#### Transcript Lecture_3_Heat and t..

Example Problems A gas undergoes a thermodynamic cycle consisting of three processes: Process 1 - 2: Compression with pV = constant from p1 = 105 Pa, V1 = 1.6 m3, V2 = 0.2 m3, (U2 –U1) = 0. Process 2 – 3: Constant pressure expansion from V2 to V3. Process 3 – 1: Constant volume, (U3 –U1) = -3549 kJ. (a) Sketch the cycle. (b) Determine the heat transfer and the work for process 2 – 3. (a) p 2 3 Notes: For 1 - 2: Since U12 = 0, Q12 = W12 1 V For 3 – 1: Since dV31 = 0, W31 = 0. Ucycle = 0. Example Problems (b) U cycle U12 U 23 U 31 0 0 ΔU 23 -3549 0 U 23 3549 J The work in the process 2 – 3 is given by, dW23 VV3 pdV p2 (V3 V2 ) 2 In order to determine this work, we need V2 since we know that V3 = V1. We can get V2 from the relation, pV = constant . p1V1 constant p2V2 ; p2 p1V1 / V2 8x105 Pa. W23 p2 (V3 V2 ) 8x105 (1.6 0.2) 1120 kJ Q23 U 23 W23 3549 1120 4669 kJ. Also note that since U12 0, Q12 W12 and, W12 pdV V2 V1 V2 dV constant dV p1V1 1.6 x105 ln 0.2 /1.6 332.7 kJ V1 V V Example Problems Summary in kJ, note that U for the cycle = 0. U12 = 0 W12 = -332.7 Q12 = -332.7 U23 = 3549 W23 = 1120 Q23 = 4669 U31 = -3549 W31 = 0 Q31 = -3549 Wcycle = Qcycle = 787.3 kJ A gas in a piston-cylinder assembly undergoes an expansion process for which the relationship between pressure and volume is given by PVn = constant = A The initial pressure is 3 bars, the initial volume is 0.1m3 and the final volume is 0.2 m3. Determine the work for the process in kJ if (a) n = 1.5, (b) n = 1.0 and (c) n = 0. In each case the work is given by, p pdV A Vn V21 n V11 n dV W A n A 1 n V V1 V2 The constant A can be evaluated at either end state V21 n V11 n dV W A n A 1 n V1 V V2 p2V2n p1V1n A p2V2 p1V1 1 n This expression is valid for all n except n =1. (a) To evaluate W, the pressure at state 2 is required. n V1 0.1 p2 p1 3bar V 0.2 2 1.5 1.06bar 1.06bar 0.2m3 3bar 0.1m3 105 N / m2 W 1 1.5 1bar 17.6kJ 1kJ 3 10 N / m (b) For n = 1 we need to consider this special case, pV A V2 W p1V1 ln V1 W 20.79kJ (c) For n = 0, the pressure volume relation is just p = A W pdV p V2 V1 W 30kJ Example Problems A gas is enclosed in a cylinder with a movable piston. Under conditions that the walls are Adiabatic, a quasi-static increase in volume results in a decrease in pressure according to, PA, VA PV 5 / 3 constant for Q = 0. WAB A D 105 P (Pa) Find the quasi-static work done on the system and the net heat transfer to the system in each of the three processes, ADB, ACB and the linear AB. In the process ADB the gas is heated at constant pressure (P = 105 Pa) until its volume increases from 10-3 to 8 x 10-3 m3. The gas is then cooled at constant volume until its pressure decreases to 105/32 Pa. The other processes (ACB and AB) can be similarly interpreted according to the figure. 105/32 C B 10-3 8 x 10-3 V (m3) Example Problems U = Q – W From the first Law: For the adiabatic process: U = – W U U B U A WAB PA VB VA 5/3 VA V dV 3 = PAVA5 / 3 VB 2 / 3 VA 2 / 3 2 3 = 25 100 112.5 J 2 For the process ADB: WADB PdV PA VD VA 700 J but, U B U A QADB WADB and from the result of the adiabatic process QADB U B U A WADB 112.5 700 587.5 J Example Problems Similarly for the process ACB: we find that WACB = -21.9 J and QACB = -90.6 J For the process AB: W can also calculated from the area: WAB 360.9 J and from the first law, QAB = 248.4 J Note that while we can calculate QADB and QACB , we can not calculate QAD , QDB , QAC and QCB , separately for we do not know (UD – UA) or (UC – UA). Given the standard enthalpies of formation, H f ,find the enthalpy for the following reaction at 298K and 1 atmosphere pressure. o H of kJ / mol MnSiO3 MnO SiO2 -246 -384 -910 MnSiO3 (s, 298) MnO(s,298) + SiO2 (quartz,298) H o products H of H of reac tan ts 910 384 (246) 1048 KJ Given the specific heat functionality find the heat of reaction at 800K. Heat capacity Constants (J/mol-K) MnSiO3 MnO SiO2 c x 10-5 b x 103 a 110 46.0 46.9 16.2 8.2 34.2 -25.8 -3.7 -11.3 Temp range (K) 298-1500 298-900 298-1000 First bring each of the components in the reaction from 298K to 800K dH C p dT 800 298 dH 800 298 C p dT H (800) H (298) 800 298 H (800) H of 800 H (800) H 800 298 o f 298 C p dT C p dT a bT cT dT 2 Then evaluate H for the reaction at 800K H (800) H o f 800 298 a bT cT dT 2 800 b 2 c H of aT T 2 T 298 a =-17.1 b = 26.2 x 10-3 c = 10.8 x 105 26.2 x 10-3 2 2 H (800) H 17.1 800 298 800 298 2 1 5 1 10.8 x 10 800 298 o f H (800) 1048kJ 1.0kJ 1049kJ P,V, T Relations and Thermodynamic Properties Graphical equation of state P, v, T surface projected on the P-T plane. Phase diagram P,v, T surface projected on the P- v plane. P,v, T surface for a substance that expands on freezing. v = V/m, specific volume P,V, T Relations and Thermodynamic Properties Graphical equation of state P,v, T surface for a substance that contracts on freezing. Phase diagram and P-v surface for a substance that contracts on freezing. P,V, T Relations and Thermodynamic Properties Phase change: consider a container of liquid water heated at constant pressure (1 Atm ~ 105 Pa) (a) The temperature of the water rises to 100 C. (b) At 100 C the heat energy goes into converting the liquid water to water vapor. The volume of the system increases. The temperature of the two – phase, liquid/vapor system stays constant (100 C) until the last drop of liquid disappears. (c) After all the fluid is converted to vapor, the temperature and volume rise as heat is added to the vapor. P,V, T Relations and Thermodynamic Properties Saturated liquid Saturated vapor x 1-x g f vf v P,v, T surface projected on the T- v plane for water. Lever rule: x(vg v) (1 x)(v v f ) v (1 x)v f xvg rule of mixtures x v vf vg v f vg-v v-vf vg vf v vg x is the mass fraction of the vapor called the quality. x mvap mliq mvap P,V, T Relations and Thermodynamic Properties Specific internal energy and enthalpy • u = U/unit mass • h = H/unit mass (J/kg) (J/kg) u = Q/mass – W/mass h = u + pv The simple rule of mixtures can always be used for any of the specific quantities (v, u, h): h = (1-x)hf + xhg u = (1-x)uf + xug v = (1-x)vf + xvg h, u, v are the specific enthalpy, internal energy, and volume in the two-phase fluid/gas region. Example Water in a piston cylinder assembly undergoes two processes from an initial state where the pressure is 106 Pa and the temperature is 400 C. Process 1-2: Process 2-3: The water is cooled as it compressed at constant pressure to the saturated vapor state at 106 Pa. The water is cooled at constant volume to 150 ºC. Sketch the processes on p-v and T-v diagrams. For the overall process determine the work and heat transfer in kJ/kg. Water 1 10 2 1 400 C 179.9 C 4.758 3 v (m3/kg) 150 C T (ºC) P (105 Pa) Boundary 106 Pa 400 C 179.9 C 2 4.758 x 105 Pa 150 C 3 v (m3/kg) The only work done in this process is the constant pressure compression, process 1-2, since process 2-3 is constant volume. W pdV p V2 V1 V2 V1 W V V p 2 1 p v2 v1 m m m Since for state 1 we know P and T, we can get the specific volume from a data Table, (properties of superheated water vapor) v1 = 0.3066 m3/kg. The specific volume at state 2 is the saturated value at 10 bar v2 = 0.1944 m3/kg. W/m = 106 (0.1944 – 0.3066) = -112.2 kJ/kg The minus sign indicates work done ON the system From the 1st Law, Q = U + W, and dividing through by the mass, Q U W W u3 u1 m m m m u1 is the specific internal energy in state 1, which we can get also get from a Table as 2957.3 kJ/kg. P (105 Pa) To get u3 note that we are in a 2-phase fluid/vapor region, so u3 will be a linear combination of uf and ug determined from the quantity of fluid and vapor present at v3 = v2. The mass fraction of vapor present or “quality” is given by, Table 2 10 1 400 C 179.9 C 4.758 3 v (m3/kg) vf v3 v g 150 C v3 v f 0.1994 1.0905 x103 x3 0.494 3 vg v f 0.3928 1.0905 x10 Then using the rule of mixtures we can evaluate u3. u3 1 x u f xu g 1583.9 kJ/kg 631.68 2559.5 Finally, Q U W W u3 u1 1583.9 2957.3 112.2 1485.6 kJ/kg m m m m The minus sign indicates that heat is transferred OUT of the system A two-phase liquid-vapor mixture is initially at 5 bar in a closed container of volume 0.2 m3 and “quality”, x = 0.10. The system is heated until only saturated water remains. Determine the mass of the water in the tank and the final pressure. Given Data V = 0.2 m3 P1= 5 bar x1 = 0.10 x2 = 1.0 (saturated water) H2O Using the Table: State 1: vf1= 1.0296 x 10-3 m3/kg vg1= 0.3749 m3/kg The water is in a closed system and the total volume and mass are constant. P (bar) v1 xvg1 (1 x)v f 1 0.03842m3 / kg 1 m v V V 0.2m3 m 5.21kg v 0.03842m3 / kg ? P2 =? 2 5.0 P1 1 v v2 v1 P2 Psat @ v2 51.45bar linearly extrapolated between vf1 v v g1 50 and 60 bar.