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. R. Shanthini 09 Feb 2012 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. R. Shanthini 09 Feb 2012 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 27 o C throughout. The average Fanning friction factor may be taken as 0.0066. Problem 4 from Problem Set 1 in Compressible Fluid Flow: p = 600 kPa p L = ? L = 11.5 m D = 15 mm = 1.5 mol/s; T = 300 K = 0.0066γ = 1.4; molecular mass = 28;

4. R. Shanthini 09 Feb 2012 (1.3) Design equation: = 0.0066; L = 11.5 m; D = 15 mm = 0.015 m; = 20.240 unit? p = 600 kPa p L = ? L = 11.5 m D = 15 mm T = 300 K = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

5. R. Shanthini 09 Feb 2012 p = 600 kPa = 600,000 Pa; 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; T = 300 K; A = πD 2 /4 = π(15 mm) 2 /4 = π(0.015 m) 2 /4; = 71.544 unit? (1.3) Design equation: p = 600 kPa p L = ? L = 11.5 m D = 15 mm T = 300 K = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

6. R. Shanthini 09 Feb 2012 = 71.54420.240 p = 600 kPa = 600,000 Pa p L = ? (1.3) Design equation: Solve the nonlinear equation above to determine p L p = 600 kPa p L = ? L = 11.5 m D = 15 mm T = 300 K = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

7. R. Shanthini 09 Feb 2012 = 71.54420.240 p = 600 kPa = 600,000 Pa This value is small when compared to 20.240. And therefore p L = 508.1 kPa is a good first approximation. Determine the approximate solution by ignoring the ln-term: p L = p (1-20.240/71.544) 0.5 = 508.1 kPa Check the value of the ln-term using p L = 508.1 kPa: ln[(p L /p) 2 ] = ln[(508.1 /600) 2 ]= -0.3325

8. R. Shanthini 09 Feb 2012 = 71.54420.240 p = 600 kPa = 600,000 Pa p L kPaLHS of the above equation RHS of the above equation 51020.24019.528 50920.24019.727 508.120.24019.905 50720.24020.123 506.520.24020.222 50620.24020.320 Now, solve the nonlinear equation for p L values close to 508.1 kPa:

9. R. Shanthini 09 Feb 2012 Rework the problem in terms of Mach number and determine M L. Problem 4 continued : Design equation: (1.4) = 20.240 (already calculated in Problem 4) M = ? p = 600 kPa M L = ? L = 11.5 m D = 15 mm T = 300 K = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

10. R. Shanthini 09 Feb 2012 M = u c = u = 1 = 1 Ap RT p = 600 kPa M L = ? L = 11.5 m D = 15 mm T = 300 K = 4 (1.5x 28/1000 kg/s) π(15/1000 m) 2 (600,000 Pa) () (8314/28)(300) J/kg 1.4 0.5 = 0.1 = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

11. R. Shanthini 09 Feb 2012 Design equation: (1.4) M L = ? p = 600 kPa L = 11.5 m D = 15 mm T = 300 K Solve the nonlinear equation above to determine M L M L = ? 20.240 = 1.5 mol/s;= 0.0066γ = 1.4; molecular mass = 28;

12. R. Shanthini 09 Feb 2012 20.240 This value is small when compared to 20.240. And therefore M L = 0.118 is a good first approximation. Determine the approximate solution by ignoring the ln-term: M L = 0.1 / (1-20.240 x 1.4 x 0.1 2 ) 0.5 = 0.118 Check the value of the ln-term using M L = 0.118: ln[(0.1/M L ) 2 ] = ln[(0.1 /0.118) 2 ]= -0.3310

13. R. Shanthini 09 Feb 2012 p L kPaLHS of the above equation RHS of the above equation 0.11620.24018.049 0.11720.240 0.11820.24019.798 0.118520.24020.222 0.11920.24020.64 Now, solve the nonlinear equation for M L values close to 0.118: 20.240

14. R. Shanthini 09 Feb 2012 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.

15. R. Shanthini 09 Feb 2012 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 satisfies the following condition: in flows where p decreases along the flow direction, and in flows where p increases along the flow direction. (1.5) (1.6)

16. R. Shanthini 09 Feb 2012 Differential equation of Problem 2: (1.2) In flows where p decreases along the flow direction can be rearranged to give (1.5)

17. R. Shanthini 09 Feb 2012 Differential equation of Problem 2: (1.2) In flows where p increases along the flow direction can be rearranged to give (1.6)

18. R. Shanthini 09 Feb 2012 Problem 7 from Problem Set 1 in Compressible Fluid Flow: Air enters a horizontal constant-area pipe at 40 atm and 97 o C 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. L 40 atm 97 o C 500 m/s p*=? Air: γ = 1.4; molecular mass = 29;

19. R. Shanthini 09 Feb 2012 L Limiting pressure: 40 atm 97 o C 500 m/s p*=? Air: γ = 1.4; molecular mass = 29;

20. R. Shanthini 09 Feb 2012 L 40 atm 97 o C 500 m/s p*=? = (40 atm) (500 m/s) [(8314/29)(273+97) J/kg] 0.5 = 61.4 atm Air: γ = 1.4; molecular mass = 29;

21. R. Shanthini 09 Feb 2012 L 40 atm 97 o C 500 m/s p*=61.4 atm 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. Air: γ = 1.4; molecular mass = 29;

22. R. Shanthini 09 Feb 2012 Problem 8 from Problem Set 1 in Compressible Fluid Flow: Show that the equations in Problem 6 are equivalent to the following: (1.7) (1.8) in flows where p decreases along the flow direction in flows where p increases along the flow direction

23. R. Shanthini 09 Feb 2012 In flows where p decreases along the flow direction (1.5) Sincewe get (1.7)

24. R. Shanthini 09 Feb 2012 In flows where p increases along the flow direction (1.6) Sincewe get (1.8)

25. R. Shanthini 09 Feb 2012 x and Limiting pressure: Limiting Mach number: Summary:

26. R. Shanthini 09 Feb 2012 For air, γ = 1.4 = 0.845 If M < 0.845 then pressure decreases in the flow direction. That is, the pressure gradient causes the flow. x 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. Limiting Mach number for air:

27. R. Shanthini 09 Feb 2012 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 L max, shall satisfy the following equation: where pressure p and Mach number M are the conditions of the flow at the entrance of the pipe.

28. R. Shanthini 09 Feb 2012 L max p; Mp*; M* (1.3) Start with the following: Substitute L = L max and p L = in (1.3) to get (part of 1.9)

29. R. Shanthini 09 Feb 2012 L max p; Mp*; M* Start with the following: Substitute L = L max and M L = in (1.4) to get (part of 1.9) (1.4) Therefore, (1.9)