Week 5. Energy Analysis of Closed Systems. GENESYS Laboratory

Transcription

1 Week 5. Energy Analysis of Closed Systems

2 Objectives 1. Examine the moving boundary work or PdV work commonly encountered in reciprocating devices such as automotive engines and compressors 2. Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems 3. Develop the general energy balance applied to closed systems 4. Define the specific heat at constant volume and the specific heat at constant pressure 5. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases 6. Describe incompressible substances and determine the changes in their internal energy and enthalpy 7. Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances

3 1 st law of Thermodynamics Control mass (Closed System) Control volume (Opened System) Moving boundary (closed System) E mass =PV D W + D Q = DU + D KE + DPE If your system is a stationary system D W + D Q + DE = DU + D KE + DPE mass D W + D Q = D H + DKE + DPE D W + D Q = D U D W + D Q + DE ( DE ) = DU DW + DQ = DH mass boundary

4 Moving Boundary Work Moving boundary work: the expansion and compression work d W = Fds = PAds = PdV b W b = ò 1 2 PdV (kj) dv > 0 : expansion W > 0 dv < 0 : compression W<0 The area under the process curve on a P-v diagram represents the boundary work Area = A 1 2 = ò da = ò 1 2 PdV (kj)

5 Summary - Work + Work

6 Moving Boundary Work II The boundary work done during a process depends on the path followed as well as the end states Quasi-equilibrium (Quasi-static) process - a process during which the system remains nearly in equilibrium at all times - reversible process - idealized process and is not a true representation of actual process The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used

7 Quasi-equilibrium vs. Non-quasi equilibrium (Compression) Fast change Slow change

8 Quasi-equilibrium vs. Non-quasi equilibrium (Expansion) Fast change Slow change

9 Polytropic Process Work is dependent on detailed process In polytropic process, Pv n =constant ü n=1 ; isothermal process (T=const.) ü n=0 ; isobaric process (P=const.) ü n= ; isovolumetric process (V=const.) ü n=κ=c p /C v ; isentropic process (s=const.)

10 Summary Here is a tip! 1) Check your system (A piston-cylinder device vs. a rigid tank) - A piston-cylinder device à moving boundary (W b ) - A rigid tank à fixed boundary (No W b ) 2) Note the type of fluid (Water, steam, R-134 vs. Other gases) - Water, steam, R-134 à Use property table - Other gases, air à Use an ideal gas equation

11 Energy Balance for Closed Systems Energy balance (or the first law) E - E = DE (kj) in out system The rate form E& in - E& out = desystem dt (kw) For a closed system undergoing a cycle E - E = DE = 0 E in out system in = E out The energy balance in terms of heat and work interactions Q -W = DE or Q -W = DE net, in net, out system where Q = Q - Q, W = W -W net, in in out net, out out in

12 Specific Heats A property that enables to explain the energy storage capabilities of various substances The energy required to raise the temperature of a unit mass of a substance by one degree - Specific heat at constant volume (C v ) as the volume is maintained constant -Specific heat at constant pressure (C p ) as the pressure is maintained constant

13 Specific Heats (Continue) d e - d e = du C V d e C P in in out dt = du dt - d e = du + PdV out = dh

14 Different Specific Heat I am burnt 21 C What the heck!! Piece of cake 20 C Oh, Come on Temperature

15 Internal Energy, Enthalpy, And Specific Heats of Ideal Gases For ideal gas, u is a function of temperature only according to Joule h = u + Pv where Pv = h = u + RT RT Since R is constant and h = h( T ) du = C ( T ) dt V dh = C ( T ) dt P u = u( T )

16 Specific Heats (Continue) Three ways to determine the Δu and Δh of ideal gas -Using the tabulated u and h data -Using the C v or C p relations as a function of temperature and performing the integrations. -Using average specific heats if ΔT is small ò ò D u = C ( T ) C ( T -T ) v v, av 2 1 D h = C ( T ) C ( T -T ) P P, av 2 1

17 Specific Heats (Continue) Specific heat relations of ideal gases h = u + Pv where Pv = RT h = u + RT dh = du + RdT dh du = + R dt dt C p = C v + R Specific heat ratio k = C p /C v for monatomic gases for diatomic gases

18 u, h, and c of incompressible substance A substance whose specific volume is constant during a process (e.g. solid, liquid) The constant-volume and constant-pressure specific heats are identical for incompressible substances The specific heat of incompressible substances depend on temperature only Internal Energy Changes du = c dt = c( T ) dt v Du = u c - u avg 1 ( T 2 = ò 1 2 -T ) 1 c( T ) dt c p = cv = c

19 u, h, and c of incompressible substance II Enthalpy Changes dh = du + vdp + Pdv Dh = Du + c avg DT + vdp = du + vdp Solid D h = D c avg D T Liquid (if D P = 0) : D h = D c DT (if D T = 0) : D h = vdp avg For a process between states 1 and 2, the last relation can be expressed as By taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same temperature, the enthalpy of the compressed liquid can be expressed as h + v( P - P ) h + v ( P - P P, T T T T Since ( P - P ) << 0 \ h T T