Chapter5_B - Chapter 5 - Section B - Non-Numerical...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 5 - Section B - Non-Numerical Solutions 5.1 Shown to the right is a PV diagram with two adi- abatic lines 1 2 and 2 3, assumed to inter- sect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce work. For the cycle ± U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is there- fore false. 5.5 The energy balance for the over-all process is written: Q = ± U t + ± E K + ± E P Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q . Thus heat is transferred to the surroundings. The total entropy change of the process is: ± S total = ± S t + ± S t surr Just as ± U t for the egg is zero, so is ± S t . Therefore, ± S total = ± S t surr = Q surr T σ = Q T σ Since Q is negative, ± S total is positive, and the process is irreversible. 5.6 By Eq. (5.8) the thermal efficiency of a Carnot engine is: η = 1 T C T H Differentiate: ± ∂η T C ² T H =− 1 T H and ± T H ² T C = T C T H 2 = T C T H 1 T H Since T C / T H is less unity, the efficiency changes more rapidly with T C than with T H . So in theory it is more effective to decrease T C . In practice, however, T C is fixed by the environment, and is not subject to control. The practical way to increase η is to increase T H . Of course, there are limits to this too. 5.11 For an ideal gas with constant heat capacities, and for the changes T 1 T 2 and P 1 P 2 , Eq. (5.14) can be rewritten as: ± S = C P ln ± T 2 T 1 ² R ln ± P 2 P 1 ² ( a )I f P 2 = P 1 , ± S P = C P ln ± T 2 T 1 ² If V 2 = V 1 , P 2 P 1 = T 2 T 1 Whence, ± S V = C P ln ± T 2 T 1 ² R ln ± T 2 T 1 ² = C V ln ± T 2 T 1 ² 642
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Since C P > C V , this demonstrates that ± S P S V . ( b )I f T 2 = T 1 , ± S T =− R ln ± P 2 P 1 ² If V 2 = V 1 , T 2 T 1 = P 2 P 1 Whence, ± S V = C P ln ± P 2 P 1 ² R ln ± P 2 P 1 ² = C V ln ± P 2 P 1 ² This demonstrates that the signs for ± S T and ± S V are opposite. 5.12 Start with the equation just preceding Eq. (5.14) on p. 170: dS R = C ig P R dT T d ln P = C P R T dP P Foranidealgas PV = RT ,a n dl n P + ln V = ln R + ln T . Therefore, P + dV V = T or P = T V Whence, R = C P R T T + V = ³ C P R 1 ´ T + d ln V Because ( C P / R ) 1 = C V / R , this reduces to: R = C V R T + d ln V Integration yields: ± S R = µ T T 0 C V R T + ln V V 0 ********************** As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute / T = / P + / V : R = C P R P + C P R V P = C V R P + C P R V Integration yields: ± S R = C V R ln P P 0 + C P R ln V V 0 5.13 As indicated in the problem statement the basic differential equations are: dW
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

Page1 / 9

Chapter5_B - Chapter 5 - Section B - Non-Numerical...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online