Wallace and Hobbs 2.8 – Second Law and Entropy

by Joe Archive on August 7, 2008

2.8.1 Carnot Cyclebr /br /In a thermodynamic context, a cycle refers a series of changes undergone by a “working substance” which does external work and transfers heat energy while eventually returning to its initial conditions, ready to make another cycle.br /br /As the internal energy at the beginning and end of the cycle is the same, the net heat taken in by the system must equal the net work done.br /br /Efficiency is defined as:br /br /eta = (Work done)/(Heat taken in)br /br /Ideal heat engine – a working substance in a cylinder with insulating walls and a conducting base, fitted with a frictionless, insulated piston to which varying loads can be applied. Also, an insulating stand that the cylinder can rest upon and become totally thermally isolated, as well as two infinite reservoirs of heat one “hot” @ T_1 and the other “cold” at T_2 (T_1T_2). By infinite, we mean that heat can flow into or out of wither reservoir into the working substance with out changing the temperature of the reservoir.br /br /Cycle as follows:br /1) Start in state “A” at temperature T2. The system is places on the Stand and compressed (adiabatically) via the piston until it reaches T1 – state “B”.br /2) The system is then transferred to the warm reservoir, where it extracts heat (in quantity Q1), expanding isothermally to state “C”.br /3) The cylinder is returned to the stand and expands adiabatically until the temperatrue returns to T2, state “D”.br /4) The cylinder is placed on the cold reservoir and compressed isothermally to its original volume, expelling heat Q2 in the process, returning to state “A”.br /br /This is best visualized in PV space – the net work done in the process is the area contained within the cyclic path of the system.br /br /Example 2.14: Show that, for a Carnot engine, the ratio of heat absorbed to heat expelled is equal to the ratio of the reservoir temperatures.br /br /the heat absorbed is given by:br /br /Q1 = int_B^C pdVbr / = int_B^C RT_1dV/Vbr / = RT_1 ln(V_C/V_B)br /br /similarly,br /br /Q2 = RT_2 ln(V_D/V_A)br /br /(note change of sign, as heat flows in other direction here)br /br /br /For the adiabatic paths, we have:br /br /p_AV_A^gamma = p_BV_B^gammabr /p_CV_C^gamma = p_DV_D^gammabr /br /for the isothermal paths, we havebr /br /p_BV_B = p_CV_Cbr /p_DV_D = p_AV_Abr /br /combining, we get:br /br /V_C/V_B = V_D/V_Abr /br /Thus,br /br /Q1/Q2 = T_1/T_2 br /br /br /Following Carnot’s cycle in reverse tranfers heat from the cold reservoir to the hot reservior while taking in external energy in the form of work – this is a refrigeration machine.br /br /2.8.2 – Entropybr /Passing reversibly from one adiabat to another along an isotherm (temp T), heat goes in or out of a system (Q_rev). However, it can be shown that Q_rev/T is an constant for a given pair of adiabats. This quantity is the “difference” between the two adiabats and is called the entropy.br /br /More generally, when a system passes from one state to another, the entropy changes:br /br /s_2 – s_1 = int_1^2 dq_rev/Tbr /br /We can also see that br /br /ds = dq/T = c_p d theta/thetabr /br /so, br /br /s = c_p ln(theta) + constbr /br /Example 2.15: Calculate the change in entropy when 5g of water at 0C is raised to 100C and converted to steam at this temperature.br /br /S_373 – S_273 = int_{273}^{373} dQ/Tbr /br /where dQ = m*c(T)*dT for the liquid water part of the integration, while the system remains at a constant temperature during the vaporization.br /br /assuming c is constant, we have:br /br /DeltaS_{liquid} = 20.9 * ln(373/273) = 6.5J/degbr /br /the additional entropy flow during vaporization is:br /br /m*L/373 = 30.2 J/degbr /br /so the total entropy gain by the water isbr /br /36.7J/degbr /br /br /Consider the change in entropy during the Carnot cycle. There is no entropy flow during the adiabatic paths. For BC, the entropy of the working fluid increases by Q_1/T_1. Similarly, during DA, the entropy decreases by Q_2/T_2. As we have shown, these tow amounts are equal (but of opposite sign) so the entropy of the working fluid is unchanged, consistent with the definition of a cycle. Entropy is transferred between the reservoirs, but a negligible amount compared to their infinite extent.br /br /2.8.3 Clausius-Clapeyronbr /br /CC eqn can be derived from the Carnot cycle. Describes the change in saturation vapor pressure of a liquid with temperature (or melting point of a solid with pressure).br /br /Consider a working substance made up a liquid in equilibrium with saturated vapor.br /Let state A be the initial state with vapor pressure e_s-de_s at temperature T-dT. Adiabatic compression to B, with e_s, T, can be achieved with an infinitesimal compression. br /Then, place the system in contact with a reservior at T and expand it until a unit mass of liquid has evaporated. In this process, the pressure remains constant at e_s and the system passes to state C. The change in volume can be expressed in terms of the specific volumes of the gas and liquid:br /br /DeltaV = (alpha_g – alpha_L)br /br /and the heat absorbed is equal to L, the latent heat of vaporization.br /br /Now, we return tot he insulating stand and make an infinitesimal expansion back to pressure e_s-de_s, causing the temperature to fall again to T-dT. Finally, place the system on the heat sink at temperature T-dT and isothermally compress back to the initial state, condensing a unit mass of liquid.br /br /From earlier results:br /br /Q1/T1 = Q2/T2 = (Q1-Q2)/(T1-T2)br /br /Q1 – Q2 is the net work done by the system, and thus is given by:br /br /Q1-Q2 = (alpha_g-alpha_l)de_sbr /br /and we also have Q1 = L and T1-T2 = dTbr /br /so we havebr /br /L/T = (alpha_g-alpha_l)de_s/dTbr /br /orbr /br /de_s/dT = L/[T(alpha_g-alpha_l)]br /br /The calculation can be repeated for a mixture of solid and liquid, for T being the melting point at pressure p and we get an equivalent equationbr /br /dT/dp = T(alpha_l – alpha_s)/Lbr /br /Example 2.16 Calculate the change in the melting point of ice when the pressure chanes from 1atm to 2atm (alpha_Ice = 1.0903×10-3m^3/kg, alpha_Water = 1.0010×10^-3m^3/kg)br /br /dT = T(alpha_W -alpha_I) dp/L = -0.007degbr /br /Ice’s melting point reduced for increases in pressure – unusual, related to the fact the ice is less dense than waterbr /br /Example 2.17 Derive an expression of L(T) in terms of the specific heats of the liquid and the vapor.br /br /When a unit mass of liquid vaporized, the entropy is increased:br /br /delta s = L/Tbr /br /taking d/dT of both sides:br /br /ds_v/dT – ds_l/dT = 1/T dL/dT – L/T^2br /br /Tds_v/dT – Tds_l/dT = dL/dT – L/Tbr /br /dq_v/dT – dq_l/dT = dL/dT – L/Tbr /br /thus:br /br /dL/dT – L/T – c_v – c_lbr /br /2.8.5 Generalized Second Lawbr /br /Tds = du + p dalphabr /br /equality applies to reversible, equilibrium transformations.br /inequality for irreversible.

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