PERFORMANCE OF BOILERS

Evaporation Rate: It is steam generation rate of boilers which may be expressed in terms of kg of steam per unit heating surface area or kg of steam per cubic metre of furnace volume or kg of steam per kg fuel burnt.

Equivalent Evaporation: It is equivalent of evaporation of 1 kg of water at 100°C to dry and saturated steam at 100°C, standard atmospheric pressure of 1.013 bar. Hence, the equivalent evaporation of 1 kg of water at 100°C needs 2,257 kJ.

Factor of Evaporation: It is ratio of heat absorbed by 1 kg of feed water under working conditions to latent heat of steam at atmospheric pressure.

Boiler Efficiency: It is ratio of heat absorbed by water in boiler to heat supplied to boiler per unit time.

Example 4.15: In a boiler trial observations made are as follows:

Calculate the evaporation factor and equivalent evaporation at 100°C in kg/kg of coal. Specific heat of feed water = 4.18 kJ/kg K.

Solution:

At 15 bar (from steam table)

Example 4.16: In a boiler observations made were as follows:

Calculate:

1. Boiler efficiency.
2. Percentage of heat loss in flue gas.
3. Percentage of heat loss to ash and unburnt coal.
4. Percentage of heat loss unaccounted for.

Solution:

From steam table. At 10 bar,

 

h_f = 762 kJ/kg;    h_fg = 2,015.3 kJ/kg

 

Enthalpy of feed water at 30°C, h_w = 125.79 kJ/kg

1. 
2. Heat carried away by flue gas, Q = m_f C_f (t_f – t_a) = 20 × 1.03 (300 – 20) = 5,768kJ/kg of coalHence, loss 
3. Quantity of ash and unburnt coal = 10kg/hHeat lost in ash and unburnt coal 
4. Total heat accounted for = 74.24 + 17.47 + 0.9 = 92.61%Heat unaccounted for = (100 – 92.62) = 7.39%

Example 4.17: A boiler plant delivers steam at 20 bar and 360°C to an engine developing 1,350 kW at the rate of 10 kg/kWh. Temperature of feed water = 80°C. Calorific value of fuel = 28,000 kJ/kg. The grate is to be designed to burn 400 kg of coal per m² per hour. Find the grate area required for the above duty assuming the combustion efficiency of 90% and boiler efficiency including superheater as 75%.

Solution:

Enthalpy of steam at 20 bar and 360°C (from steam table),

h_g = 3,159 kJ/kg (use interpolation)

h_f = Enthalpy of feed water at 80°C = 334.91kJ/kg

Mass of steam required to produce, 350 kW power = 1,350 × 10 = 13,500 kg/h

Heat required = 13,500 (h_g – h_f) = 13,500 (3,159 – 334.91) = 38,125.215 × 10³ kJ

Calorific value of fuel = 28,000 kJ/kg

But, combustion efficiency given as 90%