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Boiler efficiency with superheated steam

Short answer

Superheated steam at 15 bar, 300°C: h=3038.3 kJ/kg; feedwater at 50°C: 209.4 kJ/kg. Heat supplied = 2000*(3038.3-209.4)=5.66x10^6 kJ/h. Efficiency = 5.66x10^6/(250*32000)=70.7%. Factor = (3038.3-209.4)/2257=1.253. Equivalent evaporation = 2000*1.253=2507 kg/h.

A boiler produces 2000 kg/h of superheated steam at 15 bar and 300 °C from feed water at 50 °C. The fuel consumption is 250 kg/h with calorific value 32000 kJ/kg. Find (a) the heat supplied to the water and steam per hour, (b) the boiler efficiency, (c) the factor of evaporation and the equivalent evaporation from and at 100 °C (latent heat 2257 kJ/kg).

Enthalpies (steam tables)

  • Superheated steam at 15 bar, 300 °C: h=3038.3h = 3038.3 kJ/kg.
  • Feed water at 50 °C: hfw=209.4h_{fw} = 209.4 kJ/kg.

(a) Heat supplied to water and steam

Q=m˙s(h−hfw)=2000×(3038.3−209.4)=5657800 kJ/hQ = \dot m_s (h - h_{fw}) = 2000 \times (3038.3 - 209.4) = 5657800\ \text{kJ/h}

(b) Efficiency

Heat released by fuel =250×32000=8000000= 250 \times 32000 = 8000000 kJ/h.

η=56578008000000=0.707\eta = \frac{5657800}{8000000} = 0.707

(c) Factor and equivalent evaporation

Fe=h−hfw2257=3038.3−209.42257=1.253F_e = \frac{h - h_{fw}}{2257} = \frac{3038.3 - 209.4}{2257} = 1.253 E=Fe×m˙s=1.253×2000=2507 kg/hE = F_e \times \dot m_s = 1.253 \times 2000 = 2507\ \text{kg/h}

Answer: (a) 5.66×1065.66 \times 10^{6} kJ/h; (b) 70.7 %; (c) Fe=1.253F_e = 1.253, E≈2507E \approx 2507 kg/h.

✓ Answer checked by Claude Opus 5.5 (high effort)

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