MAE 5350 Gas Turbines Integral Forms of Mass

MAE 5350: Gas Turbines Integral Forms of Mass and Momentum Equations Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk

Kinematic Properties: Two ‘Views’ of Motion 1. Lagrangian Description – Follow individual particle trajectories – Choice in solid mechanics – Control mass analyses – Mass, momentum, and energy usually formulated for particles or systems of fixed identity • ex. , F=d/dt(m. V) is Lagrangian in nature 2. Eulerian Description – Study field as a function of position and time; not follow any specific particle paths – Usually choice in fluid mechanics – Control volume analyses – Eulerian velocity vector field: – Knowing scalars u, v, w as f(x, y, z, t) is a solution

CONSERVATION OF MASS Relative to CS Inertial • • This is a single scalar equation – Velocity doted with normal unit vector results in a scalar 1 st Term: Rate of change of mass inside CV – If steady d/dt( ) = 0 – Velocity, density, etc. at any point in space do not change with time, but may vary from point to point 2 nd Term: Rate of convection of mass into and out of CV through bounding surface, S 3 rd Term (=0): Production or source terms

Integral vs. Differential Form • Integral form of mass conservation • Apply Divergence (Gauss’) Theorem • Transform both terms to volume integrals • Results in continuity equation in the form of a partial differential equation • Applies to a fixed point in the flow • Only assumption is that fluid is a continuum – Steady vs. unsteady – Viscous vs. inviscid – Compressible vs. incompressible

Summary: Incompressible vs. Constant Density • Two equivalent statements of conservation of mass in differential form • In an incompressible flow • Says particles are constant volume, but not necessarily constant shape • Density of a fluid particle does not change as it moves through the flow field • Incompressible: Density may change within the flow field but may not change along a particle path • Constant Density: Density is the same everywhere in the flow field

MOMENTUM EQUATION: NEWTONS 2 nd LAW Inertial • • • Relative to CS This is a vector equation in 3 directions 1 st Term: Rate of change of momentum inside CV or Total (vector sum) of the momentum of all parts of the CV at any one instant of time – If steady d/dt( ) = 0 – Velocity, density, etc. at any point in space do not change with time, but may vary from point to point 2 nd Term: Rate of convection of momentum into and out of CV through bounding surface, S or Net rate of flow of momentum out of the control surface (outflow minus inflow) 3 rd Term: – Notice that sign on pressure, pressure always acts inward – Shear stress tensor, t, drag – Body forces, gravity, are volumetric phenomena – External forces, for example reaction force on an engine test stand Application of a set of forces to a control volume has two possible consequences 1. Changing the total momentum instantaneously contained within the control volume, and/or 2. Changing the net flow rate of momentum leaving the control volume

Application to Rocket Engines Chemical Energy Thermal Energy F Rocket Propulsion (class of jet propulsion) that produces thrust by ejecting stored matter • Propellants are combined in a combustion chamber where chemically react to form high T&P gases • Gases accelerated and ejected at high velocity through nozzle, imparting momentum to engine • Thrust force of rocket motor is reaction experienced by structure due to ejection of high velocity matter • Same phenomenon which pushes a garden hose backward as water flows from nozzle, gun recoil Kinetic Energy

Application to Airbreathing Engines Chemical Energy Thermal Energy • Flow through engine is conventionally called THRUST – Composed of net change in momentum of inlet and exit air • Fluid that passes around engine is conventionally called DRAG Kinetic Energy

Thrust Definitions • Use of conservation of mass and momentum in control volume form to derive governing equation for “net uninstalled thrust” – See Section 3. 1 – Textbook notation: Fn)uninstalled – Pay close attention to control volume choices • Control volume option 1: pages 113 – 119 • Control volume option 2: pages 121 - 124 – Understand each term, including the ram drag and pressure mismatch • Can divide into nozzle contribution and inlet contribution – Nozzle contribution is called “gross thrust” – Textbook notation: Fg • Uninstalled Thrust: thrust produced by engine if it had zero external losses • Installed Thrust: actual propulsive force transmitted to aircraft by engine

Efficiency Summary • Overall Efficiency – What you get / What you pay for – Propulsive Power / Fuel Power – Propulsive Power = TUo – Fuel Power = (fuel mass flow rate) x (fuel energy per unit mass) • Thermal Efficiency – Rate of production of propulsive kinetic energy / fuel power – This is cycle efficiency • Propulsive Efficiency – Propulsive Power / Rate of production of propulsive kinetic energy, or – Power to airplane / Power in Jet

MOMENTUM EQUATION: NEWTONS 2 nd LAW Inertial • • • Relative to CS This is a vector equation in 3 directions 1 st Term: Rate of change of momentum inside CV or Total (vector sum) of the momentum of all parts of the CV at any one instant of time – If steady d/dt( ) = 0 – Velocity, density, etc. at any point in space do not change with time, but may vary from point to point 2 nd Term: Rate of convection of momentum into and out of CV through bounding surface, S or Net rate of flow of momentum out of the control surface (outflow minus inflow) 3 rd Term: – Notice that sign on pressure, pressure always acts inward – Shear stress tensor, t, drag – Body forces, gravity, are volumetric phenomena – External forces, for example reaction force on an engine test stand Application of a set of forces to a control volume has two possible consequences 1. Changing the total momentum instantaneously contained within the control volume, and/or 2. Changing the net flow rate of momentum leaving the control volume

Momentum Equation: Mid-Air Refueling • • • An F-4 Phantom is being refueled in mid-air Refueling boom enters at an angle of 30° from the F-4 flight path Fuel flow rate through the boom is 20 kg/s at a velocity of 30 m/s relative to the two aircraft Density of jet fuel is less than water, and is about 700 kg/m 3 What additional lift force is necessary to overcome force on F-4 fighter due to momentum transfer during refueling?