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Unformatted text preview: Chapter 5 The First Law of Thermodynamics Mixing Chambers and Heat Exchangers 5101C Yes, if the mixing chamber is losing heat to the surrounding medium. 5102C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium. 5103C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium. 5104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is wellinsulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < T sat @ 250 kPa = 127.44C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h 1 2245 h f @ 80 C = 334.91 kJ/kg h 2 2245 h f @ 20 C = 83.96 kJ/kg h 3 2245 h f @ 42 C = 175.92 kJ/kg Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance: m m E m m m in out = = + = system (steady) 1 2 3 Energy balance: 0) pe ke (since 3 3 2 2 1 1 energies etc. potential, kinetic, internal, in change of Rate (steady) system mass and work, heat, by nsfer energy tra net of Rate 2245 2245 = = = + = = = W Q h m h m h m E E E E E out in out in Combining the two relations and solving for m 2 gives ( 29 3 2 1 2 2 1 1 h m m h m h m + = + m h h h h m 2 1 3 3 2 1 = Substituting, the mass flow rate of cold water stream is determined to be ( 29 ( 29 ( 29 kg/s 0.864 = = kg/s 0.5 kJ/kg 83.96 175.92 kJ/kg 175.92 334.91 2 m 5105 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. 590 H 2 O (P = 250 kPa) T 3 = 42 C T 1 = 80 C m 1 = 0.5 kg/s T 2 = 20 C m 2 Chapter 5 The First Law of Thermodynamics Properties Noting that T < T sat @ 300 kPa = 133.55 C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h 1 2245 h f @ 20 C = 83.96 kJ/kg h 3 2245 h f @ 60 C = 251.13 kJ/kg and kJ/kg . 3069 3 C 300 kPa 300 2 2 2 = = = h T P Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the...
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
 Spring '08
 Chung

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