CP502 Advanced Fluid Mechanics Compressible Flow Part 01_Set 01: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction Incompressible flow assumption is not valid if Mach number > 0.3 What is a Mach number? Definition of Mach number (M): M≡ For an ideal gas, Speed of the flow (u) Speed of sound (c) in the fluid at the flow temperature c RT specific heat ratio specific gas constant (in J/kg.K) R. Shanthini 02 Feb 2012 absolute temperature of the flow at the point concerned (in K) For an ideal gas, M= u c = u RT Unit of u = m/s Unit of c = [(J/kg.K)(K)]0.5 R. Shanthini 02 Feb 2012 = [J/kg]0.5 = (N.m/kg)0.5 = [m2/s2]0.5 = m/s = [kg.(m/s2).m/kg]0.5 constant area duct Diameter (D) quasi one-dimensional flow speed (u) A D 2 / 4 is a constant u varies only in x-direction x compressible flow Au is a constant Mass flow rate m steady flow isothermal flow ideal gas wall friction R. Shanthini 02 Feb 2012 Density (ρ) is NOT a constant Temperature (T) is a constant Obeys the Ideal Gas equation w fu2 / 2 is the shear stress acting on the wall where f is the average Fanning friction factor Friction factor: f 16/ Re For laminar flow in circular pipes: where Re is the Reynolds number of the flow defined as follows: uD m D Re A 4m D D 2 For lamina flow in a square channel: For the turbulent flow regime: R. Shanthini 02 Feb 2012 4m D f 14.227/ Re 3.7 D 4.0 log10 f 1 Quasi one-dimensional flow is closer to turbulent velocity profile than to laminar velocity profile. Ideal Gas equation of state: pV mRT temperature pressure specific gas constant (not universal gas constant) volume mass Ideal Gas equation of state can be rearranged to give m p RT V R. Shanthini 02 Feb 2012 p RT K Pa = N/m2 kg/m3 J/(kg.K) Problem 1 from Problem Set 1 in Compressible Fluid Flow: Starting from the mass and momentum balances, show that the differential equation describing the quasi one-dimensional, compressible, isothermal, steady flow of an ideal gas through a constant area pipe of diameter D and average Fanning friction factor f shall be written as follows: 4f 2 2 dx 2 dp du 0 D u u (1.1) where p, ρ and u are the respective pressure, density and velocity at distance x from the entrance of the pipe. R. Shanthini 02 Feb 2012 w x p p+dp u u+du D dx Write the momentum balance over the differential volume chosen. u ( p dp) A m (u du) wdAw pA m (1) steady mass flow rate cross-sectional area shear stress acting on the wall dAw Ddx is the wetted area on which shear is acting R. Shanthini 02 Feb 2012 w p p+dp u u+du dx x Equation (1) can be reduced to Au Substituting m Since A D 2 / 4 , du wdAw 0 Adp m A(dp udu) w dAw 0 w fu / 2 2 dp udu R. Shanthini 02 Feb 2012 D and dAw Ddx , we get u 4 f 2 2 D dx 0 4f 2 2 dx 2 dp du 0 D u u (1.1) Problem 2 from Problem Set 1 in Compressible Fluid Flow: Show that the differential equation of Problem (1) can be converted into 4f 2 2 dx pdp dp 2 D RT (m / A) p (1.2) which in turn can be integrated to yield the following design equation: 4f L p2 D RT (m / A) 2 pL2 pL2 1 2 ln 2 p p (1.3) where p is the pressure at the entrance of the pipe, pL is the pressure at length L from the entrance of the pipe, R is the gas constant, T is the is the mass flow rate of the gas flowing temperature of the gas, m through the pipe, and A is the cross-sectional area of the pipe. R. Shanthini 02 Feb 2012 The differential equation of problem (1) is 4f 2 2 dx 2 dp du 0 D u u (1.1) in which the variables ρ and u must be replaced by the variable p. Let us use the mass flow rate equation m Au and the ideal gas equation p RT to obtain the following: p RT m m RT and u A Ap m RT and therefore pu A It is a constant for steady, isothermal flow in a constant area duct R. Shanthini 02 Feb 2012 d ( pu) pdu udp 0 du dp u p p m m RT , u Using RT A Ap and du dp u p 4 f 2 2 in dx 2 dp du 0 D u u (1.1) we get 4f 2 2 dx pdp dp 2 D RT (m / A) p R. Shanthini 02 Feb 2012 (1.2) p pL L Integrating (1.2) from 0 to L, we get 4f D L 2 0 dx RT (m / A) 2 pL pL p p pdp 2 dp p which becomes 4f L p2 D RT (m / A) 2 R. Shanthini 02 Feb 2012 pL2 pL2 1 2 ln 2 p p (1.3) Problem 3 from Problem Set 1 in Compressible Fluid Flow: Show that the design equation of Problem (2) is equivalent to 4f L 1 D M2 M2 M2 1 2 ln 2 ML ML (1.4) where M is the Mach number at the entry and ML is the Mach number at length L from the entry. R. Shanthini 02 Feb 2012 Design equation of Problem (2) is 4f L p2 D RT (m / A) 2 pL2 pL2 1 2 ln 2 p p (1.3) which should be shown to be equivalent to 4f L 1 D M2 M2 M2 1 2 ln 2 ML ML (1.4) where p and M are the pressure and Mach number at the entry and pL and ML are the pressure and Mach number at length L from the entry. We need to relate p to M! R. Shanthini 02 Feb 2012 We need to relate p to M! m m RT 1 m RT 1 m p RT RT Au A u A M RT AM RT which gives m RT pM A = constant for steady, isothermal flow in a constant area duct Substituting the above in (1.3), we get 4f L 1 D M2 R. Shanthini 02 Feb 2012 M2 M2 1 2 ln 2 ML ML (1.4) Summary Design equations for steady, quasi one-dimensional, isothermal,compressible flow of an ideal gas in a constant area duct with wall friction 4f 2 2 dx 2 dp du 0 D u u 4f 2 2 dx pdp dp 2 D RT (m / A) p 4f L p2 D RT (m / A) 2 R. Shanthini 02 Feb 2012 4f L 1 D M2 (1.1) (1.2) pL2 pL2 1 2 ln 2 p p M2 M2 1 2 ln 2 ML ML (1.3) (1.4)