!Required informationProblem 09.015 - 3-Step Air-Standard Cycle with Constant Specific Heats - DEPENDENT MULTI-PARTPROBLEM - ASSIGN ALL PARTSAn air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of airand consists of the following three processes:1-2 v= Constant heat addition from 95 kPa and 17°C to 380 kPa2-3 Isentropic expansion to 95 kPa3-1 P=Constant heat rejection to initial stateThe properties of air at room temperature are cp=1.005 kJ/kg-K, cv=0.718 kJ/kg-K, and k=1.4.Problem 09.015.a - P-v and T-s Diagrams for Constant Heat Capacity Air-Standard CycleShow the cycle on P-v and T-s diagrams. (Please upload your response/solution using the controls provided below.) (You must provide an answer before moving on to the next part.)upload a response file (15MB max)Browse... No file selected.save

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Answered: Required information Problem 09.015 -…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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16-4M A proposed air-standard cycle begins at 1.1 bars, 37°C, and 0.808 m³/kg. Process 1-2 is isentro- pic compression to 3.2 bars: process 2-3 is constant-pressure heating to 1050 K and 0.940 m³/kg; process 3-4 is constant-volume cooling to 1.1 bars; process 4-1 is constant-pressure cooling to the initial state, (4) Sketch Pv and Ts diagrams of the cycle. (b) Determine the amount of heat added and the net work output, in kJ/kg. (c) Compute the thermal efficiency. (d) What is the Carnot efficiency for a heat engine with the same maximum and minimum temperatures as the above cycle? Use air-table data.
Required information Problem 09.015 - 3-Step Air-Standard Cycle with Constant Specific Heats - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of air and consists of the following three processes: 1-2 v= Constant heat addition from 95 kPa and 17°C to 380 kPa 2-3 Isentropic expansion to 95 kPa 3-1 P=Constant heat rejection to initial state The properties of air at room temperature are cp=1.005 kJ/kg-K, cv=0.718 kJ/kg-K, and k=1.4. Problem 09.015.c - Cycle Efficiency for Constant Heat Capacity Air-Standard Cycle Determine the thermal efficiency. The thermal efficiency is %.
Required information Problem 09.015 - 3-Step Air-Standard Cycle with Constant Specific Heats - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of air and consists of the following three processes: 1-2 v= Constant heat addition from 95 kPa and 17°C to 380 kPa 2-3 Isentropic expansion to 95 kPa 3-1 P = Constant heat rejection to initial state The properties of air at room temperature are cp = 1.005 kJ/kg-K, cy= 0.718 kJ/kg-K, and k = 1.4. Problem 09.015.b - Net Work for Constant Heat Capacity Air-Standard Cycle Calculate the net work per cycle, in kJ. (You must provide an answer before moving on to the next part.) The net work per cycle is kJ.
Chapter 1 Gas Power Cycles Q1.1 (v6q9-15) An air-standard cycle with variable specific heats is executed in a closed system and is composed of the following four processes: 1-2 isentropic compression from 100kPa and 27C to 800 kPa; 2-3 v=constant heat addition to 1800K; 3-4 isentropic expansion to 100kPa; 4-1 p=constant heat rejection to initial state. (a) Show the cycle on p-v and T-s diagrams. (b) Calculate the net work output per unit mass. (c) Determine the thermal efficiency. Ans: 570.1 kJ/kg, 51.9%
Required information Problem 09.014 - 3-Step Air-Standard Cycle with Variable Specific Heats - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS An air-standard cycle with variable specific heats is executed in a closed system with 0.0025 kg of air and consists of the following three processes: 1-2 v= Constant heat addition from 95 kPa and 17°C to 380 kPa 2-3 Isentropic expansion to 95 kPa 3-1 P = Constant heat rejection to initial state Use data from tables. Problem 09.014.c - Cycle Efficiency for Variable Heat Capacity Air-Standard Cycle Determine the thermal efficiency. The thermal efficiency is %.
Q2:A constant pressure heat addition air standard cycle af a compression ratio of (17.0) .Heat is added at a rate of (1500 kJ/kg) and the atmospheric conditions are (1.0 bar) and (35 °C). Show the cycle on both P-V &T-S diagrams and calculate : 1-the temperature and pressure at each point in the cycle. 2-the thermal efficiency 3-mean effective pressure. [take y=1.4,Cv=0.715 kJ/kg.k & R=0.287kJ/kg.k]
The pressure ratio of a Brayton cycle with an air-powered regenerator is 8. The lowest and highest temperatures of the cycle are 310 K and 1150 K. The adiabatic efficiencies of the compressor and turbine are 75% and 82%, respectively, and the regenerator efficiency is 65%. Show the cycle in the T-sdiagram. Consider the variation of specific heats with temperature. a) The temperature of the air at the exit of the turbine,b) The completion of the cycle,c) Calculate the thermal efficiency of the cycle.
16-3M A proposed air-standard cycle begins at 1 bar, 27°C, and 0.860 m³/kg. Process 1-2 is isentropic compression to 2.75 bar; process 2-3 is constant-pressure heating to 1200°K and 1.25 m³/kg: process 3-4 is constant-volume cooling to 1 bar; process 4-1 is constant-pressure cooling to the initial state. (a) Sketch Pu and Ts diagrams of the cycle. (b) Determine the amount of heat added and the net work output, in kJ/kg. (c) Compute the thermal efficiency. and wong wadi atau m hogy (d) What is the Carnot efficiency for a heat engine with the same maximum and minimum
2. For the gas system power illustrated in Figure 2 (with the air-standard analysis assumptions), calculate... 8 T₁ - 2100 K S=C SEC DMC UNC T₁ - 310 K T₂ - 2100 K Figure 2. Operating condition of a gas power system cycle (The P-V and T-S diagram) a. Thermal efficiency and mean effective pressure of the system b. What happens to the system if the compression ratio increases twice from the initial value? Explain and provide proof in the form of calculation!
"It is important to take your measurements of temperature, pressure and piston height as quickly as possible after each transition stabilizes since " A) the ice melts easily and we want to get the isothermal process for the whole cycle the apparatus does leak air slightly and it is desired that the completed cycle to return to its initial conditions providing a closed PV diagram to analyze (c) the pressure changes drastically and we want to get an isobaric cycle D the hot bath could have a higher temperature and we want to get a lower temperature
Air - Standard Cycles Exercises : Q1: An Otto air standard cycle of a compression ratio of (9.0),the ambient condition are (1.013 bar) and (30°C).The maximum temperature in the cycle is (1700K).Show the cycle on both P-V and T-S diagrams and calculate: 1-the thermal efficiency 2-the mea effective pressure 3- the amount of heat rejected. [take y=1.4 ,Cv=0.715 kJ/kg.k & R=0.287KJ/kg.k] Q2:A constant pressure heat addition air standard cycle af a compression ratio of (17.0) .Heat is added at a rate of (1500 kJ/kg) and the atmospheric conditions are (1.0 bar) and (35 °C). Show the cycle on both P-V &T-S diagrams and calculate : 1-the temperature and pressure at each point in the cycle. 2-the thermal efficiency 3-mean effective pressure. [take y=1.4,Cv=0.715 kJ/kg.k & R=0.287KJ/kg.k] Q3: An air standard Dual cycle has a compression ratio of (20),and the atmospheric conditions are (1.01 bar) and (28 C).If maximum temperature in the cycle is (2200K) while the maximum pressure is (60 bar).…
Required information An air-standard cycle with variable specific heats is executed in a closed system with 0.0015 kg of air and consists of the following three processes: 1-2 v= Constant heat addition from 95 kPa and 17°C to 380 kPa 2-3 Isentropic expansion to 95 kPa 3-1 P= Constant heat rejection to initial state Use data from tables. Determine the thermal efficiency. (Enter the percentage value but without the percentage sign.) The thermal efficiency is %.

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