FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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Q1. Heat engines convert internal energy to mechanical energy. Describe the operation of this
reversible engine on a PV diagram and show how to determine the efficiency of the cycle.
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.
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- Figure 5.15 in the text gives a schematic of a Carnot cycle operating with a H2O liquid/vapor with a steady flow (constant mass flow rate) through each component. From the properties given below your cycle may or may not be a Carnot cycle. Kinetic energy and potential energy changes can be ignored in this problem. The cycle conditions are as follows: Process 4 – 1: constant pressure at 300 kPa from saturated liquid to saturated vapor Process 2 – 3: constant pressure at 30 kPa from x2 = 87.9% to x3 = 10.9% a) Determine the thermal efficiency using steam table data b) Compare the result of part a) with the Carnot efficiency using the boiler and condenser temperatures. c) State if the cycle is internally reversible, irreversible or impossiblearrow_forwardAn open feedwater heater is a direct-contact heat exchanger used in vapor power plants. Shown in the figure below are feedwater heater with H20 operating data for an open 31 °C as the working fluid operating at steady state, where T1 P2=3 bar = 0.92 +2 P3 3 bar Saturated liquid m380 kg/s T, PT=3 bar Open feedwater 3 heater Ignoring stray heat transfer from the outside of the heat exchanger to its surroundings and kinetic and potential energy effects, determine the rate of entropy production, in kW/K.arrow_forward6.110 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. wwwarrow_forward
- A closed system undergoes a thermodynamic cycle with 2 steps: process 1-2 (from state 1 to state 2), process 2-1 (from state 2 to state 1). During process 1-2, the system received energy by heat transfer of 25J. During process 2-1, energy was transferred from the system to its surrounding by heat transfer of 15J. This is a power cycle. True or false?arrow_forwardFigure P6.90 shows a simple vapor power cycle operating at steady state with water as the working fluid. Data at key locations are given on the figure. Flow through the turbine and pump occurs isentropically. Flow through the steam generator and condenser occurs at constant pressure. Stray heat transfer and kinetic and potential energy effects are negligible. Sketch the four processes of this cycle in series on a T-s diagram. Determine the thermal efficiency.arrow_forwardAt steady state, Refrigerant 22 enters (1) the compressor at 40C, 5.5bar and is compressed to 60C, 13.8bar. R-22 exiting (2) the compressor enters a heat exchanger where energy transfer to air as a separate stream occurs and the refrigerant exits (3) as a liquid at 13.5bar, 32C. Air enters (4) the condenser at 27C, 1.0bar with a volumetric flow rate of 21.2m3/min and exits (5) at 43C. Assuming ideal gas behavior for the air and stray heat transfer and kinetic and potential energy effects are negligible, determine the compressor powerarrow_forward
- Define the second-law efficiency.arrow_forward10. Carnot cycle thermal efficiency is increased by reducing lower(sink) temperature 11. A system with higher exergy has more capacity to do work than a system with lower exergy. 12. A diffuser has low pressure at inlet and high pressure at outlet 13. Carnot efficiency equation can also be used to find efficiency of conversion of electrical and/or magnetic energy to work. 14. Quality factor of 0.75 indicates supersaturated vapor 15. Law of conservation of Mass and energy applied to Nozzles and Diffusers indicates that under ideal conditions, the change in kinetic energy of the fluid results in a complimentary change in enthalpy of the fluidarrow_forwardThermodynamicsarrow_forward
- 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.arrow_forwardPlease helparrow_forwardExplain the heat transferred from a furnace (Q) to convert the boiler feed water at 25 ° C into superheated steam at 17 bar and 250 ° C.arrow_forward
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