Review: Problem Solving in IsometricProcess 03

(Last Updated On: December 18, 2017)

Review: Problem Solving in IsometricProcess 03

Problem Statement:

There are 2.33 kg of gas for which R = 377 J/kg.°K and k = 1.25, that undergo a nonflow constant volume process from P1 = 657.8 kPa and T1 = 66°C to P2 = 1745 kPa. During the process the gas is internally stirred and there are also added 108.25 kJ of heat. Determine (a) T2 (b) the work input and (c) the change of entropy.

Course: Thermodynamics

TOPIC: Processes of Ideal Gas

SUBTOPIC: Isometric Process

DEFINITION: An Isometric Process is a reversible constant volume process. A constant volume process maybe reversible or irreversible.

Problem Answer:

a.) T2 = 907.97°K b.) Wn = -1890.38 kJ c.) ∆S = 3.46 kJ/°K

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Review: Problem Solving in IsometricProcess 03
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