Lesson 6. Performance evaluation of boilers

Boilers

Boiler is a system which transfers the energy from the fuel to water.  The boiler fuel will be combusted and the thermal energy generated will be transferred to the feed water.  The water gets heated up and hot water or steam will be generated.  The maximum transfer of energy from fuel to steam shows better performance of boiler. 

If a boiler has good combustion system, which converted the fuel energy into thermal energy more thermal energy will be produced from the available fuel energy.  If the boiler’s heat transfer system is efficient, a maximum portion of generated heat will be absorbed by water during the generation of steam.         

The performance of boiler can be assessed by the following procedure.

Performance assessment

The efficiency of boiler is found out by calculating fuel energy input to the thermal energy available in the form of steam.  Fuel energy can be calculated by knowing the calorific value of fuel used in the boiler.  The thermal energy used for the production of steam can be calculated from the enthalpy of the steam based on the initial and final temperatures.

Observation

The boiler will be allowed to operate for a specific period to reach stabilization to avoid the fluctuations in the energy consumption during initial period.  After stabilization period, the quantity of fuel consumed will be calculated and its average calorific value has to be assessed per unit time.  The energy content of the input water and the output steam can be calculated based on the inlet and out let temperature.  Based on the temperature the enthalpy will be found out and the thermal energy consumed can be calculated.

            Quantity of fuel consumed                 -           F, kg h-1

            Calorific value of fuel                        -           CVF, kCal

            Quantity of steam generated               -           M, kg h-1

            Initial temperature of water                 -           t1, °C

            Enthalpy of water                               -           h, kCal kg-1

            Pressure of steam                               -           p2, ksc

            Temperature of steam                         -           t2, °C

            Enthalpy of steam based on the

pressure and temperature                     -           H, kCal kg-1

Boiler efficiency                                  =          Heat in steam / Heat in Fuel

Heat in steam                                      =          M (H-h)          

Heat in fuel                                         =          F x CVF 

 

Problem

A 10 TPH boiler is fed with water at temperature 75°C and steam is produced at 10 ksc with temperature of 180°C.  The sugarcane bagasse is used as fuel and its fuel consumption rate is 3 TPH with calorific value of 2800 kCal kg-1.  Find out the efficiency of boiler for the given condition.

Fuel consumption rate, F                     -           3000 kg h-1

Calorific value of fuel, CVF                 -           2800 kCal kg-1

Steam production rate, M                    -           10 000 kg h-1

Inlet water temperature                       -           75°C

Enthalpy of feed water, h                    -           75 kCal kg

Steam pressure, p2                             -           10 ksc

Steam temperature, t2                         -           180°C

Enthalpy of steam, H                          -           665 kCal kg-1

Heat in steam                                     -           M (H-h)

   -           10 000 x (665-75)

   -           59 00 000 kCal

Heat in fuel                                        -           F x CVF

   -           3000 x 2800

   -           84 00 000                      

                                                                             Heat in steam

Boiler efficiency                                 =          -----------------

                                                                             Heat in fuel

 

                                                                             59 00 000

                                                               =          -------------

                                                                              84 00 000

                                                               =          70%

Last modified: Wednesday, 9 April 2014, 5:49 AM