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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