**Hi Friends,**

In previous post, we have briefly described the

'As is' Hydrate feed rate to Kiln = 1720 kg/hr and

Calcined alumina coming out from Kiln = 1000 kg/hr.

**material balance calculations**for a typical 3000 tpa Specialty grade alumina production unit. Here, we are presenting the sample energy balance calculations across all associated facilities like Kiln, Air cooler and Water cooler installed for calcination as well as cooling of alumina for production of 3000 tpa Refractory grade calcined alumina. On the basis of this basic calculations, it becomes very easy to arrive at estimated fuel requirement for calcination, air requirement for combustion of fuel and generation rate of flue gases released to atmosphere through stack. In the proposed system, the feed hydrate with about 10% physical moisture will be entering to Rotary Kiln at about 30^{o}C and calcined alumina will be leaving the Kiln at about 1400^{o}C. From hot end side of the Kiln, preheated air and furnace oil will be entering to the Kiln at around 95^{o}C and 80^{o}C respectively. The furnace oil temperature will be very close to its flash point while entering the Kiln through the burner.'As is' Hydrate feed rate to Kiln = 1720 kg/hr and

Calcined alumina coming out from Kiln = 1000 kg/hr.

**Energy Balance across Rotary Kiln:**Assumed composition of Heavy furnace oil -

C = 87.0% H = 10.5% S = 1.0%.

O = 0.10% N = 0.20% Moisture = 1.20%.

Calorific value = 9600 k.cal/kg.

Specific gravity = 0.92

Specific heat = 0.56 k.cal/kg.

Taking the basis of 1 kg of HFO-

Element Composition Moles O

_{2}RequirementC 0.87 kg 0.0725 kg. mole 0.0725 kg. mole

H 0.105 kg 0.0525 kg. mole 0.0262 kg. mole

S 0.010 kg 0.0003 kg. mole 0.0003 kg. mole.

In order to ensure complete combustion of fuel oil inside the Rotary Kiln, 10% excess O

_{2}will be required.Thus O

_{2}requirement = 1.10*0.0990 kg. mole = 0.1089 kg. moleConsidering 21% O

_{2}in air,Air requirement = 0.1089/0.21 = 0.5186 kg. mole.

N

_{2}in air = 0.79*0.5186 kg. mole = 0.4097 kg. mole.Now the composition of Flue gas may be derived as-

CO

_{2}= 0.0725 kg. mole = 0.0725 * 44 = 3.19 kg.SO

_{2}= 0.0003 kg. mole = 0.0003 * 64 = 0.0192 kg.O

_{2 }= 0.01 kg. mole = 0.01 * 32 =0.320 kg.N

_{2}= 0.4097 kg. mole = 0.4097 *28 = 11.472 kg.Thus total dry flue gas= 3.19+0.0192+0.32+11.472 = 15.00 kg.

Water vapour generated on combustion of Hydrogen= 0.0525*18 = 0.945 kg.

Mass of vapor due to moisture in HFO = 0.012 kg.

Hence total water vapour = 0.945 kg + 0.012 kg = 0.957 kg of water vapour per kg of HFO burnt.

Let x kg of HFO is required per tonne of product alumina.

Hence air requirement = 0.5186*x kg.

Mass of Flue gas = 15.00*x kg.

Mass of water vapour=0.957*x kg.

Total heat value of HFO per tonne of product =9600*x k.cal.

Sensible heat of HFO = x*0.56*(30-0) k.cal. = 16.8* x k.cal.

Mass of water with HFO = 0.012*x kg.

Enthalpy of water vapour = 0.012*x*539 k.cal. = 6.5* x k.cal.

Enthalpy of air = 15*x*0.24*(95-0) k.cal = 342* x k.cal.

Enthalpy of input feed hydrate = 1720 * 0.20*(30-0) k.cal = 10320 k.cal.

Therefore, total heat input to Kiln = 9600*x + 16.8*x + 6.5*x + 342*x + 10320.

Thus total heat input to Kiln = 9965.3*x + 10320 k.cals.

Heat with dry stack gas = 15.0*0.24*(250-0) k.cal = 901 k.cal.

Enthalpy of calcined alumina = 1000*0.27*(1400-0) k.cal = 378000 k.cals.

Heat of vaporization of free moisture=170*1.08(100-30)+170*0.45*(250-100).

= 11900 + 91630 + 11475 k.cal. = 115005 k.cals.

Heat of vaporization of combined water = 0.55*1000*0.45*(1400-250).

= 284625 k.cal.

Heat of vaporization of water formed during combustion of Fuel oil

= 0.945*1.0*(100-30)+0.945*539+0.945*0.45*(250-100) k.cal. = 639 k.cal.

During operation of Rotary Kiln, substantial thermal energy is lost by radiation.

Assuming 20% of heat given by HFO is lost by radiation,

Heat loss by radiation = 0.20*9600*x k.cal. = 1920*x k.cals.

Therefore, total heat output=901+378000+115005+639+284625+1920*x.

Thus, Total heat out = 779170 + 1920x k.cals.

Since Heat Input = Heat Output.

Therefore, 9956.3*x + 10320 = 779170 + 1920*x.

or, 9956.3 *x - 1920*x = 779170 - 10320.

or, 8036.3 *x = 768850.

or, x = 768850 / 8036.3 = 95.67 kg.

Taking design margin as 15%,

Specific HFO requirement=1.15*95.67 kg/t=110 kg HFO per tonne alumina.

Air requirement = 110 * 15 kg per hour. = 1650 kg per hour.

Flue gas generation rate = 110*15.0+110*0.957 = 1755 kg per hour.

Specific HFO requirement = 110 kg per tonne of alumina.

Air requirement for calcination = 1650 kg per hour.

Flue gas generation rate = 1755 kg per hour.

Trust, all associated technical aspects have been covered systematically. In case you have some better method for carrying out energy balance calculations, please feel free to share with us. Your valued suggestions / remarks / comments, if any, shall be welcomed.

Thanks and regards.

Let x kg of HFO is required per tonne of product alumina.

Hence air requirement = 0.5186*x kg.

Mass of Flue gas = 15.00*x kg.

Mass of water vapour=0.957*x kg.

**Heat Input to Rotary Kiln:-**Total heat value of HFO per tonne of product =9600*x k.cal.

Sensible heat of HFO = x*0.56*(30-0) k.cal. = 16.8* x k.cal.

Mass of water with HFO = 0.012*x kg.

Enthalpy of water vapour = 0.012*x*539 k.cal. = 6.5* x k.cal.

Enthalpy of air = 15*x*0.24*(95-0) k.cal = 342* x k.cal.

Enthalpy of input feed hydrate = 1720 * 0.20*(30-0) k.cal = 10320 k.cal.

Therefore, total heat input to Kiln = 9600*x + 16.8*x + 6.5*x + 342*x + 10320.

Thus total heat input to Kiln = 9965.3*x + 10320 k.cals.

**Heat Output from Rotary Kiln:-**Heat with dry stack gas = 15.0*0.24*(250-0) k.cal = 901 k.cal.

Enthalpy of calcined alumina = 1000*0.27*(1400-0) k.cal = 378000 k.cals.

Heat of vaporization of free moisture=170*1.08(100-30)+170*0.45*(250-100).

= 11900 + 91630 + 11475 k.cal. = 115005 k.cals.

Heat of vaporization of combined water = 0.55*1000*0.45*(1400-250).

= 284625 k.cal.

Heat of vaporization of water formed during combustion of Fuel oil

= 0.945*1.0*(100-30)+0.945*539+0.945*0.45*(250-100) k.cal. = 639 k.cal.

During operation of Rotary Kiln, substantial thermal energy is lost by radiation.

Assuming 20% of heat given by HFO is lost by radiation,

Heat loss by radiation = 0.20*9600*x k.cal. = 1920*x k.cals.

Therefore, total heat output=901+378000+115005+639+284625+1920*x.

Thus, Total heat out = 779170 + 1920x k.cals.

Since Heat Input = Heat Output.

Therefore, 9956.3*x + 10320 = 779170 + 1920*x.

or, 9956.3 *x - 1920*x = 779170 - 10320.

or, 8036.3 *x = 768850.

or, x = 768850 / 8036.3 = 95.67 kg.

Taking design margin as 15%,

Specific HFO requirement=1.15*95.67 kg/t=110 kg HFO per tonne alumina.

Air requirement = 110 * 15 kg per hour. = 1650 kg per hour.

Flue gas generation rate = 110*15.0+110*0.957 = 1755 kg per hour.

**Conclusions:**Specific HFO requirement = 110 kg per tonne of alumina.

Air requirement for calcination = 1650 kg per hour.

Flue gas generation rate = 1755 kg per hour.

Trust, all associated technical aspects have been covered systematically. In case you have some better method for carrying out energy balance calculations, please feel free to share with us. Your valued suggestions / remarks / comments, if any, shall be welcomed.

**If you like this article, then please press your rating as +1 appearing at the footer of this page.**

Thanks and regards.

**Kunwar Rajendra**
How can the heat and mass balance for precipitation circuit be calculated manually other than using SysCad? Can the steps shown above be followed?

ReplyDeleteDear Win Tinha,

ReplyDeleteThe methodology for material and heat balance calculations for all unit operations is the same. Thus, For carrying out material and heat balance calculations for Precipitation circuit, you should be well aware about the process circuit as well as characteristics, heat capacities and associated properties of each stream of precipitation circuit.

Thanks and regards.

Kunwar Rajendra

Hi Win Thiha,

ReplyDeleteI have requested my team to develop the material and thermal energy balance calculations for Precipitation circuit using manual method as presented above. On Excel spreadsheet, we have detailed calculations which will be published on this platform in simplified manner shortly.

Regards

Rajendra Kunwar

www.ceti.co.in