{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW21s - Spring 2006 Thermodynamics Homework#21 Solution 1...

This preview shows pages 1–3. Sign up to view the full content.

Spring 2006 1 ME 201 Thermodynamics Homework #21 Solution 1. Consider a jet aircraft flying at 300 m/s at an altitude of 3,000 m (use Table A-16 in the text to determine the pressure and temperature). The jet operates with a simple, ideal turbojet engine. The burner of the engine operates at 1300 kPa and 1800 K. Determine a. specific thrust produced b. propulsive efficiency Solution: We begin with a block diagram of the system. Now we set up our table. Node T(K) P(kPa) h(kJ/kg) φ (kJ/kg K) ° v (m/s) 1 268.5 70.12 268.62 1.5907 300 2 313.4 120.4 313.4 1.7458 0 3 611.3 1300 618.93 2.4287 0 4 1800 1300 2003.3 3.6684 0 5 1548.6 683.4 1697.77 3.4838 0 6 885.1 70.12 916.23 2.829812 1250 Italicized values form air tables Bold values are calculated We begin by entering our operating information. Since state 1 and state 4 are fixed from the operating information we can go to the air tables and find h 1 = 268.62 kJ/kg and φ 1 = 1.5907 kJ/(kg K) h 4 = 2003.3 kJ/kg and φ 4 =3.6684 kJ/(kg K) We now traverse the cycle. Diffuser Compressor Burner Turbine Nozzle 1 2 3 4 6 5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ME 201 Thermodynamics Spring 1999 2 1-2 Diffuser: Isentropic The 1 st law gives kJ/kg 313.62 = ) (10 2 (300) + 268.62 = 2 v
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}