Question

Question 3 - 70 Marks In this question we are considering a scramjet engine. A scramjet engine is a high-speed propulsion engine which begins working at approximately Mach 5. Similar to a regular jet engine, the air is compressed in the inlet, fuel is added and combustion occurs in the combustion chamber and the higher pressure and temperature mixture is then exhausted through a nozzle to provide thrust. Unlike a conventional jet engine the air is compressed using the shock wave phenomena rather than a compressor, and the air stays supersonic throughout the engine. Researchers in South East Queensland are amongst the world leaders in researching Scramjet technology. We are going to analyse the scramjet shown in Figure 2. The scramjet is flying at an altitude of 30km, where the air temperature is 50 °C and pressure is 1.1 kPa, and at a speed of 2.5 km/s. In the combustor hydrogen fuel is added at a rate of 3 kg of hydrogen per 100 kg of air. As a result of the combustion process, the mixture at the exit of the combustor is now at 2500K, 300 kPa with a velocity of 1600 m/s. From this point the flow is expanded through the nozzle to the exit plane with a density ratio of 11.35 (condutor/Ped = 11.35). The scramjet shown in Figure 2 is a 2D geometry, that is, from front on it looks rectangular with a width of 500 mm. You can assume that the ideal gas relations can be used at each point in the flow and use the specific gas constant for air. 2.5 km/s -50°C 1.1 kPa 0.2 m 0.08 m Exit plane 1600 m/s 2500K 300 kPa Figure 2: Schematic of scramjet engine. Part (a) - 20 Marks What is the mass flow entering the engine? Part (b) 40 Marks What is the nozzle exit flow velocity?

Answer 1

Solution 25 km/s - 50°C likPa 0.2m 0108m - Buit plane 1600 ml 2500k 300tlal the Part (9) air mass flow rate (und = Situ entering engine where Si = density of air at section B-6 Cooss sectional area of scramjet at serta meno of the passage air at section 6 - ① Uz Velocity of entering P = SRT so at section ①0 Piz S, PaTi Rra Gas constant for air = 0,287 kilky k 1.1X10? fi (0, 287-X103) x (273-50) Si = 0.017 kg/m3 Aiz I Love ㅈ Ai = 0.1257 m2 v z 2500 mls (Giren) na = Siti = Lina = 5.34 kg/s I (2X0.2) 2 So 0.017% 0.1257* 2500 Ausner

Part (6) at section (2) (2) P2 = 2₂ R2 TL fi= combustor R₂ = Rmixture (mixture of air and Hydrogen) Rumsture for ideal gases, ma B tmh Rh Rmixture = me tmh Ph + Malmy Pa cha Thiba) Ph z (M.W.) W Ra= 0.287 kJ/1gok T 2 Omversal gas constant (M.W) w= Molendar & weight of Hydrogen Phe 8.314 4. 157 kJ / 1gok 2 4,157 +(!) x 0.287 2 so I+ (10) = 0.3997 kJ/ 19.6 P2 I combustar Roisture & Tr 2 P2 z 300 kle (given) Tr = 2500k (given)

300x103 combus for (1.39977113) (2500) I combusfor = 0.30 kg/m3 It is given that I combustor 2 11:35 remit Scombustar 1.35 So Penit - 0.30 11.35 Senit = 0.026 kg/m3 La man flow rate of Hydrogen enit mass flow rate (meant) na timu man flow ante of air menit = in Cit with my ana (It The ina (It FAR) Fuel Air Ratio = 5:34 (It too menit 25,5002 kals and, ment a fenit tz Vz 5.500 2 0026 % * COVET vs Lits- 132 10,521. 44 m/s taber Dz2260 nozzle entreterity

Raste) According to your assumption as given in problem Fa menit x Menit - infreestream freestream we have data Men't = 5,5002 legls. Menit 10, 512194 m/s in frestream = 5.34 kgle "freestream = = 2500 ml SO F: (5.5002 x 10,512044) - (5,3 4x 2500) F = 44, 470.52 Hewton Answer