Equivalent evaporation and boiler efficiency

Data from steam tables

At 10 bar: $h_f = 762.7$ kJ/kg, $h_{fg} = 2014.4$ kJ/kg. Feed water at 40 °C: $h_{fw} = 167.6$ kJ/kg.

Enthalpy of wet steam:

$$h = h_f + x\,h_{fg} = 762.7 + 0.95 \times 2014.4 = 2676.4\ \text{kJ/kg}$$

Heat added per kg of steam $= h - h_{fw} = 2676.4 - 167.6 = 2508.8$ kJ/kg.

(a) Equivalent evaporation

$$E = \frac{\dot m_s\,(h - h_{fw})}{2257} = \frac{4500 \times 2508.8}{2257} = 5002\ \text{kg/h}$$

(b) Factor of evaporation

$$F_e = \frac{h - h_{fw}}{2257} = \frac{2508.8}{2257} = 1.112$$

(c) Boiler efficiency

$$\begin{aligned} \eta &= \frac{\dot m_s (h - h_{fw})}{\dot m_f \times CV} = \frac{4500 \times 2508.8}{600 \times 29000}\\ &= \frac{11289600}{17400000} = 0.649 \end{aligned}$$

(d) Equivalent evaporation per kg of coal

$$\frac{E}{\dot m_f} = \frac{5002}{600} = 8.34\ \text{kg steam/kg coal}$$

(Actual evaporation is $4500/600 = 7.5$ kg/kg coal.)

Answer: (a) about 5002 kg/h; (b) 1.112; (c) about 64.9 %; (d) about 8.34 kg/kg of coal.