Q1: The Clausius inequality holds good for [GATE ME 2022, SET-1]
(a) any process
(b) any cycle
(c) only reversible process
(d) only reversible cycle
Ans: (b)
The Clausius inequality holds good for any cycle.
= 0 ⇒ Reversible cycle
< 0⇒ Irreversible cycle
>0⇒ Impossible cycle
Q1: Keeping all other parameters identical, the Compression Ratio (CR) of an air standard diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle rc = 2
The difference between the new and the old efficiency values, in percentage,
(ηnew|CR = 21) - (ηold |CR = 15) = _______ %. (round off to one decimal place) [GATE ME 2020, SET-2]
Ans: (4.6 to 4.9)

= 4.8%
Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself: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]
Explanation
Entropy change for hot reservoir

Energy gain in cold reservoir

Loss in availability = T0 [ ΔSc + ΔSh ]
⇒ 300(30 – 10)
= 300(20)
= 6000 kJ
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Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself:Availability of a system at any given state is
[2000]
Explanation
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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Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself: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]
Explanation
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.
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Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself: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]
Explanation
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
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Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself: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]
Explanation
For flow stream,
A.E. =(h2 – h1) + K.E. – T0 (S2 – S1)

= 187 kJ/kg
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Question for GATE Past Year Questions: Availability & Irreversibility
Try yourself:The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called
[2014]
Explanation
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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