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Unformatted text preview: Chapter 7 Entropy Reversible SteadyFlow Work 778C The work associated with steadyflow devices is proportional to the specific volume of the gas. Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the compressor. 779C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the work output of the turbine. Therefore, this is not a good proposal. 780C We would not support this proposal since the steadyflow work input to the pump is proportional to the specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly. 752 Chapter 7 Entropy 781 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to the pump is to be determined. Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. Properties The specific volume of saturated liquid water at 20 kPa is v 1 = v f @ 20 kPa = 0.001017 m 3 /kg (Table A5). Analysis The power input to the pump can be determined directly from the steadyflow work relation for a liquid, ( 29 1 2 1 2 1 P P v m pe ke vdP m W in = + + = Substituting, kW 274 =  = 3 3 m kPa 1 kJ 1 kPa ) 20 6000 )( /kg m 0.001017 )( kg/s 45 ( in W 782 Liquid water is to be pumped by a 10kW pump at a specified rate. The highest pressure the water can be pumped to is to be determined. Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is assumed to be reversible since we will determine the limiting case. Properties The specific volume of saturated liquid water at 20 kPa is v 1 = v f @ 20 kPa = 0.001017 m 3 /kg (Table A5). Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversible steadyflow work relation for a liquid, Thus, ( 29  = = + + = 3 2 3 1 2 1 2 1 m kPa 1 kJ 1 Pa k ) 100 )( /kg m 0.001017 )( kg/s 5 ( kJ/s 10 P P P v m pe ke vdP m W in It yields P 2 = 2100 kPa 753 PUMP P 2 100 kPa 10 kW H 2 O 2 1 45 kg/s Chapter 7 Entropy 783E Saturated refrigerant134a vapor is to be compressed reversibly to a specified pressure. The power input to the compressor is to be determined, and it is also to be compared to the work input for the liquid case. Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The compressor is adiabatic....
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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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