CP502 Advanced Fluid Mechanics

1. CP502 Advanced Fluid Mechanics Compressible Flow Part 01_Set 03: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction (continued)

2. Problem 10 from Problem Set 1 in Compressible Fluid Flow: Determine the isothermal mass flow rate of air in a pipe of 10-mm-i.d. and 1m long with upstream condition of 1 MPa and 300 K with a exit pressure low enough to choke the flow in the pipe assuming an average Fanning friction factor of 0.0075. Determine also the exit pressure. Given μ= 2.17 x 10-5 kg/m.s, calculate the Reynolds number of the flow to check if the given flow were turbulent. Air: γ = 1.4; molecular mass = 29; = ?; = 0.0075 D = 10 mm flow is choked p = 1 MPa T = 300 K L = 1 m At choking condition, pL = p*,ML= and L = Lmax

3. Air: γ = 1.4; molecular mass = 29; = ?; = 0.0075 D = 10 mm p = 1 MPa T = 300 K L = Lmax = 1 m 0.5 ( ) (π) (10/1000 m)2 (1000,000 Pa) M 1.4 = 4 (8314/29)(300) J/kg = 0.317 M M at the entrance could be determined using (1.9)

4. Air: γ = 1.4; molecular mass = 29; = ?; = 0.0075 D = 10 mm p = 1 MPa T = 300 K L = Lmax = 1 m Use (part of 1.9) Solving the nonlinear equation above gives M = 0.352 at the entrance = 0.317 M = 0.317 x 0.352 = 0.1116 kg/s

5. Air: γ = 1.4; molecular mass = 29; = ?; = 0.0075 D = 10 mm p = 1 MPa T = 300 K L = Lmax = 1 m Determine the exit pressure. Since (pM)entrance = (pM)exit (1 MPa) (0.352)= pexit () 0.5 pexit = (1 MPa) (0.352) (1.4) = 0.417 MPa

6. Air: γ = 1.4; molecular mass = 29; = ?; = 0.0075 D = 10 mm p = 1 MPa T = 300 K L = Lmax = 1 m Reynolds Number: 4 (0.1116 kg/s) = π(10/1000 m) (2.17 x 10-5kg/m.s) = 6.5 x 105 Therefore, flow is turbulent

7. Problem 11 from Problem Set 1 in Compressible Fluid Flow: Air flows at a mass flow rate of 9.0 kg/s isothermally at 300 K through a straight rough duct of constant cross-sectional area 1.5 x 10-3 m2. At one end A the pressure is 6.5 bar and at the other end B the pressure is 8.5 bar. Determine the following: (i) Velocities uA and uB (ii) Force acting on the duct wall (iii) Rate of heat transfer through the duct wall In which direction is the gas flowing?

8. Air: γ = 1.4; molecular mass = 29; = 9.0 kg/s; A = 1.5x10-3 m2 pA= 6.5 bar T = 300 K pB= 8.5 bar (i) Velocities uA and uB = ? (9 kg/s) (8314/29 J/kg.K) (300 K) = (1.5x10-3 m2) (6.5 bar) (100,000 Pa/bar) = 794 m/s 6.5 bar = 794 m/s = 607 m/s 8.5 bar

9. Air: γ = 1.4; molecular mass = 29; = 9.0 kg/s; A = 1.5x10-3 m2 pA= 6.5 bar T = 300 K pB= 8.5 bar (ii) Force acting on the duct wall = ? Force balance on the entire duct gives the following: pA A + uA = pB A + uB+ Force acting on the duct wall Force acting on the duct wall = (pA – pB ) A + (uA– uB ) = (6.5 –8.5) bar x 100,000 Pa/bar x 1.5 x 10-3 m2 + (9.0 kg/s)(794 –607) m/s = -300 Pa.m2 + 1683 kg.m/s2 = -300 N + 1683 N = 1383 N

10. Air: γ = 1.4; molecular mass = 29; = 9.0 kg/s; A = 1.5x10-3 m2 pA= 6.5 bar T = 300 K pB= 8.5 bar (iii) Rate of heat transfer through the duct wall = ? Energy balance on the entire duct gives the following: Rate of heat transfer through the duct wall from the surroundings + hA + uA2/2 = hB + uB2/2 Enthalpy at A Enthalpy at B Kinetic energy at B Kinetic energy at A

11. Air: γ = 1.4; molecular mass = 29; = 9.0 kg/s; A = 1.5x10-3 m2 pA= 6.5 bar T = 300 K pB= 8.5 bar Rate of heat transfer through the duct wall from the surroundings = (hB – hA) +(uB 2 – uA 2)/2 Since (hB – hA) = cp(TB – TA) = 0 for isothermal flow of an ideal gas Rate of heat transfer through the duct wall from the surroundings = (uB 2 – uA 2)/2 = (6072 – 7942)/2 m2/s2 = (-130993.5 m2/s2)= (-130993.5 J/kg) = (-130993.5 J/kg) (9.0 kg/s) = -1178942 J/s = -1179 kW Heat is lost to the surroundings

12. Air: γ = 1.4; molecular mass = 29; = 9.0 kg/s; A = 1.5x10-3 m2 pA= 6.5 bar T = 300 K pB= 8.5 bar Direction of the gas flow: Determine first the limiting pressure as follows: = (9.0/1.5x10-3) kg/m2.s (8314*300/29 J/kg)0.5 = (9.0/1.5x10-3) kg/m2.s (8314*300/29 J/kg)0.5 = 17.6 bar Since pAand pBare lower than the limiting pressure, p increases along the flow direction (see Problem 6). Therefore, gas is flowing from A to B.

13. Summarizing the results of Problem 11: = 9.0 kg/s; pB= 8.5 bar pA= 6.5 bar T = 300 K uB= 607 m/s uA= 794 m/s Pressure increases in the flow direction and therefore velocity decreases according to the following equation: Force acting on the entire duct wall is 1383 N Velocity decreases and therefore kinetic energy is lost across the duct. The lost energy is transferred from the duct to the surroundings through the duct wall.

14. Problem 12 from Problem Set 1 in Compressible Fluid Flow: Gas produced in a coal gasification plant (molecular weight = 0.013 kg/mol, μ = 10-5 kg/m.s, γ = 1.36) is sent to neighbouring industrial users through a bare 15-cm-i.d. commercial steel pipe 100 m long. The pressure gauge at one end of the pipe reads 1 MPa absolute. At the other end it reads 500 kPa. The temperature is 87oC. Estimate the flow rate of coal gas through the pipe? Additional data: ε= 0.046 mm for commercial steel. For fully developed turbulent flow in rough pipes, the average Fanning friction factor can be found by use of the following:

15. Properties of gas produced: Molecular weight = 0.013 kg/mol; μ= 10-5 kg/m.s; γ = 1.36 pL = 500 kPa D = 15 cm p = 1 MPa T = (273+87) K L = 100 m What is the flow rate through the pipe?

16. Design equation to be used: (1.3) = 1/[4 log(3.7x15x10/0.046)] = 0.0613 = 0.0038 = 4 x 0.0038 x 100 m / (15 cm) = 10.1333 = (500/1000)2 = 0.25 Using the above in (1.3), we get 10.1333 – ln(0.25) = = 15.3595 1 – 0.25

17. = 15.3595 p = 1 MPa = 1,000,000 Pa; R = 8.314 J/mol.K = 8.314/0.013 J/kg.K; T = 360 K; A = πD2/4 = π(15 cm)2/4 = π(0.15 m)2/4; Therefore, = 9.4 kg/s; Check the Reynolds number: Re = uDρ/μ = D/μ = (9.4 kg/s)(15/100 m)/(10-5 kg/m.s) = 1.4x105 Therefore, flow is turbulent

18. Governing equation for incompressible flow: Starting from the mass and momentum balances, obtain the differential equation describing the quasi one-dimensional, incompressible, isothermal, steady flow of an ideal gas through a constant area pipe of diameter D and average Fanning friction factor. Density (ρ) is a constant Incompressible flow Mass flow rate is a constant Steady flow Constant area pipe A is a constant Therefore, u is a constant for a steady, quasi one-dimensional, compressible flow in a constant area pipe.

19. p p+dp D u u+du dx x Write the momentum balance over the differential volume chosen. Since , and , we get

20. Therefore, we get (2) Rearranging (2) gives It means p decreases in the flow direction. Since ρand u are constants, integrating the above gives pressure at the exit pressure at the entrance

21. Rework Problem 12 assuming incompressible flow: Molecular weight = 0.013 kg/mol; = 0.0038 pL = 500 kPa D = 15 cm p = 1 MPa T = (273+87) K L = 100 m What is the flow rate through the pipe? Design equation used for compressible flow (1.3) Design equation to be used with incompressible flow

22. Molecular weight = 0.013 kg/mol; = 0.0038 pL = 500 kPa D = 15 cm p = 1 MPa T = (273+87) K L = 100 m Substitute in the above, we get

23. Molecular weight = 0.013 kg/mol; = 0.0038 pL = 500 kPa D = 15 cm p = 1 MPa T = (273+87) K L = 100 m What is ρ? ρ = (ρentrance + ρexit) / 2 = [(p/RT)entrance + (p/RT)exit)] / 2 = (pentrance + pexit) / 2RT = (1,000,000 + 500,000) Pa / [2 x (8.314/0.013) x 300 J/kg] = 1.9545 kg/m3 Compare 7.76 kg/s with the 9.4 kg/s obtained considering the flow to be compressible. Therefore, = 7.76 kg/s

24. Important Note: Problems (13) and (14) from Problem Set 1 in Compressible Fluid Flow are assignments to be worked out by the students themselves in preparation to the mid-semester examination.