Question

Questions 21-28 The following will allow you to demonstrate competency in. (i) applying the First Law of Thermodynamics to a closed system; (ii) un ders tanding the concept of boundary work; and (ii) deriving thermo-physical data for water states from tables. The solutions provided should also be done in a way that allows you to practice systematic problem solving. You have been asked to use your knowledge of thermodynamics to help analyse the compression of water in a closed system. The process is simulated as a well insulated piston-cylinder device which initially contains 5 litres of saturated liquid water (H2O) at 175kPa. The water is now slowly heated with a resistance heater (at the base of this piston-cylinder device) for a period of time. Whilst the water is being heated it is also stirred using a small (mechanically operated) paddle. This helps homogenise the temperature inside the liquid during heating but also supplies 300KJ of energy to the water during the time it is being heated. At the end of the combined stirring and heating, it is found that half the water has evaporated Q21 [Marks=1%) Identify the initial system temperature (units: °C) before the heater and paddle are activated Answering Space Q22 (Marks 1 %) Identify the enthalpy (h) for H2O (units: kJ/kg), before the heater and paddle Answering Space are activated. Q23 [Marks=1% Identify the specific volume (v) for H20 (units: m3/kg), before the heater and paddle are activated Answering Space

Answer 1

Given 5x10'in (saturated State) Initial volume of water slitres Pressure P = 175KP a Heat supplied a= 300kJ @ Initial temperature Ti from Steam table at P = 175 kPa., Saturated water Tscut = 116c hg - 25 au kilkg, he: 487 kJ/kg. Vg = 1.00 Milkg, I = 0.00105 m²/kg. How Tsat = T, = 1160 Initial Enthalpy Hi = h = 487 kJ/kg a Initial specific volume 9 = 9: 0.00105 milkg Day Total mass of H₂O = Initial volume of water - 5xion 8p. volume v 0.00105 mass = 4.76 kg /

eas since ginen hay of wath is operated i mass of Vapour = 112 of total initial mass of watu mw = 112x4.76= 2.38 kg 926 final speiche volume of Heo How dryness raition (x) = mass of Valour - 2.38 = 0.5. 476 Total man .: Final. Sp. volume Up + x (ug-24) : 0.00105 + 0.5*(1.00 -0.00105) U2 0.5005 m/kg Qot Enthalpy at End (Ho) He hy + & Chgeht): 487+0.5 (a5a4-887) H₂ 1505.5 kilka Because (te in kilky Bag Power supplied - Total mass x (H2 - H.) x1000 time time = 4.76 * (150s.S-487/x 100 = 1353.45 sec 3582