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Past Year Questions: Availability And Irreversibility - Mechanical Engineering MCQ


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6 Questions MCQ Test Thermodynamics - Past Year Questions: Availability And Irreversibility

Past Year Questions: Availability And Irreversibility for Mechanical Engineering 2024 is part of Thermodynamics preparation. The Past Year Questions: Availability And Irreversibility questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Past Year Questions: Availability And Irreversibility MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Past Year Questions: Availability And Irreversibility below.
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Past Year Questions: Availability And Irreversibility - Question 1

A heat reservoir at 900 K is brought into contact with the ambient at 300 K for a short time. During this period 9000 kJ of heat is lost by the heat reservoir. The total loss in availability due to this process is

[1995]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 1

Entropy change for hot reservoir

Energy gain in cold reservoir

Loss in availability = T0 [ ΔSc + ΔSh ]
⇒ 300(30 – 10)
= 300(20)
= 6000 kJ

Past Year Questions: Availability And Irreversibility - Question 2

Availability of a system at any given state is

[2000]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 2

Availability of system of any given state is when no maximum useful work obtainable as the system goes to dead state.

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Past Year Questions: Availability And Irreversibility - Question 3

Considering the relationship TdS = dU + pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which of the following statements is correct?

[2003]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 3

TdS=dU+pdV
This equation holds for for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of path.

Past Year Questions: Availability And Irreversibility - Question 4

A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during this process is to be used as a source of energy. The ambient temperature is 303 K and specific heat of steel is 0.5 kJ/kgK. The available energy of this billet is

[2004]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 4

Heat lost by steel = (0.5)(2000)(450 – 1250)
Gained = 800MJ
ΔSLost by steel = (2000) (0.5)) ln 
⇒ –1.021 MJ
(ΔS) gained = 1.021MJ
AE (W) = a – T0dS
= 800 – (303) (1.021)
= 490.7 MJ

Past Year Questions: Availability And Irreversibility - Question 5

The pressure, temperature and velocity of air flowing in pipe are 5 bar, 500 K and 50 m/s, respectively. The specific heats of air at a constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively.Neglect potential energy. If the pressure and temperature of the surroundings are 1 bar and 300 K, respectively, the available energy in kJ/ kg of the air stream is

[2013]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 5

For flow stream,
A.E. =(h2 – h1) + K.E. – T0 (S2 – S1)

= 187 kJ/kg

Past Year Questions: Availability And Irreversibility - Question 6

The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called

[2014]

Detailed Solution for Past Year Questions: Availability And Irreversibility - Question 6

Exergy (or) Available Energy :
The maximum portion of energy which could be converted into useful work by ideal processes which reduce the system to dead state (a state in equilibrium with the earth and its atmosphere).

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