Entropy guide for teaching entropy-April 2020 PDF

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son apuntes de clase del profesor Carlos Acevedo...

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P_1=100 T_1=800 X_1= F(T_1 y P_1) s_1= F(T_1 y P_1) v_1= F(T_1 y P_1) h_1= F(T_1 y P_1) u_1= F(T_1 y P_1) If v_2=v_1 X_2= F(v_2 y P_2) s_2= F(v_2 y P_2) u_2= F(v_2 y P_2) h_2= F(v_2y P_2)

s2-s1= (0.8524-1.845) Btu/lb R=-0.9926 ) Btu/lb R v2=v1= 7.446 pies3/lb h2-h1=-878.5 Btu/lb u2-u1=-768.1 Btu/lb x1=1 x2=0.37 If 𝑄 = (𝑢2 − 𝑢1 ) + 𝑤 = −768.1 volumen=0).

𝐵𝑡𝑢 𝑙𝑏

; w=0 ( the process undergoes at constant

𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 ∆𝑆𝑡𝑜𝑡𝑎𝑙 𝑖𝑛 𝑙𝑎𝑠𝑡 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 𝑖𝑓 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑠𝑖𝑛𝑘 𝑖𝑠 95°𝐹 𝐵𝑡𝑢 𝑙𝑏 𝐵𝑡𝑢 𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠= 768.1 𝑙𝑏 𝐵𝑡𝑢 ∆𝑠𝑠𝑦𝑠𝑡𝑒𝑚 = −0.9926 𝐿𝑏 𝑅 𝑄𝑠𝑦𝑠𝑡𝑒𝑚 = −768.1

∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠

𝐵𝑡𝑢 768.1 𝑙𝑏 𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = = (95 + 459.6)𝑅 𝑇𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠

∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = 1.38

𝐵𝑡𝑢 𝐿𝑏 𝑅

∆𝑠𝑡𝑜𝑡𝑎𝑙 = ∆𝑠𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ∆𝑠𝑡𝑜𝑡𝑎𝑙 == −0.9926

𝐵𝑡𝑢 𝐵𝑡𝑢 + 1.38 𝐿𝑏 𝑅 𝐿𝑏 𝑅

∆𝑠𝑡𝑜𝑡𝑎𝑙 = 0.3874

𝐵𝑡𝑢 𝐿𝑏 𝑅

s2-s1= (0.5772-0.341) kJ/kg K=0.2362 ) kJ/kg K v2=v1= 0.03382 m3/ kg u2-u1=64.09 kJ/kg x1=0.4 x2=0.7793 If 𝑄 = (𝑢2 − 𝑢1 ) + 𝑤 = 64.09

𝑘𝐽 𝑘𝑔

; w=0 ( the process undergoes at constant volumen=0).

𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 ∆𝑆𝑡𝑜𝑡𝑎𝑙 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 𝑖𝑓 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑠𝑖𝑛𝑘 𝑖𝑠 35°𝐶 𝑄𝑠𝑦𝑠𝑡𝑒𝑚 = 64.09

𝑘𝐽 𝑘𝑔

𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠= − 64.09 ∆𝑠𝑠𝑦𝑠𝑡𝑒𝑚 = 0.2362

𝑘𝐽 𝑘𝑔

𝑘𝐽 𝑘𝑔𝐾

𝑘𝐽 𝑘𝑔 − 64 . 09 𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = ∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = 𝑇𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 (35 + 273.16)𝐾 𝑘𝐽 ∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = −0.2079 𝑘𝑔 𝐾 ∆𝑠𝑡𝑜𝑡𝑎𝑙 = ∆𝑠𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑠𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ∆𝑠𝑡𝑜𝑡𝑎𝑙 == 0.2362

∆𝑠𝑡𝑜𝑡𝑎𝑙

𝑘𝐽 𝑘𝐽 − 0.2079 𝑘𝑔 𝐾 𝑘𝑔𝐾

𝑘𝐽 = 0.0283 𝑘𝑔𝐾

A rigid tank contains an ideal gas that is being stirred by a paddle wheel. The temperature of the gas remains constant as a result of heat transfer out. The entropy change of the gas is to be determined. Assumptions The gas in the tank is given to be an ideal gas. Analysis The temperature and the specific volume of the gas remain constant during this process. Therefore, the initial and the final states of the gas are the same. Then s2 = s1 since

∆𝑠𝑦𝑠𝑡𝑒𝑚=0

entropy is a property. Therefore,...