Hi, can you please help me with this entropy problem? An initially open, rigid, 2-m³ tank is sitting in a large room. The room is filled with air at 300 K and 1 bar.A paddlewheel is then inserted into the tank and the tank is sealed. Some amount of work is then doneby the paddlewheel, under various conditions.a) Accounting for the variability of specific heat with temperature, determine the entropygenerated during this process for W = 600 kJ if the tank is insulated.b) Accounting for the variability of specific heat with temperature, determine the entropygenerated during this process for W = 600 kJ if the tank is not insulated. In this scenario, thetank eventually reestablishes thermal equilibrium with the large room.

(a) Entropy generation is 0 .(b) Entropy generation is −2kJ/K .Explanation:Initial condition,Volume…

Answered: An initially open, rigid, 2-m³ tank is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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1.2 I kg of a fluid is compressed reversibly according to a law pu = 0.25, where p is in bar and r is in m'/kg. The final volume is į of the initial volume. Calculate the work done on the fluid and sketch the process on a p-r diagram.
1. Air initially at 400 K expands in a constant pressure non-flow process until the initial volume is doubled. Find the work in kJ/kg 2. The actual velocity of gas entering in a chimney is 8 m/s the gas is 25°C with the gas constant of 0.287 kJ/kg-K. determine the gas pressure for a mass of a gas is 50,000 kg/hr and chimney diameter of 1.39 m
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 fluid is heated reversibly at a constant pressure of 1.05 bar until it has a specific volume of 0.1 m'/kg. It is then compressed reversibly according to a law pu = constant to a pressure of 4.2 bar, then allowed to expand reversibly according to a law pel.- constant, and is finally heated at constant volume back to the initial conditions. The work done in the constant pressure process is -515 N m, and the mass of fluid present is 0.2 kg. Calculate the net work of the cycle and sketch the cycle on a p-v diagram.
b) A steam turbine operates with 1.6 MPa and 350°C steam at its inlet and saturated vapor at 30°C at its exit as shown in Fig. Q2(b). The mass flow rate of the steam is 22 kg/s, and the turbine produces 12,350 kW of power. By applying the first law of thermodynamics on the steady flow device compute the rate at which heat is lost through the casing of this turbine. 1.6 MPa 350°C 22 kyls Heat Turbine 30°C sat. vapor Fig. Q2(b)
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A closed system undergoes a process in which work is done on the system and heat transfer occurs only at a location on the boundary where the temperature is Tb. Considering an adiabatic an internally reversible process, the system entropy change is O positive negative equal to zero O is not predictable from the info given.
psl answer both i iwill give thumbs up
Consider a container with two chambers connected by a valve. The first chamber has volume, V1, is held at a constant temperature, T, and is filled with an ideal gas. The second chamber has volume, V2, is held at the same temperature, T, and is under vacuum. a) Sketch a picture of this container and its contents. b) Mathematically, how does entropy change as the valve is opened? c) When is the change in entropy maximized during this process?
Hi, pls help me solve this problem. Thank you so much.
In a closed system contains R-134a and it undergoes an isothermal process from state 1 to state 2. The initial tempetature and pressure of R-134a are 320 kPa and 400C respectively. The final quality of R-134a is 50 percent. Find the specific work and heat transfer during the isothermal process. a) W = -27.65KJ/kg and q = -130.075 KJ/kg b) W = -47.65KJ/kg and q = -13.075 KJ/kg c) W = -47.65KJ/kg and q = -130.075 KJ/kg d) W = -147.65KJ/kg and q = -10.075 KJ/kg do it fast so that, I can rate you..

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