FTFS Chap05 P101 - Chapter 5 The First Law of...

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Chapter 5 The First Law of Thermodynamics Mixing Chambers and Heat Exchangers 5-101C Yes, if the mixing chamber is losing heat to the surrounding medium. 5-102C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium. 5-103C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium. 5-104 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 well-insulated 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 = 127.44°C, 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 h f @ 80 ° C = 334.91 kJ/kg h 2 h f @ 20 ° C = 83.96 kJ/kg h 3 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 steady-flow system can be expressed in the rate form as Mass balance: && & & mm E m in out −= =⎯ ⎯+ = Δ system (steady) Ê 0 12 0 3 Energy balance: 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 · · 0) pe ke (since 0 3 3 2 2 1 1 energies etc. potential, kinetic, internal, in change of Rate (steady) 0 system mass and work, heat, by nsfer energy tra net of Rate Δ Δ = = = + = = Δ = W Q h m h m h m E E E E E out in out in & & & & & & & 4 43 4 42 1 & 1 & & © Combining the two relations and solving for gives & m 2 ( ) 3 2 1 2 2 1 1 h m m h m h m & & & & + = + m hh m 2 13 32 1 = Substituting, the mass flow rate of cold water stream is determined to be () kg/s 0.864 = = kg/s 0.5 kJ/kg 83.96 175.92 kJ/kg 175.92 334.91 2 m & 5-105 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. 5-90
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Chapter 5 The First Law of Thermodynamics Assumptions 1 This is a steady-flow 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. 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 h f @ 20 ° C = 83.96 kJ/kg h 3 h f @ 60 ° C = 251.13 kJ/kg and kJ/kg 3069.3 C 300 kPa 300 2 2 2 = = = h T P o Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: && & & mm m m in out in out −= =→ = →+ = Δ system (steady) Ê 0 12 0 3 Energy balance: & & & & EE E mh Q W in out in out = = += Rate of net energy transfer by heat, work, and mass system (steady)
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This note was uploaded on 04/14/2008 for the course ME 320 taught by Professor Dr.kinne during the Spring '07 term at University of Texas at Austin.

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FTFS Chap05 P101 - Chapter 5 The First Law of...

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