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)

R. Shanthini 22 Dec 2010 Show that the steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats can be described by the following equations: Problem 6 from Problem Set #2 in Compressible Fluid Flow: (2.6) (2.5) (2.7) (2.8) where p 0, T 0 and ρ 0 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.

R. Shanthini 22 Dec 2010 T0T0 p0p0 u0=0u0=0 Stagnation properties T p u M Ideal gas satisfies Isentropic flow of an ideal gas satisfies Using the above two equations, we can easily prove (2.5)

R. Shanthini 22 Dec 2010 Start with the energy balance for steady, adiabatic, inviscid, quasi one-dimensional, compressible flow: dh + udu = 0 T0T0 p0p0 Stagnation properties Using dh = c p dT, which is applicable for ideal gas, in the above, we get c p dT + udu = 0 Integrating the above between the two cross-sections, we get c p (T – T 0 ) + (u 2 – 0)/2 = 0 (2.2) T p u M u0=0u0=0

R. Shanthini 22 Dec 2010 Using the definition of M, we get c p (T – T 0 ) + M 2 γRT/2 = 0 Since c p = γR/( γ -1) for an ideal gas, the above can be written as γR (T – T 0 ) /(γ -1) + M 2 γRT/2 = 0 which can be rearranged to give the following: 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. (2.6)

R. Shanthini 22 Dec 2010 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)

R. Shanthini 22 Dec 2010 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)

R. Shanthini 22 Dec 2010 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 mm 2. 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. Problem 8 from Problem Set #2 in Compressible Fluid Flow: T 0 = 400 K p 0 = 600 kPa Air A e = 100 mm 2 p b kPa is given Assumptions: Steady, isentropic flow Determine the following at the exit of the converging duct: Mach Number M e = ? Temperature T e = ? Mass flow rate = ? p b is known as the back pressure

R. Shanthini 22 Dec 2010 (i) For p b = 600 kPa, there will be no flow since p 0 = 600 kPa as well (ii) For p b = 500 kPa, assume p e is the same as p b. Using (2.5), we get = K = (100 x10 -6 m 2 ) (0.517) (500,000 Pa) (8314/29)(379.7) J/kg 0.5 () 1.4 = kg/s = Using (2.7), we get Using (2.9), we get

R. Shanthini 22 Dec 2010 Results summarized: Back pressure, p b (in kPa) Exit pressure, p e (in kPa) Exit Mach number, M e Exit temperature, T e (in K) Mass flow rate (in kg/s) T 0 = 400 K p 0 = 600 kPa Air A e = 100 mm 2 p b kPa (given) Is there a problem with the above results?

R. Shanthini 22 Dec 2010 Results summarized: Back pressure, p b (in kPa) Exit pressure, p e (in kPa) Exit Mach number, M e Exit temperature, T e (in K) Mass flow rate (in kg/s) > > Yes, there is. Supersonic Mach numbers have been reached from stagnation condition in a converging duct? It can not happen! T 0 = 400 K p 0 = 600 kPa Air A e = 100 mm 2 p b kPa (given)

R. Shanthini 22 Dec 2010 Maximum Mach number at the exit of a converging duct must be M e = 1. Corresponding p e (denoted by p*) could be calculated using (2.7) as follows: = 317 kPa p* = 600 Using (2.5), we get= K Using (2.9), we get the mass flow rate as kg/s.

R. Shanthini 22 Dec 2010 Results summarized: Back pressure, p b (in kPa) Exit pressure, p e (in kPa) Exit Mach number, M e Exit temperature, T e (in K) Mass flow rate (in kg/s) Flow chocks at the throat of the converging duct at an exit pressure of 317 kPa at a maximum flow rate of kg/s? T 0 = 400 K p 0 = 600 kPa Air A e = 100 mm 2 p b kPa (given)

R. Shanthini 22 Dec 2010 T 0 = 400 K p 0 = 600 kPa Air A e = 100 mm 2 p e kPa (given) 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 ( kg/s in this case) even if we increase the driving force by decreasing the back pressure in the second chamber.

R. Shanthini 22 Dec 2010 Air at 900 kPa and 400 K enters a converging nozzle with a negligible velocity. The throat area of the nozzle is 10 cm 2. 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 p b for 900 ≤ p b ≤ 100 kPa. Problem 9 from Problem Set #2 in Compressible Fluid Flow:

R. Shanthini 22 Dec 2010 Limiting pressure = kPa Results:

R. Shanthini 22 Dec 2010 Sonic condition M = 1 Results (continued):

R. Shanthini 22 Dec 2010 Maximum velocity m/s Results (continued):

R. Shanthini 22 Dec 2010 Maximum mass flow rate kg/s Results (continued):

R. Shanthini 22 Dec 2010 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 200 o C. 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. Problem 10 from Problem Set #2 in Compressible Fluid Flow: p 0 = 1400 kPa T 0 = ( ) K M 0 ≈ 0 = 3 kg/s p e = 100 kPa

R. Shanthini 22 Dec 2010 (2.6) (2.7) p 0 and T 0 are known. Use the equations below to calculate p and T for different values of M. Use (2.9) to calculate the exit are of the duct as follows:

R. Shanthini 22 Dec 2010

Enlarged version

R. Shanthini 22 Dec 2010 Shape of the converging-diverging duct

R. Shanthini 22 Dec 2010 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 m 2. 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. Problem 11 from Problem Set #2 in Compressible Fluid Flow: p 0 = 1000 kPa T 0 = 480 K M 0 ≈ 0 A throat = m 2 p e = 31.7 kPa M e = ? A e = ?

R. Shanthini 22 Dec 2010 (i) Rearrange (2.7) to determine M e as follows: = 2.9 (ii) Rearrange (2.5) to determine T e as follows: = 179 K

R. Shanthini 22 Dec 2010 (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 M t = 1 in the above expression. Therefore, we get = m 2 Since M e = 2.9 and A t = m 2, the above expression gives

R. Shanthini 22 Dec 2010 (iv) Use (2.9) to get the mass flow rate through the duct as follows: = kg/s