Question

I) Boiler superheater [3 marks) B) A steam boiler fired with oil of calorific value of 42 MJ/kg generates steam at a rate of 2500 kg/h and 1.5 MPa. The generated steam is further heated in a superheater to a temperature of 300°C. Feed water at 30°c is preheated to 90°C through passage in the economizer. The boiler unit has an efficiency of 78% i) Calculate the fuel/oil consumption rate (in kg/h) ii) Calculate the percentage increase in fuel consumption when the economizer is not utilized energy. Calculate the overall (boiler and turbine) plant efficiency [Data: Specific heat capacity of water4.187 kJ/kg °c] [4 marks) 17 marks] 13 marks] ii) The generated superheated steam drives a turbine that develops 1800 kW work

Answer 1

a) The steam boiler generates steam by taking heat energy from the oil, so the heat from oil is sensible heat given by

mass flowrate × calorific value= m × 42 ×10^6

But it is given that the efficiency of boiler is 78% that means that only 78% of total heat is available for generating steam.

Steam is generated by taking this heat as latent heat of vapourisation = mass flowrate of steam × latent heat per kg

= 2500 × 2256 × 10^3

Equating these two expressions will give us the value of mass flowrate of oil in kg/h which comes out to be 134.285 kg/h.

b) Economizer preheats the water from 30°C to 90°C and then it is sent to the boiler.

So for preheating water

mass flowrate × specific heat × temperature difference

This much amount of heat is required.

If economizer is not used then this much heat is also to be given by the boiler itself.

Hence the fuel consumption will increase.

Calculations:

More heat that will be required by the boiler now

= (2500) kg/h × (4.187 × 10^3) KJ/ kg °C × (100-30)°C

= 732,725,000 J/h

Total heat consumption = 6.37 × 10^9 J/h

Since fuel consumption is proportional to heat consumption we can directly find fuel consumption percentage by dividing the more heat required by boiler to total heat required

=(7.32 × 10^8 / 6.37× 10^9) × 100%

= 0.1149 × 100%

11.49%

So there is 11.49% increase in fuel consumption.

c) The overall plant efficiency is given by

= work output from plant / total heat input to plant

Total heat input = heat from boiler + heat from superheater +

heat from economizer

Heat required in superheater = mass flowrate × specific enthalpy of steam at 300 ° C and 1.5 MPa - specific enthalpy at.100 °C and 1.5 MPa.

= 2500 × ( 3073.31 - 2668.45 ) × 10^3 J/h

Enthalpy values are taken from superheated steam tables.

= 1.012 × 10^9 J /h

Total heat input = 7.28 × 10^9 J/h

= 7.28 × 10^9 / 3600 J/s

= 2.022 × 10^6 W

Overall plant efficiency% = work done / total heat input × 100%

= 1800 kW / (2.022 ×1000) kW

= 1800/2022 ×100%

= 0.89 × 100%

= 89%