00:01
Hello, here we have to solve the following problem.
00:03
The efficiency of the cycle 1 -2 -3 -1 is 20 % and the efficiency of the cycle 1 -3 -4 -1 is 10%.
00:19
We have to determine the efficiency of the cycle which is 1 -2 -3 -4 -1.
00:29
So first of all let's look at the cycle 1 -2 -3 -1.
00:33
Its efficiency is work is actually a q of 1 2 plus q of 2 3 minus q of 3 1 divided by q12 plus q 2 3 times 100 % on the other hand for the cycle 13 414 its efficiency is q13 plus q13, actually q13 minus q34.
01:55
And actually, you know, for the simplicity, let's label all the q's as absolute values, because it will help us to avoid the errors with signs.
02:15
And it means that q1, 2, 3, 4, the one which we have to calculate is q12 plus q2 -2 minus q -24 and minus q41.
02:43
That's divided by q12 plus q22 plus q -2 3 -3 times 100.
02:52
Percent so it basically means that we have to combine equation one and two to get somehow equation three let's see how this could have worked so here we will rewrite the first equation so we will simplify this one minus q13 divided by q1 2 plus q 2 3 and that is all multiplied by 100 % in the second equation that is 1 minus q34 over q13 minus q41 or q14 divided by q13 times 100 % and in the third equation that is 1 plus q34 actually 1 minus yeah q34 plus q34 plus q14 divided by q12 plus q2 plus q23 times 100%.
04:30
So now let's see how then this information could have helped us so from the first equation we can express q1 2 plus q2 3 let's do this so efficiency of 1 2 3 1 over 100 % which is 0 .2 equals to 1 minus q13 over q12 plus q2 therefore q13 over this denominator equals to 0 .8 and thereby q12 plus q23 equals to q13 over 0 .8 so now let's look at, yeah, let's keep this ratio in mind.
06:12
And let's look at the second equation.
06:15
So from the second equation, 0 .1 equals to 1 minus q31, or q34 plus q41, q14 plus q14, q14 divided by q13 and here we have to express q34 plus q14...