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This post was last edited by cloudyangel on 2021-9-5 at 10:58. The topic involves a turbine driven by superheated steam at 735K and 1520 kPa (https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-712169d492a112d06720dd0be8327ee7_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-031591b8619af8147b5e5f37fe1be862_l3.svg). While driving the turbine to perform work, each 1 kg of steam loses 7.1 kJ of heat to the surroundings. The ambient temperature is 293K. The steam pressure after work is done is 71 kPa. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-c0e1701e41b1e39724dd0c58de9b093f_l3.svg; the energy output per kilogram is 2770.5 kJ/kg. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-2270b21541bc5cd6d9b0d8570a7b178b_l3.svg. What is the thermodynamic efficiency η (%) for this process? (A92.3 (B) 88.8 (C) 79.6 (D) 76.6 The question tests knowledge of energy conservation and isentropic expansion processes. The loss of enthalpy is used to drive the turbine to perform work, as well as to dissipate heat into the environment. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-bc1ae8c0588a142ec042d7b868fbe744_l3.svg The actual work done is W = (3426.7–2770.5) – 7.1 = 649.1 kJ/kg. In an ideal process, there is no heat exchange with the environment and entropy remains constant; as a result, enthalpy decreases more, leading to greater work output in such an ideal process. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-bc6276183ce78b8f570d9b17f5d5574e_l3.svg The ideal total work is W_i = 649.1 + 7.1 + 74.8 = 731 kJ. The thermodynamic efficiency is given by https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-d70e6221be2f1a1cb1396738b313c5d7_l3.svg. Option B is chosen; it assumes that the indoor refrigerator operates ideally. The indoor temperature is 28 ℃, while the temperature inside the refrigerator is maintained at 10 ℃. Now, 30 kg of glue, whose specific heat capacity is 4.2 kJ/kg·K, needs to be cooled from 35 °C to 10 °C. What are the values of the electrical energy consumed by the refrigerator (in kJ) and the heat released by the refrigerator into the room (also in kJ)? The correct options are: (A) 188.28, 3338.28; (B) 188.28, 2961.72; (C) 200.25, 3350.25; (D) 200.25, 2949.75. The solution involves applying the reverse Carnot refrigeration cycle. The heat absorbed by the glue is given by https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-8e0de11a06241ecbc1315fe6d2c24a66_l3.svg = 304.2 × (35 – 10) = 3150 kJ. The electrical energy consumed by the refrigerator, plus the heat absorbed by the glue, equals the heat released by the refrigerator into the room. BD can be ruled out immediately; therefore, the focus is on determining the electrical work consumed. An ideal refrigerator operates according to the reverse Carnot cycle: https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-fb8478646b29d85411afc12ca761561a_l3.svg. Here, T1 should be set to the ambient temperature of 28°C. The value of w is calculated as follows: w = 3150 / (273.15 + 28) * (28 – 10) = 188.28 kJ. For question A, a reaction heat recovery system in a coal chemical plant can generate saturated steam at 4.0 MPa using the heat released from reactions; 20 tons of such steam correspond to what amount of standard coal saved? (The enthalpy difference between 4.0 MPa saturated steam and boiler feedwater is 2800 kJ/kg) (A) 1.911 (B) 7.714 (C) 1843 (D) 18.43. The solution involves a simple heat calculation. Heat released by saturated steam = Heat released upon burning standard coal; standard coal refers to coal with a calorific value of 7000 kcal per kilogram. 2028001000/(7000*4.18)/1000 = 1.914 t