00:01
There are supplied 317 kilojoule per cycle of the diesel engine operating on 2 to 7 gram of air at the pressure p1 which is 97 .91 kilo pascal and t1 value is 48 .9 degrees celsius at the end of compression we have pressure p2 equals 3930 kilo pascal assuming that air and product within the air within the cycle have air properties then determine the compression ratio the cutoff ratio net work done the thermal efficiency and mean effective pressure the process in the diesel cycle is 1 to 2 is the reversible adiative compression.
01:08
2 .3 is constant pressure heat addition.
01:12
3 .4 is the reversible adiabetic expansion and 4 to 1 is again the constant volume heat projection.
01:19
The heat supplied qs is 317 kilojou per cycle.
01:26
Mass given is 2 to 7 gram or in kj is 0 .27 .7.
01:35
The compression ratio are is the ratio of the volume before compression and after compression that is v1 by v2 the process one to two is reversible adiabetic then we will use the relation p1 v1 v1 to the power gamma is equal to p2 v2 v2 from here we can write v1 by v2 is equal to p2 by p2 by p1 raised to the power gamma 1 by gamma where gamma is 1 .4 then substitute the value and we get the value for v1 by v2 is equals to 3930 divided by 97 .91 raised to the power 1 by 1 .4 this will give us the value for the compression ratio r is equal to 13 .97 next to find the cutoff ratio cutoff ratio cutoff ratio is donated by rc which is v3 by v2 that is the volume after heat addition to the volume before the addition v3 by v2 is equal to t3 by t2 for 2 3 3 process which is the constant pressure heat addition p2 is equals to p3 then we can write q23 is equals to qs which is h3 minus h2 that is change in enthalpy which is mcp t3 minus t2 then from here we have to find the value for t3 t3 t3 minus t2 is equals to 1389 .5 calvin substitute the value of t2 which is 924 .7 138 9 .52 from here we get the value for t3 is 2314 .2 to calvin substituted into the equation for air we have used here cp equals 1 .0 .05 kilojoule per kg kelvin the value of t2 is find out by t1 v1 raised to the power gamma minus 1 is equal to t2 v2 raised to the power gamma minus 1.
04:07
From here, t2 by t1 is equal to v2 by v1 raised to the power gamma minus 1 or compression ratio raised to the power gamma minus 1.
04:17
We will substitute the value of r and gamma and thus we get to know the value of t2 from here 9 to 4 .7 kelvin, which is substituted in this equation to get the value of t3.
04:31
Now then we have t3 and t2 values.
04:35
The cutoff ratio is equal to 2314 .22 by 924 .7.
04:44
The cutoff ratio comes out to be 2 .5.
04:49
Rc is 2 .5...