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Unformatted text preview: Department of Mechanical Science and Engineering ME 300 Thermodynamics Section D Fall 2010 Homework #1  Solution 1 Problem 1.1 To get familiar with the SIunit system convert the following temperatures, pressures and velocities. (a) Convert the following temperatures first into ◦ C and then into Kelvin.40 ◦ F, 0 ◦ F, 32 ◦ F, 98.6 ◦ F, 212 ◦ F, 425 ◦ F, 2066 ◦ F Analysis : T ( ◦ C ) = ( T ( ◦ F ) 32) · 5 9 (1) T ( K ) = T ( ◦ C ) + 273 . 15 (2) Using equation 1 and 2 we get:40 ◦ F = 0 ◦ C = 233.15 K (only at 0 ◦ F are the numerical values between ◦ F and ◦ C identical!) ◦ F = 17.78 ◦ C = 255.37 K (a cold day in Illinois) 32 ◦ F = 0 ◦ C = 273.15 K (ice point) 98.6 ◦ F = 37 ◦ C = 310.15 K (approximately body temperature of a human) 212 ◦ F = 100 ◦ C = 373.15 K (steam point) 425 ◦ F = 218.3 ◦ C = 491.48 K (baking temperature for pizzas) 2066 ◦ F = 1130 ◦ C = 1403.15 K (the lowest temperature at which a plain carbon steel can begin to melt) (b) Convert the following pressures into kPa. 14.5 lbf/in. 2 , 34 lbf/in. 2 , 110 lbf/in. 2 , 3190.8 lbf/in. 2 Analysis : P ( kPa ) = P ( lbf/in. 2 ) · 6 . 8948 kPa lbf/in. 2 (3) Using equation 3 we get: 14.5 lbf/in. 2 = 100 kPa (roughly atmospheric pressure) 34 lbf/in....
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 Spring '08
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 Thermodynamics, Energy, Kinetic Energy, Mass, kg

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