A solar thermal power plane with a point-focus concentrating collector comprisin

Question

A solar thermal power plane with a point-focus concentrating collector comprising a heliostat Held and central receiver and a steam turbine generator unit is operating in the following conditions: Power plant electric power output P_el = 75 MW; Single heliostat surface area A_h = 125 m^2; Beam solar radiation flux incident ont the heliostat I_b, N= 835 W/m^2; Heliostat-receiver concentration ratio C = 510; Efficiency of solar collector, that is, heliostat field and central receiver eta_th-g= 0.36: HTF specific heat c- 1.03 kJ/kg.K; HTF temperature rises in the absorber/receiver from 298 degree C to 563 degree C. Calculate: The overall (electrical) efficiency of the power plant; The rate of useful heat output of the central receiver/absorber; The total reflecting surface area of the heliostats field; The number of heliostats and absorber surface area in the central receiver The HTF mass flow rate.

Explanation / Answer

The concentration ratio C=510=Area of the collector aperture / Surface area of the receiver

Therefore area of the collector aperture = 510 x 125 sq.m

Solar radiation flux incident of the heliostat = 835 W/sq.m

Total solar power from one heliostat = 835 x 510 x 125 W

Efficiency of the solar collector i..e, heliostat field and central receiver = 0.36

The power output from each heliostat considering efficiency = 835 x 510 x 125 x 0.36 = 19.16 MW

Let n be the no. of helistats.

Then the total power generated from n heliostats would be = 19.16n MW ---- 1

Given the electric power output from the plant = 75 MW.

The smallest value of n in the equation 1 above, which would give >= 75 is 4.

Therefor the no. of heliostats would be 4

Now, 4 heliostats would produce 19.16 X 4 = 76.64 MW

But the power plant output is 75MW.

Therefore the loss due to the process after collector would be : 76.64 – 75 = 1.64 MW

The efficiency of the subsequent processes = 75 / 76.64 = 0.98

The overall efficiency of the power plant = 0.98 x 0.36 = 0.35

a.      The overall efficiency of the power plant = 0.35

b.      The rate of useful heat output of the central receiver = 835 x 510 x 125 W = 53 MW

c.      The total reflecting surface area of the heliostat field : 510 x 125 sq.m = 63750 sqm

d.      The no. of heliostats = 4, the absorber surface area = 63750 x 4 = 255000 sqm

e.      Let The HTF mass flow rate be m kg/sec

Then m x 1.03 x ( 563-298 ) = 76.64 MW

m=280.78 Kg/sec

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