# Ex7_6 - First with ef3 available we only recover ef2-ef1...

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Example 7.6 Parameters: R Cp p0 T0 mdot 0.287 1 100 293 90 kJ/kg,K kJ/kg,K kPa K kg/s Below, bold fonts indicate given values, regular font for calculated values Note that, once the formulae for h, s and ef have been entered with proper anchoring they can be copied to the next state. p T h s ef State 0 100 293 0 0 0 State 1 1000 610 317 0.0724444 295.77378 State 2 970 860 567 0.4246597 442.57472 Heat recovery: 22500 kW Exergy recovery: 13212.08 kW State 3 110 1020 727 1.2200313 369.53084 heat exchanger equation gives T4 State 4 100 770 477 0.9662179 193.89815 From that perspective, the efficiency of the heat exchanger can be measured in two ways:
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Unformatted text preview: First, with ef3 available, we only recover ef2-ef1, i.e. an efficiency of 0.397263 Alternatively, without prejudging of further recovery past point 4, comparing ef2-ef1 to ef3-ef4 also makes sense 0.835841 Exergy destruction: 2594.857 kW Organization of a spreadsheet for exergy analysis: constant Cp The ef's measure the local `ability to do work' per unit mass flow rate of air....
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