CHAPTER7_SECTION3 - Chapter 7 Entropy Reversible...

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Unformatted text preview: Chapter 7 Entropy Reversible Steady-Flow Work 7-78C The work associated with steady-flow 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. 7-79C 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. 7-80C We would not support this proposal since the steady-flow 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. 7-52 Chapter 7 Entropy 7-81 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 A-5). Analysis The power input to the pump can be determined directly from the steady-flow 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 7-82 Liquid water is to be pumped by a 10-kW 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 A-5). Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversible steady-flow 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 7-53 PUMP P 2 100 kPa 10 kW H 2 O 2 1 45 kg/s Chapter 7 Entropy 7-83E Saturated refrigerant-134a 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.

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CHAPTER7_SECTION3 - Chapter 7 Entropy Reversible...

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