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a modern state-of-the-art biomass power plant with heat recovery feature has the capacity to combust up to 1.4 x 106 kg/day of pellets to generate energy. the pellets have an average lhv of 18.1 mj/kg and hhv of 19.1 mj/kg. with a capacity factor of 57% and conversion efficiency of 38%, how much energy does this plant generate on an annual basis?



Answer :

Energy in terms of Joules = 1956.50 GJ is the answer.

Energy generation by the Power plant (on annual basis) is given as: -

A.  When heat recovery feature is Present

E = (M * HHV * C * ηc * 365) / 3600 watt – hr         ------------- equation (i)

B.  When heat recovery feature is not present

E = (M * LHV * C * ηc * 365) / 3600 watt – hr        ------------- equation (ii)

Here,

 M = Mass of Pellets consumed per day (Kg/day)

 HHV = Higher Heating Value of Biomass (J/Kg)

 LHV = Lower Heating Value of Biomass (J/Kg)

 C = Capacity factor {Plant running this % time in a day}

 ηc = Conversion efficiency

 Joule to watt – hr conversion factor

 1 Joule = 1 /3600 watt – hr

Given:

 M = 1.3 * 106 Kg/day;

 HHV = 19.9 * 106 J/Kg;

 LHV = 17.4 * 106 J/Kg

 C = 56% = 0.56

 ηc = 37% = 0.37

Heating Recovery feature ---> Present

As heating recovery feature is present so eq. 1 will be used

Now, putting corresponding given values in Eq. 1 to get the value of E.

E = (M * HHV * C * ηc * 365) / 3600

E = (1.3 * 106 * 19.9 * 106 * 0.56 * 0.37 * 365) / 3600

E = (1956.496 * 1012) / 3600

E = 543.47 Gwh

Energy in terms of Joules = 543.47 * 3600

Energy in terms of Joules = 1956.50 GJ

Energy Generation (E) by the power plant (on annual basis) = 543.47 Gwh (Giga Watt – hour)

Energy Generation (E) by the power plant (on annual basis) = 1956.50 GJ (Giga Joule)

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