In the previous section we saw the details about cyclic process. We saw the need to maximize the area enclosed by the curves. In this section, we will see how this can be achieved
Some basics can be written in 14 steps:
1. Consider the last fig.12.14 that we saw in the previous section
• It shows a cyclic process A-B-A
♦ The forward process A-B is along the red curve
♦ The backward process B-A is along the green curve
• This is shown in fig.12.15 below also
2. In the forward process, volume of the gas increases from VA to VB
♦ We can consider this as the upward motion of the piston
♦ So work is done by the gas during the forward process
3. In the backward process, volume of the gas decreases from VB to VA
♦ We can consider this as the downward motion of the piston
♦ So work is done on the gas during the backward process
4. We can write:
♦ The piston starts from the extreme low point A
♦ Rises up to the extreme high point B
♦ Returns to the extreme low point A
• This completes one cycle
5. Consider the forward process A-B
♦ Let the heat absorbed by the gas during this process be QAB
♦ Let the work done by the gas during this process be WAB
• This is shown in fig.12.15 below:
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| Fig.12.15 |
• Then we get: UB = UA + QAB - WAB
⇒ UB - UA = QAB - WAB
6. Consider the backward process B-A
♦ Let the heat released by the gas during this process be QBA
♦ Let the work done on the gas during this process be WBA
• Then we get: UA = UB - QBA + WBA
⇒ UA - UB = -QBA + WBA
7. Consider the results in (5) and (6)
• The left sides are numerically equal. Only difference is in the signs
8. Let us multiply both sides of (6) by '-1'
• We get: UB - UA = QBA - WBA
9. Now the left sides in (5) and (8) are the same
• So the right sides must be equal. We get:
QAB - WAB = QBA - WBA
⇒ WAB - WBA = QAB - QBA
10. Consider the left side of the result in (9):
♦ WAB is the work done by the gas when the piston moves up
♦ WBA is the work done on the gas when the piston moves down
• So (WAB - WBA) is the net work done by the gas
♦ We will denote this net work as W
♦ So we can write: W = WAB - WBA
11. Consider the right side of the result in (9):
♦ QAB is the heat absorbed by the gas when the piston moves up
♦ QBA is the heat released by the gas when the piston moves down
• So (QAB - QBA) is the net heat absorbed by the gas
12. From (9), we get Eq.12.8: W = QAB - QBA
• W is the net work
• So it is clear that, the net heat absorbed is converted into net work
13. Now, W is the area enclosed by the curves
• We want to maximize this area
• It is obvious from Eq.12.8 that:
For a given QAB, the net work W will be maximum when QBA is minimum
14. So we need to ensure that, the gas releases minimum possible heat
• Lesser the 'released heat', greater is the 'net work'
Heat engines
• We saw that, the gas receives heat and gives us a net work
• The net work is obtained when the piston completes one cycle
• If the cycle repeats continuously, we will get continuous work
• Such an arrangement which can produce continuous work by absorbing heat is called a heat engine
The main features of the heat engine can be written in 6 steps:
1. In the heat engine, all the molecules of the gas together form the ‘system’
2. We need to supply heat to this system. It can be done in any one of the two methods given below:
(i) Put the cylinder in contact with a heat reservoir at a high temperature T1
(ii) If the gas is combustible, ignite it using an electric circuit
3. As a result of the heat (QAB), the gas expands and does 'useful work' on the piston
• This piston can be connected to the wheels of an automobile
4. When the upward motion of the piston is complete, ‘half of the first cycle’ is complete
• Next, we need to bring the piston to it’s original lower position. Only then can the gas expand in the next cycle
5. For that, we push the piston downwards
• During this process, the cylinder is kept in contact with a cold reservoir at temperature T2
♦ So some heat (QBA) flows out of the system
• When the piston reaches the lowermost point, the ‘second half of the first cycle’ is complete
• Now the piston is ready for the second cycle
6. The above steps can be schematically represented as shown in fig.12.16 below:
 |
| Fig.12.16 |
• We see that:
♦ one item enters the system. It is: QAB
♦ two items leave the system they are: W and QBA
• So QAB must be equal to (W + QBA)
• Indeed, by rearranging Eq.12.8, we get: W + QBA = QAB
• This implies that
♦ part of QAB is converted into useful work
♦ the remaining part is released into the surroundings
Efficiency of a heat engine
This can be explained in 3 steps:
1. If we can obtain 'more useful work (W)' by supplying a 'less quantity of heat (QAB)', we can say that, the efficiency of the engine is high
2. Efficiency is denoted using the symbol 𝜼
• Mathematically, we can write Eq.12.9: $\mathbf\small{\rm{\eta=\frac{W}{Q_{AB}}}}$
• It is clear that, if W is high and QAB is low, 𝜼 will be high
3. Let us see the possible values of 𝜼:
It can be written in 4 steps:
(i) We have Eq.12.8: W = QAB – QBA
• So it is clear that W will be always less than QAB
• So $\mathbf\small{\rm{\frac{W}{Q_{AB}}}}$ will be always less than 1
• It will be a proper fraction. Engineers will be trying to obtain a high fraction like 9/10 or 95/100
(ii) We can derive another expression for 𝜼:
• Dividing both sides of Eq.12.8 by QAB, we get:
$\mathbf\small{\rm{\frac{W}{Q_{AB}}=\frac{Q_{AB}}{Q_{AB}}-\frac{Q_{BA}}{Q_{AB}}}}$
• Thus we get Eq.12.10: $\mathbf\small{\rm{\eta=1-\frac{Q_{BA}}{Q_{AB}}}}$
(iii) So it is clear that, if QBA is equal to zero, W will be equal to the absorbed heat
• That means, all the heat is converted into useful work
We will get 𝜼 = 1
(iv) This can be achieved if the 'heat released (QBA)' can be made zero
• But such an engine with 'efficiency 1' is not possible. We will see the reason in a later section
In the next section, we will see the reverse of the heat engine cycle
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