Consider a carnot cycle heat pump cycle with r410a
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Of the plotted technologies, four are commercially available: (1,2) single-effect H 2O/LiBr or NH 4/H 2O machines, (3) double-effect H 2O/LiBr machines, and (4) adsorption machines with various working pairs. regeneration temperature for various HACTįigure 11.16 shows the typical ranges of firing temperature and COP for absorption, adsorption and ejector cycles. System-wide approaches to measuring efficiency are addressed in Section 11.8.ġ1.16. This is shown for typical system COPs in Fig. 11.16. Another perspective on thermal COP is provided by comparing it with the Carnot limit for a given regeneration temperature. This highlights the need to take a system-wide approach to system performance, rather than simply focusing on the COP values for different cooling technologies. Depending on factors such as equipment utilization rate, the assumed conversion efficiency of the electric grid, climate, and many others, facility ‘A’ might turn out to use less source fuel energy than facility ‘B’ to meet the same loads throughout the year (it will likely also eliminate expensive demand charges on its electric bill which are usually incurred during the hottest days of the year).
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For example, imagine that facility ‘A’ utilizes a CCHP system which incorporates a waste-heat-driven HACD with a COP of 0.5, while facility ‘B’ utilizes grid electricity and a VCC heat pump with a COP of 4. In the case of waste-heat-activated cooling technologies, the accounting is even more favorable towards the HACT. high-temperature heat) are required to produce one unit of electricity, then an electrically-driven heat pump with a COP of 3 will have equivalent source fuel utilization rate to a fuel-fired, heat-activated heat pump with a COP of only 1. In this regard, if it is assumed that three units of fuel (i.e. the COP.It is important to remember the more thermodynamically valuable nature of electricity compared with fuel or heat when comparing VCC COPs to thermal COPs. the heat transfer to the R-410a in the evaporator, and d. the heat transfer from the R-410a in the condenser, c. the heat transfer from the compressor, b. With the data given in the table, calculate: a. An R-410a heat pump cycle as shown in the Figure has an R-410a flow rate of 0.05 kg/s with 5 kW into a non-adiabatic compressor. Condenser -lcomp -Ocong to room 5 State 4 comp Compressor P, kPa 3100 3050 3000 420 400 390 Expansion valve T, ☌ h, kJ/kg -5 120 110 45 -10 367 377 134 280 284 Evaporator Oevap from cold outside air 2. Engineering Mechanical Engineering Q&A Library 2.