Engineering Thermodynamics: Unit III: Availability and Applications of II Law

Expressions for energy of an open systems in terms of availability and Thermodynamic second law efficiency

We know that the steady flow energy equation for any kind of open system follows the energy balance concept.

EXPRESSIONS FOR ENERGY OF AN OPEN SYSTEMS IN TERMS OF AVAILABILITY AND SECOND LAW EFFICIENCY

We know that the steady flow energy equation for any kind of open system follows the energy balance concept.


Case (a): Turbine

The following assumptions are made:

(i) Potential energies are negligible.

(ii) Inlet velocity of the turbine is negligible when compared to exit velocity.

Therefore, Steady Flow Energy Equation (SFEE) reduces to

m (h1) + Q = m (h2) + W

Actual work output, Wact = m (h1 - h2) + Q

[⸫ For air, h1 - h2 = Cp (T1 - T2), Q = Negative when the heat is transferred to the surrounding]

Reversible work (or) maximum work or availability change,

Wmax = (h1 - h2) - To (S1 - S2)

⸫ Irreversibility, I = To ΔS

Second law efficiency,

Case (b): Compressor/pump

The following assumptions are made:

(i) Potential energies are negligible.

(ii) Kinetic energies are negligible.

Therefore, SFEE reduces to

m (h1) + Q = m (h2) + W

Actual work output, W = m (h1 - h2) + Q

Compressor work is negative because it is supplied to perform the action.

[⸪ Compressor/ pumps are work absorbing devices] 

Maximum work input, Wmax = (h1 - h2) - To (S1 - S2)


Case (c): Heat exchanger

Consider my kg of hot fluid with a specific heat of C1 exchanging heat with a cold fluid of m2 kg with a specific of C2.

By heat balance,

⸫ Heat gained by cold fluid = Heat given out by hot fluid

m2 C2 (T4 - T3) = m1 C1 (T1 - T2)

Availability of cold fluid, Bc = m2 C2 (T4 - T3) = To C1 (S4S3)

Availability, of Hot fluid, Bh = m1 C1 (T1 - T2) - To (S1 - S2)

Irreversibility, I = Bh - Bc


Case (d): Throttling process

For throttling process,

Enthalpy before throttling = Enthalpy after throttling

h1 = h2

For throttling process, both work output and heat transfer are zero.

⸫ Maximum work, Wmax = (h1 - h2) - To (S1 - S2)

So, maximum work, Wmax = To (S2 - S1) = ψ1 – ψ2

where

ψ1 = Availability at inlet and 

ψ2 = Availability at outlet


Engineering Thermodynamics: Unit III: Availability and Applications of II Law : Tag: : - Expressions for energy of an open systems in terms of availability and Thermodynamic second law efficiency


Engineering Thermodynamics: Unit III: Availability and Applications of II Law



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Engineering Thermodynamics

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