Solucionario Van Wylen Termodinámica 6ª Edición Capitulo 7

CHAPTER 7. 6th edition Sonntag/Borgnakke/Wylen
This problem set compared to the fifth edition chapter 7 set.
Study guide problems 7.1-7.17 are all new
New
5th

7.1
Electrical appliances (TV, stereo) use electric power as input. What happens to the power? Are those heat engines? What does the second law say about those devices?
Most electric appliances such as TV, VCR, stereo and clocksdissipate power in electrical circuits into internal energy (they get warm) some power goes into light and some power into mechanical energy. The light is absorbed by the room walls, furniture etc. and the mechanical energy is dissipated by friction so all the power eventually ends up as internal energy in the room mass of air and other substances.
These are not heat engines, just the oppositehappens, namely electrical power is turned into internal energy and redistributed by heat transfer. These are irreversible processes.
7.2
A gasoline engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency and how much power is rejected to the ambient?
Conversion Table A.1: 20 hp = 20 × 0.7457 kW = 14.91 kW
Efficiency: ηTH = W.out/Q.H = 14.9135 = 0.43Energy equation: Q.L = Q.H - W.out = 35 – 14.91 = 20.1 kW
Q.H
⇒
Q.L
⇒
W.out
⇒
Sonntag, Borgnakke and van Wylen
7.3
A refrigerator removes 1.5 kJ from the cold space using 1 kJ work input. How much energy goes into the kitchen and what is its coefficient of performance?
C.V. Refrigerator. The energy QH goes into the kitchen air.
Energy Eq.: QH = W + QL = 1 + 1.5 = 2.5 kJ
COP: β = QLW =1.5 / 1 = 1.5
The back side of the refrigerator has a black grille that heats the kitchen air. Other models have that at the bottom with a fan to drive the air over it.
12Air in, 3Air out, 4
7.4
Assume we have a refrigerator operating at steady state using 500 W of electric power with a COP of 2.5. What is the net effect on the kitchen air?
Take a C.V. around the whole kitchen. The only energyterm that crosses the control surface is the work input W. apart from energy exchanged with the kitchen surroundings. That is the kitchen is being heated with a rate of W..
Remark: The two heat transfer rates are both internal to the kitchen. Q.H goes into the kitchen air and Q.L actually leaks from the kitchen into the refrigerated space, which is the reason we need to drive it out again.Sonntag, Borgnakke and van Wylen
7.5
A window air-conditioner unit is placed on a laboratory bench and tested in cooling mode using 750 W of electric power with a COP of 1.75. What is the cooling power capacity and what is the net effect on the laboratory?
Definition of COP: β = Q.L / W.
Cooling capacity: Q.L = β W. = 1.75 × 750 = 1313 W
For steady state operation the Q.L comes from thelaboratory and Q.H goes to the laboratory giving a net to the lab of W. = Q.H - Q.L = 750 W, that is heating it.
7.6
Geothermal underground hot water or steam can be used to generate electric power. Does that violate the second law?
No.
Since the earth is not uniform we consider the hot water or steam supply as coming from one energy source (the high T) and we must reject heat to a low temperaturereservoir as the ocean, a lake or the atmosphere which is another energy reservoir.
Iceland uses a significant amount of steam to heat buildings and to generate electricity.
Sonntag, Borgnakke and van Wylen
7.7
A car engine takes atmospheric air in at 20oC, no fuel, and exhausts the air at –20oC producing work in the process. What do the first and the second laws say about that?
Energy Eq.: W =QH − QL = change in energy of air. OK
2nd law: Exchange energy with only one reservoir. NOT OK.
This is a violation of the statement of Kelvin-Planck.
Remark: You cannot create and maintain your own energy reservoir.
7.8
A windmill produces power on a shaft taking kinetic energy out of the wind. Is it a heat engine? Is it a perpetual machine? Explain.
Since the wind is generated by a...