EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 1 Air flows into a constant area insulated duct with a Mach number of 0.20. For duct diameter of 1 cm and friction coefficient of 0.02, determine the duct length required to reach Mach 0.6. Determine the length required to attain Mach 1. Finally, if an additional 75 cm is added to the duct length needed to reach Mach 1, while the initial stagnation conditions are maintained, determine the reduction in flow rate that would occur. Solution Note, you will often find two different notations for the Fanno parameter. One is often written as: 4ππΏ∗ π· While the book uses: ππΏ∗ π· Where the constant of 4 is incorporated into the fanning friction coefficient. Then for a Mach number π1 = 0.2 and πΎ = 1.4: ππΏ∗ ( ) = 14.5333 π· 1 Where total ratios were obtained from Mach-isentropic equations. Similar to PM expansion waves, the length required for the flow to achieve some condition ππ can be obtained from the difference between Fanno parameters. For π2 = 0.6: ( ππΏ∗ ) = 0.4908 π· 2 Then: ππΏ ππΏ∗ ππΏ∗ = ( ) − ( ) = 14.0425 π· π· 1 π· 2 Thus, to reach π2 = 0.6, πΏ = 702.1250 cm. For the flow to be choked, π3 = π∗ = 1: ππΏ∗ ( ) =0 π· 2 Then, we are finding πΏ∗1 = 726.6650 cm. EAS4132/5714 FALL 2023 HW08 SOLUTIONS Now, for the case were length is added to the tube, recall the form of the Fanno line: The maximum point the flow can achieve is a Mach number of 1. Even if length is added to the tube, the flow will still be choked at the point where area is most reduced due to the formation of the boundary layer. The new Fanno parameter will be: ππΏ∗ ( ) = 16.0333 π· 4 Then, from equations/flow calculators/tables: π4 = 0.1917 π4 = 0.9747 ππ,1 π4 = 0.9927 ππ,1 Note that ππ,1 was noted for stagnation conditions being equal to the initial flow as stated in the problem. While no numeric values are given for pressure, the ratio of flow rates can be found where from: π πΜ = π΄π√πΎπ π π π The ratio can be found where: π4 π1 π1 πΜ4 (ππ,1 ) (ππ,1 ) π4 (ππ,1 ) = ( )√ = 0.9604 πΜ1 ( π1 ) ( π4 ) π1 ( π4 ) ππ,1 ππ,1 ππ,1 Thus the reduction is 3.9622% EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 2 Air flows from a large tank, in which the pressure and temperature are 100 kPa and 30°C respectively, through a 1.6 m long pipe with a diameter of 2.5 cm. The pipe is connected to a converging nozzle with an exit diameter of 2.1 cm. The air from the nozzle is discharged into a large tank in which the pressure is maintained at 35 kPa. Assuming that the friction is equal to 0.008, find the mass flow rate through the system. The flow in the nozzle can be assumed to be isentropic and the pipe is heavily insulated. Solution Let (3) be the exit plane of the nozzle, (2) the entrance of the nozzle and (1) the entrance of the pipe. Assume that the flow at the exit of the converging flow is choked. The area ratio will be: π΄ 2 π΄2 = = 1.4172 π΄ 3 π΄∗ Then π2 = 0.4408, which has a Fanno parameter of 1.681. The Fanno parameter can be found from the duct properties such that: ππΏ∗ ππΏ∗ ππΏππ’ππ‘ ( ) =( ) + = 2.193 π· 1 π· 2 π·ππ’ππ‘ Which corresponds to a Mach number of π1 = 0.4065. By propagation of pressure ratios, the pressures from inlet to exit can be found from: π3 π3 ππ,2 π2 π ∗ π1 = ππ,1 ππ,2 π2 π ∗ π1 ππ,1 Where the first two ratios come from isentropic-Mach relations, ratios relative to chocking can be found from Fanno relations, and the last ratio comes from isentropic-Mach relations again. Then π3 = 49.5 kPa, validating the choked assumption. Then, using the given total conditions for a known Mach number of 0.4065: π1 = 0.4065 π1 = 89.2459 kPa π1 = 293.3 K πΜ = 0.073 kg/s EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 3 Air flows through a 0.15 by 0.25 m rectangular duct. At a given section of the duct: π = 2; π = 75 kPa; π = 5°πΆ Assume a friction factor of 0.024, find the maximum length of the duct that can be installed downstream of this section if no shock wave is to occur in the duct. Also, find the exit pressure and temperature that will occur for this maximum length of duct. Solution Assume that the duct will be slowed down smoothly such that the exit plane (2) has π2 = π∗ = 1. For the rectangular duct, the equivalent diameter for the duct can be found from the ratio: ππππ β∗π€ π·=4 =4 = 0.1875 πππππππ‘ππ 2(β + π€) Where β is the height of the duct and π€ the width of the duct. Then, the length of the duct can be found from the Fanno parameter of the inlet where: ππΏπππ₯ ππΏ∗ = ( ) → πΏπππ₯ = 2.383 m π· π· 1 Where the Fanno parameter for the choked flow is 0. As the pressure and temperature ratios for Fanno flow are computed as a ratio of choked properties, such that only the state property ratios at (1) are needed. Then for (2) = (*) : π2 = 183.7 kPa π2 = 417.0 K EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 4 A rocket nozzle is operating with a stretched out throat (πΏ = 50 cm and π· = 10 cm) as shown in the figure below. If the inlet stagnation conditions are ππ,1 = 1 MPa and ππ,1 = 1500 K. determine the nozzle exit velocity and mass flow for a back pressure of 30 kPa. The diameter of the nozzle at the exit station is the same as at the inlet station 30 cm. Treat the exhaust gas as a perfect gas with πΎ = 1.4 and π = 500 J/kg β K. Assume isentropic flow in variable area sections and Fanno flow in constant-area sections with π = 0.22. Solution Assume the system is choked at (2) such that π2 = π∗ = 1. Then at (2) the Fanno parameter will be 0, and the state properties will correspond to choked conditions. Using pipe parameters: ππΏ ππΏ∗ = ( ) = 0.110 ∴ π1 = 0.7637 π· π· 1 Additionally, for an area ratio of 9, the exit Mach number will be ππ = 3.8061. Then isentropicMach equations can be used to find state properties at (1), as well as at the exit from propagation of ratios where: ππ ππ,2 π ∗ π1 ππ = ( )( )( )( ) = 8.1108 kPa ππ,2 π2 π1 ππ,1 ππ πππ‘πππππ ππ − ππ πππ‘πππππ π2 − πΉππππ π1 − ππ πππ‘πππππ π1 From NSW relations, the back pressure required to cause a normal shot at the exit of the nozzle is 135.7230 kPa. Then no wave forms immediately after the exit plane, instead OSW will form to negotiate the gradient to the back pressure. Note that as there is no heat transfer to the flow, the total temperature will be constant across the flow. So from isentropic-Mach equations: ππ = 384.9 K ∴ ππ is known ππ = 1975.6189 m/s πΜ = ( ππ ) π΄ π = 5.8854 kg/s π ππ π π EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 5 Air is expanded from a large reservoir in which the pressure and temperature are 200 kPa and 30°C respectively through a convergent nozzle which gives an exit Mach number of 0.2. The air then flows down a pipe with diameter of 25 mm, the Mach number at the end of the pipe being 0.8. Assuming the flow in the nozzle is isentropic and the flow in the pipe is adiabatic, find the length of pipe and the pressure at the exit of the pipe. The friction factor can be assumed to be 0.2. Solution From Fanno relationships, for π1 = 0.2: ππΏ∗ ( ) = 14.5333 π· 1 π1 = 5.4554 π∗ And π1 can be found from isentropic-Mach relations given that ππ,1 is known. For π2 = 0.8: ππΏ∗ ( ) = 0.0722 π· 2 π2 = 1.2893 π∗ The length of the pipe will be πΏ = 18.07 m π2 = 46.0 kPa EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 6 Air enters a pipe having a diameter of 0.1 m and a length of 1 m with a Mach number of 2 and a pressure of 90 kPa. Assuming the flow is adiabatic and the mean friction factor of the pipe is 0.02, plot a graph of the pressure variation along the length of the duct: Solution Note that initial conditions are known where: π1 ππΏ∗ π1 = 2; ∗ = 0.4082; ( ) = 0.3050 π π· 1 At any π₯ location along the pipe: ππΏ∗ ππΏ∗ ππ₯ ( ) =( ) − π· π₯ π· 1 π· The Mach number can be found from the Fanno parameter where: (πΎ + 1)π2 ππΏ∗ 1 − π2 πΎ + 1 = + ln [ ] π· πΎπ2 2πΎ 2 + (πΎ − 1)π2 And the pressure can be found from the critical pressure π∗ = 220.4082 kPa as a function of Mach number where: 1 2 π 1 πΎ+1 = [ ] π ∗ π 2 + (πΎ + 1)π2 EAS4132/5714 FALL 2023 HW08 SOLUTIONS EAS4132/5714 FALL 2023 HW08 SOLUTIONS Problem 7 Air enters a pipe at Mach 2.5, a temperature of 40 °C and a pressure of 70 kPa. The pipe has a diameter of 2.0 cm and the flow can be assumed to be adiabatic. A shock occurs in the pipe at a location where the Mach number is 2. If the Mach number at the exit of the pipe is 0.8 and if the average friction factor is 0.02, find the distance to the shock from the entrance to the pipe and the total length of pipe. Also find the pressure at the exit of the pipe. Solution For π1 = 2.5 at the pipe inlet: Before the shock where π2 = 2.0: ππΏ∗1 ( ) = 0.4320 π· 1 π1 = 0.2921 π∗ ππΏ∗1 ( ) = 0.3050 π· 2 π2 = 0.4082 π∗ Then the distance between (1) and (2) can be found from: ππΏ1→2 ππΏ∗1 ππΏ∗1 =( ) −( ) π· π· 1 π· 2 Then πΏ1→2 = 0.127 m and π2 = 97.82 kPa. Note that the distance πΏ1→2 represents 0.127 m represents the distance from the inlet of the tube to the location of the shock. The pressure π2 represents the static pressure of the flow before the shock. π Across the normal shock, π3 = 0.5774 and 3⁄π2 = 4.50 therefore π3 = 440.2 kPa. The normal shock will form to negotiate the pressure between (1) and (4). By definition of a shock, the pressure must increase across a shock such that π3 > π2 (be careful when using notation from a table or calculator). The Fanno parameter will be: ( ππΏ∗3 ) = 0.5876 π· 3 π3 = 1.837 π∗ EAS4132/5714 FALL 2023 HW08 SOLUTIONS As the exit Mach number is known to be 0.8, for π4 = 0.8 ( ππΏ∗3 ) = 0.07229 π· 4 π4 = 1.284 π∗ Then π4 at the exit can be found to be 308.4 kPa from a propagation of ratios as shown below: π4 π ∗ π3 π2 π ∗ π4 = ( ∗ ) ( ) ( ) ( ∗ ) ( ) π1 π π3 π2 π π1 The length segment πΏ3→4 = 0.5153 m must be found from the Fanno parameters after the shock for (3) and (4): ππΏ3→4 ππΏ∗3 ππΏ∗3 =( ) −( ) π· π· 3 π· 4 For a total pipe length of πΏπ‘ππ‘ππ = 0.6423 m.
