Steam Generation Calculations Of Boiler cp54

Steam Generation Calculations of Boiler Steam Requirement = 13.5 ton/hr Feed water temperature = 28oC Steam Pressure = 25 kg/cm2 Steam Temperature = 350oC Boiling point at 25 kg/cm2 = 221.8oC Specific Heat of Water = 4.62 KJ/kg For the production of super-heated steam three phases are involved. In first phase sensible heat will be calculated then in 2 nd one latent heat of vaporization will be calculated for saturated steam then in 3rd phase have to calculate latent heat for superheated steam.

1st Phase: For the calculation of sensible heat feed water temperature should be known i.e 28oC and for final temperature steam pressure would be required i.e 25kg/cm2. At this pressure boiling point of water is 221.8oC. So now, Q1 = m∆T M = 13.5 ton = 13500kg Q1 = 13500*4.62*(221.8-28) = 12,087,306KJ/Hr

2nd Phase:

Then heat of vaporization will be calculated from steam table at 25kg/cm2. there will be 2 values there, 1st will be total enthalpy which includes sensible heat while 2nd value tells us just enthalpy of vapors. Here we are needed with only heat of vaporization so we will consider 2nd one i.e ∆Hv = 1848.5 KJ/kg Q2 = mλ1 = 13500*1848.5 = 24954750KJ/hr

3rd phase: Then enthalpy of super-heated will be calculated, procedure is as follows as the enthalpy value at given temperature and pressure is not available in steam table. Required Pressure = 348lb/in2 Required temperature = 662oF Values available in steam table are: Temp=600 H=1311.4 H=1310.6

P=340 P=350

Calculations for 348lb/in2 at 600oF 350-348/1310.6-X1 =

348-340/X1-1311.4

2X1-2622.8

=

X1

=

10484.8-8X1

1310.76Btu/lb

Calculations for 348lb/in2 at 700oF

Temp=700 H=1365.2 H=1364.7

350-348/1364.7-X2 =

348-340/X2-1365.2

2X2-2730.4

=

X2

=

10917.6-8X2

1364.8Btu/lb

Now calculations at 662oF 700-662/1364.8-X

=

662-600/X-1310.76

38X-49808.8

=

9461736-62X

X

=

1344.264Btu/lb

= 3126.8KJ/kg

This value of latent heat is total enthalpy of super heated steam with saturated and sensible steam, to calculate value of latent heat for super heated steam we have to subtract total latent heat of saturated steam (from steam table) to total enthalpy of superheated steam i.e ∆Hg (Total enthalpy of saturated steam) = 2800.4KJ/kg ∆H = 3126.8 – 2800.4 = 326.4KJ/kg Q3 = mλ2 = 13500*326.4 =4406400KJ/kg Total Heat Energy = 41448456KJ/kg Assume Boiler Efficiency is 85% Then, Q = 4876288.9KJ/kg = 11654330.5Kcal/hr

Sui Gas Industrial: Connection Capacity = 25MCF/hr = 25000Cf/hr Calorific value of Sui gas = 935Btu/Cf So energy produced = 25000*935 = 23375000Btu/hr 1Btu = 1.055KJ So, = 24660625KJ/hr 1KJ = 0.239Kcal So now, Heat Energy produced = 5893889 Kcal/hr As 11654330.5Kcal/hr energy is required to produce = 13.5Ton steam So 5893889Kcal/hr energy will produce = 5893889*13.5/11654330.5 = 6.87ton/hr Cf= Cubic feet SCf = Standard Cubic feet Cubic feet MCf = Metric Cubic Feet

NCf = Normal

MMCf = Million Cubic feet

Calorific value tells us about the heat energy released by gas per cubic feet. Its value differs as composition so diff for diff gas fields.

Bio Gas: Some standards are given below: COD reduction = 65% Bio Gas production = 0.5m3/kg Methane Content = 50-55% Apent wash COD = 110,000mg/lit Spent wash quantity = 58m3/hr = 58000lit/hr Total COD = 58000*110,000 = 6380kg/hr COD reduction = 6380*.65 = 4147kg/hr Bio Gas Production = 0.5*4147 = 2073.5 m3/hr Methane content = 2073.5*0.53 = 1098.9m3/hr Calorific Value = 5400 Calorific Value of Bio gas = 59,34,357 Kcal/hr Steam Produced = 5934357*13.5/11654330.5 = 6.88 ton/hr