FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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Steam at 44 bar and a dryness fraction, x = 0.9 is throttled to a pressure of 12 bar. Calculate thedifference in power output in kilowatts between the following two expansion processes:a) Steam at the initial pressure of 44 bar and x = 0.9 at State 1 is expanded in a turbine to State 3 at 0.12 bar.b) Steam at the reduced pressure of 12 bar after throttling at State 2 is expanded in another turbine to State 4 at the same exhaust pressure of 0.12 bar.The mass flow rate of steam is 8 kg/sec in both cases and the expansion in both turbines can be assumed to be reversible and adiabatic. Sketch both expansion processes on the same T-s diagram using the respective initial and final state points as described above.Explain the reason for the difference in power output.Calculate the mass flow rate of steam for the turbine operating at the throttled/reduced pressure to generate the same output as the turbine operating at the pressure before throttling.NOTE: You are required to number the state…
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- An oxygen gas R = 0.2598 KJ/kg°k and k = 1.395. If 4 kg of oxygen undergo a reversible non flow constant pressure process from initial volume =1.2 cubic meter and initial pressure = 690 kPa to a state where final temperature = 600°C. 1. Determine the Change in Internal Energy. choices: a.200.60 KJ. b.198.45 KJ. c.99.54 KJ. d.200.55 KJ 2. Determine the constant pressure-specific heat. choices: a.0.9865 KJ/kg-°K. b.0.9175 KJ/kg-°K. c.0.8580 KJ/Kg-°K. d.0.7843 KJ/kg-°K need complete solution, cancellation and symbol:)arrow_forwardThe following processes occur in a reversible thermodynamic cycle: 1-2: 0.2 kg heating at constant pressure 1.05 bar at specific volume 0.1 m/kg and work done -515 J. 2-3: Isothermal compression to 4.2 bar. 3-4: Expansion according to law pv1./= constant. 4-1: heating at constant volume back to the iniüal conditions. Calculate the work done for the constant volume heating process? Moving to another question will save this response. «< Questio acar A s P I R Earrow_forwardExample 5.23. When a closed system executes a certain non-flow process, the work and heat interactions per degree rise in temperature at each temperature attained are given by SW = (4 0.08 T) kJ/kg and 8Q = 1.00 kJ/K dT dT Make calculations for the increase or decrease in the internal energy of the system if it is to operate between the temperature limits of 200 °C and 400 °C.arrow_forward
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