Chapter 6, Problem 6.118P

Rankine cycle/ engine (Thermodynamics) Show complete and step by step solution and the diagram. Answer as soon as possible. Note it is 750°K

Problem 5.060 SI The figure shows the schematic of a vapor power plant in which 100 kg/s of water circulates through the four components operating at steady state. The water flows through the boiler and condenser at constant pressure and through the turbine and pump adiabatically. Kinetic and potential energy effects can be ignored. Process data follow: Process 4-1: constant-pressure at 4000 kPa from saturated liquid to saturated vapor. Process 2-3: constant-pressure at 20 kPa from x2 = 88% to x3 = 18%. m = 100 kg/s Boiler X2 = 88% X3 = 18% Water Pump Turbine Condenser 2 Determine thermal efficiency. Determine ởrvle, in kW/K. Determine if the cycle is internally reversible, irreversible, or impossible. v Step 1 Determine the cycle thermal efficiency. % the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.…

4. In part of a Carnot cycle, water undergoes two internally reversible processes shown in the image. Determine the heat transfer (in kJ/kg) for each process. pi =1 MPa, T2 = 400 °C, p2 = 2 MPa. %3D %3D 3.

Thermodynamics subject power cycles. Show the ts diagram and complete and step by step solution. Bix the final answer.

Q.1 A piston-cylinder device operates 1 kg of fluid at 20 bar pressure. The initial volume is 0.1 (the last digit of your id.no). The fluid is expand reversibly following the process pv¹.3 = constant. at point B the volume becomes 1.5 m³. The fluid is then cooled at a constant temperature. Calculate the work done in the cycle. p (bar) 0.1 last digit of your id. no 0.1 v cu.m Py ¹.3 B 1.5

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…

The Figure shows a simple vapor power plant operating at steady state with water circulating through the components. Relevant data at key locations are given on the figure. The mass flow rate of the water is 90 kg/s. Kinetic and potential energy effects are negligible as are all stray heat transfers. Determine a. The heat added in boiler to the water b. If the combustion efficiency is 85%, find the mass of diesel fuel combustion rate in kg/day if CVDiesel =52 MJ/kg. c. The output turbine power in kW.

6.110 Figure P6.110 shows a simple vapor power plant operating at steady state with water as the working fluid. Data at key locations are given on the figure. The mass flow rate of the water circulating through the components is 109 kg/s. Stray heat transfer and kinetic and potential energy effects can be ignored. Determine a. the net power developed, in MW. b. the thermal efficiency. c. the isentropic turbine efficiency. t2 d. the isentropic pump efficiency. e. the mass flow rate of the cooling water, in kg/s. f. the rates of entropy production, each in kW/K, for the turbine, condenser, and pump. P = 100 bar T = 520°C %3D Power out Turbine P2 = 0.08 bar 2 = 90% %3D Steam Cooling water in at 20°C generator Condenser Pa= 100 bar T= 43°C Cooling water out at 35°C 4. Pump 3 P3 0.08 bar Saturated liquid Power in FIGURE P6.110 2. www

The first step of a thermodynamic cycle is an isobaric process with increasing volume. The second is an isochoric process, with decreasing pressure. The last step may be either an isothermal or adiabatic process, ending at the starting point of the isobaric process. Sketch a graph of these two possibilities, and comment on which will have greater net work per cycle.