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ENGR5003 / Week 1
Lecture 3 Example 1 - energy vs exergy losses in a steam plantTutorialCurrent specThird10 min

ENGR5003 Lecture 3 Example 1 (slide 21)

Two different analyses of the same steam power plant give the following loss breakdowns (energy = exergy + anergy).

EquipmentEnergy loss / energy input (%)Exergy loss / exergy input (%)
Boiler and super-heater949 (combustion 29.7, heat transfer 14.9, flue 0.68, others 3.72)
Turbineapprox. 04
Condenser471.5
Pumpapprox. 01.0
Others35.5
In total5961
Formulas you may need
  • Exergy of heat qq supplied at mean temperature TmHT_{mH}: ex,Q=q(1−T0TmH)=q−T0Δse_{x,Q} = q\left(1 - \dfrac{T_0}{T_{mH}}\right) = q - T_0\Delta s (learn this, Lecture 3 slide 4)
  • Maximum (Carnot) work from heat: wmax=q(1−T0T)w_{max} = q\left(1 - \dfrac{T_0}{T}\right) (learn this, Lecture 3 slide 20)
  • Entropy increase principle: ΔSiso=Sg≥0\Delta S_{iso} = S_g \ge 0 (on the formula sheet)
  1. (a)
    Explain why the condenser accounts for most of the ENERGY loss but almost none of the EXERGY loss, while the boiler is the other way round.
  2. (b)
    Use the totals to estimate the plant's energy (first-law) efficiency and its exergy (second-law) efficiency.
  3. (c)
    What are strategies for reducing the exergy and energy losses?