Chapter 1Introduction and basic conceptsNewtons second lawF = Ma (N)WeightW = mg (N) 1J = 1NmDensity = m V kg m3Specic volumev =V m=1 Specic weights =g N m3Kelvin to CelciusT(K) = T(C)+273.15T(K) =T(C)Rankine to FahrenheitT(R) = T(F)+459.67T(R) =T(F)(R) = 1.8T(K) T(F) = 1.8T(C)+32 1Pa = 1 N m2 1bar = 105Pa = 0.1MPa = 100kPaAbsolute, gage and vacuum pressurePgage = PabsPatm Pvac = PatmPabs1The pressure at depth h from the free surface isP = Patm+gh or Pgage =ghRelation for the variation of pressure with elevationdP dz =g P = P2P1 =2 R 1gdzThe atmospheric pressure is measured by a barometer and is given byPatm =ghChapter 2Energy, Energy Transfer and General Energy AnalysisThe total energy of a system on a unit mass basise =E m kJ kgKinetic EnergyKE = mV2 2(kJ)Kinetic Energy on a unit mass basiske =V2 2 kJ kgPotentional EnergyPE = mgz (kJ) Potentional Energy on a unit mass basis pe = gz kJ kg Total Energy of a system E =U +KE+PE =U +m V2 2+mgzTotal Energy of a system on a unit mass basise = u+ke+pe = u+V2 2+gzMass ow rate m = V =AcVavg kg sEnergy ow rate E = me kJ sor kW3Mechanical Energy of a owing uid on a unit mass basisemech =P +V2 2+gzMechanical Energy of a owing uid expressed in rate form Emech = memech = mP +V2 2+gz Mechanical Energy change of a uid during incompressible owemech = P2P1 +V2 2V2 1 2 +g(z2z1) kJ kgAnd Emech = memech = mP2P1 + V2 2V2 1 2 +g(z2z1) (kW) Heat transfer per unit mass of a systemq =Q m kJ kgAmount of heat transfer during a processQ =t2 R t1 Qdt (kJ)When Q remains constantQ = Qt (kJ)Work done per unit mass of a systemw =W m kJ kg The total volume change during a process between states 1 and 2 2Z 1 dV =V2V1 =V The total work done during process 1-2 2Z 1 W =W12 (NotW) Electrical work (where N is the amount of coulombs andV V V is a potentional difference)We =V V VNElectrical work expressed in rate form (Electrical Power) We =V V VI (W) Electrical work done during time intervaltWe =2Z 1V V VIdt (kJ)Work done by a constant forceW = Fs (kJ)Work done by a not constant forceW =2Z 1Fds (kJ)TorqueT = FrF =T r This force acts through a distance s, which is related to the radius r bys = (2r)nShaft WorkWsh = Fs =T r(2rn) = 2nT (kJ) Power transmitted through the shaft Wsh = 2nT (kW)Spring WorkWspring = FdxTotal spring workWspring =1 2k(x2 2x2 1) (kJ) Work associated with the expansion or contraction of a solid barWelastic =2Z 1Fdx =2Z 1nAdx (kJ)Work associated with the stretching of a lm (also called surface tension work)Wsurface =2Z 1sAdx (kJ)Energy balanceEinEout =Esystem The change in the total energy of a system during a process (in the absence of electric, magnetic and surface tension effects)E =U +KE+PEWhereU = m(u2u1) KE = 1 2m(V2 2 V2 1 ) PE = mg(z2z1)Energy balance more explicitly EinEout = (QinQout)+(WinWout)+(Emass,inEmass,out) =Esystem Energy balance for any system undergoing any kind of process EinEout =Esystem (kJ) Or in the rate form Ein Eout = dEsystem dt (kW) For constant rates, the total quantities during a time intervalt are related to the quantities per unit time as Q = Qt (kJ) W = Wt (kJ) E = dE dt t (kJ) Energy balance on a unit mass basis eineout =esystem kJ kg Energy balance in differential form EinEout = dEsystem or eineout = desystem The Energy balance for a cycle Wnet,out = Qnet,in or Wnet,out = Qnet,inPerformance or efciencyPerformance =Desiredoutput RequiredinputCombustion efciency combustion = Q HV=Amount of heat released during combustion Heating value of the fuel burnedMechanical efciencymech =Emech,out Emech,in= 1Emech,loss Emech,inPump efciencypump = Emech,fluid Wshaft,in= Wpump,u WpumpTurbine efciencyturbine = Wshaft,out | Emech,fluid|= Wturbine Wturbine,eMotor efciencymotor = Wshaft,out Welect,inGenerator efciencygenerator = Welect,out Wshaft,in Combined efciency of a pump-motor combinationpumpmotor =pumpmotor = Wpump,u Welect,in= Emech,fluid Welect,inCombined efciency of a turbine-generator combinationturbinegenerator =turbinegenerator = Welect,out Wturb,in= Welect,out | Emech,fluid|Rate of heat conduction Qcond = ktAT xIn the limiting case ofx0 (Fouriers law) Qcond =ktAdT dx(W)Rate of heat transfer by convection Qconv = hA(TsTf) (W)Maximum rate of radiation Qemit,max =AT4 sRadiation emitted by a real surface Qemit =AT4 s (W)Rate at which a surface absorbs radiation Qabs = Qincident (W)Net rate of radiation heat transfer Qrad =A(T4 s T4 s ) (W)Chapter 3Properties of Pure SubstancesThe quality x as the ratio of the mass of vapor to the total mass of the mixture (for saturated mixtures only) x = mvapor mtotal Where mtotal = mliquid +mvapor = mf +mgThe total volume in a tank containing a saturated liquid-vapor mixture isV =Vf +Vg V = mvmtvavg = mfvf +mgvg mf = mtmg mtvavg = (mtmg)vf +mgvgDividing my mt yieldsvavg = (1x)vf +xvg Since x = mg/mt. This relation can also be expressed as vavg = vf +xvfg m3 kg Where vfg = vgvg. Solving for quality we obtain x = vavgvf vfg Theanalysisgivenabovecanberepeatedforinternalenergyandenthalpywiththefollowingresults uavg = uf +xufg kJ kgh avg = hf +xhfg kJ kg All the results are of the same format, and they can be summarized in a single equation asyavg = yf +xyfgWhere y is v, u or h.9Ideal-gas Equation of StateP = RT vPv = RT The gasconstant R is determinded fromR =Ru M kJ kgKorkPam3 kgK Where Ru is the universal gas constant The mass of a systemm = MN (kg)The ideal-gas Equation of State can be written in several different forms V = mv PV = mRT mR = (MN)R = NRu PV = NRuT V = N v P vRuT The properties of an ideal gas at two different states are related to each other by P1V1 T1 = P2V2 T2 Compressibility factor Z = Pv RT or Pv = ZRTGases behave differently at a given temperature and pressure, but they behave very much the same at temperatures and pressure normalized with respect to their critical temperatures and pressures. The normalization is done as PR = P Pcr and TR = T Tcr Pseudeo-reduced specic volume vR = vactual RTcr/Pcr Van der Waals Equation of State P+ a v2(vb) = RT The determination of the two constants appearing in this equation is based on the observation that the critical isotherm on a Pv diagram has a horizontal inection point of the cricital point. Thus, the rst and the second derivatives of P with respect to v at the critical point must be zero. That is P vT=Tcr=const = 0 and 2P v2T=Tcr=const = 0By performing the differentiations and eliminating vcr, the constants a and b are determined to bea =27R2T2 cr 64Pcrand b =RTcr 8PcrBeattie-Bridgeman Equation of StateP = RuT v2 1 c vT3( v+B) A v2whereA = A01a v and B = B01b vBenedict-Webb-Rubin Equation of StateP =RuT v+B0RuTA0C0 T21 v2 +bRuTa v3 +a v6 +c v3T21+ v2e v2Virial Equation of StateP =RT v+a(T) v2 +b(T) v3 +c(T) v4 +d(T) v5 +Vapor PressurePatm = Pa+Pv