CP 502 Advanced Fluid Mechanics Compressible Flow Part
CP 502 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)
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 1 m 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. = ? ; p = 1 MPa Air: γ = 1. 4; molecular mass = 29; D = 10 mm T = 300 K = 0. 0075 flow is choked L=1 m At choking condition, p. L = p*, ML = R. Shanthini 19 July 2019 and L = Lmax
Air: γ = 1. 4; molecular mass = 29; = ? ; p = 1 MPa D = 10 mm = 0. 0075 T = 300 K L = Lmax = 1 m = (π) (10/1000 m)2 (1000, 000 4 Pa) M ( 1. 4 (8314/29)(300) J/kg 0. 5 ) = 0. 317 M M at the entrance could be determined using (1. 9) R. Shanthini 19 July 2019
= ? ; p = 1 MPa Air: γ = 1. 4; molecular mass = 29; D = 10 mm = 0. 0075 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 R. Shanthini 19 July 2019
= ? ; p = 1 MPa Air: γ = 1. 4; molecular mass = 29; D = 10 mm = 0. 0075 T = 300 K L = Lmax = 1 m Determine the exit pressure. Since (p. M)entrance = (p. M)exit (1 MPa) (0. 352) = pexit ( ) 0. 5 pexit = (1 MPa) (0. 352) (1. 4) R. Shanthini 19 July 2019 = 0. 417 MPa
= ? ; p = 1 MPa Air: γ = 1. 4; molecular mass = 29; D = 10 mm = 0. 0075 T = 300 K L = Lmax = 1 m Reynolds Number: 4 (0. 1116 kg/s) = π (10/1000 m) (2. 17 x 10 -5 kg/m. s) = 6. 5 x 105 Therefore, flow is turbulent R. Shanthini 19 July 2019
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 m 2. 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 u. A and u. B (ii) Force acting on the duct wall (iii) Rate of heat transfer through the duct wall In which direction is the gas flowing? R. Shanthini 19 July 2019
Air: γ = 1. 4; molecular mass = 29; p. A = 6. 5 bar A = 1. 5 x 10 -3 m 2 = 9. 0 kg/s; T = 300 K p. B = 8. 5 bar (i) Velocities u. A and u. B = ? = (9 kg/s) (8314/29 J/kg. K) (300 K) (1. 5 x 10 -3 m 2) (6. 5 bar) (100, 000 Pa/bar) = 794 m/s 6. 5 bar = 8. 5 bar 794 m/s R. Shanthini 19 July 2019 = 607 m/s
Air: γ = 1. 4; molecular mass = 29; p. A = 6. 5 bar (ii) Force A = 1. 5 x 10 -3 m 2 = 9. 0 kg/s; T = 300 K p. B = 8. 5 bar acting on the duct wall = ? Force balance on the entire duct gives the following: p. A A + u A = p. B A + u. B + Force acting on the duct wall = (p. A – p. B ) A + (u. A – u. B ) = (6. 5 – 8. 5) bar x 100, 000 Pa/bar x 1. 5 x 10 -3 m 2 + (9. 0 kg/s) (794 – 607) m/s = -300 Pa. m 2 + 1683 kg. m/s 2 R. Shanthini 19 July 2019 = -300 N + 1683 N = 1383 N
Air: γ = 1. 4; molecular mass = 29; p. A = 6. 5 bar (iii) Rate A = 1. 5 x 10 -3 m 2 = 9. 0 kg/s; T = 300 K p. B = 8. 5 bar 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 + h. A + u. A 2/2 Enthalpy at A = Enthalpy at B Kinetic energy at A R. Shanthini 19 July 2019 h. B + u. B 2/2 Kinetic energy at B
Air: γ = 1. 4; molecular mass = 29; p. A = 6. 5 bar A = 1. 5 x 10 -3 m 2 = 9. 0 kg/s; T = 300 K p. B = 8. 5 bar Rate of heat transfer through the duct wall from the surroundings = (h. B – h. A) + (u. B 2 – u. A 2)/2 Since (h. B – h. A) = cp (TB – TA) = 0 for isothermal flow of an ideal gas Rate of heat transfer through the duct wall from the surroundings = (u. B 2 – u. A 2)/2 = (607 2 – 794 2)/2 m 2/s 2 = (-130993. 5 m 2/s 2) = (-130993. 5 J/kg) (9. 0 kg/s) = -1178942 J/s = -1179 k. W R. Shanthini 19 July 2019 Heat is lost to the surroundings
Air: γ = 1. 4; molecular mass = 29; p. A = 6. 5 bar A = 1. 5 x 10 -3 m 2 = 9. 0 kg/s; T = 300 K p. B = 8. 5 bar Direction of the gas flow: Determine first the limiting pressure as follows: = (9. 0/1. 5 x 10 -3) kg/m 2. s (8314*300/29 J/kg)0. 5 = 17. 6 bar Since p. A and p. B are lower than the limiting pressure, p increases along the flow direction (see Problem 6). R. Shanthini 19 July 2019 Therefore, gas is flowing from A to B.
Summarizing the results of Problem 11: = 9. 0 kg/s; p. A = 6. 5 bar u. A = 794 m/s T = 300 K p. B = 8. 5 bar u. B = 607 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. R. Shanthini 19 July 2019
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 k. Pa. The temperature is 87 o. C. 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: R. Shanthini 19 July 2019
Properties of gas produced: Molecular weight = 0. 013 kg/mol; μ = 10 -5 kg/m. s; γ = 1. 36 p = 1 MPa D = 15 cm T = (273+87) K L = 100 m What is the flow rate through the pipe? R. Shanthini 19 July 2019 p. L = 500 k. Pa
Design equation to be used: (1. 3) = 1/[4 log(3. 7 x 15 x 10/0. 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 = R. Shanthini 19 July 2019 10. 1333 – ln(0. 25) 1 – 0. 25 = 15. 3595
= 15. 3595 p = 1 MPa = 1, 000 Pa; R = 8. 314 J/mol. K = 8. 314/0. 013 J/kg. K; T = 360 K; A = πD 2/4 = π(15 cm)2/4 = π(0. 15 m)2/4; Therefore, = 9. 4 kg/s; Check the Reynolds number: Re = u. Dρ/μ = D/μ = (9. 4 kg/s)(15/100 m)/(10 -5 kg/m. s) = 1. 4 x 105 R. Shanthini 19 July 2019 Therefore, flow is turbulent
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. Incompressible flow Steady flow Density (ρ) is a constant Mass flow rate Constant area pipe is a constant A is a constant Therefore, u is a constant for a steady, quasi one-dimensional, compressible flow in a constant area pipe. R. Shanthini 19 July 2019
p p+dp u u+du D dx x Write the momentum balance over the differential volume chosen. Since R. Shanthini 19 July 2019 , and , we get
Therefore, we get Rearranging (2) gives It means p decreases in the flow direction. Since ρ and u are constants, integrating the above gives R. Shanthini 19 July 2019 pressure at the exit pressure at the entrance (2)
Rework Problem 12 assuming incompressible flow: = 0. 0038 Molecular weight = 0. 013 kg/mol; p = 1 MPa D = 15 cm T = (273+87) K p. L = 500 k. Pa 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 R. Shanthini 19 July 2019
Molecular weight = 0. 013 kg/mol; p = 1 MPa D = 15 cm T = (273+87) K L = 100 m Substitute R. Shanthini 19 July 2019 = 0. 0038 in the above, we get p. L = 500 k. Pa
Molecular weight = 0. 013 kg/mol; p = 1 MPa D = 15 cm = 0. 0038 T = (273+87) K p. L = 500 k. Pa L = 100 m What is ρ? ρ = (ρentrance + ρexit ) / 2 = [(p/RT)entrance + (p/RT)exit )] / 2 = (pentrance + pexit ) / 2 RT = (1, 000 + 500, 000) Pa / [2 x (8. 314/0. 013) x 300 J/kg] = 1. 9545 kg/m 3 Therefore, R. Shanthini 19 July 2019 = 7. 76 kg/s Compare 7. 76 kg/s with the 9. 4 kg/s obtained considering the flow to be compressible.
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 examinations. R. Shanthini 19 July 2019
Ignore this d 1 = 0. 20 m p. A = 3. 88 bar A p=? 3 km 7 km m 2 = 0. 63 kg/s d 2 = 0. 15 m 5 km R. Shanthini 19 July 2019 p. B = 3. 69 bar B
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