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(Throughout, we will assume the processes involve an ideal gas with constant n.)
In a cyclic process, the state variables all return to their initial values after one cycle.
Assuming this PV diagram describes a process involving an ideal gas, we can deduce the following:
as a→b | P_{a} = P_{b} | ΔE_{a→b} = n c_{v} (T_{b} - T_{a}) | ΔQ_{a→b} = n c_{p} (T_{b} - T_{a}) | W_{a→b} = -P_{a} (V_{b} - V_{a}) |
at b | T_{b} = P_{b} V_{b} / (n R) | T_{b} > T_{a} | ΔQ_{a→b} > 0 | W_{a→b} < 0 |
as b→c | T_{b} = T_{c} | ΔE_{b→c} = 0 | ΔQ_{b→c} = -W_{b→c} | W_{b→c} = -n R T_{b} ln (V_{c} / V_{b}) |
at c | T_{c} = P_{c} V_{c} / (n R) | ΔQ_{b→c} > 0 | W_{b→c} < 0 | |
as c→d | V_{c} = V_{d} | ΔE_{c→d} = n c_{v} (T_{d} - T_{c}) | ΔQ_{c→d} = n c_{v} (T_{d} - T_{c}) | W_{c→d} = 0 |
at d | T_{d} = P_{d} V_{d} / (n R) | T_{d} < T_{c} | ΔQ_{c→d} < 0 | |
as d→a | P_{d} V_{d}^{cp/cv} = P_{a} V_{a}^{cp/cv} | ΔE_{d→a} = n c_{v} (T_{a} - T_{d}) | ΔQ_{d→a} = 0 | W_{d→a} = n c_{v} (T_{a} - T_{d}) |
at a | T_{a} = P_{a} V_{a} / (n R) | T_{a} > T_{d} | W_{d→a} > 0 |
In addition, we know that ΔE_{a→b} + ΔE_{c→d} + ΔE_{d→a} = 0, but in this case it only tells us that T_{b} = T_{c}.
From this we can compute the total heat added during the cycle:
n c_{p} (T_{b} - T_{a}) + n R T_{b} ln (V_{c} / V_{b})and the total work done by the gas:
P_{a} (V_{b} - V_{a}) + n R T_{b} ln (V_{c} / V_{b}) - n c_{v} (T_{a} - T_{d}),corresponding to the shaded area in the PV diagram.
Note that heat enters this "engine" in steps a→b and b→c, and exits it in step c→d; the total work done by the engine is positive. If the cycle were operated in reverse, it would be a "refrigerator", taking heat from a cooler environment during step d→c and exhausting it to a warmer environment from c→a; the total work done by the refrigerator is negative. The meaning of this is that it requires external work to move heat from a cooler to a warmer place.
Suppose we are given P_{a} = 200,000 Pa, P_{b} = 300,000 Pa, P_{c} = 100,000 Pa and V_{a} = 65 L.
We know that P_{d} = P_{c}, V_{b} = V_{a}, and T_{b} = T_{c}. We can use the ideal gas law to compute the temperatures at points a and b from the pressure and volumes at those points: 390.907 K and 586.36 K, respectively. From the latter we can compute the volume at point c, which is 195 L.
We can use the adiabatic equation for an ideal gas:
P_{d} V_{d}^{γ} = P_{a} V_{a}^{γ},to compute V_{d} = 98.5216 L. Finally, we can use the ideal gas law to find that T_{d} = 296.252 K.
Since b to c is an isothermal expansion, the first law tells us that ΔQ_{b→c} = - ΔW_{b→c}. Using the ideal gas law, we find
ΔW_{b→c} = - ∫_{b}^{c} P dVUsing the equations for isovolumetric and isobaric processes, we have= - n R T_{b} ∫_{b}^{c} dV / V(Note the sign: the gas does external work.)= - n R T_{b} ln (V_{c} / V_{b}) = -21,422.9 J.
Finally, consistent with the fact that ΔE = 0 for a cycle, we compute
ΔQ_{a→b} = n c_{v} (T_{b} - T_{a}) ΔQ_{b→c} = n R T_{b} ln (V_{c} / V_{b}) ΔQ_{c→d} = n c_{p} (T_{d} - T_{c}) ΔQ_{d→a} = 0 = 9750 J = 21,422.9 J =-24,119.6 J
ΔW_{a→b} = 0 ΔW_{b→c} = - n R T_{b} ln (V_{c} / V_{b}) ΔW_{c→d} = P_{c} (V_{c} - V_{d}) ΔW_{d→a} = c_{v} (P_{a} V_{a} - P_{d} V_{d}) / R = -21,422.9 J = 9647.84 J = 4721.76 J
ε = W / Q_{h}In the example above, Q_{h} = 31,172.9 J, W = 7053.33 J, Q_{c} = 24,119.6 J and ε = 22.6265%.= (Q_{h} - Q_{c}) / Q_{h}Here, energy is taken in as Q_{h}, work W is done by the engine, and the waste heat is discharged as Q_{c}.= 1 - Q_{c} / Q_{h}.
This calculation obscures some interesting facts:For an air conditioner, heat pump in air conditioning mode, or refrigerator, ε = Q_{transferred} / W_{required}. The efficiency of a refrigerator can be greater than 1, since
- for this cycle, as the number of degrees of freedom of the gas χ→∞ (with all other values unchanged), ε→0 as 1/χ (more heat must be removed during isobaric cooling to lower the temperature the same amount, because more of the heat comes from non-translational degrees of freedom which do not affect temperature);
- ε does not depend on n; increasing n will only lower the temperatures (via the ideal gas law; with more molecules, less energy is available per molecule);
- while the efficiency is a complicated function of the given values, it is turns out that, holding all other values constant, increasing P_{b} or decreasing P_{c} will increase ε, while the value of V_{a} does not alter ε at all and the chosen value of P_{a} is almost optimal.
Of course, for another cycle, the results can be very different.
ε = Q_{c} / (Q_{h} - Q_{c})where Q_{c} is the heat removed from the cooler environment and Q_{h} is the heat exhausted into the warmer environment. Note that Q_{h} must be greater than Q_{c}.= 1 / (Q_{h} / Q_{c} - 1),
ΔS = - ΔQ_{h} / T_{h} + ΔQ_{c} / T_{c}.In general, ΔS = ∫ dQ / T.
For the example above, ΔS_{b→c} = 36.5355 J/K, and of course ΔS_{d→a} = 0. To compute ΔS_{a→b} and ΔS_{c→d}, we use the ideal gas law to express T as a function of P and V:
ΔS_{a→b} = ∫ n c_{v} dT / Tand= ∫ n c_{v} (V dP / (n R)) / (P V / (n R))= n c_{v} ln (P_{b} / P_{a}) = 20.2262 J/K
ΔS_{c→d} = ∫ n c_{p} dT / TThe changes in entropy of the reservoirs are of course dependent on their temperatures. Assuming that the isobaric compression lost heat to a reservoir at room temperature (< T_{d}), and that the isochoric and isothermal processes gained heat from a reservoir at 320 C (> T_{b}),= ∫ n c_{p} (P dV / (n R)) / (P V / (n R))= n c_{p} ln (V_{d} / V_{c}) = -56.7617 J/K
ΔS_{cold reservoir} = 24119.6 / 293.15 J/K = 82.277 J/K andso thatΔS_{hot reservoir} = -31172.9 / 593.15 J/K = -52.555 J/K,
ΔS_{reservoirs} = 29.722 J/K,which is greater than zero, as required by the second law of thermodynamics.
If we did not know T as a function of P and V, it would be necessary to connect each pair of endpoints with a combination of adiabatic and isothermal processes, and compute the change in entropy using those new processes. For example:
Process a→e is an adiabatic process, and e→b is isothermal. P_{e} = 551,135 Pa and V_{e} = 35.3815 L, from
P_{a} V_{a}^{γ} = P_{e} V_{e}^{γ}This gives us (using the same equations as above)= n R T_{b} V_{e}^{γ - 1}
ΔQ_{a→e} = 0 andSince entropy is a state variable,ΔQ_{e→b} = 11,859.9 J.
ΔS_{a→b} = ΔS_{e→b} = 20.2262 J/K,as before. Either way, the entropy change of the engine is zero, because it is a cycle: all state variables, including entropy, end at the same state in which they began. Therefore ΔS_{c→d} = -56.7617 J/K.
What makes it so is that each step of the Carnot cycle is reversible: all changes of state are infinitesimally close to equilibrium.
Reversible changes of state are infinitely long; all real changes of state are irreversible.Since it is reversible, the total entropy change of the Carnot cycle reservoirs is zero. This gives usIn isothermal processes, by definition the gas is in thermal equilibrium with the reservoir. An adiabatic process can be irreversible:
Consider a closed, thermally insulated volume with a partition in the middle, such that nearly all of the gas molecules are on one side of the partition; the other side is a nearly perfect vacuum. With removal of the partition, the gas adiabatically but irreversibly doubles in volume.A reversible adiabatic process would be a change in volume by infinitesimal steps while the gas is thermally insulated.
0 = - Q_{h} / T_{h} + Q_{c} / T_{c}or
Q_{c} / Q_{h} = T_{c} / T_{h}.If we attempt to compute the entropy change for c→d by finding a point f on the PV diagram such that c→f is adiabatic and f→d is isothermal, we find that we have constructed a Carnot cycle:
Hence the total entropy change is zero including the reservoirs, and its efficiency is
1 - T_{d} / T_{b} = 49.4761%.The efficiency of a Carnot Refrigerator is T_{c} / (T_{h} - T_{c}).
ε = 1 - (T_{d} - T_{a}) / ((1 + 2 / χ) (T_{c} - T_{b}))
ε = 1 - (V_{small} / V_{large})^{(2 / χ)}
ε = 1 / (1 + T_{lower} / |Δ T| + χ / (2 ln (V_{large} / V_{small})))
In each case, ε was computed for an ideal gas. Note that in each case, it is independent of n, and increases with decreasing χ. This generalizes what we saw above: if we think of work as the result of increased temperature, less heat input is required to change the temperature of a gas with fewer degrees of freedom.
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©2012, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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