CLOSED system, i.e. non-flow processMaximum useful work (Wu)max = Wmax - Watm = Wmax - Po (V2 - V1)where, Po = surrounding or atmospheric pressure, Wmax = the max work that can be extracted when the system is expandingand Watm = Po(V2 - V1) = the work DONE by the system against the surrounding having a flexible BOUNDARY, for a rigid boundary Watm = 0.For a very slow expansion i.e. an isothermal expansion Q = Wmax = P (V2 - V1) as ΔU = 0. (Wu)max = P (V2 - V1) - Po (V2 - V1) = (P - Po) ΔV Calculation:Given, Po = 100 kPa, P = 300 kPa, ΔV = 0.01 m3 from, (Wu)max = (P - Po) ΔV (Wu)max = ( 300-100) x 0.01 = 2kJ For a closed system, maximum useful work is the availability, available energy or exergy.

"> CLOSED system, i.e. non-flow processMaximum useful work (Wu)max = Wmax - Watm = Wmax - Po (V2 - V1)where, Po = surrounding or atmospheric pressure, Wmax = the max work that can be extracted when the system is expandingand Watm = Po(V2 - V1) = the work DONE by the system against the surrounding having a flexible BOUNDARY, for a rigid boundary Watm = 0.For a very slow expansion i.e. an isothermal expansion Q = Wmax = P (V2 - V1) as ΔU = 0. (Wu)max = P (V2 - V1) - Po (V2 - V1) = (P - Po) ΔV Calculation:Given, Po = 100 kPa, P = 300 kPa, ΔV = 0.01 m3 from, (Wu)max = (P - Po) ΔV (Wu)max = ( 300-100) x 0.01 = 2kJ For a closed system, maximum useful work is the availability, available energy or exergy.

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A gas expands in a frictionless piston – cylinder arrangement. The expansion process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work that could be utilized from the above process is

Fluid Mechanics First Law Thermodynamics in Fluid Mechanics 7 months ago

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Concept:For a CLOSED system, i.e. non-flow processMaximum useful work (Wu)max = Wmax - Watm = Wmax - Po (V2 - V1)where, Po = surrounding or atmospheric pressure, Wmax = the max work that can be extracted when the system is expandingand Watm = Po(V2 - V1) = the work DONE by the system against the surrounding having a flexible BOUNDARY, for a rigid boundary Watm = 0.For a very slow expansion i.e. an isothermal expansion Q = Wmax = P (V2 - V1) as ΔU = 0. (Wu)max = P (V2 - V1) - Po (V2 - V1) = (P - Po) ΔV Calculation:Given, Po = 100 kPa, P = 300 kPa, ΔV = 0.01 m3 from, (Wu)max = (P - Po) ΔV (Wu)max = ( 300-100) x 0.01 = 2kJ For a closed system, maximum useful work is the availability, available energy or exergy.

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Posted on 22 Nov 2024, this text provides information on Fluid Mechanics related to First Law Thermodynamics in Fluid Mechanics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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