Solutions to Exercises in Applied Thermodynamics Set 3 1. General equation of continuity for an extensive property of an open system with n streams transporting i into the system is n = i i =1 of the extensive property is given by The rate of change, n = i i =1 Applications i) Conservation of mass The T-elbow of hot and cold water shower serves as the mixing chamber for the hot and cold water streams. Under steady flow conditions the net mass flow will be zero. Warm water to shower Hot water in Cold water in Open system (ii) Conservation of matter in a combustion chamber In a combustion chamber, under steady flow conditions, net inflow of matter is zero. Also conservation can be applied to each chemical element. Fuel in Products of combustion out Air in iii) Conservation of electric charge The junction of an electric circuit cannot store electric charge and conservation of electric charge is stated as Kirchoff’s law (which, in this case, is i1 + i2 + i3 + i4 = 0). i1 i4 i2 i3 2. Properties common to two systems in equilibrium i) Simple thermodynamic systems: temperature ii) Electrically charged bodies: electric potential iii) Fluids in separate vessels connected by a valve: pressure 3 pi = 300 kPa Vi= 150 l po=105 kPa Given that theprocess is fully rtesisted, 2 W1 2 p dV 1 450 400 ABSOLUTE PRESSURE/kPa 350 300 250 200 150 100 50 0 0 100 200 300 400 500 600 700 VOLUME/ l FIG.Q3:ABSOLUTE PRESSURE VS VOLUME By graphical interpolation of Figure 1 to give 6 equal intervals of V at 100 litres each, Table Q3a: Absolute pressure-volume relationship p/ kPa V/l 405 150 245 250 185 350 155 450 115 550 85 650 70 750 800 Using Simpson’s rule, for any even number (2n) of equal width strips width x, area A is given by, A = x (y1+ 4y2+2y3+ . . . + 4y2n+ y2n+1) / 3 Work done, W = area under p-V curve in Figure Q3 = –100 [405+ 4(245+ 155+ 85) + 2(185+115) + 70] / 3 = –100 500 J = – 100 kJ The result could be obtained with the tabulated values, ignoring V = 200 litres to give Table 3b and very nearly the same result as above Table Q3b: Absolute pressure-volume relationship p/ kPa V/l 405 150 210 300 155 450 95 600 70 750 Work done, W = area under p-V curve in Figure Q3 = –150 [405+ 4(210+ 95) + 2(155) + 70] / 3 = –100 250 J = – 100 kJ Shaft work = Ws = W – po (V1– V2) = – 100 kJ – 105 (150-750)/1000 = – 37 kJ 4. The expression is valid only for fully resisted expansion of a closed system. (a) i) The expression cannot be used for a partly resisted expansion. ii) The expression cannot be used for an un-resisted expansion. iii) The expression can be used for a fully resisted expansion. (b) i) Friction does not affect the process within the closed system comprising the gas. Thus, the expression can be used to find the work done on the system. ii) The expression cannot be used to find the work done on the system comprising the surroundings, as friction occurs within the surroundings and affects the work transmitted across the piston. (c) 5 The expression is not for open systems and therefore cannot be used for any flow process. For an ideal gas, pV = mRT For the process, pVn = K 2 Therefore, work done in process 1-2 = W1 2 pdV 1 n = KV dV = (p2V2 – p1V1) / (n-1) For a constant pressure process, p = constant. Therefore n = 0. p2 = p1= p W1-2 = (p2V2 – p1V1) / (n-1) = p (V1-V2) For a constant volume process, V = constant. Therefore n = . W1-2 = 0 For constant temperature process, n = 1. The formula, W1-2 = (p2V2 – p1V1) / (n-1) cannot be used, and the result is obtained by integration as W1-2 = mRT ln (V1/V2) 6 Cyclic process: There is no net change in the system boundary during a cycle. So there is no work done in displacing the matter in the surroundings. Thus the use of either gauge pressure or absolute pressure will give the same result. Non-cyclic process: Here the use of gauge pressure will give the shaft work or useful work and the use of absolute pressure the work interaction across the system boundary. Set 5 1 Qc Ws Qin = 68 kW Wd Qx System boundary The boundary interactions including the unaccounted heat interaction rate Qx are as identified in the figure above. Ws and Ws are, respectively the work output rates to the shaft and the dynamo, Qin is the heat supply rate, and Qc is the heat removal rate by the cooling water. Rate of work transfer through shaft = Ws Power delivered to dynamo = Total work interaction rate Wd Ratio of shaft power to heat input = = 40*2*4000/60 W = 16.8 kW = 0.5 kW (by data) = 16.8 kW + 0.5 kW = 17.3 kW = 17.3 / 68 = 0.254 (or 25.4%) Applying the first law of thermodynamics to the system, Qin – Qc – Qx – Ws- Wd =0 Qx = 68 - 12.5 -17.3 = 38.2 kW 2 partition system boundary acid water Before breaking of partition mixture After breaking of partition Consider the rigid insulated sealed vessel shown above. There is no boundary interaction Therefore there is no change energy content. When the system is cooled there is a heat interaction with heat flow to the surroundings, but no work. Therefore the energy content will decrease. p 3 Data: m = 0.2 kg R = 287 J/kg K = 1.4 p1 = 800 kPa, T1 = 1000 K Using ideal gas equation, V1 = m R T1/ p1 = 0.0717 m3 = 71.7 litres 1 pV1.2 = constant 2 Table of properties of the air State p/kPa V / l T/K 1 800 71.7 1000 2 100 407 709 (Numbers in red are directly obtained from the data. Numbers in green are calculated values). Cv = R / (–1) = 717 J/kg K Cp = R / (–1) = 1004 J/kg K V1 = m R T1/ p1 = 0.0717 m3 = 71.7 litres According to the law of expansion, p1V11.2 = p2V21.2 V2 = (100/800) 1/1.2 V1 = 407 l Using ideal gas equation, T2 = p2V2 T1 / p1V1 = 709 K Work done on the system, W1-2 = (p2V2 – p1V1) / (n –1) = m R (T2 – T1) / (n –1) = 0.2*287* (709 - 1000) / 0.2 = – 83.3 kJ Change in internal energy U = m Cv (T2 – T1) = 0.2* 717 (709-1000) J = – 41.7 kJ By first law of thermodynamics,U = Q – W Heat interaction, Q = – 41.7 – (–83.3) = 41.6 kJ V 4 Data: position 2 position 1 V = 1 litre R = 287 J/kg K = 1.4 p1 = 100 kPa, T1 = 303 K 50 mm p Using ideal gas equation, m = p1 V1 / R T1 = 1.15*10 -3 kg = 1.15 g Cv = R / (–1) = 717 J/kg K Cp = R / (–1) = 1004 J/kg K 2 pV = constant Table of properties of the air State p/kPa V / l T/K 1 1 100 1.0 303 2 201 0.607 370 (Numbers in red are directly obtained from the data. Numbers in green are calculated values). V1 – V2 = ( d 2 /4)* x, where x is the inward displacement of the piston. V2 = 1 – (*0.1*0.1*0.05 / 4 * 103) litre = 0.607 l According to law of expansion, p1V1 = p2V2 p2 = 100* (1/0.607) = 201 kPa Using ideal gas equation, T2 = p2V2 T1 / p1V1 = 367 K V Work done on the system, W1-2 = (p2V2 – p1V1) / (n –1) = m R (T2 – T1) / (n –1) = 1.15*10 -3 *287* (370 - 303) / 0.4 = 55 J Change in internal energy U2 – U1 = m Cv (T2 – T1) = 1.15*10 -3 * 717 (370-303) J = 55 J By first law of thermodynamics, U2 – U1 = Q1-2 + W1-2 Heat interaction, Q1-2 = 0 kJ (A small difference between W1-2 and U2 – U1 is possible because of rounding errors in the arithmetic.)