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120 100 80 superheat horn excess work 2 in .NET Print barcode 3 of 9 in .NET 120 100 80 superheat horn excess work 2




How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
140 120 100 80 superheat horn excess work 2 use .net framework 3 of 9 barcode encoder topaint barcode 3 of 9 for .net JasperReports temperature, T ( C). 60 3 40 20 0 -20 -40 -60 -1 0 1 a b 4 1 throttling excess work d c specific entropy, s (kJ k Code-39 for .NET g-1 K-1 ). Figure 2.8: T S diagram for the idealized ammonia vapor-compression cycle. 23 Cool Thermodynamics Mecha Visual Studio .NET USS Code 39 nochemistry of Materials From standard thermodynamic tables [Mayhew & Rogers 1971, ASHRAE 1998], we find the following specific (per unit mass) properties for ammonia: at 20 C at 40 C h 1 = 1420 kJ kg 1 h a = 89.8 kJ kg 1 h d = 1473.

3 kJ kg 1 s 1 = s 2 = 5.623 kJ kg 1 K 1 s a = 0.368 kJ kg 1 K 1 s d = 4.

877 kJ kg 1 K 1.. We also note that h3 = 37 1.9 kJ kg 1 and s3 = s b = 1.360 kJ kg 1 K 1.

The values of s provided at 50 and 100 K degrees of superheating are: s (at 50 K superheating) = 5.321 kJ kg 1 K 1 s (at 100 K superheating) = 5.655 kJ kg 1 K 1.

The actual degree of superheating is calculated by interpolation between values available in the tables:. degree of superheating = 50 + (100 - 50) (5.623 - 5.321 ) = 95.

2 K 5.655 - 5.321.

i.e., 45.

2 K above the ta VS .NET ANSI/AIM Code 39 bulated value The specific enthalpy at the top of a knowledge of h tabulated at a 50 K of the additional 45.2 K superheating.

h2 = h ( at 50 K superheating) + of 50 K superheating. the Code 39 for .NET superheat horn h2 is obtained from degree of superheating, plus the effect just calculated:.

45.2(1751.9 - 1622.

4) = 1 739.46 kJ kg -1. 50.

Armed with these data, we are now ready to generate the figures requested. Carnot cycle: Recall Equa tion (2.5) for the Carnot COP. In the reversible limit, corresponding refrigerant and reservoir temperatures are the same.

Now introduce the given temperatures to obtain. COPCarnot = Tevap Tcond - Tevap 253 = 4.216. 313 - 253.

Next,. W rev = (h2 hd) T 3 (s c s d). = (1739.46 1473.3) 319 (5.623 4.877) = 28.216 kJ kg 1. Thermodynamic and Operational Fundamentals rev Finally, Qevap = T 3 (s 1 s 3) = 253 (5.623 1.360) = 1078.54 kJ kg 1. = 4.216 , which of course is Wrev the same as the Carnot COP calculated at the beginning of the exercise..

As a quick consistency ch 3 of 9 barcode for .NET eck for COP:. rev Qevap Idealized vapor compressi 39 barcode for .NET on (theoretical) cycle: Relative to the reversible Carnot cycle, this irreversible cycle incurs excess work due to: (1) the superheat horn, and (2) throttling. In addition it suffers from a loss of cooling capacity.

Each of these is readily calculated as follows. Excess work due to the superheat horn = (h 2 h d) T 3 (s c s d) = (1739.46 1473.

3) 319 (5.623 4.877) = 28.

216 kJ kg 1. Excess work from throttling = (h3 ha) T 4 (sb s a) = (371.9 89.

8) 253 (1.360 0.368) = 31.

12 kJ kg 1. Loss of cooling capacity = h4 hb = T 4 (s 4 s b). We take advantage of the following 3 thermodynamic equivalences for this particular cycle h3 = h4 (since throttling is isenthalpic) sb = s 3 and.

hb ha = T4 (s b s a) to obtain that the loss of cooling capacity = (h4 h a) T4 (s 3 sa) = (371.9 89.8) 253 (1.

360 0.368) = 31.12 kJ kg 1.

The irreversible cycle s COP can now be expressed in terms of the values of Qevap and W for the reversible cycle, modified by the losses just calculated:. Cool Thermodynamics Mechanochemistry of Materials rev Qevap - ( loss of coo ling capacity). COP = Wrev + ( two excess work contributions). 1078.54 - 3112 . = 3.324 251.78 + 3112 + 28.216 . which is 79% of the Carno t COP. This exceptionally high figure derives from the idealized nature of the irreversible cycle, i.e.

, key loss mechanisms having been omitted. (At this stage, Qevap refers to cooling capacity in kJ kg 1. In later chapters, where we will be examining cooling power, as the product of cooling capacity and refrigerant mass flow rate, we will also be using the symbol Qevap to denote cooling rates in kW.

) _________________________________________________________________________. Gas cycles (e.g., reverse .

NET Code-39 Brayton and reverse Otto cycles with isobaric or isochoric heat transfer) are possible, and in fact constitute a small fraction of special-application refrigeration systems. They suffer, however, from the heat transfer branches being substantially non-isothermal. This translates into a considerable increase in the required work input for the same cooling capacity, in other words, lower COP.

Other factors militate against adopting the idealized processes in the Carnot refrigeration cycle. For example, cycles must be performed in finite time in order to attain non-zero cooling rates. Heat exchangers must be finite in extent, so a thermal bottleneck develops in removing and rejecting heat.

Compressors and throttlers incur significant fluid friction losses. Losses derive from superheating in the evaporator and de-superheating in the condenser (to ensure pure vapor or pure liquid at their exits). There are also heat leaks in each component and in refrigerant lines, as well as pressure drops, and mechanical friction losses in the compressor shaft.

Hence actual COPs are far below the Carnot limit (a point to which we ll return at the beginning of 4). B3. Real vapor-compression cycles Figure 2.

9 is a schematic of a real vapor compression chiller. The branches of the cycle are depicted in Figure 2.10: Two-phase refrigerant exits the evaporator and is superheated prior to being sucked into the compressor (1 2).

Refrigerant vapor is compressed and discharged to the condenser (2 3). De-superheating in the condenser (3 4). Condensation/heat rejection (4 5 6).

Throttling (expansion) (6 7). Evaporation/cooling effect/heat removal (7 1)..

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