ME 200 L23: Clausius Inequality and Control Volume Example Problems Kim See’s Office ME Gatewood Wing Room 2172 Please check your HW Grades on Blackboard Please return all previously graded Homework(s) today with HW8 ThermoMentor © Program Launched Quiz 2 performance suggests it is making a difference, Let Examination 2 prove that too. Spring 2014 MWF AM J. P. Gore Gatewood Wing 3166, Office Hours: MWF TAs: Robert Kapaku Dong Han

Clausius Inequality ►The Clausius inequality is developed from the Kelvin-Planck as: ∫ (Eq. 5.13)  cycle = 0 no irreversibilities present within the system  cycle > 0 irreversibilities present within the system  cycle < 0 impossible Eq. 5.14

Example: Use of Clausius Inequality Q H =1000 kJ, T H =500 K and Q L =600 kJ at (a) 200 K, (b) 300 K, (c) 400 K. Find if each cycle is reversible, irreversible or ideal. Solution: Use the given QH, QL values to find work and ensure that the work produced does not result in a negative value for  cycle ∫

Example: Use of Clausius Inequality (b)  cycle = 0 kJ/K = 0 (a)  cycle = +1 kJ/K > 0 Irreversibilities present within systemNo irreversibilities present within system (c)  cycle = –0.5 kJ/K < 0 Impossible

Review for Examination 2 5 –Control Mass and Control Volume –Simple Compressible Substance: State Principle –Conservation of Mass –First Law of Thermodynamics or Conservation of Energy –Property Relations Subcooled or Saturated Solid, Subcooled or compressed liquid and Saturated Liquid, Saturated Liquid Vapor Mixture, Superheated Vapor, Ideal Gases p-V-T, p-v-T, and p-V-Z-T relations Internal Energy, Enthalpy, Entropy –SI and British System of Units. Make Sure lbm versus lbf is managed with the 32 ft/s 2 factor properly –Boundary Work versus Shaft Work –Heat Transfer and Entropy relationship –Reversible processes: Internally reversible, Externally reversible

Review for Examination 2 6 –High Temperature Reservoir, Low Temperature Reservoir –Second Law of Thermodynamics –Efficiency and Coefficient of Performance –Carnot Engine and Carnot Heat Pump –Control Volumes Nozzles Diffusers Compressors Pumps Turbines Heaters Heat Exchangers –Integrated Control Volumes such as a Pump feeding into a Boiler which feeds into a Turbine which feeds into a Condenser

Example Problem: Nozzle 7 Given: Air at 800 K expands to an exit temperature of 660 K. The inlet velocity is sufficiently low to not contribute significantly to the total energy. Find: The exit velocity. Assumptions: Change in PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved.

Example Problem: Diffuser 8 Given: Steam at 100 o C, 1 bar is pressurized through a diffuser to 1.5 bars, 120 o C and negligible velocity. Find the inlet velocity. Find: The inlet velocity. Assumptions: Change in PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved.

Example Problem: Compressor 9 Given: Air is compressed from 1 bar, 300 K to 10 bars, 800 K by a compressor using 550 kJ/kg of electrical work input. Find: Heat transferred to the cooling fluid. Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved.

Example Problem: Heater/Combustor 10 Given: Air at 800 K and 10 bars is heated to 1400 K by heat addition from a combustor. Find: Find the heat added by the combustor. Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, No work done.

Example Problem: Turbine 11 Given: Air at 1400 K and 10 bars is expanded to 900 K by a turbine. Find: Find the work output per unit mass of air if the process is adiabatic. Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, Heat Transfer is negligible. Consider the compressor, combustor, and turbine on slides 11, 12, and 13 as a system: Net work =

Example Problem: Heat Exchanger 12 Given: 0.6 kg/s of air at 2000 K flows through a counter-flow HX and exits at 1000 K. On the other side of the HX, 1 kg/s of air is heated from 800 K to 1400 K. Find: The heat loss to the surroundings. Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, No work done.