1. ME 300: Thermodynamics II Lecture 9 A. Udupa

2. 2 nd 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 Q s at the source temperature T s and delivers Q u at the use temperature T u. ►Energy is lost by heat transfer at the rate Q l at temperature T l. ►There is no work and operation is at steady state. ∙ ∙ ∙

3. Exergetic Efficiency ►Applying closed system energy and exergy rate balances at steady-state ►These equations can be rewritten as (Eq. 1a) (Eq. 1b)

4. Exergetic Efficiency ► Equation 1a shows that the energy carried in by heat transfer, Q s, is either used, Q u, or lost, Q l. This can be described by an energy-based efficiency in the form product/input as (Eq. 2) The value of  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%). ∙ ∙∙

5. Exergetic Efficiency ► Equation 1b shows that the exergy carried in accompanying heat transfer Q s is either transferred from the system accompanying the heat transfers Q u and Q l or destroyed by irreversibilities within the system. This can be described by an exergy- based efficiency in the form product/input as ∙∙ ∙ (Eq. 3)

6. 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,  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  to as close to unity as practical. ►Increase the use temperature, T u, so it better matches the source temperature, T s.

7. Exergetic Efficiencies of Typical Systems Turbine Compressor Counterflow Heat- Exchanger Direct Heat-Exchanger

8. Example 1 Find a.Work per unit mass of flow b.Exergy Destruction per unit mass c.Isentropic Work per unit mass d.2 nd Law Efficiency and isentropic efficiency

10. 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.

11. 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))