Thermodynamics problem, seems pretty straight forward however I don\'t think I g

Question

Thermodynamics problem, seems pretty straight forward however I don't think I got the right answers.

Thank you again!

6-A.) Draw a complete schematic diagram (show arrows, symbols & numeric values for each w & Ø). Consider the steady operation of a self-contained refrigeration "package" for a shipping container. Instead of directly using an engine or electric motor the cyclic device is thermally activated by heat transfer from a steady 1373 K external source. The external thermal source is due to complete external combustion of propane fuel. Assume all of the fuel energy is captured by the device. The device provides a cooling capacity of 22 kJ/s of heat transfer from a 260 K shipping container and rejects 55 kJ/s of waste heat to the +37 atmosphere Notice external work is zero, and for a reversible cycle, (QT)REy2 0 (using sign convention Q1>0, Qord) Draw a complete schematic diagram (show arrows, symbols & numeric values for each 0). For the actual cycle Find the fuel input rate kws, the cycle COP-_-, and rate of entropy creation production. cycle kJ/s-K For a completely reversible cycle operating at the same temperatures and the same cooling capacity Find the minimum possible fuel input rate - kJ/s and the maximum possible COP = Combustion of Propane Fuel Maintains Hot Reservoir Waste Heat to Atmosphere = 55 kJ/s To ) = 310 K 1373 K-TH OH CYCLE No External Work Cooling Capacity 22 kJ/s Tc ) = 260 K

Explanation / Answer

Ans

Refrigeration cycle can be considered as the reversed Carnot cycle. Typically a heat pump mechanism is

used to transfer heat from low temperature to high temerature.

Hence to drag heat out of the refrigerator at 22KJ/s work has to be done. Assuming all the heat is absorbed by the sytem , the net heat flow = 22 kJ/s + 55 KJ/s = 77KJ/s

The coefficient of performance COP = Tc / (To-Tc ) = 260/ (370-260) = 5.2

The rate of entropy creation = (dQo/dt) / To - (dQc/ dt) /Tc = (55 Kj/s/ 310K) - (22 kJ/s)/(260) =0.092Kj/sK

For reversible process:

waste heat =0. Heat suppied to ambient should be added to the source.

hence minimum possible fuel input rate = 22KJ/s

As the source is at 1373 k , COP will be changing accordingly.

However, if the ambient tempearture can be kept at 310 K, it will be maximum COP=5.5

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