00:01
In this question it has been given that there is a dam of, let's say, given area a equals to 3 x10 to the power 6 meter square, that is the base area.
00:22
Now, further it has been given that in order to produce 100 kilowatt hours of electrical energy, how much volume of water must pass through the dam.
00:36
Now, e equals to 100 kilowatt hour will be, sorry, 1000 kilowatt hour will be 1000 into 10 to the power 3 into 3 ,600 jule.
00:54
When we convert 1 ,000 kilowatt hour into jule, this will be the net magnitude.
01:05
Now let's look this diagram.
01:08
So water must be let's suppose field width up to this height.
01:13
So when we open the base or the tap at the bottom, the water level will get on decrease.
01:21
So in order to find that how much energy it will take to produce this much amount of energy e, or this much amount of energy in joules, one need to also find that in the upper layer, let's say at given height h, how this much amount of energy must be stored in this level, in this layers of water in the form of gravitational potential energy.
02:01
Now we know that let's say gravitational potential energy pe equals to mgh.
02:09
Now in differential form we can write that in the mass can be written as row into, that is row is here density into area of cross section into volume, sorry, row into volume, that is density into volume into gh and further it can also be written as row that is the density and volume can be written as area of cross section into height h, into height dh into g h because here i have only taken let's suppose that this is the water this is the water so a thin layer of this i have taken dh so for this the gravitational potential energy is stored in this thin layer of water due to its height from the datum line that is it the base can be written as let's say dpe that is small fractional amount of potential energy stored in this water level or this water layer can be written as row into a into g into h into d.
03:40
So this is the basically height from the base and this is the width of this layer.
03:47
Now integrating that we will get if we do the integration both sides we will get potential energy stored let's suppose if we do the integration with limits h1 and h2 that is the upper limit is that's say h1 and the lower limit is h2 so we will have p equals to row a g into h1 into h2 row ag is the constant so i have taken out of the integration sign and here it will be h d h so solving this we will get p e equals to row a g into h2 square minus h1 squared divided by 2 where h sorry yes h2 is the upper limit and h1 is the lower limit one need to correct here that h2 is the maximum height and h1 is the minimum height from the base which will be enough to produce this much amount of energy when water starts falling down or coming out of the tap so this must this amount of energy stored sorry must be equals to the given amount of the required amount of electrical energy produced by the turbine.
05:28
So let's write the equation, row a, g into h2 square minus h1 square divided by 2 into efficiency.
05:40
Because all the energy has not been converted into electrical energy, so into efficiency that is 0 .9 must be equals to the amount of energy here.
05:52
So here it is 10 to the power 3.
05:58
6 into 10 to the power 9 jule.
06:05
Now in this question it has been given that the water level in the lake was initially at initially at the height of 150 meter.
06:15
So h2 is 150 meter.
06:18
So one can write here row into a let's start putting the value also.
06:24
So row is here 10 to the power 3.
06:27
Area of cross section is 3.
06:29
3 .10 to the power 6 meter square.
06:34
G let's take 10 meter per second square in spite of 9 .8 meter per second square and h2 will be 150 square minus h1 square that is the lower limit of, sorry, the lower height of the water level when the water level starts decreasing into 2 into g.
06:59
0 .9 equals to 3 .6 into 10 to the power 9.
07:05
Now we all required to solve h1...