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Physics Topics can be both theoretical and experimental, with scientists using a range of tools and techniques to understand the phenomena they investigate.
What is the Formula for the Coefficient of Performance of a Refrigerator?
According to the second law of thermodynamics. no heat engine can have 100% efficiency. In 1824 french scientist N.L Sadi Carnot, developed a hypothetical, idealised heat engine which has the maximum efficiency (given two heat reservoirs at temperatures T1 and T2) consistent with the second law of thermodynamics. The cycle over which this engine operates is called the Carnot cycle.
Standard cylinder-piston arrangement: In Fig. (a), C is a cylinder. The piston P is an airtight one, capable of sliding within C. Thus, expansion or contraction of a substance (gas in most cases) kept inside the cylinder may occur. B is the bottom surface of the cylinder through which exchange of heat between the cylinder and the surroundings can take place. This surface has been drawn by a single line.
On the other hand, the other surfaces of the cylinder and the piston are insulating. No heat exchange can take place through them. Each of these surfaces has been drawn by using two lines. The whole arrangement is called a standard cylinder-piston arrangement.
In Fig.(b) A is an insulating plate. If a cylinder filled with some quantity of a gas is placed on the plate A, the gas exchanges no heat with the surroundings through any wall. So upward or downward movement of the piston causes adiabatic expansion or adiabatic contraction, respectively.
On the other hand, if the cylinder is placed on a heat reservoir at temperature T, the equilibrium temperature of the gas inside the cylinder also becomes T [Fig.(c)]. Now, if the piston is moved upward or downward, the gas takes heat from or rejects heat to the heat reservoir, respectively, and all along the temperature of the gas remains at T. In this way, Isothermal expansion or isothermal contraction takes place. Of course, the piston is to move up and down very slowly, so that sufficient time is obtained for heat exchange between the cylinder and the heat reservoir.
Description of a Carnot cycle on a pV-diagram:
Two isothermal curves at temperatures T1 and T2 are shown on a pV-diagram [Fig.]. In addition, two adiabatic curves are also shown. The points of intersection of these four curves are a, b, c, d.
Initially the system (in most cases a gas) is taken in a cylinder fitted with a piston in such a way that it is in the state a. In this condition pressure, volume and temperature of the gas are Pa, Va, T1 respectively. Next the following four reversible processes are performed on the system.
i) Reversible isothermal expansion a → b: At the state b the properties of the gas are Pb, Vb, T1. According to the figure, since Vb > Va it is an expansion. Since this expansion takes place along an isothermal curve, the temperature T1 remains constant.
ii) Reversible adiabatic expansion b → c: At the state c the properties of the gas are pc, Vc, T2. In this case also, since Vc > Vb, it is an expansion. Due to this expansion, the temperature comes down to T2 from T1.
iii) Reversible Isothermal contraction c → d: The point d is situated on the adiabatic curve through the point a. The properties of the gas at the state d are pd, Vd, T2.
iv) Reversible adiabatic contraction d → a: Due to this contraction, the gas returns to its initial state a, i.e., a cycle becomes complete. This cycle is a Carnot cycle. So, a cycle enclosed by four reversible processes, two isothermals and two adiabatics, is called a Carnot cycle.
Action of a Carnot cycle: The Carnot cycle shown in Fig. is a clockwise cycle. So the work done due to this cycle is positive and the area abcd of the cycle on the p-V diagram indicates this work (W).
The two process b → c and d → a are adiabatic, i.e., in these two processes no heat exchange takes place with the surroundings. Again, the process a → b is an isothermal expansion. So the gas takes up heat Q1 from the source at higher temperature T1. On the other hand, the process c → d is an isothermal contraction. So an amount of heat Q2 is rejected to the sink at lower temperature T2. This action of Carnot cycle can be expressed by Fig.
Obviously, in a complete cycle, the Q1 amount of heat is taken from the source at higher temperature T1. A part of this heat is converted into work W, and the remaining part Q2 is rejected to the sink at lower temperature T2. This means that this clockwise carnot cycle acts as a heat engine. This is called a Carnot engine. Since each process of the cycle is reversible, the engine E [Fig.] is a reversible heat engine working between two temperatures.
Efficiency of a Carnot engine using an ideal gas:
Suppose, n mol of an ideal gas is used as the working substance of the Carnot cycle, in this case, temperature of the source = T1 and temperature of the sink = T2 ; T2 < T1.
In isothermal expansion a → b [Fig.],
heat taken, Q1 = nRT1ln\(\frac{V_b}{V_a}\) …… (1)
Again, in isothermal compression c → d,
heat rejected, -Q2 = nRT2ln\(\frac{V_d}{V_c}\);
∴ Q2 = nRT2ln\(\frac{V_c}{V_d}\) ….. (2)
So, efficiency of Carnot engine,
η = 1 – \(\frac{Q_2}{Q_1}\) = 1 – \(\frac{T_2}{T_1} \cdot \frac{\ln \left(\frac{V_c}{V_d}\right)}{\ln \left(\frac{V_b}{V_a}\right)}\) ….. (3)
Now, in case of an adiabatic process of an ideal gas,
TVγ-1 = constant
where γ is the ratio of the two specific heats Cp and Cv of the gas.
So, for the adiabatic process b → c,
From this equation it is evident that the efficiency of a Carnot engine depends only on the temperatures of the source and the sink. Again, η = 1 = 100% if T2 = 0. Since absolute zero is attainable, Carnot engine with 100% efficiency is a practical impossibility.
Use of other materials in a Carnot engine: It is not that only ideal gas is always used as working substance in heat engines. Substances like a real gas, or a mixture of a liquid and vapour (for example, mixture of steam and water in a steam engine) are used in many cases as working substances. Using the second law of thermodynamics, Carnot showed that
- all reversible engines working between the same two temperatures have the same efficiency and
- no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures.
This is Carnot’s theorem. Proof of this theorem is beyond our syllabus.
Carnot engine is a reversible engine. Its efficiency between the temperatures T1 and T2 is η = 1 – \(\frac{T_2}{T_1}\), when the working substance is an ideal gas. According to Carnot’s theorem, it is not that this expression is only applicable to ideal gases. Whatever the working substance used in a Carnot engine may be, if it works between the same two temperatures, its efficiency will be the same.
In general, a Carnot cycle can be performed with any thermodynamic system as a working substance e.g. hydrostatic, chemical, electrical, magnetic or otherwise.
So the expression of efficiency of Carnot engine, i.e., equation (8) will be valid not only for ideal gases but also for any working substance.
Carnot refrigerator: A clockwise Carnot cycle behaves as a heat engine. This is called a Carnot engine. If now this Carnot cycle operates in an anticlockwise direction adcba [Fig.], then work done and exchange of heat in the four reversible processes will be equal but oppostie to the corresponding quantities of the Carnot cycle will act as an arrangement such that in each cycle-
i) some amount of heat Q2 will be taken [Fig.] from lower temperature T2;
ii) some amount of external work (W) is to be supplied;
iii) some amount of heat Q1 will be rejected to higher temperature T1.
It is obvious that the arrangement R shown in Fig. acts as a refrigerator. In this case, the anticlockwise Carnot cycle bounded by two reversible isothermal processes and two reversible adiabatic processes is called a Carnot refrigerator.
Inter conversion between Carnot engines and refrigerators: Both Carnot engine and Carnot refrigerator are reversible. So it is possible to use Carnot engine as Carnot refrigerator, and vice versa. For this, only the processes of a Carnot cycle should be reversed. But no process is reversible in reality. All real processes are irreversible. So Carnot engine or Carnot refrigerator is only an ideal arrangement. They cannot be prepared practically. All practical engines or refrigerators are irreversible. These cannot be operated in reverse. So no heat engine can be converted into a refrigerator, and vice versa.
Coefficient of performance of a Carnot refrigerator: The coefficient of performance of a Carnot refrigerator,
e = \(\frac{\text { output }}{\text { input }}\) = \(\frac{\text { heat taken from cold body }}{\text { work supplied from outside }}\)
= \(\frac{Q_2}{W}\) [according to Fig’s]
= \(\frac{Q_2}{Q_1-Q_2}\) = \(\frac{1}{\frac{Q_1-Q_2}{Q_2}}\) = \(\frac{1}{\frac{Q_1}{Q_2}-1}\)
Now, in case of a Carnot refrigerator, work done and heat exchange are equal and opposite to the corresponding qunatities of a Carnot engine. So, following the calculations for the determination of efficiency of Carnot engine, it is evident that in case of a Carnot refrigerator,
\(\frac{Q_1}{Q_2}\) = \(\frac{T_1}{T_2}\)
Therefore, e = \(\frac{1}{\frac{T_1}{T_2}-1}\) = \(\frac{1}{\frac{T_1-T_2}{T_2}}\) = \(\frac{T_2}{T_1-T_2}\)
Numerical Examples
Example 1.
A Carnot engine is being operated by taking heat from a source at temperature 527°C. If the surrounding temperature is 27°C, what is the efficiency of the engine? If the source supplies heat at the rate of 109J per minute, how much usable work is obtained per minute?
Solution:
Temperature of the source,
T1 = 527°C = (527 + 273) K = 800K
Temperature of the sink,
T2 = 27°C = (27 + 273)K = 300K
So, efficiency η = 1 – \(\frac{T_2}{T_1}\) = 1 – \(\frac{300}{800}\) = \(\frac{5}{8}\) = \(\frac{5}{8}\) × 100% = 62.5%
Now, η = \(\frac{W}{Q_1}\)
∴ W = ηQ1 = \(\frac{5}{8}\) × 109 = 6.25 × 108 J ᐧ min-1.
Example 2.
A Carnot engine takes up 200 cal of heat per cycle from a source at temperature 500K and rejects 150 cal of heat to the sink. What is the temperature of the sink and the efficiency of the engine?
Solution:
\(\frac{Q_1}{Q_2}\) = \(\frac{T_1}{T_2}\) or, T2 = T1 ᐧ \(\frac{Q_2}{Q_1}\) = 500 × \(\frac{150}{200}\) = 375 K
Efficiency, η = 1 – \(\frac{Q_2}{Q_1}\)
= 1 – \(\frac{T_2}{T_1}\) = 1 – \(\frac{375}{500}\) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
= \(\frac{1}{4}\) × 100% = 25%
Example 3.
The temperature of the source and the sink of a Carnot engine are 500K and 300K respectively. If the temperature of the source diminishes to 450K, what will be the percentage change in efficiency?
Solution:
In the first case, efficiency
η1 = 1 – \(\frac{T_2}{T_1}\) = 1 – \(\frac{300}{500}\) = \(\frac{2}{5}\)
In the second case, efficiency
η2 = 1 – \(\frac{T_2}{T_1}\) = 1 – \(\frac{300}{450}\) = \(\frac{1}{3}\)
∴ Fractional change in efficiency
= \(\frac{\eta_2-\eta_1}{\eta_1}\) = \(\frac{\eta_2}{\eta_1}\) – 1 = \(\frac{1 / 3}{2 / 5}\) – 1 = \(\frac{5}{6}\) – 1 = –\(\frac{1}{6}\)
∴ Percentage change = –\(\frac{1}{6}\) × 100 = -16.67%
So, efficiency decrease by 16.67%
Example 4.
The temperature of the ice-box of a refrigerator is -7°C and the room temperature is 31°C. To maintain the temperature of the ice-box, the refrigerator rejects 2000 cal of heat per minute. If the refrigerator approximates a Carnot refrigerator, what is the power consumption?
Solution:
Temperature of the ice-box = (-7 + 273) K = 266 K
Temperature of the surrounding = (31 + 273)K = 304 K
∴ Coefficient of performance,
e = \(\frac{T_2}{T_1-T_2}\) = \(\frac{266}{304-266}\) = \(\frac{266}{38}\) = 7
Again, e = \(\frac{Q_2}{W}\)
∴ W = \(\frac{Q_2}{e}\) = \(\frac{20000}{7}\) cal ᐧ min-1
= \(\frac{20000 \times 4.2}{7 \times 60}\) J ᐧ s-1 = 200 J ᐧ s-1
= 200 W.
Example 5.
An ideal gas is kept in a closed, rigid and heat insulating vessel. A coil of resistance 100 ohm, carrying a current of 1 ampere, is supplying heat to the gas. What will be the change of internal energy of the gas?
Solution:
Since the vessel is insulated, exchange of heat Q = 0
Work done by electrical energy in 5 minutes, i.e., 300 seconds, W = I2Rt = 12 × 100 × 300 = 30000 J
This work is suppplied to the gas from outside. So it is negative, i.e., W = -30000J
From the first law of the thermodynamics,
Q = (Uf – Ui) + W
or, 0 = ΔU + (-30000)
[ΔU = Uf – Ui = change of internal energy]
or ΔU = 30000 J.
Example 6.
A Carnot engine takes 3000 kcal of heat from a reservoir at 627°C. The sink is at 27°C. What is the amount of work done by the engine? [AIEEE ’03]
Solution:
Temperature of the reservoir,
T1 = (627 + 273)K = 900K
Temperature of the sink,
T2 = (27 + 273)K = 300K
∴ Efficiency, η = 1 – \(\frac{T_2}{T_1}\) = 1 – \(\frac{300}{900}\) = \(\frac{2}{3}\)
Again, η = \(\frac{W}{Q_1}\),
or, W = ηQ1 = \(\frac{2}{3}\) × 3000
= 2000 kcal = 2000 × 4.2 × 103J
= 8.4 × 106J
Example 7.
An ideal gas is taken through the cycle A → B → C → A as shown in Fig.
If the net heat supplied to the gas in the cycle is 5J, determine the work done by the gas in the process C → A. [‘IIT ’02]
Solution:
Heat taken by the gas, Q = 5 J
Since in the cycle A → B → C → A, the initial and the final states are both A, the change of internal energy,
ΔU = UA – UA = 0
Now, WAB = area of ABED = AD × ED = 10 × (2 – 1) = 10 J
WBC = 0
So, W = WAB + WBC + WCA = 10 + WCA
From the first law of thermodynamics we have,
Q = ΔU + W
or, 5 = 0 + (10 + WCA) or, WCA = 5 – 10 = -5 J.
Example 8.
If a system is taken from i to f by the process i → a → f [Fig.], Q is 50 cal and W is 20 cal. If in the process i → b → f, Q is 36 cal, what is the value of W?
Solution:
In both the processes, initial state i and final state f are the same. So change of internal energy Uf – Ui is also equal.
In process i → a → f, Uf – Ui = Q – W = 50 – 20 = 30 cal
∴ In process i → b → f, Uf – Ui = Q – W
or, W = Q – (Uf – Ui) = 36 – 30 = 6 cal.
Example 9.
A thermodynamic process is shown in Fig. The pressures and volumes corresponding to some points in the figure are pA = 3 × 104Pa; pB = 8 × 104Pa; VA = 2 × 10-3m3; VD = 5 × 10-3m3. In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added. What would be the change in internal energy of the system in the process AC?
Solution:
Since AB is an isochronic process, no work is done.
∴ From the first law of thermodynamics,
Q = ΔU + W or, 600 = Δ U + 0 or, ΔU = 600 J
Again, work done in the process BC
= area under the line BC
= pB(Vc – VB) = pB(VD – VA)
= 8 × 104 × (5 × 10-3 – 2 × 10-3) = 240 J
∴ Q = ΔU + W or, ΔU = Q – W = 200 – 240 = -40 J
So, change of internal energy in the process ABC
= 600 – 400 = 560 J
Since internal energy is independent of the path, the change of internal energy in the process AC = 560 J.
Example 10.
What is the amount of heat energy absorbed by a system going through a cyclic process shown in Fig.?
Solution:
The system is shown on a p-V diagram. In the diagram, the cycle is circular.
Radius of the cycle = \(\frac{30-10}{2}\) = 10 unit
∴ Work done, W = area of the cycle
= π × 102 (litre × kPa)
= 100π × 10-3m3 × 103 Pa
= 100πJ = 100 × 3.14 J = 314 J
In a cyclic process, change of internal energy = 0
So according to the first law of thermodynamics,
Q = ΔU + W = 0 + 314 = 314 J