressed as

ρ1 V1 An1 = ρ 2 V2 An2

Alternatively, for a flow that has more than one inlet andexit points, then for all inlet and exit points, theconservation of mass is given by:

Ʃ ρi Vi Ani = Ʃ ρe Ve

Ane--------------------------------------- (1.1.1)

Example 1.1.1

Gas flows at the rate m = 0.20 kg/s through each nozzle in thebank of nozzles shown in figure. The inlet specific volume is0.60 m3/kg and at the outlet it is 1.00 m3/kg. Spacing of thenozzles is s = 5.0 cm, wall thickness at the inlet is t1 = 2.5mm, and at the outlet it is t2 = 2.0 mm. Blade height is b = 4.0cm. Nozzle angle is a2 = 70°. Find the steam velocity at theinlet and at the outlet.


Figure 1.1.1 Turning of flow by steam nozzles

Solution

The Area of the inlet is A1 = b(s- t1) = 4(5-0.25) = 19 cm2

Velocity at the inlet is given by,

ṁ = ρ1V1A1 =V1A1

v1

thus V1 = ṁv1

A1 = 0.20∗0.60∗100

2

19 = 63.15 m /sec.

The Area of the outlet is A2 = b(s cosa2- t2) = 4(5 cos(70⁰)-0.2) =6.04 cm2 .

Hence the velocity at outlet is

V2 = ṁv2

A2 = 0.20∗1∗100

2

6.04 = 331.25 m /sec.


1.2 First Law of Thermodynamics

The first law of thermodynamics for a uniform steady flow in achannel can be expressed as:

ṁ (u1 + p1 v1 + 12 V21 + gz1) + Q = ṁ (u2 + p2 v2 + 12 V 2

2 + gz2) + Ẇ-------------

(1.2.1)

Here,

u + 12 V2+ gz = e is the specific energy of the fluid which is

the sum of internal energy (u), kinetic energy (v22

), and

potential energy gz, where, g is the acceleration of gravityand z is the height.

p1 v 1 is the energy flow into the control volume also known asflow work.

p2 v 2 is the energy transfer as work leaving the controlvolume.

h = u + pv is the Enthalpy which is the sum of internal energyand flow work.

Q is the heat transfer rate into control volume and Ẇ is thework deliver rate.

On dividing the Eq. 1.2.1 withṁ, and substituting q = Q /ṁ andw = Ẇ/ṁ as heat transfer and work done per unit mass, we get amodified form of First law of thermodynamics as:

h1 + 12 v21 + gz1 + q = h2 + 12 v2

2 + gz2 + w


The sum of enthalpy, kinetic energy, and potential energy is

called the stagnation enthalpy, denoted by, h0 = h + 12 v

2 + gz,

which give us another modified form of 1st law ofthermodynamics as:

h01 + q = h02 + w

In case some water turbines there is a need to retain thepotential energy terms. Otherwise, for flow of gases thepotential energy terms can be neglected as they are small.Similarly, in case of pumps the potential energy differencecan be neglected as the changes in elevation is small.

On neglecting the change in potential energy, the first lawreduces to:

h1 + 12 v

21 + q = h2 +

12 v

22 + w

This equation is further simplified because even if velocityis large and the difference in kinetic energy between theinlet and exit may be small. We now have:

h1 + q = h2 + w

Since the turbomachinery flows are almost adiabatic, we candrop q from the above equation, which will give us the value ofwork delivered by a turbine as

w = h01 — h02 = ∆h0 (Stagnation Enthalpy Change)

And the work done on the fluid in a compressor as

w = h02- h0i

It should be observed the equations are written in a form thatgives a positive value for work, for both a turbine and acompressor. The convention of thermodynamics of denoting workout from a system as positive and work in as negative isignored. (SAME PARAGRAPH) Thus, we assume that during flow inPrinciple of Thermodynamics and Fluid Flow

a turbomachine there will be kinetic energy (high velocity)and stagnation energy change will be considered under dynamicconditions. In case of power generating turbo machines,stagnation enthalpy decreases from inlet to outlet of a turbomachine rotor for which ‘w’ will be positive and ∆h0 will benegative. While in case of a power absorbing turbo machine,stagnation enthalpy will be increasing from inlet to outlet ofa turbo machine rotor, and hence ‘w’ will be negative and ∆h0

will be positive.

It is critical to note that the work done or Enthalpy changeoccurs only during transfer of energy through rotors orimpellers. In case of stators or fixed passages only pressurechange, kinetic or potential energy change occurs depending onshape during the dynamic action of flow in the turbo machine.

Example 1.2.1

Gas flows adiabatically at a rate m = 0.02 kg/s through adiffuser, shown in figure, with inlet diameter D1 = 2.0 cm.Specific volume at the inlet v1 = 2.40 m3/kg. Exit diameter is D2

= 2.5 cm, with specific volume at the outlet v2 = 3.20m3/kg. Findthe change in enthalpy neglecting any change in the potentialenergy.


Figure 1.2.1 Flow through a diffuser

Solution:

The area of inlet is A1 = πr2 = 3.14 * (0.01)2

=3.14 * 10-4 m2

The area of outlet is A2 = πr2 = 3.14 * (0.0125)2

=4.90 * 10-4 m2

The velocity at inlet is V1 = ṁv1

A1 = 0.02∗2.403.14∗10−4 = 152.86 m

/sec, and at the outlet is

V2 = ṁv2

A2 = 0.02∗3.204.9∗10−4 = 130.62 m /sec.

Since no work is done and the process is adiabatic so thestagnation enthalpy remains constant. With negligible changein potential energy, this equation reduces to

h2 - h1 = 12V1

2- 12V22 = 12(152.86

2 – 130.622)


= 3.15 kJ/kg

1.3 Entropy

Entropy is classically defined as a measure of the unavailableenergy in a closed thermodynamic system that is also usuallyconsidered to be a measure of the system's disorder, and thatvaries directly with any reversible change in heat in thesystem and inversely with the temperature of the system.Entropy is the function of the state of the system. Thisimplies that the change in its value is independent of thepaths between initial and last states. Entropy (S) is athermodynamic property defined by the relation:

ds = (δQT )reversible ------------------------------------------------------------------------------------------

(1.3.1)

ds = δQT , for an infinitesimal reversible process, for unit

mass

ds > δQT , for an infinitesimal irreversible (spontaneous)

process, for unit mass

Equation 1.3.1 applies only for a reversible process, hencethe entropy change must be calculated from values of heat andtemperature along a reversible path between initial and laststate.

1.3.1 Isentropic Relations

This section will derive two extremely important relationsinvolving entropy for a pure substance. Leaving out the


kinetic energy and potential energy for a closed system, thedifferential form of first law of thermodynamics is writtenas:

δq = du + δw

For a reversible process the work done is given by theequation

δw = p dv

thus we get, T ds = du + p dv.----------------------------------------------------------------------- (1.3.2)

Equation 1.3.2 is deprived of any path function and ispresented only in terms of properties. Thus, this equation isapplicable for a state change due to flow across systemboundaries and hence can be used for both reversible andirreversible process. From the enthalpy equation we canfurther add that

h = u + pv, or

dh = du + p dv + v dp.

The above equation 1.3.1 gets transformed to

T ds = dh – v dp.

Thus on mass basis the equations are transformed to

T ds = du + p dv

T ds = dh - v dp.

Thus the ideal gas equation takes the form:

dh = cp dT

pv = RT.

Therefore, ds = cpdTT

−Rdpp .


Assuming constant specific heats, the equation yields

s2 – s1 = cplnT2

T1 – R ln

p2p1

--------------------------------------------------------------(1.3.3)

s2 – s1 = cv lnT2

T1 + R ln

v2v1

But the process is an adiabatic reversible process, for which

s2 – s1 = 0

Therefore equation (1.3.3) is transformed to

cplnT2

T1 = R ln

p2p1

cplnT2

T1 = (cp−cv) ln

p2p1, or

T2T1 = ¿) (ɤ-1)/k

-------------------------------------------------------------------------------(1.3.4)

or, T2T1 = ¿) (ɤ-1)

--------------------------------------------------------------(1.3.5)

or,p2p1 = ¿) ɤ

--------------------------------------------------------------------------------------(1.3.6)

The important thing to note here is that equations 1.3.4,1.3.5 and 1.3.6 are only applicable to an ideal gas withPrinciple of Thermodynamics and Fluid Flow

constant specific heat and undergoing a reversible adiabaticprocess.

1.3.2 Gas Tables

When accounting for higher temperature changes, variation ofspecific heat is not negligible and thus should be taken intoaccount when calculating the change in properties for an idealgas. A method for calculating the change in enthalpy andvariations of specific heat is to use tables that have alreadybeen printed such as the Gas Tables. Column 1 of the tablegives temperature in Kelvin, column 2 is standard-stateinternal energy in J/mol, column 3 gives standard-stateenthalpy, column 4 gives standard-state entropy, column 5 and6 list the relative pressure and relative volume respectively.

Table 1.3.1 Ideal gas entropy table for Air

The change in entropy can be easily calculated by integratingthe equation

ds=cpdTT

−Rdpp


For 1 mole of gas

ds =s2-s1 = ∫cpdTT - ∫ Rdp

P

By using the standard state entropy values from the tableabove

s2 –s1 = sOT2 - sOT1 – R lnp2p1

As for an adiabatic process s2 –s1 =0

Thus for an adiabatic reversible process the equation gettransformed to

sOT2 - sOT1 = R lnp2p1

or

p2p1 = e

soT2−soT1R

As it is clear from this equation that soT2 and so

T1 arefunctions of temperature only R is a constant, thus right sideof the equation is a function of temperature only.

Thus a new quantity is defined as Pr called the relativepressure as

Pr = esoTR

therefore, p2p1 = Pr2Pr1

The values of Pr have been provided in the table 1.3.1 incolumn 5.


1.4 Equations of State

The thermodynamic state for a compressible substance isentirely determined by specifying two independentthermodynamic properties and all other properties are just thefunctions of these two independent variables. This functionalrelation is known as ‘Equations of State”.

1.4.1 Ideal Gases

In an Ideal gas model, we assume that the internal energy isonly a function of temperature u = u (T). The relation betweenpressure and specific volume to temperature is given by

pv = RT; or p = ρ RT,

here R is an ideal gas constant and R =R /Μ, where R = 8.3145J mol-1 K-1,and Μ is the molecular mass.

For an ideal gas at constant volume and constant pressure,specific heat is given by:

cv(T )= ( δuδT)v = dudT , which implies du = cv(T) dT.

This represents the dimensionless heat capacity at constantvolume; it is generally a function of temperature due tointermolecular forces.

cp(T )= ( δhδT)p =

dhdT , which implies du = cp(T) dT

The heat capacity ratio or adiabatic index is the ratio of theheat capacity at constant pressure to heat capacity atconstant volume. It is sometimes also known as the isentropicexpansion factor:

γ=¿ cp / cv , which implies that,


cv = R

γ−1 ; and cp = γRγ−1 ; the values are depicted in the figure

below.

Figure 1.4.1 Specific heat for air and their ratio

1.5 Efficiency

Efficiency in general, describes the extent to which time,effort or cost is well used for the intended task or purpose.It is often used with the specific purpose of relaying thecapability of a specific application of effort to produce aspecific outcome effectively with a minimum amount or quantityof waste, expense, or unnecessary effort. This part is aims atfinding the various measures of efficiency for turbo-machineryflows and their relation to thermodynamic losses as discussed.

To measure how well a turbine is performing we can look at itsisentropic efficiency. This compares the actual performance ofthe turbine with the performance that would be achieved by anideal, isentropic, turbine. When calculating this efficiency,heat lost to the surroundings is assumed to be zero. The


starting pressure and temperature is the same for both theactual and the ideal turbines, but at turbine exit the energycontent ('specific enthalpy') for the actual turbine isgreater than that for the ideal turbine because ofirreversibility in the actual turbine. The specific enthalpyis evaluated at the same pressure for the actual and idealturbines in order to give a good comparison between the two.

A quantity called total-to-total efficiency is defined as theratio of actual work done to the isentropic work.

ŋtt = wws

= h01−h03

h01−h03s

---------------------------------------------------------------(1.5.1)

The label 1 is the inlet to a nozzle and label 3 is the exitstate from rotor.


Figure 1.5.1 Thermodynamic states used to define turbineefficiency

Figure 1.5.1 shows the adiabatic expansion between staticstate h1 and h3. The process lines have also been shown betweenthe stagnation states h01 and h03. In addition to the constantpressure lines corresponding to these states a line ofconstant stagnation pressure po3i is drawn.

A measure of the irreversibility, loss of stagnation pressureis given by,

Δp0 = p03i – p03.

But if the kinetic energy at the exit state being wasted andthe exit velocity is not diffused then a total-to-static


efficiency is used to get a measure of the efficiency of thesystem.

ŋts = h01−h03h01−h3s

----------------------------------------------------------------------(1.5.2)

The denominator is larger than in previous case, thusefficiency is reduced because of the wasted Kinetic energy.

The total-to-total efficiency can be written in the terms ofvelocity as

ŋtt = wws

=

h1+¿

12V12−

h3−¿ 1

2V32

h1+12V12−h3s−

12V3s2

¿

¿

-------------------------------------------------(1.5.3)

If the different between the inlet and outlet kinetic energiesare small

12V12=

12V32 then

ŋt = h1−h3h1−h3s

-----------------------------------------------------------------------(1.5.4)

For a compressor the process lines between the stagnationstates and the corresponding static states are shown in figure1.5.2 and the compressor efficiency is


ŋt = isentropicworkacutalwork

=h2s−h1

h2−h1-----------------------------------------------(1.5.5)

Figure 1.5.2 Thermodynamic states used to define compressorefficiency

Example 1.5.1

At a velocity of 50 m/sec steam enters a turbine at a staticpressure of 80 bars and temperature 520°C. It leaves theturbine at a pressure 0.35 bar, temperature 80°C and velocity200 m/sec. Find the total temperature and pressure at the


inlet, total temperature and pressure at the exit, total-to-total efficiency, total-to static efficiency.

Solution

Using steam tables static enthalpy and entropy of steam at theinlet and exit are

h1 = 3447.8 kJ/kg S1 = 6.7873 kJ/(kg • K)

he = 2645.0 kJ/kg Se = 7.7553 kJ/(kg • K)

Stagnation enthalpies are

h01 = h1 + 12V1

2

= 3447.8 + 5022∗1000

=3449.1 kJ/kg

h0e = he + 12V2

2

=2645 + 2002

2∗1000

=2665.0 kJ/kg

Since the flow is isentropic

xes = ses−sf

sg−sf

= 6.78−0.987.71−0.98

=0.8621

The enthalpy at this state

hes = hf + xes(hg-hf)


=304.20 + 0.8621( 2630.7 – 304.20)

= 2309.9 kJ/kg

Assuming Ves = Ve

h0es = hes + 12Ve

2

=2309.9 + 2002

2∗1000

=2329.9 kJ/kg

Total-to-total efficiency is

ŋtt = h01−hoeh01−hoes

= 3449.1−2665.03449.1−2329.9

=0.7006

Total-to-static efficiency is

ŋts = h01−hoe

h01−hes

= 3449.1−2665.03449.1−2309.9 =0.6883

And the efficiency when Kinetic energy is neglected is

ŋt = h1−he

h1−hes

= 3447.8−2645.03447.8−2309.9


=0.7055

Example 1.5.2

A fan with a radius R = 50 cm draws air from atmosphere atpressure 101.325 kPa and temperature 288 K. The volumetricflow rate is Q = 4.72 m3/s of standard air, and the power tothe fan is W= 2.52 kW. The total-to-total efficiency of fan is0.8.

Find

a) Total-to-Static efficiency

b) The stagnation pressure rise across the fan.

Solution

It is a simple reversible adiabatic process as air is drawnfrom atmosphere without any other constraints.

Therefore p01 = 101.325 kPa and T01= 288 K

Density of air ρ01 = 1.225 kg/m3

The mass flow rate thus can be calculated as

ṁ = ρ01 * Q

=1.225 * 4.72

=5.78 kg/s

The fan flow area and velocity are

A=πr2

= 3.14 * 0.50 * 0.50

=0.785 m2

V = QA


= 4.720.785

= 6.012 m/sec

The specific work is given by

w = Wṁ

= 25205.78 = 435.9 J/kg

Then the isentropic work is

ws = ŋtt * w

=0.8 * 435.8

= 348.7 J/kg

Total-to- static efficiency can be calculated as

ŋts = T3s−To1

T03−T01 = ŋtt –V3/2w

= 0.8 – 6.6∗6.62∗436

=0.75

(b) The stagnation pressure rise can be calculated from

P03-P01 = pws

= 1.225-348.7

= 427Pa

= 43.6 mm H2 0


Example 1.5.3

At a pressure 1 bar and temperature 300 K air blows into amultistage compressor. The ratio of total pressure acrosscompressor is 30 and its total-to-total efficiency is 0.85.Find the loss of stagnation pressure during this compressionprocess, assuming specific heats to be constant

Solution

The inlet is assumed to be state 1 and outlet as e . Thestagnation temperature is given by isentropic compression as

T0es = T01* (p0ep01

)(γ-1)/γ

= 300 * 30 1/3.5

=792.8 K

Efficiency is given by

ŋtt = T02s−To1

T02−T01

The exit temperature is given by

T0e = T01 + (ŋtt)-1* (T0es – T01)

= 879.76 K

If the same process had been done isentropically , thenpressure ratio would be

p0eip01

=(T0eT01

)γ/γ-1

=(901300)3.5

= 46.94Principle of Thermodynamics and Fluid Flow

Hence loss of stagnation pressure is

= poei –poe

= 46.94 – 30

=16.94 bar

1.5.1 Thermodynamic Losses

Generally speaking, losses occur in turbo-machines due to:

(a) Bearing friction, windage, etc., all of which may beclassified as mechanical losses,

(b) Unsteady flow, friction between the blade and the fluid,etc., which are internal to the system and may be classifiedas fluid-rotor losses.

There are other losses like leakage across blades, labyrinthleakage, etc. in addition to the above losses. These arecovered under fluid-rotor losses. Usually, mechanical lossesin large turbo-machines do not exceed 1%. In very largemachines dealing with hundreds of megawatts of power, (thekind used in large power stations), these losses may besmaller than 0.5%.

The effect of thermodynamic loss can be observed in thecompressor process line. From the Gibbs equation we have

T ds = dh – v dp---------------------------------------------------------------------------------- (1.5.4)

with the slope

(δhδs

)p = T which gives us the idea that the slope is equal

to the absolute temperature on an enthalpy- entropy diagram.


The enthalpy change between 2 states is made of two parts

a) The irreversible change in enthalpy is dhf = T ds

b) The isentropic change by setting ds =0 which gives dhs=vdp .Substituting this in equation (1) we get

dh = dhs + dhf

Figure 1.5.3 An infinitesimal irreversible process in turbineand compressor

Points to remember:

1) For a compressor all 3 values dh, dhs and dhf are positive

2) For a turbine dhf is positive but dh and dhs are negative.

The nature of the irreversibility may be further illustratedby considering a flow channel that extends from an inlet atlocation l1 to some general location l. The first law of

Thermodynamics for this control volume is


Q + m(u1+ p1v1 +12V12 + gz1) = m(u+ pv +

12V22 + gz) + W

Differentiating the above equation with respect to l we get,

m dudl = Q' - m¿ + 12

dV2

dl + gdzdl ] - W '

where

Q'=d Qdl and W'=

d Wdl

Thus in parts where there is no heat interaction Q'=0and partswith no work interaction have W'=0.

Differentiating the second law of thermodynamics we get,

m(s - s1) = ∫l1

l Q'

T dl + ∫l1

l˙sp'dl

which gives,

m dsdl =

Q'T + ˙sp'

1.5.2 Small Stage or Polytrophic Efficiency

While dealing with multistage compressors and turbines it isuseful to introduce an efficiency defined slightly differentlyto the isentropic or adiabatic efficiency. The small-stage orpolytrophic efficiency (Ƞp) removes a bias in the isentropicefficiency when comparing turbomachines of different pressureratio.

Compression Process

Figure 1.5.4 represents an enthalpy-entropy diagram. Point 1to 2 in the figure represents adiabatic compression between


pressures from p1 to p2 and point 1 to 2s represents theisentropic line for reversible process. For the small stageefficiency (Ƞp) the actual work is denoted by δW and idealwork by δWmin.

Figure 1.5.4 Compression process by small stages

Taking notations from the figure 1.5.4, the polytropicefficiency can is given as:

Ƞp = δWminδW =

hxs−h1

hx−h1 =

hys−hx

hy−hx = …

Also, Ƞp = ∑ δWminδW since the small stages have equal

efficiency.

Since ∑δW= {(hx−h1) + hy−hx + …} = (h2 – h1)

this implies ƞp = [(hxs−h1) + hys−hx + …] / (h2 – h1).

Whereas, the adiabatic efficiency for compression process isgiven as:


Ƞc = h2s−h1

h2−h1 ,

due to the divergence of constant pressure lines

[(hxs−h1) + hys−hx + …} > h2s−h1

which implies that, ƩδWmin > Wmin.

Therefore, Ƞp > Ƞc .

Thus, taking into consideration the divergence of constantpressure lines, we can conclude that small stage efficiency isgreater than the isentropic efficiency for a compressionprocess.

Perfect Gas

In case of a perfect gas, where Cp is considered constant, thepolytropic efficiency for small stage is given by:

Ƞp = dhisdh

= vdpCpdT

------------------------------------------------------------------- (1.5.5)

where,

dp = incremental change in pressure

dh = incremental enthalpy rise

dhis = ideal enthalpy rise, the notations are as represented infigure 1.5.5 below


Figure 1.5.5 Incremental change of state in a compressionprocess

If v = RT/ p, then equation 1.5.5 becomes

Ƞp = RCp Tp

dpdT

And therefore we have

dTT =

γ−1γȠp

dpp

------------------------------------------------------------------------------------ (1.5.6)

As Cp = γRγ−1 .

Taking equal efficiency for small stage and integratingequation 1.5.6, we get

T2T2 = (

p2p1

)γ−1γȠp

Isentropic efficiency for compression process is


Ƞc = T2s−T1

T2−T1 , for equal velocity at inlet and outlet.

Also, T2sT1 = (

p2p1

)γ−1γ , as Ƞp = 1 for ideal compression process.

Thus we have,

Ƞc = [(p2

p1)γ−1γ −1 ]

[(p2

p1)γ−1γȠp−1 ]

------------------------------------------------ (1.5.7)

Figure 1.5.6 Isentropic efficiency, Polytropic efficiency andpressure ratio for compressor

The figure 1.5.6 provides us with the relation betweenisentropic efficiency, polytropic efficiency and pressureratio for a compressor at γ=1.4


In case of an adiabatic turbine, the polytropic efficiency canbe derived similar to equation 1.5.7 as,

Ƞt = [1−(p2

p1)Ƞp (γ−1 )

γ ][1−(p2p1 )

(γ−1)γ ]

Unlike in a compression process, isentropic efficiency exceedspolytropic efficiency for an expansion. The figure 1.5.7 showsturbine isentropic efficiency against pressure ration forvarious small stage efficiencies (γ=1.4)

Figure 1.5.7 Turbine Isentropic efficiency

Nozzle Efficiency

The function of the nozzle is to transform the high pressuretemperature energy (enthalpy) of the gases at the inletposition into kinetic energy. This is achieved by decreasingthe pressure and temperature of the gases in the nozzle.


Fig. 1.5.8 Mollier Diagram (a) Nozzle (b) Diffuser

The flow process through a nozzle and a diffuser is shown infigure 1.5.8 (a) and (b) respectively. For a nozzle theprocess assumed to be steady and adiabatic where h01 = h02. Nozzleefficiency (ȠN) is given as:

ȠN = (12C22) / (12

C2s2 ) =

h01−h2

h01−h2s

---------------------------------------------------------------- (1.5.8)

An enthalpy loss coefficient for nozzle is defined as

ζN = h2−h2s

12C22

and a velocity coefficient as

KN = C2

C2s

and the above coefficients are related as


ŋN = 1

1+ςN = KN2

------------------------------------------------------------------------------------(1.5.9)

Example 2.2

Gas is blown into the nozzle of turbine at a stagnationtemperature and pressure of 1000K and 5.0 bar and leaves witha velocity 600m/s at a pressure of 2.36 bar. Determine nozzleefficiency.

Solution

ŋN = 1−

T2T01

1−T2s

T01

= 1−

T2

T01

1−p2

p01

(γ−1)/γ

Assuming adiabatic flow T02 = T01

T2= T02- 12C22/Cp

= 1000 – 155.17

= 844.83 K

and thus we can solve for ŋN

ŋN = 1−

T2T01

1−T2s

T01

= 0.1560.18

=0.8667

Diffusers

The main aim of a diffuser is to increase the pressure atoutlet by reducing the flow velocity. Mostly all turbomachines


have a diffuser to fulfill the stated purpose. The flows in aturbo-machine are subsonic (M<1). A diffuser is a long hollowchannel which diverges along the flow thereby decreasing theflow velocity of the fluids.(figure 1.5.9)

Fig 1.5.9 Subsonic diffusers a) 2 Dimensional b) Conical c)annular


For steady and adiabatic flow h01 = h02

So h2-h1 = 12 (C1

2−C22)

For state point 1 to state point 2s

h2s – h1 = 12 (C1

2−C2s2 )

Diffuser efficiency ŋD or diffuser effectiveness is defined as

ŋD = h2s−h1

h2−h1=c12−c2s

2

c12−c22

---------------------------------------------------------------------------------(1.5.10)

In a low speed flow density ρ can be considered constant

h2s – h1 = p2−p1

ρ------------------------------------------------------------------------------------------- (1.5.11)

Now the diffuser efficiency can be written as

ŋD = p2−p1

ρ(c¿¿12−c22)¿

Equation 1.5.10 can be written in terms of pressure as

h2 – h2s = (h2-h1) – (h2s-h1)

= p01−p02

ρ

So ŋD = h2s−h1

(h2s−h1 )−(h2s−h2)


= 1

1+(p01−p02)/(p2−p1)

1.6 Newton’s 2nd Law of Motion

According to Newton’s second law of motion, in linear motionforce is equal to the rate of change of momentum of the fluidacross the control volume in a particular direction. If theuniform velocity of the fluid at entry is C1 and at outlet isC2, then

∑F=¿ṁ ¿ (C1 – C2)

Since turbomachines have rotational system, the mostdescriptive form of Newton’s law second law of motion is“torque is equal to the rate of change of angular momentum.

1.7 Summary and Key Points

1. Adiabatic energy equation

2. Specific-Heat Ratio

3. Isentropic flow is an adiabatic, reversible andfrictionless flow so that

4. Isentropic relations


w=h02−h01h0=h+ 1

2 v2

γ=cpcv

dS=0

P2P1

=(ρ2ρ1 )γ

=(T2T1)γ

γ−1T2T1

=(V2V1

)γ−1 T2T1

=(p2p1

)γ-1γ

5. Turbine efficiency

6. Compressor efficiency

7. Polytrophic efficiency

Last modified at 16:02 on 06 May


ηt=h1−h2h1−h2s

ηc=h2s−h1h2−h1

principle of thermodynamics and fluid flow

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