1. CP502 Advanced Fluid Mechanics Compressible Flow Lectures 1 & 2 Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction

2. R. Shanthini 08 Dec 2010 What is a Mach number? Definition of Mach number (M): M ≡M ≡ Speed of the flow (u) Speed of sound (c) in the fluid at the flow temperature Incompressible flow assumption is not valid if Mach number > 0.3 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. R. Shanthini 08 Dec 2010 For an ideal gas, Unit of c = [(J/kg.K)(K)] 0.5 = [m 2 /s 2 ] 0.5 = m/s = [kg.(m/s 2 ).m/kg] 0.5 M =M = u u c = Unit of u = m/s = [J/kg] 0.5 = (N.m/kg) 0.5

4. R. Shanthini 08 Dec 2010 constant area duct quasi one-dimensional flow compressible flow steady flow isothermal flow ideal gas wall friction Diameter (D) is a constant speed (u) x u varies only in x-direction Density (ρ) is NOT 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 Mass flow rate is a constant

5. R. Shanthini 08 Dec 2010 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. R. Shanthini 08 Dec 2010 Ideal Gas equation of state: pressure volume mass specific gas constant (not universal gas constant) temperature Ideal Gas equation of state can be rearranged to give kg/m 3 J/(kg.K) K Pa = N/m 2

7. R. Shanthini 08 Dec 2010 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. Problem 1 from Problem Set 1 in Compressible Fluid Flow: (1.1)

8. R. Shanthini 08 Dec 2010 p p+dp u u+du x dx D Write the momentum balance over the differential volume chosen. cross-sectional area is the wetted area on which shear is acting steady mass flow rate shear stress acting on the wall (1)

9. R. Shanthini 08 Dec 2010 Substituting Equation (1) can be reduced to Since, and, we get (1.1) p p+dp u u+du x dx D

10. R. Shanthini 08 Dec 2010 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, 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. Problem 2 from Problem Set 1 in Compressible Fluid Flow: (1.2) (1.3)

11. R. Shanthini 08 Dec 2010 The differential equation of problem (1) is in which the variables ρ and u must 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: andand therefore It is a constant for steady, isothermal flow in a constant area duct

12. R. Shanthini 08 Dec 2010 (1.1), Usingand (1.2) in we get

13. R. Shanthini 08 Dec 2010 (1.3) ppLpL L Integrating (1.2) from 0 to L, we get which becomes

14. R. Shanthini 08 Dec 2010 Show that the design equation of Problem (2) is equivalent to where M is the Mach number at the entry and M L is the Mach number at length L from the entry. Problem 3 from Problem Set 1 in Compressible Fluid Flow: (1.4)

15. R. Shanthini 08 Dec 2010 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 p L and M L are the pressure and Mach number at length L from the entry. (1.4) (1.3) We need to relate p to M!

16. R. Shanthini 08 Dec 2010 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. R. Shanthini 08 Dec 2010 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)

18. R. Shanthini 08 Dec 2010 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;

19. R. Shanthini 08 Dec 2010 (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;

20. R. Shanthini 08 Dec 2010 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;

21. R. Shanthini 08 Dec 2010 = 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;

22. R. Shanthini 08 Dec 2010 = 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

23. R. Shanthini 08 Dec 2010 = 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:

24. R. Shanthini 08 Dec 2010 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;

25. R. Shanthini 08 Dec 2010 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;

26. R. Shanthini 08 Dec 2010 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;

27. R. Shanthini 08 Dec 2010 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

28. R. Shanthini 08 Dec 2010 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

29. R. Shanthini 08 Dec 2010 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.

30. R. Shanthini 08 Dec 2010 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)

31. R. Shanthini 08 Dec 2010 Differential equation of Problem 2: (1.2) In flows where p decreases along the flow direction can be rearranged to give (1.5)

32. R. Shanthini 08 Dec 2010 Differential equation of Problem 2: (1.2) In flows where p increases along the flow direction can be rearranged to give (1.6)

33. R. Shanthini 08 Dec 2010 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;

34. R. Shanthini 08 Dec 2010 L Limiting pressure: 40 atm 97 o C 500 m/s p*=? Air: γ = 1.4; molecular mass = 29;

35. R. Shanthini 08 Dec 2010 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;

36. R. Shanthini 08 Dec 2010 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;