I need the answer quickly ExExample 8.14 A compressor operating at steady state takes in 1 kg/s of air at 1bar and 25°C and compresses it to 8 bar and 160°C. Heat transfer from the compres-sor to its surroundings occurs at a rate of 100 kW. (a) Determine the power input inkW. (b) Evaluate the second law efficiency for the compressor. Neglect K.E. and P.E.changes. Take To = 25°C and Po = 1 bar.Sc%3D

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Answered: EX Example 8.14 bar and 25°C and…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A cycle works with one kg of air and is compressed through a polytropic process (pv12 = C) from state-l at 120 kPa, 350K to state-2 of 600kPa, following a constant pressure cooling process to state-3, then constant temperature heating from state-3 to state-1. Represent the cycle 3 to: on P-V diagram, find the temperature, pressure and volume at the end of each process of the cycle, and find also the net heat and work done.
1:Heat flows from a hot reservoir at 800 K to another reservoir at 250 K. H entropy change of the overall process is 4.25 kJ/K, make calculations for the heat flowinger of the high temperature reservoir.
A turbine, operating under steady- flow conditions, reccives 1000 kg/min of stcam. At the inlet, the pressure is 30 bar, the temperature is 400°C, the velocity At the exit, the pressure is 0.7 bar, the quality is (100%), and the velocity is 100 m/s. If the turbine produced a power output of 9300 KW. By using the energy balance of open system with sutable tables, answer the following: (a) What are the main assumptions ? (b) Calculate dh, AKe ? (c) Calculate the rate of heat transfer between the turbine and surroundings, in kW.
A cycle works with one kg of air and is compressed through a polytropic process (pv12 = C) from state-1 at 120 kPa, 350K to state-2 of 600kPa, following a constant pressure cooling process to state-3, then constant temperature heating from state-3 to state-1. Represent the cycle on P-V diagram, find the temperature, pressure and volume at the end of each process of the cycle, and find also the net heat and work done.
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Moodle in the Cloud 0% In turbochargers a turbine is directly connected to a compressor. The power harvested by the turbine is used to power the compressor. In the figure below a turbine runs on steam and powers a compressor on air. Air can be considered a perfect gas with k = 1.34, R = 0.28 kJ/kgk and cp = 1.02 kJ/kgK. The air enters the compressor at p₁ 1bar, T₁ = 300K and a mass flow rate of mdot = 5.9 kg/s and exits at p2 = 5.2 bars. The steam enters the isentropic turbine at P3 = 2 MPa, T3 = 320°C and exits at p4 = 100 kPa. = State 2 Isentropic Compressor State 1 Air - perfect gas T₁ = 300K P₁= 1 bar Get Help Answer: W = ? tubine State 4 Isentropic Turbine State 3 Steam T,=320°C P3= 2MPa What is the power produced by the turbine? Please enter your answers in kW.
There are required 1902.3 kW of compressor power to handle air adiabatically from 1 atm, 26.7 C to 304.06 kPaa. The initial velocity air velocity is 21 m/s and the final velocity is 85 m/s. a). If the process is isentropic, find the volume of air handled, m3/min measured at inlet conditions. b). If the compression is an irreversible adiabatic to a temperature of 157.2 C, with the capacity found in (a), find the power input, kW
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Thermodynamics: Please show step by step solution for the following Practice problem
200Kg/min superheated steam at 40 bar and 350C enters a turbine through a 7.5cm ID pipe. It exits at 5bar as saturated steam through a 5cm ID pipe neglect the change in potential energy of the system. What is the temperature of the outlet saturated system? How much energy is transferred to or from the turbine?
1. Two kilograms of air at a pressure of 2.25 MPa and temperature of 27C is heated atconstant volume to a final pressure of 2.5 MPa. Compute for the:a. Change in internal energyb. Change in enthalpyc. Change in entropyd. Heat transferrede. Non-flow work 2. In a constant pressure process, 325 KJ of heat are added to 6.55 kg of ideal gas(R=208 J/kg-K and k = 1.65) initially at 290 K. Calculate:a. Final Temperatureb. Change in enthalpyc. Change in internal energyd. Change in entropye. Non-flow work
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