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Unformatted text preview: 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 = 127.44C, 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 steady-flow 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 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. 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. 5-90 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