Sol termodinamica cengel cap 1

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Chapter 1 INTRODUCTION AND BASIC CONCEPTS
Thermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-2C 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 ofenergy, and thus no violation of the conservation of energy principle. 1-3C There is no truth to his claim. It violates the second law of thermodynamics.

Mass, Force, and Units 1-4C 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/s2. In other words, the weight of a 1-lbmmass at sea level is 1 lbf. 1-5C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1 kg-force. 1-6C There is no acceleration, thus the net force is zero in both cases.

1-7 A plastic tank is filled with water. The weight of the combinedsystem is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The mass of the water in the tank and the total mass are
V =0.2 m
3

mtank = 3 kg

mw =ρV =(1000 kg/m )(0.2 m ) = 200 kg
3 3

mtotal = mw + mtank = 200 + 3 = 203 kg Thus,
 1N   W = mg = (203 kg)(9.81 m/s 2 ) 2   1 kg ⋅ m/s = 1991 N 

H2O

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1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. Analysis The mass of the air in the room is
m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg
3 3

ROOM AIR 6X6X8 m3

Thus,
 1N W =mg = (334.1 kg)(9.81 m/s 2 )  1 kg ⋅ m/s 2    = 3277 N  

1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 1% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as
W = mg = m(9.807 − 3.32 × 10−6 z )

In our case,
W = 0.99Ws = 0.99mgs =0.99(m)(9.807)

Substituting,
0.99(9.81) = (9.81 − 3.32 × 10 −6 z)  → z = 29,539 m

0 Sea level

1-10E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the spring and beam scales in space. Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body:
 1 lbf W = mg = (150 lbm)(5.48 ft/s 2 )  32.2lbm ⋅ ft/s 2    = 25.5 lbf  

(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scale will read what it reads on earth,
W = 150 lbf

1-11 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
 1NF = ma = m(6 g) = (90 kg)(6 × 9.81 m/s 2 )  1 kg ⋅ m/s 2    = 5297 N  

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1-12 [Also solved by EES on enclosed CD] 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
 1N W = mg = (5 kg)(9.79 m/s 2 )  1 kg ⋅ m/s 2    = 48.95 N  

Then the net force that acts on the rock is
Fnet = Fup −Fdown = 150 − 48.95 = 101.05 N

From the Newton's second law, the acceleration of the rock becomes
a= F 101.05 N  1 kg ⋅ m/s  = m 5 kg  1N 
2

Stone

  = 20.2 m/s 2  

1-13 EES Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below....