00:01
Now in this question we are given that as we drill down to the earth, the temperature increases by 3 degrees celsius for every 100 meter of depth.
00:13
And oil wells can be drilled to a depth of 1 ,830 meter.
00:20
And from this height below the earth, water is pumped into the shaft and this steam is, and this steam is, is used as is used in a heat engine and the atmospheric temperature or the surrounding temperature we are given to it with 20 degrees celsius and we have to find two things in the in this question in the first part we have to find the maximum efficiency and in the second part we have to calculate that how much heat is absorbed from the interior of the earth each day when the work produce basically the power produce w dot is 2 .5 megawatt so we have to calculate this qh for a single day for single for one day we can write for one day now first we'll be calculating the maximum temperature up to a depth of 1830 meter now at the surface the temperature is 20 degrees celsius so th will be 20 degrees celsius plus now we are the hole is drilled up to 1830 meter and in every one meter three degrees raised so up to this much of meter how much degree would be raised so three divided by hundred this is three degrees celsius and this is hundred meter so from here we'll get this value to be 74 .9 degrees celsius which will be equals to 348 kelvin therefore the temperature up to a depth of 1 ,8 130 meter would be around 348 kelvin and tc we are given to be in 20 degrees celsius means 293 kelvin now we can easily calculate the maximum efficiency so the maximum efficiency is given by 1 minus t l divide with th 2.
02:35
2 .93 divide with th we have 3 48 so that will be equals to 0 .158 or we can write 15 .8 percentage.
02:54
So the maximum efficiency comes out to be as let me write here.
02:59
This is 15 .8 percentage.
03:04
Now next we're going to solve the b part of the problem where we have to calculate that how much energy is absorbed from the interior of the earth each day.
03:16
So qh we have to calculate.
03:20
So b part we're going to solve now.
03:24
Now efficiency, we know it is also equals to the power produce or the work produce divide with the heat input.
03:32
Qh.
03:33
This is w dot.
03:34
This is qh dot because you are finding the rate first.
03:38
So efficiency we have calculated to be 0 .158 and power we are given as 2 .5 megawatts.
03:46
So mega what means 10 to the power 6 watt or you could write.
03:51
Jule per second...