CP 502 Advanced Fluid Mechanics Compressible Flow Part

CP 502 Advanced Fluid Mechanics Compressible Flow Part 02_Set 02: Steady, quasi one-dimensional, isentropic compressible flow of an ideal gas in a variable area duct (continued)

Problem 6 from Problem Set #2 in Compressible Fluid Flow: Show that the steady, one-dimensional, isentropic, compressible flow of an ideal gas with constant specific heats can be described by the following equations: (2. 5) (2. 6) (2. 7) (2. 8) where p 0, T 0 and ρ0 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 21 July 2019

T 0 p 0 u 0=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 21 July 2019

T 0 p 0 u 0=0 Stagnation properties T p u M Start with the energy balance for steady, adiabatic, inviscid, quasi one-dimensional, compressible flow: (2. 2) dh + udu = 0 Using dh = cp d. T, which is applicable for ideal gas, in the above, we get cp d. T + udu = 0 Integrating the above between the two cross-sections, we get cp (T – T 0) + (u 2 – 0)/2 = 0 R. Shanthini 21 July 2019

Using the definition of M, we get cp (T – T 0) + M 2 γRT/2 = 0 Since cp = 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: (2. 6) The above equation relates the stagnation temperature (at near zero velocity) to a temperature at Mach number M for steady, invicid, one-dimensional, compressible flow of an ideal gas in a variable area duct. R. Shanthini 21 July 2019

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 21 July 2019

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 21 July 2019

Problem 8 from Problem Set #2 in Compressible Fluid Flow: A large air reservoir contains air at a temperature of 400 K and a pressure of 600 k. Pa. 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 k. Pa, (ii) 500 k. Pa, (iii) 400 k. Pa, (iv) 300 k. Pa and (v) 200 k. Pa. Assumptions: Steady, isentropic flow Ae = 100 mm 2 Air T 0 = 400 K p 0 = 600 k. Pa R. Shanthini 21 July 2019 pb k. Pa is given pb Determine the following at the exit of the converging duct: Mach Number Me = ? Temperature Te = ? Mass flow rate = ? is known as the back pressure

(i) For pb = 600 k. Pa, there will be no flow since p 0 = 600 k. Pa as well (ii) For pb = 500 k. Pa, assume pe is the same as pb. = 0. 517 Using (2. 7), we get Using (2. 5), we get = 379. 7 K Using (2. 9), we get ( = (100 x 10 -6 m 2) (0. 517) (500, 000 Pa) R. Shanthini 21 July 2019 1. 4 (8314/29)(379. 7) J/kg 0. 5 ) = 0. 0927 kg/s

Results summarized: Back pressure, Exit Mach Exit temperature, Mass flow pb (in k. Pa) pe (in k. Pa) number, Me Te (in K) rate (in kg/s) 600 0 400 0 500 0. 517 379. 7 0. 0927 400 0. 784 356. 2 0. 1161 300 1. 046 328. 1 0. 1211 200 1. 358 292. 2 0. 1110 Ae = 100 mm 2 Air T 0 = 400 K p = 600 k. Pa R. 0 Shanthini 21 July 2019 pb k. Pa (given) Is there a problem with the above results?

Results summarized: Back pressure, Exit Mach Exit temperature, Mass flow pb (in k. Pa) pe (in k. Pa) number, Me Te (in K) rate (in kg/s) 600 0 400 0 500 0. 517 379. 7 0. 0927 400 0. 784 356. 2 0. 1161 300 1. 046 > 1 328. 1 0. 1211 200 1. 358 > 1 292. 2 0. 1110 Ae = 100 mm 2 Air T 0 = 400 K p = 600 k. Pa R. 0 Shanthini 21 July 2019 pb k. Pa (given) Yes, there is. Supersonic Mach numbers have been reached from stagnation condition in a converging duct? It cannot happen!

Maximum Mach number at the exit of a converging duct must be Me = 1. Corresponding pe (denoted by p*) could be calculated using (2. 7) as follows: p* = 600 = 317 k. Pa Using (2. 5), we get Using (2. 9), we get the mass flow rate as 0. 1213 kg/s. R. Shanthini 21 July 2019 = 333. 3 K

Results summarized: Back pressure, Exit Mach Exit temperature, Mass flow pb (in k. Pa) pe (in k. Pa) number, Me Te (in K) rate (in kg/s) 600 0 400 0 500 0. 517 379. 7 0. 0927 400 0. 784 356. 2 0. 1161 300 317 1 333. 3 0. 1213 200 317 1 333. 3 0. 1213 Ae = 100 mm 2 Air T 0 = 400 K p = 600 k. Pa R. 0 Shanthini 21 July 2019 pb k. Pa (given) Flow chocks at the throat of the converging duct at an exit pressure of 317 k. Pa at a maximum flow rate of 0. 1213 kg/s?

Ae = 100 mm 2 Air T 0 = 400 K p 0 = 600 k. Pa pe k. Pa (given) Pressure at the throat cannot be less than the limiting pressure (317 k. Pa 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 ( 0. 1213 kg/s in this case) even if we increase the driving force by decreasing the back pressure in the second chamber. R. Shanthini 21 July 2019

Problem 9 from Problem Set #2 in Compressible Fluid Flow: Air at 900 k. Pa 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 pb for 900 ≤ pb ≤ 100 k. Pa. R. Shanthini 21 July 2019

Results: Limiting pressure = 475. 45 k. Pa R. Shanthini 21 July 2019

Results (continued): Sonic condition M = 1 R. Shanthini 21 July 2019

Results (continued): Maximum velocity 365. 77 m/s R. Shanthini 21 July 2019

Results (continued): Maximum mass flow rate 18. 1981 kg/s R. Shanthini 21 July 2019

Problem 10 from Problem Set #2 in Compressible Fluid Flow: 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 k. Pa and 200 o. C. Assume that the flow is isentropic and the exit pressure is 100 k. Pa. 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. p 0 = 1400 k. Pa T 0 = (273+200) K M 0 ≈ 0 pe = 100 k. Pa = 3 kg/s R. Shanthini 21 July 2019

p 0 and T 0 are known. Use the equations below to calculate p and T for different values of M. (2. 6) (2. 7) Use (2. 9) to calculate the exit are of the duct as follows: R. Shanthini 21 July 2019

R. Shanthini 21 July 2019

Enlarged version R. Shanthini 21 July 2019

Shape of the converging-diverging duct R. Shanthini 21 July 2019

Problem 11 from Problem Set #2 in Compressible Fluid Flow: Air at approximately zero velocity enters a converging-diverging duct at a stagnation pressure and a stagnation temperature of 1000 k. Pa and 480 K, respectively. Throat area of the duct is 0. 002 m 2. The flow inside the duct is isentropic, and the exit pressure is 31. 7 k. Pa. 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. p 0 = 1000 k. Pa T 0 = 480 K M 0 ≈ 0 Athroat = 0. 002 m 2 R. Shanthini 21 July 2019 pe = 31. 7 k. Pa Me = ? Ae = ?

(i) Rearrange (2. 7) to determine Me as follows: = 2. 9 (ii) Rearrange (2. 5) to determine Te as follows: = 179 K R. Shanthini 21 July 2019

(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 Mt = 1 in the above expression. Therefore, we get Since Me = 2. 9 and At = 0. 002 m 2, the above expression gives = 0. 0077 m 2 R. Shanthini 21 July 2019

(iv) Use (2. 9) to get the mass flow rate through the duct as follows: = 3. 695 kg/s R. Shanthini 21 July 2019