Example: Electric heater is often used in houses to provide heating during wintermonths. It consists of a simple duct with coiled resistance wires as shown. Considera 20 kW heating system such that the air enters at 100 kPa and 17° C with a massflow rate of 1.8 kg/s. If it is known that the air leaves the duct with an exit temperatureof 27° C (same pressure), determine the heat loss from the duct.dEg/dt=20 kWHeat loss dQ/dt=?To=27° CTo=17° CAssume: steady state, negligibleKE and PE changes, air can beconsidered as idea gas22outdQ()()0dt22dQdQ,(20000)(1.8)()dtdtFrom thermo. table A-19(874), h301.5(/),291.5(/)dQ(1.8)(301.5291.5)(1,000dtginoutgoutinoutinpoutindEdEVVmhgzmhgzdtdtdEmhmhhhdtCTkJkghkJkg=++++−++=+=−+=−====−&&&&)20,0002,000() heat lossW−=−→Transient Energy Balance (Unsteady State)&( )m up Vgzin in+ + +ρ22& ( )m upVgzout out+ + +ρ22Heat in q=dQ/dtWork out dW/dtFirst law of Thermodynamics (Energy Conservation):dEdtdEdtwhere dQdtinternal energy + mechanical energy) =dQdtdEdtThis is the same equation as (4.79), pp. 72 in Potter & Somertongg= + − + − − = −= + ++ + + − + + = − =& &( ) , ( )&&(&( )& ( ) & ( ) ( )E E q qdWdtq qE m m hVgzdEdtm hVgz m hVgzdWdtin out in out in outout in22 222 20E(t) not constant(4.19), p. 157 in Cengel’s book211212Integrate the transient equation in time from 1 to 2:22dQ[()()]()22dt22()[()()]22For uniform flow process: (a) the state of sydEVVdWmhgzmhgzdtdtoutindtdtVVEEhgzhgzdmQWoutin+++−++=−−+++−++=−∫∫∫&&21121212stem analyzed is uniform(b) the fluid flow at the inlet or exit section is uniform and steady22()()()22where E (internal energy of system)=muQ: total heat tranoutoutininVVEEmhgzmhgzQW−+++−++=−12sfer during time from 1 to 2W: total work done during time from 1 to 2Example: Steam at a pressure of 2 MPa and a temperature of 350° C is exitingout from a tank through a valve to drive a turbine as shown. The exhausting steamenters an initially evacuated tank with a volume of 1 m3. The valve is closed whenthe second tank is filled with steam at a pressure of 1.4 MPa and a temperature of 500° C. Assume no significant heat transfer and KE and PE changes are alsonegligible. Determine the amount of work developed by the turbine.Steam2 MPa350 ° CV=1 m3initiallyevacuatedcontrol volumeto be analyzedAssumptions: dQ/dt=0 adiabatic process, KE and PE are negligibleSteam properties at the first tank remains constantThe second tank reaches final equilibrium after the filling process ends22Mass conservationdm since there is no outletdtThe unsteady energy balance equation:dQ()()22dt, adiabatic, no outlet and negligible KEinoutoutininmmmdEVVdWmhgzmhgzdtdtdEdWmhdtdt=−=+++−++=−−=−&&&&&&f & PEIntegrate over time: E=-W+ Initially the second tank is evacuatedE=E,()ininininitialfffinitialffinfdmmhdtWhdtdtWhmEmummmmWmhu∆=−+=−+∆∆−=∆=−==−∫∫&00)(1.63)1.31213137(967.3)(mW)3137(kJ/kghC350T MPa,2 Pfirst tank theAlso,)/(1.3121u table,same theFrom)(967.32521.01/m 0.2521 v853) (p. 6-A tableFrom MPa1.4 and C500 are tank second theof pressure and re temperatufinal The: tablefrom properties determine
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