A course in Gas Dynamics…………………………………. …. …Lecturer: Dr A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 6 vii. Intensive properties properties which independent on mass of fluid contained in the control volume (CV), specific properties like u, h, s or naturally independent like pressure and temperature. viii. Conservation of mass law mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. Mass-energy equivalence is a concept formulated by Albert Einstein that explain the relationship between mass and energy. E=mc2, where c is light speed. 1-2 Control volume of the system Figure (1.2) shows an arbitrary mass at time t and the same mass at time t + ∆t, which composes the same mass particles at all times. At time t the given mass particles occupy regions 1 and 2 . At time t + ∆t the same mass particles occupy regions 2 and 3 . Regions 1 & 2, which originally confines of the mass, are called the control volume. X: the total amount of any extensive property in a given mass.

X: Mass-dependent (extensive) property; scalar or vector quantity. A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 7 x: the amount of X per unit mass ; thus X = x dm = xpdV (1 - 1) CV dX $ d $ This relation, known as Reynolds’s Transport Theorem, which can be interpreted in words as: The rate of change of property for a fixed mass system of fluid particles as it is moving is equal to the rate of change of inside the control volume plus the net efflux of from the control volume (flow out minus flow in across control volume boundary). Where d dt 0t CV: control volume that containing the mass. CS: control surface that surrounding the control volume. X: Mass-dependent (extensive) property; scalar or vector quantity. x: is the amount of the property per unit mass. For mass it equals one. p: Fluid density (kg/m3). dV: Infinitesimal (very small) control volume. dA: Infinitesimal control surface. V: Velocity vector. n": Outward unit vector which is perpendicular to dA. ( ) = xpdV + & xp(V. n" )dA (1 - 2) CV CS : Material or total or substantial derivative : Partial derivative with respect to time

Let X = mass (m) ⟹ x = 1 For fixed amount of mass that moves through A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 8 1.2.1 Conservation of mass: dX a CV CS Let X = mass (m) ⟹ x = 1 For fixed amount of mass that moves through the control volume: (dm] = 0 and for steady flow: a ∭ pdV = 0 , so: & p(V. n" )dA = 0 (1 - 3) CS for one-dimensional flow any fluid property will be constant over an entire cross section & p(V. n" )dA = p(V. n" )dA = pV dA = pV(Aexit - Ainlet ) (1 - 4) p2 V2 A2 = p1V1 A1 (1 - 5) m = pVA = cons. Where: m, kg/s V is the component of velocity perpendicular to the area A, m/s. ρ is the density kg/m3. A is the area m2 . 1.2.2 Conservation of energy. Let X = energy (E) ⟹ x = e , dE a ( dt ) = at xpdV + & xp(V. n" )dA dt at CV ( dt ) = at xpdV + & xp(V. n" )dA ( dt ) = at epdV + & ep(V. n" )dA (1 - 6)

velocity is perpendicular to the surface , we have , so: A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 9 For one-dimensional, steady flow the last integral is simple to evaluate, as e, p and V are constant over any given cross section. Assuming that the velocity is perpendicular to the surface , we have , so: & ep(V. n" )dA = U(pVA)e = U me (1 - 7) cs ∂ ∂t So, so = sw + a ∭ epdV + ff ep(V. n" )dA w=shaft work (ws)+ flow work (pv) For steady one-dimensional flow, 2 2 h1 + 2 + gz1 + q = h2 + 2 + gz2 + ws (1 - 8) dv2 epdV = 0 (steady state) cV dt dt at cV cs v1 v2 oq = ows + dh + + gdz (1 - 9) 2

And momentum equation simplify to: J F = & Vp(V. nu)dv A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 10 1.2.3 Conservation of momentum. d(momentum) r cV cs For steady flow the time rate of change of linear momentum stored inside the control volume is a CV And momentum equation simplify to: J F = & Vp(V. nu)dv cs The x-component of this equation would appear as: J Fx = & Vx p Vx dv Or J Fx = J m (Vout - Vin ) (1 - 11) J F = = rt VpdV + & Vp(V. nu)dv (1 - 10) dt at xpdV