CP502 Advanced Fluid Mechanics

1. CP502 Advanced Fluid Mechanics Compressible Flow Lectures 5 and 6 Steady, quasi one-dimensional, isentropic compressible flow of an ideal gas in a variable area duct (continued)

2. Problem 6 from Problem Set #2 in Compressible Fluid Flow: Show that the steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats can be described by the following equations: (2.5) (2.6) (2.7) (2.8) where p0, T0 and ρ0 are the stagnation (where fluid is assumed to be at rest) properties, p, T and ρ are the properties at Mach number M and γ is the specific heat ratio assumed to be a constant.

3. T0 T p0 p u0=0 u Stagnation properties M Ideal gas satisfies Isentropic flow of an ideal gas satisfies Using the above two equations, we can easily prove (2.5)

4. T0 T p0 p u0=0 u Stagnation properties M Start with the energy balance for steady, adiabatic, inviscid, quasi one-dimensional, compressible flow: dh + udu = 0 (2.2) Using dh = cp dT, which is applicable for ideal gas, in the above, we get cp dT + udu = 0 Integrating the above between the two cross-sections, we get cp(T – T0) + (u2 – 0)/2 = 0

5. Using the definition of M, we get cp(T – T0) + M2 γRT/2 = 0 Since cp = γR/(γ-1) for an ideal gas, the above can be written as γR(T – T0) /(γ -1) + M2 γRT/2 = 0 which can be rearranged to give the following: (2.6) The above equation relates the stagnation temperature (at near zero velocity) to a temperature at Mach number M for steady, isentropic, one-dimensional, compressible flow of an ideal gas in a variable area duct.

6. Using (2.6) in (2.5), we can easily get the following equations relating the stagnation properties (at near zero velocity) to properties at Mach number M for steady, isentropic, one-dimensional, compressible flow of an ideal gas in a variable area duct: (2.7) (2.8)

7. Problem 7 from Problem Set #2 in Compressible Fluid Flow: Show that the mass flow rate in a steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats is given by the following equations: (2.9) (2.10)

8. Problem 8 from Problem Set #2 in Compressible Fluid Flow: A large air reservoir contains air at a temperature of 400 K and a pressure of 600 kPa. The air reservoir is connected to a second chamber through a converging duct whose exit area is 100 mm2. The pressure inside the second chamber can be regulated independently. Assuming steady, isentropic flow in the duct, calculate the exit Mach number, exit temperature, and mass flow rate through the duct when the pressure in the second chamber is (i) 600 kPa, (ii) 500 kPa, (iii) 400 kPa, (iv) 300 kPa and (v) 200 kPa. Assumptions: Steady, isentropic flow Ae = 100 mm2 Air Determine the following at the exit of the converging duct: Mach Number Me = ? Temperature Te = ? Mass flow rate = ? pb kPa is given T0 = 400 K p0 = 600 kPa pbis known as the back pressure

9. (i) For pb = 600 kPa, there will be no flow since p0 = 600 kPa as well (ii) For pb = 500 kPa, assume pe is the same as pb. = 0.517 Using (2.7), we get Using (2.5), we get = 379.7 K Using (2.9), we get 0.5 ( ) 1.4 = (100 x10-6 m2) (0.517) (500,000 Pa) = 0.0927 kg/s (8314/29)(379.7) J/kg

10. Results summarized: Is there a problem with the above results? Ae = 100 mm2 Air pb kPa (given) T0 = 400 K p0 = 600 kPa

11. Results summarized: Yes, there is. Supersonic Mach numbers have been reached from stagnation condition in a converging duct? Ae = 100 mm2 Air pb kPa (given) T0 = 400 K p0 = 600 kPa It can not happen!

12. Maximum Mach number at the exit of a converging duct must be Me = 1. Corresponding pe (denoted by p*) could be calculated using (2.7) as follows: = 317 kPa p* = 600 Using (2.5), we get = 333.3 K Using (2.9), we get the mass flow rate as 0.1213 kg/s.

13. Results summarized: Flow chocks at the throat of the converging duct at an exit pressure of 317 kPa at a maximum flow rate of 0.1213 kg/s? Ae = 100 mm2 Air pb kPa (given) T0 = 400 K p0 = 600 kPa

14. Ae = 100 mm2 Air pe kPa (given) T0 = 400 K p0 = 600 kPa Pressure at the throat cannot be less than the limiting pressure (317 kPa in this case) even if we keep a lower pressure in the second chamber (known as the back pressure). Mass flow rate cannot be increased above the maximum mass flow rate ( 0.1213 kg/s in this case) even if we increase the driving force by decreasing the back pressure in the second chamber.

15. Problem 9 from Problem Set #2 in Compressible Fluid Flow: Air at 900 kPa and 400 K enters a converging nozzle with a negligible velocity. The throat area of the nozzle is 10 cm2. Assuming isentropic flow, calculate and plot the exit pressure, the exit Mach number, the exit velocity, and the mass flow rate versus the back pressure pb for 900 ≤pb≤ 100 kPa.

16. Results: Limiting pressure = 475.45 kPa

17. Results (continued): Sonic condition M = 1

18. Results (continued): Maximum velocity 365.77 m/s

19. Results (continued): Maximum mass flow rate 18.1981 kg/s

20. Problem 10 from Problem Set #2 in Compressible Fluid Flow: Consider a converging-diverging duct with a circular cross-section for a mass flow rate of 3 kg/s of air and inlet stagnation conditions of 1400 kPa and 200oC. Assume that the flow is isentropic and the exit pressure is 100 kPa. Plot the pressure and temperature of the air flow along the duct as a function of M. Plot also the diameter of the duct as a function of M. p0 = 1400 kPa T0 = (273+200) K M0≈ 0 pe = 100 kPa = 3 kg/s

21. p0 and T0 are known. Use the equations below to calculate p and T for different values of M. (2.6) (2.7) Use (2.9) to calculate the exit are of the duct as follows:

23. Enlarged version

24. Shape of the converging-diverging duct

25. Problem 11 from Problem Set #2 in Compressible Fluid Flow: Air at approximately zero velocity enters a converging-diverging duct at a stagnation pressure and a stagnation temperature of 1000 kPa and 480 K, respectively. Throat area of the duct is 0.002 m2. The flow inside the duct is isentropic, and the exit pressure is 31.7 kPa. For air, γ= 1.4 and R = 287 J/kg. Determine (i) the exit Mach number, (ii) the exit temperature, (iii) the exit area of the duct, and (iii) the mass flow rate through the duct. p0 = 1000 kPa T0 = 480 K M0≈ 0 pe = 31.7 kPa Me = ? Ae = ? Athroat = 0.002 m2

26. (i) Rearrange (2.7) to determine Meas follows: = 2.9 (ii) Rearrange (2.5) to determine Teas follows: = 179 K

27. (iii) Using (2.10), the exit area of the duct can be calculated as follows: Since the exit Mach number is 2.9, the flow is supersonic in the diverging section. It is not possible unless sonic conditions are achieved at the throat. Therefore Mt = 1 in the above expression. Therefore, we get Since Me = 2.9 and At = 0.002 m2, the above expression gives = 0.0077 m2

28. (iv) Use (2.9) to get the mass flow rate through the duct as follows: = 3.695 kg/s