Presentation on theme: "United Technologies Research Center"— Presentation transcript:
1United Technologies Research Center Turbomachinery in Propulsion Engines -- Problems Rensselaer – Hartford CampusDr. William T. CousinsUnited Technologies Research CenterText: S. Larry Dixon and C. Hall, “Fluid Mechanics and Thermodynamics”, 7th ed, Elsevier Inc., ISBN-10: ISBN-13:
2Problem 1Air in a jet engine enters a nozzle at 1800 R, 30 psia and 90 ft/s and exits the nozzle at 1500 R, 13 psia. Assuming that there is no heat loss,What is the exit velocity of the nozzle?What type of nozzle is it? (converging, diverging, or converging/diverging)Solution:
3Problem 2Air enters a diffuser at 14.7 psia, 540 R, with a velocity of 600 ft/s. The inlet cross-sectional area of the diffuser is 0.2 in2. A At the exit, the area is 1.75 in2 and the exit velocity is 60 ft/s. Determine the exit pressure and temperature of the air. Solution:
4Problem 3Air at 60 ft/s, 480 R, 11 lbf/in2 with 10 lbm/s flows into a turbojet engine and it flows out at 1500 ft/s, 1440 R, 11 psia. What is the change (power) in flow of kinetic energy? Solution:
5Problem 4In a jet engine a flow of air at 1000 K, 200 kPa and 40 m/s enters a nozzle where the air exits at 500 m/s, 90 kPa. What is the exit temperature assuming no heat losses and a frictionless flow? Solution:
6Problem 5The front of an engine acts as a diffuser receiving air at 900 km/h, -5°C, 50 kPa, bringing it to 80 m/s relative to the engine before entering the fan. If the flow area is reduced to 80% of the inlet area, find the temperature and pressure at the fan inlet. Solution:
7Problem 6In an aircraft engine at takeoff, the combustion products expand adiabatically in an exhaust nozzle. At the entrance to the nozzle, the pressure is MPa and the temperature is 1200 K. The kinetic energy entering the nozzle is very much smaller than the kinetic energy of the gas leaving the nozzle. The specific heat of the exhaust gas varies with temperature as follows:CP = x10-4T x10-8T in which the units of CP are kJ/kg K and T is the temperature in degrees Kelvin.The molecular weight of the exhaust gas is 30. Show how the exhaust gas velocity and pressure depend upon the temperature for 900 K, 1000 K, and 1100 K. At each temperature, determine also the speed of sound γ𝑅𝑇 . Determine the required nozzle types.The Universal Gas Constant is equal to: kJ/kmol-K.Solution:
8Problem 7Shown schematically in Part (a) of the figure is the blading of a single–stage axial turbomachine. What kind of machine is represented by cases 1 and 2? What would happen in case 3? Would it be desirable to build a compressor stage as shown in Part (b) of the figure? Part (a) Part (b)Hint:Draw the velocity diagrams for cases one, two, and three and examine them. Look at things like the U, ΔCθ, work done, etc. Also draw the velocity diagram for Part (b) and examine it.
9Problem 8At the mean radius (rm = 30 cm), the blade configuration is as shown in Part (b). For simplicity, assume the air angles and blade angles are identical (i.e., no deviation angle, etc.). The overall adiabatic efficiency of the stage is 90%. The hub-tip radius ratio is 0.8, which is high enough so that the conditions at the mean radius are a good average of the conditions from the hub to the tip. The axial velocity component at design flow rate is uniformly 125 m/s and the inlet air is at 1 atm and 20° C. a) What would be the power required to drive the compressor and the shaft speed under these conditions? b) What would be the stage total pressure rise? Hint: Draw the velocity triangles and use your geometric knowledge to calculate the power. Use your knowledge of the compressor map relationship (efficiency, pressure ratio, temperature ratio) to find the pressure ratio.
10Problem 9At a certain operating condition the mid-radius velocity triangles for an axial compressor stage are as shown in the figure. As usual, the subscripts 1 and 2 denote the entrance to the rotor and the stator, respectively. The stagnation temperature and pressure at the entrance to the rotor are 340 K and 185 kPa. Neglecting frictional effects, determine:The specific work in kJ/kg;The total and static temperatures between the rotor and the stator;The total and static pressures between the rotor and stator; andThe pressure coefficient at mid-radius, CP = (P2 – P1) / ½ρ1W12.XHint:In part a, use your knowledge of calculating work from geometry. Then, in part b, use your knowledge of the 1st Law and stagnation enthalpy. There are several ways to do part c, but the easiest is with your knowledge of the isentropic relationships. Remember, you are working with air (an ideal gas).