12 a hydro electric plant has a catchment are of 120 square...

'12. A hydro-electric plant has a catchment are of 120 square km. The available run off is 50% with annual rainfall of 100 cm: A head of 250 m is available on the average. Efficiency of the power plant is 70% _ Find average power produced (ii) capacity of the power plant: Assume a load factor of 0.6.'

Transcript

00:01 Hello everyone in this problem we have given the values of the area let's suppose it is noted by 8 that is equals to 120 km square so if i convert into meter that will be 120 into 10 raised to par 3 raised to 4 2 meter square so here it will become 120 into 10 raised to 4 6 meters square also we have given the value of annual rainfall that is equals to 100 cm so in meter it is equal to 1 meter also the value of the coefficient of the runoff is given that is equal to 50 % so here it will be equals to 0 .5 also the value of efficiency that is equals to 70 % and the value of load factor that is given 0 .6 now, firstly we are calculating the volume of the water that is available for power production.

01:33 So it comes out 0 .5 into 1 into 12 into 10 days to power 6.

01:40 So here we get 6 into 10 days to par 6 meter cube.

01:46 Now the weight of the water will be equals to volume into 10 into 10.

01:55 Density that is row so here volume that we have calculated 6 into 10 days to power 6 here it is also into g into row where row of the water is 10 days to part 3 that is thousand into 9 .8 so after multi -clines we get 58 .86 into 10 days to 9 newton so here we get the 8 of the water that is w which is equals to 58 .8 6 into 10 days to 5 .9 newton.

02:33 Now as we know the gross capacity of power plant is w into h where the value of w is 58 .86 into 10 days to power 9 into h that is 250.

02:59 So here after multiplying we get 14 .715 into 10 days to par 9 in low watt.

03:09 Now for the first part of the problem, we have to calculate the average power produced...

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