The efficiency of biogas technology: a case study.

Mandrea, Lucian ; Baran, Gheorghe ; Babutanu, Corina Alice 等

1. INTRODUCTION

Any new unit for biogas production is always we[l.sub.c]omed, especially when alcohol is used. The reactor uses the anaerobic digestion method to obtain bio-methane which leaves week pollutant compounds after burning.

Organic waste is also reduced and processed to obtain good materials for agriculture. The biogas reactor and the gasholder are placed close to the reservoir with mixture resulted from alcohol production. The final calculus shows how the biogas production raises the efficiency of the Agronad SME.

2. BRIEF PRESENTATION OF THE DISTILLERY

The next images presents the location of the biogas plant in the general landscape of the Agronad SME and the gravitational circulation of the mixture to the biogas reactor, the storage of the bio-methane and the final use of the biogas. The scheme is very simple and uses few compounds. So, final efficiency is increased (*** research grant, 2007).

[FIGURE 1 OMITTED]

[FIGURE 2 OMITTED]

3. ESTABLISH OF ENERGY CONSUMPTIONS

3.1 Calculus in area I

This area is situated between the two cylinders, first of radius [R.sub.1] which corresponds with the propeller diameter and the other one of radius [R.sub.2] which corresponds to the interior diameter of the tank, according with Fig. 3. (Baran, 2008).

[R.sub.1] = 0,5 m, [R.sub.2] = 1,68 m and the volume has the height: L = h + h2 = 1,73 + 0,48 = 2,21m .

Distillers wash from corn which fills the reactor has the dynamic viscosity [eta]=1,19*[10.sup.-3] Ns/[m.sup.2] and the density [rho]=1006,85 kg/[m.sup.3], for a temperature of 16[degrees]C (Stroade, 2010).

The momentum of viscous friction is computed in each of the two previous areas and then the total momentum is obtained.

For area I a linear distribution of the speed is admitted inside the distillers wash from corn.

V = [omega][R.sub.1] on the cylindrical surface situated at the extremity

of the propeller and V = 0 on the interior wall of radius [R.sub.2] of the reactor, in accordance with the adhesion property of the viscous fluids.

The tangential effort on the cylindrical vertical surface of radius [R.sub.1] is:

[tau] - [eta] V - 0/[R.sub.2][R.sub.1] = [eta] [omega][R.sub.1]/[R.sub.2] - [R.sub.1] (1)

This effort is active on the surface: S = 2[pi][R.sub.1]L

The viscous friction force becomes:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)

The necessary power to rotate the liquid is:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

3.2 Calculus in area II

This area include the upper cylinder of height h1=1,73m and the bottom cylinder of height h2=0,48 m, which are placed above and under the propeller.

The tangential speed of the fluid layers situated in the vicinity of the propeller (upper and bottom layers) is variable with the radius r [member of] [0,[R.sub.1]]: V = [omega]r

[FIGURE 3 OMITTED]

It depends of the radius r [member of] [0,[R.sub.1]]: V = [omega]r

The speed on the bases of the reservoir is null.

One obtains the tangential effort: [tau] = [eta] V/h, where h has

different values for the upper and under propeller areas.

Substituting the speed, the tangential effort as a function of

the variable radius r is obtained: [tau] = [eta] [omega]/h r

The elementary surface on which the tangential effort can

be considered constant is: dS = 2[pi]rdr

The elementary friction force becomes:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

The dispersed power through viscous friction is:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

The total elementary power on the upper and bottom areas

regarding the propeller is:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

The total power on the upper and bottom areas obtained by integrating the equation from 0 to [R.sub.1] is:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

3.3 The total necessary power used to rotate the fluid

Adding the relations of the power in both areas one obtains:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

3.4 The necessary power to rotate the propeller

The Reynolds number corresponding with the propeller rotation, using the rotating speed n expressed in rotations per second is:

Re = [nd.sup.2]/v (9)

Considering that water is the working liquid and that the propeller diameter is d = 1 m and the rotating speed n = 2,5 rot/s, one obtains:

Re = [nd.sup.2] = 2.5x1-- = 2,5x[10.sup.6] = 2500000 (10)

The necessary power during the operation regime is:

P = [K.sub.N][rho][n.sup.3][d.sup.5]; using the coefficient [K.sub.N] = 0,93, one obtains:

P = 0,93 x [10.sup.3] x 2,[5.sup.3] x [1.sup.5] = 14,531 x [10.sup.3] W [congruent to] 14,53kW (11)

Considering that the necessary power to start the electrical engine is 2,5 times higher than the necessary power during the operation stationary regime, it results:

[P.sub.p] = 2,5 x P = 2,5 x 14,53 = 36,32kW (12)

For n = 1 rot/s, the Reynolds number becomes:

Re = [nd.sup.2]/v = 1x[1.sup.2]/[10.sup.-6] = [10.sup.6] = 1000000 (13)

With [K.sub.N] = 0,92, the necessary power during the operation

regime is: P = 0,92 x [10.sup.3] x [1.sup.3] x [1.sup.5] = 920W

The necessary power to start the electrical engine is:

[P.sub.p] = 2,5 x P = 2,5 x 0,92 = 2,3kW (14)

For n = 1,5 rot/s, the Reynolds number becomes:

Re = [nd.sup.2]/v = 1,5x[1.sup.2]/-106 = 1,5x[10.sup.6] = 1500000 (15)

Using [K.sub.N] = 0,925 the necessary power during the operation regime is:

P = 0,925 x [10.sup.3] x 1,[5.sup.3] x [1.sup.5] = 3,128W [congruent to] 3,13kW (16)

The necessary power to start the electrical engine becomes:

[P.sub.p] = 2,5 x P = 2,5 x 3,13 = 7,8kW (17)

It is considered convenient the function of the propeller with the rotating speed n=1 rot/s, for which the necessary power of the engine is 2,3 kW.

4. THE EFFICIENCY OF THE BIOGAS REACTOR

The filling of the reactor is 50% and the volume of biogas obtained from 1 [m.sup.3] of tank per day is 0,6 [m.sup.3] in normal conditions. The total volume of biogas obtained in our reactor during one day is 6,275 [m.sup.3]. During one month the volume becomes 188,25 [m.sup.3] (Morin, 2010, Sasse, 1988).

The useful energy obtained considering the calorific power of the biogas 6,5 kWh/[m.sup.3] is:

E = P x V = 6,5 x 188,25 = 1223, 625kWh (18)

The price of this energy is:

[Pr.sub.1] = E x pr = 1223,625 x 0,486 = 594, 681lei (19)

With the necessary power during the operation regime of 920W, the electrical energy consumed is:

Ec = Pc x t = 920 x 30 x 24 = 662, 4kWh (20)

The price of this energy is:

[Pr.sub.2] = Ec * pr = 662,4 * 0,486 = 321,926lei (21) The profit becomes:

Pr = [Pr.sub.1]-[Pr.sub.2] = 594,681 - 321,926 = 272, 754lei (22)

5. CONCLUSION

Taking into account the profit realized, we can conclude that our biogas reactor is a very useful device for the alcohol private firm to improve its efficiency. We shall try in the future to maximize the production of biogas, changing everything possible from the physical and chemical points of view. The profit can be increased considering the theoretical perspective of improving the reactor design and the mixture composition.

6. ACKNOWLEDGEMENTS

The authors thank to NASR for the support offered by no. 31041 project.

7. REFERENCES

Baran, Gh. at al., (2008), Realizari si perspective in industria biogazului, Printed Publishing, ISBN 978-606-521-064-6 Bucuresti

Morin P., Marcos B., Moresoli, C. & Laflamme C (2010). Economic and environmental assessment on the energetic valorization of organic material for a municipality in Quebec, Canada, Available from http://www.sciencedirect.com Accessed: 2010-05-25

Sasse, L (1988) Biogas Plants, Available from http://www.gateinternational.org/energy.htm, Accessed: 2010-06-15

Stroade, J., Martin, A & Conrad, A. (2009) Distillers Grain Industry: Production, Use, Structure and Trends, Available from http://www.naiber.org/ Accessed: 2010-06-22

*** "New ecological technologies for energetic recovery of biodegradable wastes under the form of combustible gas with application at small treatment plants", research grant, 2007