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Unformatted text preview: MAE135 Solutions to Homework No. 1 Spring 2010 PROBLEM 1 (5 points) For calorically perfect gas, the internal energy e = c v T . With c v = 717 K/kg K for air and T = 290 K, e = (717)(290) = 207930 J/Kg. (Note that the units J/kg = m 2 /s 2 ). Therefore Kinetic energy Internal energy = 1 2 V 2 e = 1 2 V 2 207930 The figure below plots this ratio versus velocity V : (Kinetic Energy) / (Internal Energy) for T=290K 0.0 0.5 1.0 1.5 2.0 2.5 200 400 600 800 1000 V (m/s) KE/IE At low speed on the order of 10 m/s (the speed of many subsonic wind tunnels, for example), the kinetic energy is a negligible fraction of the total energy. As the speed increases, however, the kinetic energy occupies a larger and larger fraction of the total energy. For high supersonic speeds, most of the flow energy is kinetic energy. We will see soon that the above ratio is proportional to the square of the Mach number. PROBLEM 2 (10 points) R dR p d dF p dR R d To evaluate the work for the cylindrical system, we can break down the cylinder into a large number of tiny slices, each slice defined by radius R and the angle increment d...
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- Spring '08