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Examples #2: Heat Engines and Reversibility

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Examples #2: Heat Engines and Reversibility Examples #2: Heat engines and reversibility 1. Calculate the eciency of the Diesel cycle, consisting of two adiabats 1 ! 2 and 3 ! 4, one isobar 2 ! 3, and one constant-volume process 4 ! 1 (see Figure). Assume that heat capacities CV and Cp are constant. • Heat intake Q2!3 and outtake Q4!1 are: Q2!3 = Cp(T3 − T2) ;Q4!1 = CV (T4 − T1) so the eciency is: Q C (T − T ) 1 T − T η = 1 − 4!1 = 1 − V 4 1 = 1 − 4 1 Q2!3 Cp(T3 − T2) γ T3 − T2 We should try to express eciency in terms of the two variable parameters of the cycle, the xed pressure p2;3 and xed volume V1;4. So, let us eliminate the temperatures using the equation of state pV=T = const: 1 T − T 1 p V − p V V p − p η = 1 − 4 1 = 1 − 4 1;4 1 1;4 = 1 − 1;4 4 1 γ T3 − T2 γ p2;3V3 − p2;3V2 γp2;3 V3 − V2 On the two adiabats we have: γ γ γ γ p2;3V3 = p4 V1;4 ; p1 V1:4 = p2;3V2 from which we obtain: p2;3 γ p2;3 γ p2;3 γ γ p4 − p1 = γ V3 − γ V2 = γ (V3 − V2 ) V1;4 V1;4 V1;4 Therefore: γ γ γ γ V3 − V2 V1;4 p4 − p1 V1;4 V3 − V2 1 V1;4 V1;4 η = 1 − = 1 − γ = 1 − V V γp2;3 V3 − V2 γV V3 − V2 γ 3 − 2 1;4 V1;4 V1;4 We lost p2;3 from the formula in favor of three volumes that characterize this cycle. In fact, one can't do much better than this. η is dimensionless, so there is no way to construct a formula for it that depends only on two quantities p2;3 and V1;4 that have dierent incompatible units. 2. Consider an innitesimal quantity: dF = Adx + Bdy 1 where A and B are both functions of x and y. (a) Suppose that dF is an exact dierential so that F = F (x; y). Show that A and B must then satisfy the condition: @A @B = @y @x (b) If dF is an exact dierential, show that the integral dF evaluated along any closed path in the xy plane must vanish. ´ • Talking about dierentials of a quantity F implies that some coordinates such as x and y, which can be varied, control the changes of F . Then, the phrase dF is an exact dierential means that dF is an innitesimal change of a quantity F (x; y) that depends on the coordinates. A quantity dF that is not an exact dierential refers to a change of something that is not to be considered a function of the coordinates. In thermodynamics, state variables such as temperature T and volume V represent coordinates. Internal energy E(T;V ) is a state function, so dE is an exact dierential. Heat Q, however, is not a state variable so dQ is not an exact dierential. In fact, heat has meaning only as the amount of transferred energy ∆Q by means other than work; it cannot be regarded as a property of the system at given T and V . • (a) If F = F (x; y) then: @F @F @F @F dF = dx + dy = Adx + Bdy ) A(x; y) = ;B(x; y) = @x @y @x @y Therefore: @A @2F @B @2F = ; = @y @x@y @x @y@x The order of taking derivatives does not matter (if the function F is analytic, i.e. has nite derivatives). • (b) For any function F = F (x; y), the integral dF on a path A ! B is dened as a sum of innitely many innitesimal increments ∆F of the´ function F : B N N X Xh i dF = lim ∆Fi ≡ lim F (xi; yi) − F (xi−1; yi−1) ˆ N!1 N!1 A i=1 i=1 Here, the path A ! B goes through an innitely dense set of points (xi; yi), where the initial point is A = (x0; y0) and the nal point is B = (xN ; yN ). Any integral is dened in this manner! One just needs to notice that almost all terms in the nal sum are canceled out: F (xi; yi) from [···] for a particular value of i = k gets canceled by −F (xi−1; yi−1) from [···] at the next value of i = k + 1. Therefore, one derives the standard Newton's formula for integration of denite integrals (which applies to analytic functions of arbitrarily many variables): B dF = F (x ; y ) − F (x ; y ) = F (B) − F (A) ˆ N N 0 0 A Now consider a closed path in the xy plane. This path goes through some sequence of points (xi; yi), but comes back to the same (arbitrary) point at which it started. In other words, one chooses a random starting point A on the path and completes the circle with the ending point B = A. Therefore: A dF ≡ dF = F (A) − F (A) = 0 ˛ ˆ A 2 3. Consider the innitesimal quantity dF = (x2 − y)dx + xdy (a) Is this an exact dierential? (b) Evaluate the integral dF between the (x; y) points (1; 1) and (2; 2) along the straight line paths connecting the following points:´ (1; 1) ! (1; 2) ! (2; 2) (1; 1) ! (2; 1) ! (2; 2) (1; 1) ! (2; 2) (c) Consider the innitesimal quantity dG = dF=x2. Is this an exact dierential? (d) Evaluate the integral dG along all three paths of the part (b). ´ • (a) We ought to gure out if we could write: @F @F dF = dx + dy = (x2 − y)dx + xdy @x @y @F @F = x2 − y ; = x @x @y Integrate out the proposed partial derivatives: @F x3 dx = dx (x2 − y) = − xy + f (y) ˆ @x ˆ 3 1 @F dy = dy x = xy + f (x) ˆ @y ˆ 2 Note that both integrals have just one integration variable, while the other one is treated as a constant. The integrals are indenite, so after applying the usual integration rules one gets an unknown constant. In the rst integral, this constant is only prohibited from having any dependence on x, but it can have some dependence on y, i.e. it can be a function f1(y). Similarly, the second integration yields an integration constant f2(x). The question is now if it is possible to interpret both results as a single function: @F @F F (x; y) = dx + const = dy + const ˆ @x ˆ @y Is is possible to x the unknown functions f1 and f2 so that: x3 − xy + f (y) = xy + f (x) 3 1 2 (without loss of generality, we absorbed a const into the denition of one of the functions f1; f2)? We have: x3 f (x) − f (y) = − 2xy 2 1 3 It is not possible to rewrite the right-hand-side as a dierence of two expressions, one which de- pends only on xand the other that depends only on y. Therefore, dF is not an exact dierential. 3 There is no such function F (x; y) whose partial derivatives are @F=@x = x2 − y and @F=@y = x. However, there is nothing wrong with specifying dF = (x2 − y)dx + xdy. This simply describes some process in which (x; y) are gradually changed along some path, and the corresponding change dF of some quantity (which cannot be represented as a single-valued function of x and y). • (b) On the path (1; 1) ! (1; 2) ! (2; 2): (1;2) (2;2) 2 2 2 2 4 dF + dF = dy x + dx (x2 − y) = dy + dx (x2 − 2) = ˆ ˆ ˆ x=1 ˆ y=2 ˆ ˆ 3 (1;1) (1;2) 1 1 1 1 On the path (1; 1) ! (2; 1) ! (2; 2): (2;1) (2;2) 2 2 2 2 10 dF + dF = dx (x2 − y) + dy x = dx (x2 − 1) + 2 dy = ˆ ˆ ˆ y=1 ˆ x=2 ˆ ˆ 3 (1;1) (2;1) 1 1 1 1 On the straight diagonal path (1; 1) ! (2; 2) we have y = x, so that: dx = dy ; dF = (x2 − y)dx + xdy = (x2 − x)dx + xdx = x2dx (2;2) 2 7 dF = dx x2 = ˆ ˆ 3 (1;1) 1 We see that all three integrals are dierent, even though they have the same initial and nal points. • The quantity dF x2 − y 1 dG = = dx + dy x2 x2 x supposedly has: @G y @G y = 1 − ) dx = x + + g (y) @x x2 ˆ @x x 1 @G 1 @G y = ) dy = + g (x) @y x ˆ @y x 2 Hence, G(x; y) is up to a constant: y y x + + g (y) = + g (x) x 1 x 2 Clearly we can pick g1(y) = 0 and g2(x) = x. This means that dG is an exact dierential of the function: y G(x; y) = x + + const x Note that we don't can't nd the value of the constant from the given information. • Since dG is an exact dierential, all three integrals dG must have the same value, which depends only on the initial and nal points regardless of the´ path: (2;2) 2 1 dG = G(2; 2) − G(1; 1) = 2 + + const − 1 + + const = 1 ˆ 2 1 (1;1) 4 4. A system undergoes a cyclic quasi-static process which is represented in a p − V diagram by a closed loop. Show that the work done by the system is given by the area contained within this closed loop.
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