CP502 Advanced Fluid Mechanics

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

Speed of the flow (u) M≡ Speed of sound (c) in the fluid at the flow temperature Incompressible flow assumption is not valid if Mach number > 0.3 What is a Mach number? Definition of Mach number (M): 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)

For an ideal gas, u u = M= c Unit of u = m/s Unit of c = [(J/kg.K)(K)]0.5 = [J/kg]0.5 = (N.m/kg)0.5 = [kg.(m/s2).m/kg]0.5 = [m2/s2]0.5 = m/s

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

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.

Ideal Gas equation of state: temperature pressure specific gas constant (not universal gas constant) volume mass Ideal Gas equation of state can be rearranged to give K Pa = N/m2 kg/m3 J/(kg.K)

Problem 1 from Problem Set 1 in Compressible Fluid Flow: 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. (1.1)

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

p p+dp D u u+du dx x Equation (1) can be reduced to Substituting Since , and , we get (1.1)

Problem 2 from Problem Set 1 in Compressible Fluid Flow: 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, pL 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. (1.2) (1.3)

The differential equation of problem (1) is in which the variables ρ and umust 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: and and therefore It is a constant for steady, isothermal flow in a constant area duct

, Using and in (1.1) we get (1.2)

p pL L Integrating (1.2) from 0 to L, we get which becomes (1.3)

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

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 pL and ML are the pressure and Mach number at length L from the entry. (1.3) (1.4) We need to relate p to M!

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)

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)

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; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m

γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) = 0.0066; = 20.240 L = 11.5 m; D = 15 mm = 0.015 m; unit?

γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) p = 600 kPa = 600,000 Pa; T = 300 K; 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; A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4; = 71.544 unit?

γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 pL = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.3) 20.240 = 71.544 p = 600 kPa = 600,000 Pa Solve the nonlinear equation above to determine pL pL = ?

20.240 = 71.544 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.

Now, solve the nonlinear equation for pL values close to 508.1 kPa: 20.240 = 71.544 p = 600 kPa = 600,000 Pa

Problem 4 continued: Rework the problem in terms of Mach number and determine ML. γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.4) = 20.240 (already calculated in Problem 4) M = ?

γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m u u 1 1 RT M = = = = c A p ( ) 0.5 4 (1.5x 28/1000 kg/s) (8314/28)(300) J/kg = 1.4 π (15/1000 m)2 (600,000 Pa) = 0.1

γ = 1.4; molecular mass = 28; = 1.5 mol/s; = 0.0066 ML = ? D = 15 mm p = 600 kPa T = 300 K L = 11.5 m Design equation: (1.4) 20.240 Solve the nonlinear equation above to determine ML ML = ?

20.240 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.

Now, solve the nonlinear equation for ML values close to 0.118: 20.240

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.

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: (1.5) in flows where p decreases along the flow direction, and (1.6) in flows where p increases along the flow direction.

Differential equation of Problem 2: (1.2) can be rearranged to give In flows where p decreases along the flow direction (1.5)

Differential equation of Problem 2: (1.2) can be rearranged to give In flows where p increases along the flow direction (1.6)

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. Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L

Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L Limiting pressure:

Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s p*=? L (40 atm) (500 m/s) = = 61.4 atm 0.5 [(8314/29)(273+97) J/kg]

Air: γ = 1.4; molecular mass = 29; 40 atm 97oC 500 m/s 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.

CP502 Advanced Fluid Mechanics