301 HW_1_Solution

301 HW_1_Solution - MAE 301: Engineering Thermodynamics...

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1 MAE 301: Engineering Thermodynamics Spring 2010 HOMEWORK #1 Solutions Problem 1. Problem 1.34E from the book. Solution: Using the conversion relations between the various temperature scales, T (K] = T ( ° C) + 273 = 18 ° C + 273 = 291 K T ( ° F] = 1.8 T ( ° C) + 32 = (1.8)(18) + 32 = 64.4 ° F T (R] = T ( ° F) + 460 = 64.4 + 460 = 524.4 R Problem 2. Problem 1.45E from the book. Solution: 1 atm = 101.3 kPa = 14.7 psi, the maximum pressure can be expressed in SI units as kPa 241 = = = psi 14.7 kPa 3 . 101 ) psi 35 ( psi 35 max P Problem 3. Problem 2.13 from the book. Solution: The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and gz m for a given mass flow rate. kJ/kg
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This note was uploaded on 05/19/2010 for the course MAE 301 taught by Professor Hassan during the Spring '08 term at N.C. State.

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301 HW_1_Solution - MAE 301: Engineering Thermodynamics...

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