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) Summary Design equations for steady, quasi one-dimensional, isothermal,compressible flow of an ideal gas in a constant area duct with wall friction 4f dx D 4f dx D 4f L D u 2 dp RT ( m / A ) 2 du 0 (1.1) u 2 2 pdp 2 dp p (1.2) 2 p L2 pL 1 2 ln 2 2 RT ( m / A ) p p p 4f L R. Shanthini 09 Feb 2012 2 D 2 1 M 2 M 1 M 2 2 L M ln M 2 2 L (1.3) (1.4) Problem 4 from Problem Set 1 in Compressible Fluid Flow: 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 27oC throughout. The average Fanning friction factor may be taken as 0.0066. γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm = 0.0066 T = 300 K L = 11.5 m R. Shanthini 09 Feb 2012 f pL = ? γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm f = 0.0066 T = 300 K pL = ? L = 11.5 m Design equation: 4f L D 2 p L2 pL 1 2 ln 2 2 RT ( m / A ) p p p 2 (1.3) f = 0.0066; 4f L L = 11.5 m; D = 15 mm = 0.015 m; R. Shanthini 09 Feb 2012 D = 20.240 unit? γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm f = 0.0066 T = 300 K pL = ? L = 11.5 m Design equation: 4f L D 2 p L2 pL 1 2 ln 2 2 RT ( m / A ) p p p 2 p = 600 kPa = 600,000 Pa; (1.3) T = 300 K; R = 8.314 kJ/kmol.K = 8.314/28 kJ/kg.K = 8314/28 J/kg.K; m = 1.5 mol/s = 1.5 x 28 g/s = 1.5 x 28/1000 kg/s; A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4; p R. Shanthini 09 Feb 2012 2 RT ( m / A ) 2 = 71.544 unit? γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm f = 0.0066 T = 300 K pL = ? L = 11.5 m Design equation: 4f L D 2 p L2 pL 1 2 ln 2 2 RT ( m / A ) p p p 2 (1.3) 2 p L2 pL 20.240 = 71.544 1 2 ln 2 p p p = 600 kPa = 600,000 Pa R. Shanthini 09 Feb 2012 pL = ? Solve the nonlinear equation above to determine pL 2 p L2 pL 20.240 = 71.544 1 2 ln 2 p p p = 600 kPa = 600,000 Pa Determine the approximate solution by ignoring the ln-term: pL = p (1-20.240/71.544)0.5 = 508.1 kPa Check the value of the ln-term using pL = 508.1 kPa: ln[(pL /p)2] = ln[(508.1 /600)2] = -0.3325 This value is small when compared to 20.240. And therefore pL = 508.1 kPa is a good first approximation. R. Shanthini 09 Feb 2012 Now, solve the nonlinear equation for pL values close to 508.1 kPa: 2 p L2 pL 20.240 = 71.544 1 2 ln 2 p p p = 600 kPa = 600,000 Pa pL kPa 510 R. Shanthini 09 Feb 2012 LHS of the above RHS of the above equation equation 20.240 19.528 509 20.240 19.727 508.1 20.240 19.905 507 20.240 20.123 506.5 20.240 20.222 506 20.240 20.320 Problem 4 continued: Rework the problem in terms of Mach number and determine ML. γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm f = 0.0066 T = 300 K ML = ? L = 11.5 m Design equation: 4f L D 4f L D 1 M 2 2 L M ln M 2 2 L (1.4) = 20.240 (already calculated in Problem 4) M=? R. Shanthini 09 Feb 2012 2 M 1 M γ = 1.4; molecular mass = 28; m = 1.5 mol/s; D = 15 mm p = 600 kPa f = 0.0066 ML = ? T = 300 K L = 11.5 m u M= c = M u RT = 4 m RT D p 2 m A 1 RT m = A 4 (1.5x 28/1000 kg/s) = π (15/1000 m)2 (600,000 Pa) = 0.1 R. Shanthini 09 Feb 2012 RT p ( 1 RT m RT Ap (8314/28)(300) J/kg 1.4 0.5 ) γ = 1.4; molecular mass = 28; p = 600 kPa m = 1.5 mol/s; D = 15 mm f = 0.0066 ML = ? T = 300 K L = 11.5 m Design equation: 4f L D 20.240 1 M 2 M 1 M 1 (1.4)(0.1) 2 2 L M ln M 2 ( 0 . 1) 1 2 M L ML = ? R. Shanthini 09 Feb 2012 2 2 2 L (1.4) ( 0 . 1) 2 ln 2 M L Solve the nonlinear equation above to determine ML 20.240 1 (1.4)(0.1) 2 2 ( 0 . 1) 1 2 M L ( 0 . 1) 2 ln 2 M L Determine the approximate solution by ignoring the ln-term: ML = 0.1 / (1-20.240 x 1.4 x 0.12)0.5 = 0.118 Check the value of the ln-term using ML = 0.118: ln[(0.1/ML)2] = ln[(0.1 /0.118)2] = -0.3310 This value is small when compared to 20.240. And therefore ML = 0.118 is a good first approximation. R. Shanthini 09 Feb 2012 Now, solve the nonlinear equation for ML values close to 0.118: 20.240 1 (1.4)(0.1) pL kPa 0.116 R. Shanthini 09 Feb 2012 2 2 ( 0 . 1) 1 2 M L ( 0 . 1) 2 ln 2 M L LHS of the above RHS of the above equation equation 20.240 18.049 0.117 20.240 0.118 20.240 19.798 0.1185 20.240 20.222 0.119 20.240 20.64 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. 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: p ( m / A ) RT (1.5) in flows where p decreases along the flow direction, and p ( m / A ) RT (1.6) in flows where p increases along the flow direction. R. Shanthini 09 Feb 2012 Differential equation of Problem 2: 4f dx D 2 RT ( m / A ) 2 pdp 2 dp p (1.2) can be rearranged to give dp 4f /D dx 2 RT ( m / A ) 2 p 2 ( 2 f / D ) pRT ( m / A ) 2 2 RT ( m / A ) p p In flows where p decreases along the flow direction dp dx R. Shanthini 09 Feb 2012 0 2 2 RT ( m / A ) p 0 p ( m / A ) RT (1.5) 2 Differential equation of Problem 2: 4f dx D 2 RT ( m / A ) 2 pdp 2 dp p (1.2) can be rearranged to give dp 4f /D dx 2 RT ( m / A ) 2 p 2 ( 2 f / D ) pRT ( m / A ) 2 2 RT ( m / A ) p p In flows where p increases along the flow direction dp 0 dx R. Shanthini 09 Feb 2012 RT ( m / A ) p 0 2 2 p ( m / A ) RT (1.6) 2 Problem 7 from Problem Set 1 in Compressible Fluid Flow: Air enters a horizontal constant-area pipe at 40 atm and 97oC 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. 40 atm 97oC 500 m/s Air: γ = 1.4; molecular mass = 29; p*=? L R. Shanthini 09 Feb 2012 Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L dp 0 p ( m / A ) RT 0 p ( m / A ) RT dx dp dx Limiting pressure: p * ( m / A ) RT R. Shanthini 09 Feb 2012 Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L p * ( m / A ) RT ( m / A ) ( A u ) entrance / A ( u ) entrance p* pu entrance RT = R. Shanthini 09 Feb 2012 RT p u RT entrance pu entrance RT (40 atm) (500 m/s) [(8314/29)(273+97) J/kg]0.5 = 61.4 atm pu entrance RT 40 atm 97oC 500 m/s Air: γ = 1.4; molecular mass = 29; p*=61.4 atm L 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. Adp m du w dA w 0 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: M 1/ (1.7) in flows where p decreases along the flow direction M 1/ (1.8) in flows where p increases along the flow direction R. Shanthini 09 Feb 2012 In flows where p decreases along the flow direction dp p ( m / A ) RT 0 (1.5) dx Since p m RT AM we get m RT AM M 1/ R. Shanthini 09 Feb 2012 ( m / A ) RT (1.7) In flows where p increases along the flow direction dp 0 p ( m / A ) RT dx Since p m RT AM we get m RT AM ( m / A ) RT M 1/ R. Shanthini 09 Feb 2012 (1.6) (1.8) Summary: x dp 0 p ( m / A ) RT and M 1/ 0 p ( m / A ) RT and M 1/ Limiting Mach number: M * 1/ dx dp dx Limiting pressure: p * ( m / A ) RT R. Shanthini 09 Feb 2012 Limiting Mach number for air: x For air, γ = 1.4 dp M * 1 / 1 .4 0 is associated with = 0.845 M 1/ dx If M < 0.845 then pressure decreases in the flow direction. That is, the pressure gradient causes the flow. dp 0 is associated with M 1/ dx If M > 0.845 then pressure increases in the flow direction. That is, momentum causes the flow working against the R. Shanthini pressure gradient. 09 Feb 2012 Problem 9 from Problem Set 1 in Compressible Fluid Flow: Show that when the flow has reached the limiting pressure p * ( m / A ) RT or the limiting Mach number the length of the pipe across which such conditions are reached, denoted by Lmax, shall satisfy the following equation: M * 1/ 4 f L max D p2 p *2 1 ln 2 2 p * p 1 M 2 ln M 2 M 2 where pressure p and Mach number M are the conditions of the flow at the entrance of the pipe. R. Shanthini 09 Feb 2012 p; M p*; M* Lmax Start with the following: 4f L D 2 p L2 pL 1 2 ln 2 2 RT ( m / A ) p p p 2 (1.3) Substitute L = Lmax and pL = p * ( m / A ) RT in (1.3) to get 4 f L max D R. Shanthini 09 Feb 2012 p2 p *2 1 ln 2 2 p * p (part of 1.9) p; M p*; M* Lmax Start with the following: 4f L D 1 M 2 M 1 M 2 2 L M ln M 2 2 L (1.4) Substitute L = Lmax and ML = M * 1 / in (1.4) to get 4 f L max D 1 M M 2 2 ln M 2 (part of 1.9) Therefore, 4 f L max R. Shanthini 09 Feb 2012 D p2 p *2 1 ln 2 2 p * p 1 M 2 ln M 2 M 2 (1.9)