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Tutorial (TOC) > Daemons > Closed Steady > Examples |
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Ex-1 A Carnot heat engine with an efficiency of 60% receives heat from a source at a rate of 3000 kJ/min, and rejects the waste heat to a medium at 300 K. Determine (a) the power that is generated by the engine, (b) the source temperature, and (c) the rate of entropy generation if the actual efficiency is 40%. (d) What-if scenario: How would the answers change if the cold reservoir temperature was 275 K instead? |
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Solution The engine, exchanging only heat and work with the surroundings, is a closed system operating at steady state. For overall cycle analysis, launch the closed-steady daemons at TEST. Daemons. Systems.
Closed. SteadyState.
On the Analysis panel, choose Heat Engine , and enter the known cycle variables eta_Carnot (60%), Qdot_H (=3000 kJ/min) and T_C (300 K). A Calculate produces the desired answers Wdot_rev=30 kW and T_H=750 K . Enter eta_th (40%) and Calculate Sdot_gen as 0.033 kW/K. For the what-if study, simply change T_C to the new value and Calculate the new answers: Wdot_rev=30 kW and T_H=688 K . |
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| Fig. 1.1 Image of Daemons.Systems.Closed.SteadyState page. |
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Ex-2 A gas turbine, operating between a maximum temperature of 1500 K and a minimum temperature of 300 K, has a thermal efficiency of 21% and develops a power output of 8 MW. Determine (a) the actual and (b) ideal fuel consumption rate (for the same output) if the heating value of the fuel is 50 MJ/kg. Solution The engine, exchanging only heat and work with the surroundings, is a closed system operating at steady state. For overall cycle analysis, launch the closed-steady daemon at TEST. Daemons. Systems. Closed. SteadyState. |
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On the Analysis panel, choose
Heat Engine, and enter the known cycle variables T_H (1500 K), T_C (300 K), eta_th (21%),
and Wdot_net (=8 MW). A Calculate produces Qdot_H=38.1
MW . Therefore, the fuel consumption
rate is 38.1/50 kg/s = 0.762 kg/s.
To obtain the ideal fuel consumption rate, initialize the problem and enter T_H (1500 K), T_C (300 K) and Wdot_rev (=8 MW). A Calculate produces Qdot_H=10 MW . Therefore, the fuel consumption rate is 10/50 kg/s = 0.2 kg/s. |
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Ex-3 A refrigerator has a second-law efficiency of 45 %, and heat is removed from it at a rate of 200 kJ/min. If the refrigerator is maintained at 2 oC while the surrounding air is at 27oC, determine the power input to the refrigerator. Solution The refrigerator, exchanging only heat and work with the surroundings, is a closed system operating at steady state. For overall cycle analysis, launch the closed-steady daemon at TEST. Daemons. Systems. Closed. SteadyState. |
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On the Analysis panel, choose Refrigerator , and enter the known cycle variables eta_II (45%), T_H (300 K), T_C (275 K) and Qdot_C( 200 kJ/min) . A Calculate produces the desired answer Wdot_net=0.67 kW. |
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Ex-4 An air-conditioning system maintains a house at a temperature of 20 o C while the outside is 40oC. If the cooling load on this house is 10 tons, determine (a) the reversible power, (b) the reversible power if the interior is made 5 degree warmer. Solution The air-conditionar (see accompanying figure), exchanging only heat and work with the surroundings, is a closed system operating at steady state. For overall cycle analysis, launch the closed-steady daemon at TEST. Daemons. Systems. Closed. SteadyState. |
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On the Analysis panel, choose Refrigerator , and enter the known cycle variables T_H (313 K), T_C (293 K). The refrigerator works between these two temperature and must remove energy at a rate of 10 tons. Therefore, Qdot_C=10 tons, enter it. A Calculate produces Wdot_rev=2.4 kW. Change T_C to 25o C. 'Calculate' to produce Wdot_rev=1.75 kW , a 26% drop in the power consumption. |
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