Posted tagged ‘fluid mechanics’

12. Energy Exchange with Moving Blades

September 22, 2011


  • 12.1 Introduction
  • 12.2 Conservation of Angular Momentum
  • 12.3 The Euler Turbine Equation
  • 12.4 Multistage Axial Compressors
  • 12.5 Velocity Triangles for an Axial Compressor Stage
  • 12.6 Velocity Triangles for an Axial Flow Turbine Stage

12.1 Introduction

So far we have only looked at the thermodynamic results of compressors and turbines ($ pi$ ‘s and $ tau$ ‘s). Now we will look in more detail at how the components of a gas turbine engine produce these effects. You will learn later that without heat transfer, it is only possible to change the total enthalpy of a fluid with an unsteady process (e.g. moving blades). Still we will use many of the steady flow tools that we have discussed in thermodynamics and propulsion by considering the steady flow in and out of a component as shown in Figure 12.1.

Figure 12.1: Control volume around compressor or turbine.
Image fig9TurbomachineryCtrlVolume_web

12.2 Conservation of Angular Momentum

[This section is excerpted from Fluid Flow: A First Course in Fluid Mechanics, Macmillan Publishing Company, 1989.]

The momentum theorem developed in Chapter 10 gives the force acting on a fixed volume in terms of linear momentum flux through the surface of the volume. In many situations we are interested in the moment or torque on the volume. For this purpose we may adapt the angular momentum law of mechanics to the flow of fluids. Our starting point is the familiar law


$displaystyle vec{F} = frac{d}{dt}(mvec{v}),


where $ m$ , $ vec{F}$ , and $ vec{v}$ refer to a single particle. The torque exerted by the force $ vec{F}$ about a fixed point is

$displaystyle vec{T} = vec{r}timesvec{F},


where $ vec{r}$ is the radius vector from the fixed point to the point of application of $ vec{F}$ . The symbol, $ times$ , signifies, as usual, that the vector cross-product shall be taken. Then, from Newton’s law of motion,

$displaystyle vec{T}=vec{r}timesfrac{d}{dt}(mvec{v}).


We now define a vector $ vec{H}$ as the vector product of the radius vector to the particle and the linear momentum, that is,

$displaystyle vec{H}=vec{r}times mvec{v}.


The quantity $ vec{H}$ is called angular momentum. Upon differentiating $ vec{H}$ with respect to time, we find that

$displaystyle frac{dvec{H}}{dt}=frac{dvec{r}}{dt}times


However, $ dr/dt =vec{v}$ and the cross-product of a vector parallel to itself is zero. The first term in the right-hand side therefore vanishes and we have the result that

$displaystyle vec{T}=frac{dvec{H}}{dt}.$ (12..1)


Equation (12.1) states that the rate of change of angular momentum of a particle about a fixed point is equal to the torque applied to the particle.We now seek to modify the law as expressed by Equation (12.1) to be suitable for a fixed volume. The torque on a material volume $ V'$ is


$displaystyle vec{T}=frac{D}{Dt}int_{V'}rho vec{r}times vec{v}dV.


This is readily transformed into a control volume integral. We have, therefore,

$displaystyle vec{T} =frac{partialvec{H}}{partial t}+int_{S_0}rho



$displaystyle vec{H} = int_{V_0}rho vec{r}timesvec{v}dV$ (12..2)


is the angular momentum contained within the control volume. Equation (12.2) represents the angular momentum theorem. [For more information about angular momentum and rotational energy, see pages 246 and 558 in Hibbeler’s Engineering Dynamics.]

12.3 The Euler Turbine Equation

The Euler turbine equation relates the power added to or removed from the flow, to characteristics of a rotating blade row. The equation is based on the concepts of conservation of angular momentum and conservation of energy. We will work with the model of the blade row shown in Figure 12.2.


Figure 12.2: Control volume for Euler Turbine Equation.
Image fig9EulerTurbineEqCtrlVolume_web

Applying conservation of angular momentum, we note that the torque, $ mathcal{T}$ , must be equal to the time rate of change of angular momentum in a streamtube that flows through the device



$displaystyle mathcal{T} = dot{m}(v_c r_c - v_b r_b).$


This is true whether the blade row is rotating or not. The sign matters (i.e. angular momentum is a vector – positive means it is spinning in one direction, negative means it is spinning in the other direction). So depending on how things are defined, there can be positive and negative torques, and positive and negative angular momentum. In Figure 9.2, torque is positive when V_{textrm{tangential in}}$” width=”207″ height=”32″ align=”MIDDLE” border=”0″ /> — the same sense as the angular velocity.If the blade row is moving, then work is done on/by the fluid. The work per unit time, or power, $ P$ , is the torque multiplied by the angular velocity, $ omega$ :


$displaystyle P = mathcal{T} omega = omega dot{m}(v_c r_c - v_b r_b).$


If torque and angular velocity are of like sign, work is being done on the fluid (a compressor). If torque and angular velocity are of opposite sign work is being extracted from the fluid (a turbine). Here is another approach to the same idea:

  • If the tangential velocity increases across a blade row (where positive tangential velocity is defined in the same direction as the rotor motion) then work is added to the flow (this happens in a compressor).
  • If the tangential velocity decreases across a blade row (where positive tangential velocity is defined in the same direction as the rotor motion) then work is removed from the flow (this happens in a turbine).

From the steady flow energy equation,

$displaystyle dot{q} -dot{w}_s = dot{m} Delta h_T$



$displaystyle dot{q} = 0 quad textrm{and} quad -dot{w}_s = P,$



$displaystyle P = dot{m}(h_{Tc}-h_{Tb}).$


Then equating this expression of conservation of energy with our expression from conservation of angular momentum, we arrive at:

$displaystyle h_{Tc} - h_{Tb} = omega (r_c v_c - r_b v_b)$


or for a perfect gas with $ c_p=textrm{constant}$ ,

$displaystyle c_p (T_{Tc} - T_{Tb}) = omega (r_c v_c - r_b v_b).$ (12..3)


Equation (12.3) is called the Euler Turbine Equation. It relates the temperature ratio (and hence the pressure ratio) across a turbine or compressor to the rotational speed and the change in momentum per unit mass. Note that the velocities used in this equation are what we will later call absolute frame velocities (as opposed to relative frame velocities).

  • If angular momentum increases across a blade row, then  T_{Tb}$” width=”80″ height=”32″ align=”MIDDLE” border=”0″ /> and work was done on the fluid (a compressor).
  • If angular momentum decreases across a blade row, then <img src="; alt="$ T_{Tc}  and work was done by the fluid (a turbine).

12.4 Multistage Axial Compressors

An axial compressor is typically made up of many alternating rows of rotating and stationary blades called rotors and stators, respectively, as shown in Figures 12.3 and 12.4. The first stationary row (which comes in front of the rotor) is typically called the inlet guide vanes or IGV. Each successive rotor-stator pair is called a compressor stage. Hence compressors with many blade rows are termed multistage compressors.


Figure 12.3: A typical multistage axial flow compressor (Rolls-Royce, 1992).
Image fig9RollsMultistageSchematic_web


Figure 12.4: Schematic representation of an axial flow compressor.
Image fig9AxialFlowSchematicCompressor_web

One way to understand the workings of a compressor is to consider energy exchanges. We can get an approximate picture of this using the Bernoulli Equation, where $ P_T$ is the stagnation pressure, a measure of the total energy carried in the flow, $ p$ is the static pressure, a measure of the internal energy, and the velocity terms are a measure of the kinetic energy associated with each component of velocity ($ u$ is radial, $ v$ is tangential, $ w$ is axial).


$displaystyle P_T = p + frac{1}{2}rho(u^2 + v^2 + w^2).$


The rotor adds swirl to the flow, thus increasing the total energy carried in the flow by increasing the angular momentum (adding to the kinetic energy associated with the tangential or swirl velocity,$ rho v^2/2$ ).The stator removes swirl from the flow, but it is not a moving blade row and thus cannot add any net energy to the flow. Rather, the stator rather converts the kinetic energy associated with swirl to internal energy (raising the static pressure of the flow). Thus typical velocity and pressure profiles through a multistage axial compressor look like those shown in Figure 12.5.


Figure 12.5: Pressure and velocity profiles through a multi-stage axial compressor (Rolls-Royce, 1992).
Image fig9RollsPVelProfiles_web

Note that the IGV also adds no energy to the flow. It is designed to add swirl in the direction of rotor motion to lower the Mach number of the flow relative to the rotor blades, and thus improve the aerodynamic performance of the rotor.

12.5 Velocity Triangles for an Axial Compressor Stage

Velocity triangles are typically used to relate the flow properties and blade design parameters in the relative frame (rotating with the moving blades), to the properties in the stationary or absolute frame.

We begin by “unwrapping” the compressor. That is, we take a cutting plane at a particular radius (e.g. as shown in Figure 12.4) and unwrap it azimuthally to arrive at the diagrams shown in Figure 12.6. Here we have assumed that the area of the annulus through which the flow passes is nearly constant and the density changes are small so that the axial velocity is approximately constant.


Figure 12.6: Velocity triangles for an axial compressor stage. Primed quantities are in the relative frame, unprimed quantities are in the absolute frame.
Image fig9VelTrianglesCompressor_web

In drawing these velocity diagrams it is important to note that the flow typically leaves the trailing edges of the blades at approximately the trailing edge angle in the coordinate frame attached to the blade (i.e. relative frame for the rotor, absolute frame for the stator).

We will now write the Euler Turbine Equation in terms of stage design parameters: $ omega$ , the rotational speed, and $ beta_b$ $ beta_c'$ , the leaving angles of the blades.


$displaystyle c_p (T_{Tc} - T_{Tb}) = omega (r_c v_c - r_b v_b)$



From geometry,


$displaystyle v_b = w_b tan beta_b$





$displaystyle v_c = w_c tan beta_c = omega r_c - w_c tanbeta_c'$





$displaystyle c_p(T_{Tc} - T_{Tb}) = omega(omega r_c^2 - w_c r_c tan beta_c' - r_b w_b tan beta_b)$





$displaystyle underbrace{frac{T_{Tc}}{T_{Tb}}}_{substack{textrm{stagnation ...
...}_{substack{textrm{$beta$'s set by} textrm{blade design}}}right)right].$



So we see that the total or stagnation temperature rise across the stage increases with the tip Mach number squared, and for fixed positive blade angles, decreases with increasing mass flow. This behavior is represented schematically in Figure 12.7.

Figure 12.7: Compressor behavior
Image fig9CompressorStabilityBehavior_web

12.6 Velocity Triangles for an Axial Flow Turbine Stage

We can apply the same analysis techniques to a turbine, Figure 12.8. The stator, again does no work. It adds swirl to the flow, converting internal energy into kinetic energy. The turbine rotor then extracts work from the flow by removing the kinetic energy associated with the swirl velocity.

Figure 12.8: Schematic of an axial flow turbine.
Image fig9AxialFlowSchematicTurbine_web

The appropriate velocity triangles are shown in Figure 12.9, where again the axial velocity was assumed to be constant for purposes of illustration. As we did for the compressor, we can write the Euler Turbine Equation in terms of useful design variables:


$displaystyle 1-frac{T_{Tc}}{T_{Tb}} = frac{(omega r)^2}{c_p T_{Tb}}left[f... r}tanbeta_b + left(frac{w_c}{omega r} tanbeta_c'-1right)right].$



Figure 12.9: Velocity triangles for an axial flow turbine stage.
Image fig9VelTrianglesTurbine_web



Introduction | Fluid Mechanics | Fluid properties

September 16, 2011

Fluid Mechanics – Introduction

01-Fluid Mechanics example - Static and Turbulent Flow - Analysis -Dynamic  analysis - CFD

Fluid Mechanics is that section of applied mechanics, concerned with the statics and dynamics of liquids and gases.

A knowledge of fluid mechanics is essential for the Mechanical engineer, because the majority of Mechanical processing operations are conducted either partially or totally in the fluid phase.

The handling of liquids is much simpler, much cheaper, and much less troublesome than handling solids.

Even in many operations a solid is handled in a finely divided state so that it stays in suspension in a fluid.

Fluid Statics: Which treats fluids in the equilibrium state of no shear stress

 02-Fluid Mechanics example - Static and Turbulent Flow - Analysis -Dynamic  analysis - CFD

Fluid Mechanics: Which treats when portions of fluid are in motion relative to other parts.

03-Fluid Mechanics example - Static and Turbulent Flow - Analysis -Dynamic  analysis - CFD

Fluids and their Properties


In everyday life, we recognize three states of matter:

  • solid,
  • liquid and
  • gas.

Although different in many respects, liquids and gases have a common characteristic in which they differ from solids: they are fluids, lacking the ability of solids to offer a permanent resistance to a deforming force.

fluid is a substance which deforms continuously under the action of shearing forces, however small they may be.Conversely, it follows that:
If a fluid is at rest, there can be no shearing forces acting and, therefore, all forces in the fluid must be perpendicular to the planes upon which they act.

Shear stress in a moving fluid

Although there can be no shear stress in a fluid at rest, shear stresses are developed when the fluid is in motion, if the particles of the fluid move relative to each other so that they have different velocities, causing the original shape of the fluid to become distorted. If, on the other hand, the velocity of the fluid is same at every point, no shear stresses will be produced, since the fluid particles are at rest relative to each other.

Mech Engg / Gate Study Material / Gate Preparation / Related Papers

September 16, 2011

03-GATE-2011-Graduate Aptitude test in Engineering

Mechanical Engineering

Engineering Mechanics
  • Introduction to Mechanisms
  • Mechanisms and Simple Machines
  • More on Machines and Mechanisms
  • Basic Kinematics of Constrained Rigid Bodies
  • Planar Linkages
  • Cams
  • Gears
  • Other Mechanisms
  • Engineering Thermodynamics
  • Fluid Mechanics
  • Heat Transfer
  • Pro_Engineer
  • Robotics
  • Machines Kinematics
  • Strength of Materials
  • Machine Elements
  • Machine Design
  • Finite Elements
Engineering Mathematics
  • Linear Algebra
  • Calculus
  • Complex Analysis
  • Vector Calculus
  • Differential Equations
  • Probability – I
  • Probability (continued)
  • Numerical Methods
  • Transform Theory
  • Fourier Transform
  • Laplace Transform
  • Z-Transform

Engineering Mechanics:

  • Equivalent force systems
  • Equations of equilibrium

Strength of Materials:

  • Cold and Hot Working
  • Creep
  • Ductility
  • Engineering and Applications Factor of Safety Review
  • Hardness
  • Heat Treatment
  • Hookes Law
  • Material Strength
  • Malleability
  • Strain
  • Strength of Materials Summary and Overview
  • Stress
  • Cylinder Stress and Deflection with Applied Torsion
  • Toughness
  • Von Mises Criterion ( Maximum Distortion Energy Criterion )
  • Work ( Strain ) Hardening
  • Young’s Modulus
  • Vibrations

My Rendezvous with Mechanical Engineering

August 23, 2011

I don’ quite remember the exact moment when I decided that I was going to be a mechanical engineer.Though I had always wanted to be an engineer, i had never ever thought of specialization (but i did have a great deal of interest in Aeroplanes and Spacecrafts) until I had undergone a career counselling programme.After being made to write tests for hours that gauged my aptitude,skills,personality and discovered my interests i was advised by the counsellor to take up careers in any of these streams-Architecture,Mechanical Engineering,Industrial Design,Genetic Engineering,Interior Designing,Astronomy/Astrophysics,Bookkeeping ,etc.I was suggested to choose a course that required artistic and clerical abilities.but I wasn’t much interested in career planning coz it was too early…i was in my 9th standard and there were a jolly good 3 years to think about joining a college! All I wanted was to be associated with the world of technology that had immensely changed my nation.

As i moved into my eleventh standard I felt very comfortable with science chapters related to mechanical engineering like statics, dynamics and fluid mechanics and the thermal sciences.As i moved into class 12 I started feeling uncomfortable with electricity and electronics primarily because it became difficult to imagine and visualize the concepts of these sciences as compared to the mechanical sciences.

I still loved Computer Science because I was good at programming and my CS Mam was an excellent teacher.But then young India was too obsessed with Software and electronics ….especially the top rankers chose these courses…and i wanted to be different..hence I decided – to go for Mechanical Engineering .I brooded over the topic and came up with umpteen no. of reasons and decided my career…

and joined a UG Bachelor of Engineering(Mechanical)course .

Top 10 reasons why I chose to be a Mechanical Engineer on my next post