1 / 17

CP502 Advanced Fluid Mechanics

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. Speed of the flow ( u ). M ≡. Speed of sound ( c ) in the fluid at the flow temperature.

lagrange
Download Presentation

CP502 Advanced Fluid Mechanics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 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

  2. Speed of the flow (u) M≡ Speed of sound (c) in the fluid at the flow temperature Incompressible flow assumption is not valid if Mach number > 0.3 What is a Mach number? Definition of Mach number (M): For an ideal gas, specific heat ratio specific gas constant (in J/kg.K) absolute temperature of the flow at the point concerned (in K)

  3. For an ideal gas, u u = M= c Unit of u = m/s Unit of c = [(J/kg.K)(K)]0.5 = [J/kg]0.5 = (N.m/kg)0.5 = [kg.(m/s2).m/kg]0.5 = [m2/s2]0.5 = m/s

  4. constant area duct quasi one-dimensional flow compressible flow steady flow isothermal flow ideal gas wall friction is a constant Diameter (D) speed (u) u varies only in x-direction x Density (ρ) is NOT a constant Mass flow rate is a constant Temperature (T) is a constant Obeys the Ideal Gas equation is the shear stress acting on the wall where is the average Fanning friction factor

  5. 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: Quasi one-dimensional flow is closer to turbulent velocity profile than to laminar velocity profile.

  6. 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 Pa = N/m2 kg/m3 J/(kg.K)

  7. 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: where p, ρ and u are the respective pressure, density and velocity at distance x from the entrance of the pipe. (1.1)

  8. p p+dp D u u+du dx x 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

  9. p p+dp D u u+du dx x Equation (1) can be reduced to Substituting Since , and , we get (1.1)

  10. Problem 2 from Problem Set 1 in Compressible Fluid Flow: Show that the differential equation of Problem (1) can be converted into which in turn can be integrated to yield the following design equation: 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 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. (1.2) (1.3)

  11. The differential equation of problem (1) is in which the variables ρ and umust be replaced by the variable p. (1.1) Let us use the mass flow rate equation and the ideal gas equation to obtain the following: and and therefore It is a constant for steady, isothermal flow in a constant area duct

  12. , Using and in (1.1) we get (1.2)

  13. p pL L Integrating (1.2) from 0 to L, we get which becomes (1.3)

  14. Problem 3 from Problem Set 1 in Compressible Fluid Flow: Show that the design equation of Problem (2) is equivalent to where M is the Mach number at the entry and ML is the Mach number at length L from the entry. (1.4)

  15. Design equation of Problem (2) is which should be shown to be equivalent to 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. (1.3) (1.4) We need to relate p to M!

  16. 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)

  17. 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)

More Related