309 WK11TPCSWEB.TEX
ECE M.M. Yovanovich
Week 11
Lecture 1
Hand out Tables of Properties of Compressed water. Hand out P , v-diagram for water. Show the critical point, the saturated liquid line and the saturated vapor line. Show locations of the constant temperature, constant pressure and constant quality lines under the dome. Show location of constant temperature and pressure curves outside the dome. Discuss Tables B 1a and B 1b for saturated water. The columns are T; P; vf ; vg; uf ; ug; hf ; hf g; hg ; sf ; sf g; sg The subscripts: f and g denote properties at saturated liquid and saturated vapor states. The subscript fg denotes the change from liquid to vapor states.
vf g = vg , vf ; uf g = ug , uf ; hf g = hg , hf ; sf g = sg , sf At critical point where T = 374:136 C; P = 22:08 MPa, vf g = 0; uf g = 0; hf g = 0; sf g = 0. See Tables B 1a and B 1b Quality denoted as x is a dimensionless property of a liquid-vapor mixture. De nition of quality: Mass of Vapor x = Total Mass of Mixture = M MgM g+ f At points on the saturated liquid line x = 0; Mf = M; Mv = 0, and at points on the saturated vapor line x = 1; Mf = 0; Mv = M Under the dome where 0 x 1 the components of the mixture are: Mg = xM; Mf = 1 , xM
Lecture 2
Makeup lecture from 10:00-11:30 AM
Hand out a T , v Diagram for H2 O and Methodology for Energy Analysis. Some Thermodynamic Properties. Consider speci c heat capacities: cv ; cp Speci c heat capacity under constant volume process, cv ! ! @u dT + @u dv u = uT; v; therefore du = @T @v T v For constant volume process dv = 0, and ! ! " @u dT = c T dT; therefore c T = @u dT; kJ du = @T v v @T v kg K v Mean value of cv T over temperature range: T1 T T2 1 Z T2 cv = T , T cv T dT T1 2 1 Speci c heat capacity under constant pressure process, cp ! ! @h dT + @h dP h = hT; P ; therefore dh = @T @P T p For constant pressure process dP = 0, and ! " ! @h dT = c T dT; therefore c T = @h dT; kJ dh = @T p p @T P kg K P Mean value of cpT over temperature range: T1 T T2 1 Z T2 c T dT cp = T , T p T1 2 1 Both cv T and cpT are weak functions of T . Changes in u and h between States 1 and 2:
u2 , u1 = cv T2 , T1 and h2 , h1 = cpT2 , T1
Ideal Perfect Gas
PV = MRT or Pv = RT or
2
P = RT
See Table B:6b, p. 642, for nominal values of properties at low pressures SI Units: cv ; cv ; R; k = cp=cv Real gases behave like an ideal gas if they are dilute or when is small. The density is small when P=Pcritical 1 and T=Tcritical 1 Compressibility Factor Z , De nition Pv Z = RT = 1; For Ideal Gas Real Gases Z = Z T ?; P ? where P ? = P=Pcritical and T ? = T=Tcritical . When the value of Z is close to one, then the real gas can be modeled as an ideal gas. Can water vapor be modeled as an ideal gas given the following data: T1 = 200 F; P1 = 2 psia; v1 = 196 ft3 =lbm From Table B.2, p. 634-637, Tsat = 126 F . From Table B.6, R = 85:78ft lbf=lbm R. Calculation of Z1 2 Z1 = P1v1 = 85:78144 196 = 0:997 RT1 200 + 460 Yes, this water vapor is slightly superheated and it can be modeled as an ideal gas. Relation between speci c heat capacities for ideal gas.
h = u + Pv = u + RT; therefore dh = du + RdT; and cp = cv + R
Also
R k = cp = 1 + c c
v v
1
Under the dome we have the relations:
V = vf Mf + vg Mg = 1 , xMvf + xMvg U = uf Mf + ug Mg = 1 , xMuf + xMug and the intensive properties are obtained from v = 1 , xvf + xvg
3
u = 1 , xuf + xug
Chapter 5: Energy Analysis Principal Tools for Analysis are: 1 Conservation of Mass 2 Conservation of Energy 3 Equation of State Methodology: Control Mass Analysis: , Fixed Mass , Closed System Control Volume Analysis: , Speci ed Volume which may be moving and changing shape , Open System Control Volume Analysis Powerful technique for energy analysis of many important engineering devices such as: , turbines and pumps , compressors , boilers and condensers , pressure vessels , nozzles , thermoelectric devices , etc. Text gives several examples of the Control Mass Analysis: , Evaporation at constant pressure:
Q12 = M P v2 , v1 + u2 , u1 = M h2 , h1
, Dry-Ice Cooler A Sublimation Process, Solid ! Gas
Q12 = M P v2 , v1 + u2 , u1 = M h2 , h1 = M hg , hs = Mhsg where hsg = hg , hs is the Enthalpy of Sublimation. See Fig. 4.2 and 5.2 Illustrate the analysis procedure the for Dry-Ice cooler.
4
Lecture 3
Hand out Project 2. Due Date is Friday, July 23. Steady State, Steady Flow Systems Devices SSSF Assumptions for SSSF Devices 1 Control Volume CV is xed with respect to a coordinate frame 2 Mass ux and state of mass on control volume surface are time independent, _ _ and Min = Mout 3 Mass and its state at every point within the control volume are time independent, and dEcv = d Z e dV = 0 dt dt CV 4 Rates at which heat and work cross the control surface remain constant, and _ _ Q = 0; Wshaf t = 0 dt dt Conservation of Mass for Control Volume dMCV = M , M ; when dMCV = 0; M = M = M _ in _ out _ in _ out _ dt dt Mass Flow Rate " _ = AV kg M s where = mass density, A = ow area, and V = average velocity over ow area. First Law of Thermodynamics for Control Volume FLOT CV h i h i _ _ _ _ M e + Pv 1 + Wshaf t + Q = M e + Pv 2 where the total speci c energy is e, and the other terms represent rates of energy transfer as heat and work. _ _ e = u + ke + pe; Q = dQ ; Wshaf t = dW dt dt and the product Pv is called ow work. The subscripts 1 and 2 denote inlet and outlet respectively. Also note that u and Pv can be combined as the speci c enthalpy: h = u + Pv. 5
Speci c Kinetic and Potential Energies 2 ke = V2 ; pe = gz Uniform State, Uniform Flow Processes USUF Assumptions for USUF Processes 1 Control Volume CV is xed with respect to a coordinate frame 2 State of the mass crossing the control surface is independent of time and it is uniform over the various areas of the control surface where ow occurs. 3 State of the mass within the control volume may change with time, but at any instant in time, the state is uniform over the entire control volume. USUF is useful in the analysis of unsteady processes which involve rapid mixing within the control volume, e.g., , lling of tanks , discharging pressure vessels FLOT CV dECV = d M e = Q + W + hM e + Pvi , hM e + Pvi _ _ shaf t _ _ in out dt dt CV CV Alternative Form of FLOT CV MCV 2 , MCV 1 = Q + Wshaf t + f M e + Pv ing1!2 , f M e + Pv out g1!2 during the time interval: t1 t t2. See the example in text on pages 137-8: Charging of a high-pressure tank. The tutorial will also pertain to this topic.
Lecture 4
Hand out Mollier diagram h-s diagram for water. Simple Power Plant , Pump , Boiler , Turbine , Condenser 6
Show processes on T , v diagram Example of Steam Turbine Example of a SSSF process: a steam turbine. The mass ow rate through a steam turbine is 1:5 kg s, and the heat transfer rate from the boundaries of the turbine is 8:5 kW. Determine...