$\sum{\frac{Q}{T}}$ is zero

Question

I was going through a proof in my book and couldn't understand the last part of the proof. I mean I understand

$$\frac{Q_1^{\prime}}{T_1}-\frac{Q_2^{\prime}}{T_2}\leq 0$$

But after that how did author arrived at end ?

What i am saying is

$$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq 0$$ And hence for total cycle it should be $$\frac{Q_1}{T_1}-\frac{Q_2}{T_2}+\frac{Q_3}{T_3}-\frac{Q_4}{T_4}........\leq 0$$ not $$\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\frac{Q_3}{T_3}+\frac{Q_4}{T_4}........\leq 0$$ 

Or

$$\frac{Q_1^{\prime}}{T_1}-\frac{Q_2^{\prime}}{T_2}\leq 0$$ Let's take it as $a-b\leq 0 $ and later it says $$\sum{\frac{Q}{T}}\leq 0$$ Which means $a+b\leq 0$

Please help

Answer 1

What the author is saying is let $Q_2$ be the heat put in the system and $Q_1$ the heat taken out of the system. Then

$$ \frac{Q_1}{T_1}-\frac{Q_2}{T_2}\leq0\ . $$

Now, one could have more heat taken out of the system, say $\{Q_a\}$, $a=1,\ldots,p$ and more put in the system $\{Q_b\}$, $b=p+1,\ldots,n$ (what I mean by more is in more than one step). Here we take $Q_a>0$ and $Q_b<0$. Then what is meant by the formula is that the sum of all these must be zero or negative. In other words

$$ \underbrace{\frac{Q_1}{T_1}+\frac{Q_2}{T_2}+\cdots+\frac{Q_p}{T_p}}_{\text{Heat taken out}}+\underbrace{\frac{Q_{p+1}}{T_{p+1}}+\cdots+\frac{Q_{n-1}}{T_{n-1}}+\frac{Q_{n}}{T_{n}}}_{\text{Heat put in}}\leq0\ . $$

One can then write this in a more compact way as

$$ \sum_{i=1}^{n}\frac{Q_i}{T_i}\leq0\ , $$

or, even in a more compact form where the summation indices are implicit,

$$ \sum\frac{Q}{T}\leq0\ . $$

The compact notation is useful if the idea is clear, but the more explicit version is better to getting the hang of what is going on. Hope it helps!