In a boiler test 1250 kg of coal is consumed in 24 hours. The mass of water evaporated is 13,000 kg and the mean effective pressure is 7 bar. The feed water temperature is 40°C, the heating value of coal is 30000 kJ/kg. The enthalpy of 1 kg of steam at 7 bar is 2570.7 kJ. Equivalent evaporation per kg of coal is_____.

Concept:

Equivalent Evaporation:

- It is defined as the amount of feed water evaporated per unit of time during the conversion of feed water at 100 ºC into dry and saturated steam at 100 ºC.

Equivalent Evaporation \(m_e = \frac{m_w × (h-h_w)}{m_f × 2257}\)

Where, m_w = mass of water evaporated

h = enthalpy of steam after evaporation

h_w = enthalpy of feed water

m_f = mass of fuel

and 2257 in the denominator indicates heat required in kJ for 1 kg of feed water at 100 ºC and 1 atm pressure to convert into dry and saturated steam at the same pressure and temperature. 

Calculation:

m_w = 13000 kg

m_f = 1250 kg

h = 2570.7 kJ

h_w = C_p × (T-0) = 4.184 × 40 = 167.36 kJ

Here, the heating value of coal is given which is redundant data for this calculation.

\(m_e = \frac{13000 \times (2570.7-167.36)}{1250 \times 2257}\)

Therefore, \(m_e = 11.075\)