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CP502 Advanced Fluid Mechanics

CP502 Advanced Fluid Mechanics. Compressible Flow. Part 01_Set 02: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction (continued). Summary

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CP502 Advanced Fluid Mechanics

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  1. CP502 Advanced Fluid Mechanics Compressible Flow Part 01_Set 02: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction (continued)

  2. 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)

  3. Problem 4 from Problem Set 1 in Compressible Fluid Flow: Nitrogen (γ = 1.4; molecular mass = 28) is to be fed through a 15 mm-id commercial steel pipe 11.5 m long to a synthetic ammonia plant. Calculate the downstream pressure in the line for a flow rate of 1.5 mol/s, an upstream pressure of 600 kPa, and a temperature of 27oC throughout. The average Fanning friction factor may be taken as 0.0066. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m

  4. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) = 0.0066; = 20.240 L = 11.5 m; D = 15 mm = 0.015 m; unit?

  5. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) p = 600 kPa = 600,000 Pa; T = 300 K; R = 8.314 kJ/kmol.K = 8.314/28 kJ/kg.K = 8314/28 J/kg.K; = 1.5 mol/s = 1.5 x 28 g/s = 1.5 x 28/1000 kg/s; A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4; = 71.544 unit?

  6. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) 20.240 = 71.544 p = 600 kPa = 600,000 Pa Solve the nonlinear equation above to determine pL pL = ?

  7. To find a zero of a function using MATLAB: Function: where p = 600 kPa 20.240 = 71.544 Write an M-file called P4.m: function y=f(pL) p=600; y=20.240-71.544*(1-(pL/p)^2)-log((pL/p)^2); To find the zero near 600, enter on the command window: pL = fzero(@P4,600) pL = 506.4072 Easiest way to solve nonlinear equation R. Shanthini 13 Feb 2015

  8. To find a zero of a function using Other Ways: Function: where p = 600 kPa 20.240 = 71.544 Determine the approximate solution by ignoring the ln-term: pL = p (1-20.240/71.544)0.5 = 508.1 kPa Check the value of the ln-term using pL= 508.1 kPa: ln[(pL /p)2] = ln[(508.1 /600)2] = -0.3325 This value is small when compared to 20.240. And therefore pL= 508.1 kPa is a good first approximation.

  9. Now, solve the nonlinear equation for pL values close to 508.1 kPa: where p = 600 kPa 20.240 = 71.544

  10. Problem 4 continued: Rework the problem in terms of Mach number and determine ML. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.4) = 20.240 (already calculated in Problem 4) M = ?

  11. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m u u 1 1 RT M = = = = c A p ( ) 0.5 4 (1.5x 28/1000 kg/s) (8314/28)(300) J/kg = 1.4 π (15/1000 m)2 (600,000 Pa) = 0.1

  12. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.4) 20.240 Solution obtained using ‘fzero’ in MATLAB ML = 0.11852

  13. 20.240 Determine the approximate solution by ignoring the ln-term: ML = 0.1 / (1-20.240 x 1.4 x 0.12)0.5 = 0.118 Check the value of the ln-term using ML= 0.118: ln[(0.1/ML)2] = ln[(0.1 /0.118)2] = -0.3310 This value is small when compared to 20.240. And therefore ML= 0.118 is a good first approximation.

  14. Now, solve the nonlinear equation for ML values close to 0.118: 20.240

  15. Problem 5 from Problem Set 1 in Compressible Fluid Flow: Explain why the design equations of Problems (1), (2) and (3) are valid only for fully turbulent flow and not for laminar flow.

  16. Problem 6 from Problem Set 1 in Compressible Fluid Flow: Starting from the differential equation of Problem (2), or otherwise, prove that p, the pressure, in a quasi one-dimensional, compressible, isothermal, steady flow of an ideal gas in a pipe with wall friction should always satisfy the following condition: (1.5) in flows where p decreases along the flow direction, and (1.6) in flows where p increases along the flow direction.

  17. Differential equation of Problem 2: (1.2) can be rearranged to give In flows where p decreases along the flow direction (1.5)

  18. Differential equation of Problem 2: (1.2) can be rearranged to give In flows where p increases along the flow direction (1.6)

  19. Problem 7 from Problem Set 1 in Compressible Fluid Flow: Air enters a horizontal constant-area pipe at 40 atm and 97oC with a velocity of 500 m/s. What is the limiting pressure for isothermal flow? It can be observed that in the above case pressure increases in the direction of flow. Is such flow physically realizable? If yes, explain how the flow is driven along the pipe. Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L

  20. Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L Limiting pressure:

  21. Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L (40 atm) (500 m/s) = = 61.4 atm 0.5 [(8314/29)(273+97) J/kg]

  22. Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=61.4 atm L Pressure increases in the direction of flow. Is such flow physically realizable? YES If yes, explain how the flow is driven along the pipe. Use the momentum balance over a differential element of the flow (given below) to explain.

  23. Problem 8 from Problem Set 1 in Compressible Fluid Flow: Show that the equations in Problem 6 are equivalent to the following: (1.7) in flows where p decreases along the flow direction (1.8) in flows where p increases along the flow direction

  24. In flows where p decreases along the flow direction (1.5) Since we get (1.7)

  25. In flows where p increases along the flow direction (1.6) Since we get (1.8)

  26. Summary: x and and Limiting pressure: Limiting Mach number:

  27. Limiting Mach number for air: x For air, γ = 1.4 = 0.845 is associated with If M < 0.845 then pressure decreases in the flow direction. That is, the pressure gradient causes the flow. is associated with If M > 0.845 then pressure increases in the flow direction. That is, momentum causes the flow working against the pressure gradient.

  28. Problem 9 from Problem Set 1 in Compressible Fluid Flow: Show that when the flow has reached the limiting pressure or the limiting Mach number the length of the pipe across which such conditions are reached, denoted by Lmax, shall satisfy the following equation: where pressure p and Mach number M are the conditions of the flow at the entrance of the pipe.

  29. p; M p*; M* Lmax Start with the following: (1.3) Substitute L = Lmax and pL = in (1.3) to get (part of 1.9)

  30. p; M p*; M* Lmax Start with the following: (1.4) Substitute L = Lmax and ML = in (1.4) to get (part of 1.9) Therefore, (1.9)

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