Chapter 12.10 - Second Law of Thermodynamics

In the previous section, we saw the COP of the Carnot refrigerator. In this section, we will see the second law of thermodynamics and it's applications in Carnot engine and refrigerator

The Second law of thermodynamics

◼  The second law of thermodynamics states that:
When a spontaneous process occurs, the entropy of the universe always increases
   ♦ So a process will take place only in that direction which causes an increase in entropy
   ♦ Entropy is a measure of randomness or disorder
• Links to detailed notes on entropy and some related topics are given below:

   ♦ Spontaneity

   ♦ Entropy

   ♦ Second law of thermodynamics

   ♦ Gibbs free energy

• Based on the above discussion, we can write the reason for phenomena such as:
   ♦ Heat never flows from a cold object to a hot object
         ✰ Heat always flow from hot object to cold object
         ✰ A hot object has greater entropy than a cold object
   ♦ Water never spontaneously become ice
         ✰ Water has greater entropy than ice

• Application of second law to heat engines gives the explanation for why no heat engine can have 100 % efficiency • Application of second law to refrigerator gives the explanation for why no refrigerator can have a COP of infinity

 

Application to Carnot Engine

• Let us find the entropy changes in a Carnot engine
    ♦ For that, we consider one complete cycle of the engine
    ♦ There are two isothermal processes and two adiabatic processes in one complete cycle
• The PV diagram is shown again below:

• There is no exchange of heat during the adiabatic processes
    ♦ So there is no entropy changes during the adiabatic processes
• We need to consider the isothermal processes only. We can write it in 9 steps:
1. First isothermal process AB:
(i) In the first isothermal process, the hot reservoir gives away some heat
   ♦ Let this heat given away be Q_h
   ♦ Let Q_i(HR) be the initial heat content of the hot reservoir
   ♦ Let Q_f(HR) be the final heat content of the hot reservoir
• Then Q_h = Q_f(HR) – Q_i(HR)
• This Q_h will be a negative quantity because, Q_f(HR) will be less than Q_i(HR)
◼  So the entropy change suffered by the hot reservoir = $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$
(ii) The same heat content Q_h is gained by the gas
   ♦ Let Q_i(HG) be the initial heat content of the hot gas
   ♦ Let Q_f(HG) be the final heat content of the hot gas
   ♦ Then Q_h = Q_f(HG) – Q_i(HG)
• This Q_h will be a positive quantity because, Q_f(HG) will be greater than Q_i(HG)
◼  So the entropy change suffered by the hot gas = $\mathbf\small{\rm{+\frac{Q_h}{T_1}}}$
2. Second isothermal process AB:
(i) In the second isothermal process, the cold reservoir receives some heat
   ♦ Let this received heat be Q_c
   ♦ Let Q_i(CR) be the initial heat content of the cold reservoir
   ♦ Let Q_f(CR) be the final heat content of the cold reservoir
• Then Q_c = Q_f(CR) – Q_i(CR)
• This Qc will be a positive quantity because, Q_f(CR) will be greater than Q_i(CR)
◼  So the entropy change suffered by the cold reservoir = $\mathbf\small{\rm{+\frac{Q_c}{T_2}}}$
(ii) The same heat content Q_c is lost by the gas
   ♦ Let Q_i(CG) be the initial heat content of the cold gas
   ♦ Let Q_f(CG) be the final heat content of the cold gas
   ♦ Then Q_c = Q_f(CG) – Q_i(CG)
• This Q_c will be a negative quantity because, Q_f(CG) will be less than Q_i(CG)
◼  So the entropy change suffered by the cold gas = $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$
3. So now we have four entropy changes:
(i) Entropy change suffered by the hot reservoir, $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$. This we calculated in 1(i)
(ii) Entropy change suffered by the hot gas, $\mathbf\small{\rm{\frac{Q_h}{T_1}}}$. This we calculated in 1(ii)
(iii) Entropy change suffered by the cold reservoir, $\mathbf\small{\rm{+\frac{Q_c}{T_2}}}$. This we calculated in 2(i)
(iv) Entropy change suffered by the cold gas, $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$. This we calculated in 2(ii)
4. We have the basic equation:
Entropy change suffered by the universe
= Entropy change suffered by the system + Entropy change suffered by the surroundings
5. Let us calculate each item on the right side of the above equation:
(i) Entropy change suffered by the system
= Entropy change suffered by the gas = Item 3(ii) + Item 3(iv)
= $\mathbf\small{\rm{\frac{Q_h}{T_1}-\frac{Q_c}{T_2}}}$
(ii) Entropy change suffered by the surroundings
= Entropy change suffered by the reservoirs = Item 3(i) + Item 3(iii)
= $\mathbf\small{\rm{-\frac{Q_h}{T_1}+\frac{Q_c}{T_2}}}$
6. Consider the Eq.12.15 that we derived in section 12.8:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \frac{T_2}{T_1}}}$
This is same as: $\mathbf\small{\rm{\frac{Q_c}{Q_h} =  \frac{T_2}{T_1} }}$
⇒ $\mathbf\small{\rm{\frac{Q_c}{T_2} =  \frac{Q_h}{T_1} }}$
• Based on this result,
   ♦ 5(i) will become zero
   ♦ 5(ii) will also become zero
7. So we can write:
◼  Entropy change suffered by the system = 0
◼  Entropy change suffered by the surroundings = 0
• So the result in (4) becomes:
◼  Entropy change suffered by the universe = 0
• That means, when one cycle of the Carnot engine is complete, the universe suffers zero entropy change
8. This is an ideal situation
• It shows that, the Carnot engine is most efficient
• All real engines will cause the universe to suffer a positive entropy change
• This is because, all real engines dump some net heat (caused due to friction) into the surroundings
• The Carnot engine is a theoretical engine. We assume that, there is no friction
9. Now let us see if we can attain 100 % efficiency. It can be written in 6 steps:
(i) If there is to be 100% efficiency, Qc must be zero
• That is., there should be no heat exchange with a cold reservoir
• This condition of 'no heat exchange with cold reservoir' can be easily seen from the schematic diagram of a heat engine that we saw in a previous section. It is shown again below:

• It is clear that, for 100% efficiency, Qc must be zero
(ii) If Q_c is not to be given, the cold reservoir will be absent in the engine
• So there will not be any need for the three segments BC, CD and DA
• All the heat received from the hot reservoir will be converted into work
(iii) In such a situation,
    ♦ Entropy change suffered by the surroundings (hot reservoir) = $\mathbf\small{\rm{-\frac{Q_h}{T_1}}}$
    ♦ Entropy change suffered by the gas (system) = $\mathbf\small{\rm{\frac{Q_h}{T_1}}}$
(iv) Here also, the net entropy change of system and surroundings is equal to zero
• That means, the entropy change suffered by the universe is zero
• This does not violate the second law
(v) So we are inclined to adopt such an engine, where all heat is converted into work
• But such an engine can perform work only in one segment AB
• It will not return to it's original state
(If we try to return along the same path BA, the same work will have to be done on the gas, resulting in zero net work)
◼ So it cannot perform the next cycle. We can obtain continuous work only if the engine operates in cycles
(vi) Thus it is clear that, to obtain a cyclic process, some heat has to be definitely given to the cold reservoir
• When some heat is given to the cold reservoir, efficiency will become less than 100 %
• That is why, we cannot obtain 100 % efficiency

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Application to Carnot Refrigerator

• Let us find the entropy changes in a Carnot refrigerator
    ♦ For that, we consider one complete cycle of the refrigerator
    ♦ There are two isothermal processes and two adiabatic processes in one complete cycle
• The PV diagram is shown again below:

• There is no exchange of heat during the adiabatic processes
    ♦ So there is no entropy changes during the adiabatic processes
• We need to consider the isothermal processes only
• Those isothermal processes are just the reverse of what we saw in the engine
• So it is easy to prove that, in the case of refrigerator also, the entropy change suffered by the universe is zero
   ♦ The steps are left to the reader
• Our next aim is to check whether it is possible to make a refrigerator with COP infinity. It can be written in 6 steps:
(i) If there is to be infinite COP, W must be zero
• That is., there should be zero work requirement
    ♦ The cooling must be accomplished with out external work
• This condition can be easily seen from the schematic diagram of a refrigerator that we saw in a previous section. It is shown again below:

• It is clear that, for infinite COP, W must be zero
(ii) If W is not to be given, the segments CB and BA will be absent in the PV diagram below
   ♦ Because, W is applied during those segments
   ♦ The gas gets compressed during those segments
(iii) If those two segments are absent, entropy changes occur only during DC
    ♦ During DC, the entropy change suffered by the surroundings (hot reservoir) = $\mathbf\small{\rm{-\frac{Q_c}{T_2}}}$
    ♦ During DC, the entropy change suffered by the gas (system) = $\mathbf\small{\rm{\frac{Q_c}{T_2}}}$
(iv) Here also, the net entropy change of system and surroundings is equal to zero
• That means, the entropy change suffered by the universe is zero
• This does not violate the second law
(v) So we are inclined to adopt such a refrigerator, where no external work is required
(vi) But, if we avoid paths CB and BA, the gas cannot reach the hot temperature T₁
• It is essential to reach the hot reservoir temperature T₁ to dump the heat
• If we decide to retrace the path CD, the absorbed heat will be given back to the  cold reservoir
   ♦ This will not effect cooling
◼ We can obtain continuous extraction of heat only if the refrigerator operates in cycles
(vi) Thus it is clear that, to obtain a cyclic process, some work has to be definitely done
• When some work is done, COP will become less than infinity
• That is why, we cannot obtain infinite COP

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We have completed our present discussion on thermodynamics. In the next section, we will see some solved examples related to the various topics that we saw in this chapter



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Chapter 12.8 - Efficiency of a Carnot Engine

In the previous section, we saw the basic processes in a Carnot engine. In this section, we will see the work done during each of those processes. For easy reference, the final PV diagram is shown below:

Fig.12.23

First segment from A to B
• In this segment, the gas is subjected to an isothermal expansion
• We know that, the work done by the ideal gas during the isothermal expansion will be equal to: $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$ (see Eq.12.3 at end of section 12.2)

Second segment from B to C
• In this segment, the gas is subjected to an adiabatic expansion
• We know that, the work done by the ideal gas during the adiabatic expansion will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$ (see Eq.12.6 at end of section 12.3)

Third segment from C to D
• In this segment, the gas is subjected to an isothermal compression
• We know that, the work done on the ideal gas during the isothermal compression will be equal to: $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$

Fourth segment from D to A
• In this segment, the gas is subjected to an adiabatic compression
• We know that, the work done by the ideal gas during the adiabatic compression will be equal to: $\mathbf\small{\rm{\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}}}$

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Now we can find the efficiency of the Carnot engine. It can be written in 9 steps:
1. Net work, W = Work done by the gas - Work done on the gas
   ♦ Work done by the gas = Work done during segments one and two
   ♦ Work done on the gas = Work done during segments three and four
• Thus we get:
W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1}\right )- \left ( n R T_2\,\,ln \frac{V_C}{V_D}+\frac{nR(T_1 \, - \, T_2)}{\gamma\, - \,1} \right)}}$
⇒ W = $\mathbf\small{\rm{\left (n R T_1\,\,ln \frac{V_B}{V_A}- n R T_2\,\,ln \frac{V_C}{V_D} \right)}}$
2. We know that, efficiency of a heat engine is given by:
$\mathbf\small{\rm{\eta = \frac{Net \; Work}{Heat \; Absorbed}}}$ (see Eq.12.9 in section 12.5)
3. Also, we have derived another form of the above equation:
$\mathbf\small{\rm{\eta = 1-\frac{Heat \;Rejected}{Heat \; Absorbed}}}$ (see Eq.12.10 in section 12.5)
• It is easier to find efficiency using this equation
4. So our next task is to find the 'Heat absorbed' and the 'Heat rejected'
• It can be done in steps:
(i) We know that, heat absorption takes place only during the first segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(ii) We know that, heat rejection takes place only during the third segment
• It is an isothermal process
• In an isothermal process, Heat absorbed/rejected is equal to the work done during that process
• So we get: Heat rejected = $\mathbf\small{\rm{n R T_2\,\,ln \frac{V_C}{V_D}}}$
5. So the result in (3) becomes: $\mathbf\small{\rm{\eta = 1-\frac{n R T_2\,\,ln \frac{V_C}{V_D}}{n R T_1\,\,ln \frac{V_B}{V_A}}}}$
⇒ $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right ) \left (\frac{ln \frac{V_C}{V_D}}{ln \frac{V_B}{V_A}}\right )}}$
6. We see that, the right side of the above equation involves temperatures and volumes. So our next task is to find the relations between them
• For that, we consider the two adiabatic processes in the cycle. It can be written in two steps:
(i) For the  adiabatic process BC, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(see Eq.12.7 of section 12.3)
(ii) Similarly, for the adiabatic process DA, we have the relation:
$\mathbf\small{\rm{\left (\frac{V_D}{V_A} \right )^{\gamma - 1}=\left (\frac{T_1}{T_2} \right )}}$
(iii) The right sides in (i) and (ii) are the same. So we get:
$\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )^{\gamma - 1}=\left (\frac{V_D}{V_A} \right )^{\gamma - 1}}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_B} \right )=\left (\frac{V_D}{V_A} \right )}}$
⇒ $\mathbf\small{\rm{\left (\frac{V_C}{V_D} \right )=\left (\frac{V_B}{V_A} \right )}}$
7. We see the above result in (5)
So the result in (5) becomes:
Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
8. Two expressions for efficiency:
   ♦ The above Eq.12.14 gives us an expression for efficiency
   ♦ Step (3) also gives us an expression for efficiency
• So we can equate the two:
$\mathbf\small{\rm{1-\frac{Heat \;Rejected}{Heat \; Absorbed} = 1- \left (\frac{T_2}{T_1} \right )}}$
• Thus we get Eq.12.15:
$\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{T_2}{T_1} \right )}}$

⇒ $\mathbf\small{\rm{\frac{Heat \;Rejected}{Heat \; Absorbed} =  \left (\frac{Temperature \; of \; cold \; reservoir}{Temperature \; of \; hot \; reservoir} \right )}}$
9. Let us use try the results in (1) and (2) to find efficiency. it can be written in 3 steps:
(i) Using the result in (6), the result in (1) becomes:
W = $\mathbf\small{\rm{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}}$
(ii) Using the result in 4(i), we have:
Heat absorbed = $\mathbf\small{\rm{n R T_1\,\,ln \frac{V_B}{V_A}}}$
(iii) Applying the above two results into (2), we get:
$\mathbf\small{\rm{\eta = \frac{n R \,ln \frac{V_B}{V_A}\left ( T_1\,- T_2 \right)}{n R T_1\,\,ln \frac{V_B}{V_A}}=\frac{T_1 - T_2}{T_1}}}$
⇒ $\mathbf\small{\rm{\eta =1-\left (\frac{T_2}{T_1} \right )}}$
This is the same result as in Eq.12.14

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Let us see a solved example
Solved example 12.10
Three heat engines operate between the following temperature differences
    ♦ Engine A: T₁ = 1000 K, T₂ = 700 K
    ♦ Engine A: T₁ = 800 K, T₂ = 500 K
    ♦ Engine A: T₁ = 600 K, T₂ = 300 K
• Which is the most efficient engine?
• Which is the least efficient engine?
Solution:
1. Efficiency of a heat engine is given by Eq.12.14: $\mathbf\small{\rm{\eta = 1- \left (\frac{T_2}{T_1} \right )}}$
• So we get:
    ♦ Efficiency of Engine A = $\mathbf\small{\rm{\eta = 1- \left (\frac{700}{1000} \right )}}$ = 0.3
    ♦ Efficiency of Engine B = $\mathbf\small{\rm{\eta = 1- \left (\frac{500}{800} \right )}}$ = 0.375
    ♦ Efficiency of Engine C = $\mathbf\small{\rm{\eta = 1- \left (\frac{300}{600} \right )}}$ = 0.5
2. We can write:
    ♦ Engine C is the most efficient
    ♦ Engine A is the least efficient
3. Dependence of efficiency on temperature:
    ♦ We see that, the temperature difference is the same for all three engines
    ♦ In such a situation, the engine with the smallest T₂ has the most efficiency 

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In the next section, we will see Carnot refrigerator


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Chapter 12.6 - The Refrigerator

In the previous section we saw the details about heat engine. In this section, we will see the reverse of the heat engine cycle

• In the heat engine, we denoted the cycle as A-B-A. So the reverse cycle will be: B-A-B.
• However, to avoid confusion, we will use the letters M and N instead of A and B
• The basics of the reverse cycle can be written in 12 steps:
1. Consider fig.12.17 below
• It shows a cyclic process M-N-M
    ♦ The forward process M-N is along the green curve
    ♦ The backward process N-M is along the red curve

Fig.12.17

2. In the forward process, volume of the gas decreases from V_M to V_N
    ♦ We can consider this as the downward motion of the piston
    ♦ So work is done on the gas during the forward process
3. In the backward process, volume of the gas increases from V_N to V_M
    ♦ We can consider this as the upward motion of the piston
    ♦ So work is done by the gas during the backward process
4. We can write:
    ♦ The piston starts from the extreme high point M
    ♦ Lowers down to the extreme low point N
    ♦ Returns to the extreme high point M
• This completes one cycle
5. Consider the forward process M-N
    ♦ Let the heat released by the gas during this process be Q_MN
    ♦ Let the work done on the gas during this process be W_MN

• Then we get: U_N = U_M - Q_MN + W_MN
⇒ U_N - U_M = -Q_MN + W_MN

6. Consider the backward process N-M
    ♦ Let the heat absorbed by the gas during this process be Q_NM
    ♦ Let the work done by the gas during this process be W_NM
• Then we get: U_M = U_N + Q_NM - W_NM
⇒ U_M - U_N = Q_NM - W_NM
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 (5) by '-1'
• We get: U_M - U_N = Q_MN - W_MN
9. Now the left sides in (6) and (8) are the same
• So the right sides must be equal. We get:
Q_NM - W_NM = Q_MN - W_MN
⇒ W_NM - W_MN = Q_NM - Q_MN
• Multiplying both sides by -1, we get:
W_MN - W_NM = Q_MN - Q_NM
10. Consider the left side of the result in (9):
    ♦ W_MN is the work done on the gas when the piston moves down
    ♦ W_NM is the work done by the gas when the piston moves up
• So (W_MN - W_NM) is the net work done on the gas
    ♦ We will denote this net work as W
    ♦ So we can write: W = W_MN - W_NM
11. Consider the right side of the result in (9):
    ♦ Q_MN is the heat released by the gas when the piston moves down
    ♦ Q_NM is the heat absorbed by the gas when the piston moves up
• So (Q_MN - Q_NM) is the net heat absorbed by the gas
12. From (9), we get Eq.12.11: W = Q_MN - Q_NM
• W is the net work done
• So it is clear that, the net work done is same as the net heat absorbed
 

Refrigerators

• We saw that, the gas receives a net work (W) and absorbs a net heat
• The 'net heat absorbtion' occurs when the piston completes one cycle
• If the cycle repeats continuously, we will get 'continuous absorbtion of heat'
• Such an arrangement which can produce 'continuous heat absorbtion' by receiving external work is called a refrigerator 

The main features of the refrigerator can be written in 6 steps:
1. In the refrigerator, all the molecules of the gas together form the ‘system’
2. We need to do work on this system. It can be done by compressing the gas using the piston
3. As a result of the work, the gas gives off heat (Q_MN) and becomes a liquid-vapour mixture
    ♦ Note that a suitable gas like freon must be used
    ♦ Such gases are called refrigerants
• Q_MN is given off into a hot reservoir which is kept at a temperature T₁
(In practice, this hot reservoir is the 'space outside the refrigerator'. This space will be at room temperature)
4. When the downward motion of the piston is complete, ‘half of the first cycle’ is complete. We reach the point N
• Next we need to bring the piston to the original higher position M. Only then , can the gas be compressed in the next cycle
5. For that, the cylinder is placed in contact with the object to be cooled
• This object is called the 'cold reservoir'. It is at a temperature T₂
• Heat Q_NM flows from this cold reservoir into the gas
• As a result, the gas expands and reaches the initial point M
• Now the piston is ready for the second cycle
(It is called the cold reservoir because, objects inside the refrigerator will be at a lower temperature. Even though the objects inside the refrigerator are cool, outside heat will penetrate through the walls of the refrigerator and try to warm the objects. We have to remove that heat which reaches the interior of the refrigerator)
6. The above steps can be schematically represented as shown in fig.12.18 below:

Fig.12.18

 • We see that:
    ♦ two items enter the system. They are: W and Q_NM
    ♦ one item leaves the system. It is: Q_MN
• So Q_MN must be equal to (W + Q_NM)
• Indeed, by rearranging Eq.12.11, we get: W + Q_NM = Q_MN
• This implies that:
    ♦ The external work done (W)
    ♦ Plus
    ♦ The heat (Q_NM) removed from the 'object to be cooled'
    ♦ Gets dumped into the space surrounding the refrigerator
          ✰ When it is dumped, the system (gas) becomes ready for the next cycle

Coefficient of performance of a refrigerator

This can be explained in 3 steps:
1. If we can absorb 'more heat (Q_NM)' by doing a 'less net work (W)', we can say that, the coefficient of performance of the refrigerator is high
2. This coefficient is denoted using the symbol 𝛂
• Mathematically, we can write Eq.12.12: $\mathbf\small{\rm{\alpha=\frac{Q_{NM}}{W}}}$
• It is clear that, if Q_NM is high and W is low, 𝛂 will be high
3. Let us see the possible values of 𝛂:
It can be written in 4 steps:
(i) We have Eq.12.11: W = Q_MN - Q_NM
• This can be rearranged as: Q_NM = Q_MN - W
• So it is clear that W will be always less than Q_NM
• So $\mathbf\small{\rm{\frac{Q_{NM}}{W}}}$ will be always greater than 1
(ii) If the denominator W becomes smaller and smaller, the coefficient will become larger and larger
• If W becomes zero, the coefficient will become infinity. Such a refrigerator is not possible. We will see the reason in the next section
■ Since W cannot become zero, we can write:
It is impossible to remove heat from a body without doing external work W
(iii) We can derive another expression for 𝛂:
• Dividing both sides of Eq.12.11 by W, we get:
$\mathbf\small{\rm{1=\frac{Q_{MN}}{W}-\alpha}}$
⇒ $\mathbf\small{\rm{\alpha=\frac{Q_{MN}}{W}-1\;\;\;\;=\frac{Q_{MN}}{Q_{MN}-Q_{NM}}-1\;\;\;\;=\frac{Q_{MN}-Q_{MN}+Q_{NM}}{Q_{MN}-Q_{NM}}}}$
• Thus we get Eq.12.13: $\mathbf\small{\rm{\alpha=\frac{Q_{NM}}{Q_{MN}-Q_{NM}}}}$

Heat pump

• Heat pumps are devices used to heat up the interior of a room when the surroundings of the room is cold
• Heat pumps work in the same way as the refrigerators
• A comparison can be written in 4 steps:
1. From where the heat is received:
• In a refrigerator, system receives heat (Q_NM) from a cold body
    ♦ This cold body is the object to be cooled
• In a heat pump also, the system receives heat (Q_NM) from a cold body
    ♦ This cold body is the surroundings of the room
2. Upon which the work is done:
• In a refrigerator, external work (W) is done on the system
• In a heat pump also, external work (W) is done on the system
3. To where the heat is dumped:
• In a refrigerator, the ‘Q_NM + W’ is dumped into the surroundings of the refrigerator
• In a heat pump, the ‘Q_NM + W’ is dumped into the interior of the room
4. Calculation of 𝛂:
• In a refrigerator, the benefit that we receive, is the Q_NM removed from the body
    ♦ So we use the equation: $\mathbf\small{\rm{\alpha=\frac{Q_{NM}}{W}}}$
• In a heat pump, the benefit that we receive, is the Q_MN available to heat the room
    ♦ So we use the equation: $\mathbf\small{\rm{\alpha=\frac{Q_{MN}}{W}}}$

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In the next section, we will see the Carnot engine.

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Chapter 12.5 - The Heat Engine

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 V_A to V_B
    ♦ 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 V_B to V_A
    ♦ 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 Q_AB
    ♦ Let the work done by the gas during this process be W_AB 
• This is shown in fig.12.15 below:

Fig.12.15

• Then we get: U_B = U_A + Q_AB - W_AB
⇒ U_B - U_A = Q_AB - W_AB
6. Consider the backward process B-A
    ♦ Let the heat released by the gas during this process be Q_BA
    ♦ Let the work done on the gas during this process be W_BA
• Then we get: U_A = U_B - Q_BA + W_BA
⇒ U_A - U_B = -Q_BA + W_BA
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: U_B - U_A = Q_BA - W_BA
9. Now the left sides in (5) and (8) are the same
• So the right sides must be equal. We get:
Q_AB - W_AB = Q_BA - W_BA
⇒ W_AB - W_BA = Q_AB - Q_BA
10. Consider the left side of the result in (9):
    ♦ W_AB is the work done by the gas when the piston moves up
    ♦ W_BA is the work done on the gas when the piston moves down
• So (W_AB - W_BA) is the net work done by the gas
    ♦ We will denote this net work as W
    ♦ So we can write: W = W_AB - W_BA
11. Consider the right side of the result in (9):
    ♦ Q_AB is the heat absorbed by the gas when the piston moves up
    ♦ Q_BA is the heat released by the gas when the piston moves down
• So (Q_AB - Q_BA) is the net heat absorbed by the gas
12. From (9), we get Eq.12.8: W = Q_AB - Q_BA
• 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 Q_AB, the net work W will be maximum when Q_BA 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 (Q_AB), 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 (Q_BA) 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: Q_AB
    ♦ two items leave the system they are: W and Q_BA
• So Q_AB must be equal to (W + Q_BA)
• Indeed, by rearranging Eq.12.8, we get: W + Q_BA = Q_AB
• This implies that
    ♦ part of Q_AB 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 (Q_AB)', 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 Q_AB 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 = Q_AB – Q_BA
• So it is clear that W will be always less than Q_AB
• 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 Q_AB, 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 Q_BA 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 (Q_BA)' can be made zero
• But such an engine with 'efficiency 1' is not possible. We will see the reason in a later section

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In the next section, we will see the reverse of the heat engine cycle


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