![]() ![]() The large starred and striped cloth was tied to a light pole on the edge of a brick house’s front sidewalk. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 2-4.Yesterday, as I walked down my quiet, suburban-area street, I noticed something blowing in the spring wind. ![]() Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 3-2. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 9-4. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 3-2. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.Physics of Nuclear Kinetics. Addison-Wesley Pub. Nuclear and Particle Physics. Clarendon Press 1 edition, 1991, ISBN: 978-0198520467 Nuclear Reactor Engineering: Reactor Systems Engineering, Springer 4th edition, 1994, ISBN: 978-0412985317 Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 8-1. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983). But we have swapped reservoirs, without any impact on COP. For example, we assumed that the temperature difference (T hot – T cold) is the same for both modes. Note that, in this example we have many assumptions. From the definition: COP cooling = 3000/1500 = 2. Using the first law of thermodynamics, which states: ![]() In case of the cooling mode, the heat pump (air conditioner) with 1500 W motor can take heat Q cold from inside the house and then dump Q hot = 4500 W to the hot outside. Q hot = COP heating x W = 3 x 1500 = 4500 W or 4500 J/s The amount of heat the heat pump can add to a room is equal to: to act as an air conditioner in the summer), what would you expect its coefficient of performance to be? Assume all else stays the same and neglect all other losses. If the heat pump were turned to the cooling mode (i.e.Calculate the amount of heat ( Q hot) the heat pump can add to a room?.Its compressor consumes 1500 W of electric energy. An increase in the input temperature (T cold) means, for example, an oversized ground source of heat.Įxample – Heat Pump – Heating and Air ConditioningĪ reversible heat pump has a coefficient of performance, COP = 3.0, when operated in the heating mode. Therefore, reducing the output temperature (T hot) is very efficient, but requires very efficient heat transfer from heat pump system to surroundings (i.e. But in reality the best systems are around 4.5.Īs can be seen, the COP of a heat pump system can be improved by reducing the temperature difference (T hot – T cold). According to the above formula, the maximum achievable COP for T hot = 35 ☌ (308 K) and T cold = 0 ☌ (273 K) would be 8.8. Note that, these equations must use an absolute temperature scale (T cold, T hot) and it is only a theoretical maximum efficiency. Using the first law of thermodynamics define COP also as the heat removed from the cold reservoir plus the input work to the input work.įor an ideal heat pump (without losses and irreversibilities) can be derived that: In general, COP is highly dependent on operating conditions, especially absolute temperature and relative temperature between heat sink and system.įor heating, the COP is the ratio of the heat added to the system (hot reservoir). The COP usually exceeds 1, especially in heat pumps, because, instead of just converting work to heat, it pumps additional heat from a heat source to where the heat is required. The coefficient of performance, COP, is defined also for heat pumps, but at this point we follow the net heat added to the hot reservoir. We call this ratio the coefficient of performance, denoted by COP. The relevant ratio is therefore the larger this ratio, the better the refrigerator. From an economic point of view, the best refrigeration cycle is one that removes the greatest amount of heat from the inside of the refrigerator (cold reservoir) for the least expenditure of mechanical work or electric energy. For a refrigeration or heat pumps, thermal efficiency indicates the extent to which the energy added by work is converted to net heat output. Heat Pump, Refrigerator, Air Conditioner – basic principle of operationīut in heat pumps and refrigerators, the work is not an output. The thermal efficiency, η th, represents the fraction of heat, Q H, that is converted to work. In general, the thermal efficiency, η th, of any heat engine as the ratio of the work it does, W, to the heat input at the high temperature, Q H. Coefficient of Performance – Heat Pump, Refrigerator, Air Conditioner ![]()
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