Solutions Manual for Thermodynamics An Engineering Approach 10th Edition by Çengel & Boles All Chapters

Solutions Manual for Thermodynamics An Engineering Approach 10th Edition by Çengel & Boles All Chapters

Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill. 1-3C There is no truth to his claim. It violates the second law of thermodynamics. 1-4C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-3 Mass, Force, and Units 1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit. 1-6C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s . In other words, the weight of a 1-lbm mass at sea level is 1 lbf. 1-7C There is no acceleration, thus the net force is zero in both cases. 1-8 The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be 2 2 W mg 1920 N (200 kg)(9.6 m/s ) 1-9E The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be 1 lbf 2 2 (10 lbm)(32.0 ft/s ) 32.174 lbm ft/s W mg       9.95 lbf     1-10 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be determined. Analysis From the Newton's second law, the force applied is    1 N         2 2     (6 g) (90 kg)(6 9.81 m/s ) 1 kg m/s F ma m 5297 N   PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-4 1-11 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is given to be 9.807 m/s at an altitude of 13,000 m. Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from 2 at sea level and 9.767 m/s 2 %Reduction in weight %Reduction in 9.807 9.767 g g    0.41% 100 100 g 9.807 Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude. Discussion Note that the weight loss at cruising altitudes is negligible. 1-12 A plastic tank is filled with water. The weight of the combined system is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be r = 1000 kg/m 3 . Analysis The mass of the water in the tank and the total mass are mtank = 3 kg mw = V = (1000 kg/m 3 3 V = 0.2 m 3 )(0.2 m ) = 200 kg mtotal = mw + mtank = 200 + 3 = 203 kg H2O Thus, 1 N     2 2 W mg  (203 kg)(9.81 m/s )   1991 N     1 kg m/s  1-13 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is 1 N     W mg 2 2 (2 kg)(9.79 m/s )   19.58 N      1 kg m/s  Then the net force that acts on the rock is     F FF net up down 200 19.58 180.4 N Stone From the Newton's second law, the acceleration of the rock becomes 2   F 180.4 N 1 kg m/s 2 kg       2  90.2 m / s   1 N a m   PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.