ME 300: Thermodynamics IILecture 9

A. Udupa

2nd Law Efficiency / Exergetic Efficiency

►To distinguish exergy-based and energy-based efficiencies, consider the system shown in the figure. The system represents a range of applications where fuel is consumed to provide heating.

►Exergy-based efficiencies developed using exergy balances are called exergetic efficiencies.

►All energy transfers shown are in the direction of the arrows.►The system receives energy by heat transfer at the rate Qs at the source temperature Ts and delivers Qu at the use temperature Tu.►Energy is lost by heat transfer at the rate Ql at temperature Tl.►There is no work and operation is at steady state.

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Exergetic Efficiency

►Applying closed system energy and exergy rate balances at steady-state

►These equations can be rewritten as

(Eq. 1a)

(Eq. 1b)

Exergetic Efficiency

►Equation 1a shows that the energy carried in by heat transfer, Qs, is either used, Qu, or lost, Ql. This can be described by an energy-based efficiency in the form product/input as

(Eq. 2)

The value of η can be increased by applying insulation to reduce the loss. In the limit as the loss is eliminated, the value of η approaches 1 (100%).

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Exergetic Efficiency

►Equation 1b shows that the exergy carried in accompanying heat transfer Qs is either transferred from the system accompanying the heat transfers Qu and Ql or destroyed by irreversibilities within the system. This can be described by an exergy-based efficiency in the form product/input as

∙ ∙

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(Eq. 3)

Exergetic Efficiency►Introducing Eq. 3 gives

Still, the limit of 100% exergetic efficiency is not a practical objective. In most applications where fuel is consumed to provide heating, e is much less than 100% and is less than 10% in domestic water heaters. In such cases, typically there is not a good match between source and use temperature.

►Study of this expression shows that there are two ways to increase the value of the exergetic efficiency:

►Increase the value of η to as close to unity as practical.►Increase the use temperature, Tu, so it better matches

the source temperature, Ts.

Exergetic Efficiencies of Typical Systems

Turbine Compressor

Counterflow Heat-Exchanger

Direct Heat-Exchanger

Example 1

Finda. Work per unit mass of flowb. Exergy Destruction per unit massc. Isentropic Work per unit massd. 2nd Law Efficiency and isentropic efficiency

Example 2

• Argon enters an insulated turbine operating at steady state at 1000°C and 2 MPa and exhausts at 350 kPa. The mass flow rate is 0.5 kg/s. Plot each of the following versus the turbine exit temperature, in °C: – the power developed, in kW.– the rate of exergy destruction in the turbine, in

kW. – the exergetic turbine efficiency.

Solution in EES{Properties}

mdot = 0.5 [kg/s]

p1 = 2000 [kPa]T1 = 1273 [K]h1 = Enthalpy(Argon, T=T1,P=p1)s1 = Entropy(Argon, T=T1, p=p1)

p2 = 350 [kPa]h2 = Enthalpy(Argon, T=T2, p=p2)s2 = Entropy(Argon, T=T2, p=p2)

T0 = 293 [K]{Equations}

Wdot = mdot*(h1 - h2)Exdot = mdot*T0*(s2 - s1)exeff = (h1 - h2)/(h1 - h2 - T0*(s1 - s2))