CP 502 Advanced Fluid Mechanics Compressible Flow Part
CP 502 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≡ Speed of the flow (u) Speed of sound (c) in the fluid at the flow temperature For an ideal gas, specific heat ratio specific gas constant (in J/kg. K) R. Shanthini 19 July 2019 absolute temperature of the flow at the point concerned (in K)
For an ideal gas, M= u c = u Unit of u = m/s Unit of c = [(J/kg. K)(K)]0. 5 R. Shanthini 19 July 2019 = [J/kg]0. 5 = (N. m/kg)0. 5 = [m 2/s 2]0. 5 = m/s = [kg. (m/s 2). m/kg]0. 5
constant area duct is a constant Diameter (D) quasi one-dimensional flow speed (u) u varies only in x-direction x compressible flow steady flow Mass flow rate isothermal flow ideal gas Density (ρ) is NOT a constant Temperature (T) is a constant Obeys the Ideal Gas equation wall friction R. Shanthini 19 July 2019 is a constant is the shear stress acting on the wall where is the average Fanning friction factor
Friction factor: For laminar flow in circular pipes: where Re is the Reynolds number of the flow defined as follows: For lamina flow in a square channel: For the turbulent flow regime: R. Shanthini 19 July 2019 Quasi one-dimensional flow is closer to turbulent velocity profile than to laminar velocity profile.
Ideal Gas equation of state: temperature pressure specific gas constant (not universal gas constant) volume mass Ideal Gas equation of state can be rearranged to give K R. Shanthini 19 July 2019 Pa = N/m 2 kg/m 3 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 shall be written as follows: (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 19 July 2019
x p p+dp u u+du D dx Write the momentum balance over the differential volume chosen. (1) steady mass flow rate cross-sectional area shear stress acting on the wall is the wetted area on which shear is acting R. Shanthini 19 July 2019
p p+dp u u+du D dx x Equation (1) can be reduced to Substituting Since R. Shanthini 19 July 2019 , and , we get (1. 1)
Problem 2 from Problem Set 1 in Compressible Fluid Flow: Show that the differential equation of Problem (1) can be converted into (1. 2) which in turn can be integrated to yield the following design equation: (1. 3) where p is the pressure at the entrance of the pipe, p. L is the pressure at length L from the entrance of the pipe, R is the gas constant, T is the temperature of the gas, is the mass flow rate of the gas flowing through the pipe, and A is the cross-sectional area of the pipe. R. Shanthini 19 July 2019
The differential equation of problem (1) is (1. 1) in which the variables ρ and u must be replaced by the variable p. Let us use the mass flow rate equation to obtain the following: and the ideal gas and therefore It is a constant for steady, isothermal flow in a constant area duct R. Shanthini 19 July 2019
, Using in and (1. 1) we get (1. 2) R. Shanthini 19 July 2019
p p. L L Integrating (1. 2) from 0 to L, we get which becomes (1. 3) R. Shanthini 19 July 2019
Problem 3 from Problem Set 1 in Compressible Fluid Flow: Show that the design equation of Problem (2) is equivalent to (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 19 July 2019
Design equation of Problem (2) is (1. 3) which should be shown to be equivalent to (1. 4) where p and M are the pressure and Mach number at the entry and p. L and ML are the pressure and Mach number at length L from the entry. We need to relate p to M! R. Shanthini 19 July 2019
We need to relate p to M! which gives = constant for steady, isothermal flow in a constant area duct Substituting the above in (1. 3), we get (1. 4) R. Shanthini 19 July 2019
Summary Design equations for steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction (1. 1) (1. 2) (1. 3) (1. 4) R. Shanthini 19 July 2019
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