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# Past Year Questions: Availability And Irreversibility

## 6 Questions MCQ Test Engineering Mechanics | Past Year Questions: Availability And Irreversibility

Description
This mock test of Past Year Questions: Availability And Irreversibility for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 6 Multiple Choice Questions for Mechanical Engineering Past Year Questions: Availability And Irreversibility (mcq) to study with solutions a complete question bank. The solved questions answers in this Past Year Questions: Availability And Irreversibility quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Past Year Questions: Availability And Irreversibility exercise for a better result in the exam. You can find other Past Year Questions: Availability And Irreversibility extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
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]

Solution:

Entropy change for hot reservoir

Energy gain in cold reservoir

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

QUESTION: 2

### Availability of a system at any given state is [2000]

Solution:

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

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]

Solution:

The equation is valid for between reversible or inversible process of a closed system. Because it is the relation between properties which are independent to path.

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]

Solution:

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

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]

Solution:

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

= 187 kJ/kg

QUESTION: 6

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

[2014]

Solution:

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).