0:00
Hi there.
00:01
So for this problem, we are told that an ideal auto cycle has a compression ratio of 8.
00:07
At the beginning of the compression process, earth is at a pressure.
00:21
We're going to call this pressure, pressure, pressure 1, and that is equal to 95 kilo -pascal.
00:30
And at a temperature, temperature 1 of 27 celsius degrees.
00:40
So we know that this corresponds to 300 kelvin.
00:45
And with a heat that is also given, and that value is 750 kilojoules per kilogram.
01:03
Of heat that is transferred to earth during the constant volume heat addition process.
01:11
So taking into account the variation of specific heats with temperature, we need to determine, so for part a of this problem, the pressure and temperature at the end of the heat addition process.
01:29
So the earth standard assumptions are applicable.
01:35
So what we need to do in here is that the process from 1 to 2, as is shown in this figure, are going to be the following.
01:47
So we know that the temperature 1 is given.
01:51
And also we know that the heat 1 is going to be equal to 240 .07 kilojoules per kilogram.
02:08
And then the ratio at r1 is equal to 621 .2.
02:19
So to obtain this ratio of r2, which is going to be b2 divided by b1 times this.
02:34
So we will have that this is 1 divided by art.
02:37
So this times this.
02:40
And this will be 1 divided by 8 because remember that we given that r in this case, where r is the compression ratio is equal to 8.
02:52
So we will have this is 1 divided by 8 times 621 .2.
02:57
So this will give us a value of 77 .65.
03:02
So with this, we can now pass to a temperature 2, that is 6773 .1 kelvin to a heat 2.
03:14
Which is equal to 491 .2 kilojoules per kilogram.
03:21
So now we use the equation of an ideal gas, which is the pressure 2 times the volume 2, and this divided by the temperature 2, is equal to the pressure 1 times the volume 1, and this divided by the temperature 1.
03:37
So if we solve for the pressure or 2, we will obtain that this is the ratio between the volumes times the ratio between the temperatures 1 and 2, and this times the initial pressure p1.
03:49
So we now substitute all of these values in order to obtain the pressure 2.
03:53
So we have that this ratio in here is equal to 8.
03:59
So we will have that this is 8 times the temperature 2, that is 673 .1 kelvin, divided by the temperature 1, which is 300 kelvin and this times the pressure 1 which is 95 kilo pascal so from this we obtain 1 ,705 kilo pascal so now we pass from the process 2 to the process 3 in here as you can see in the picture now in that case we know that the now the what is going to be constant in that case is the volume.
04:49
So the volume is a constant heat addition.
04:57
So we will have that.
04:59
The heat between the points 2 and 3 is going to be the heat at 0 .3 minus the heat at 0 .2, which is the point 3 is equal to the heat of 0 .2 plus this in here.
05:17
So we will have that that is 491 .2 plus 750, which is the be given value.
05:26
So from this we'll obtain 1 ,241 .2 kilojoules per kilogram.
05:34
And from this, we now pass to the temperature 3, which is 1 ,539 kelvin, and the volume at this point 3, which is 6 .58.
05:49
Now with this, we use the same equation as before for r &l gas, which is the pressure 3 times the volume 3, and this divided by the temperature 3 is equal to the pressure 2 times the volume 2 divided by the temperature 2.
06:04
So from this, we will find that the pressure 3 is equal to the temperature 3 divided by the temperature 2 times the pressure 2.
06:15
And this is going to be 1 ,539 kelvin, and this divided by 673 .1 kelvin.
06:25
And this times 1 ,705 kilo pascal...