Rabu, 04 November 2009

Termodinamika

6.2 The Thermodynamic Temperature Scale

The considerations of Carnot cycles in this section have not mentioned the working medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium. Earlier we derived the Carnot efficiency with an ideal gas as a medium and the temperature definition used in the ideal gas equation was not essential to the thermodynamic arguments. More specifically, we can define a thermodynamic temperature scale that is independent of the working medium. To see this, consider the situation shown below in Figure 6.2, which has three reversible cycles. There is a high temperature heat reservoir at and a low temperature heat reservoir at . For any two temperatures , , the ratio of the magnitudes of the heat absorbed and rejected in a Carnot cycle has the same value for all systems.

**Figure 6.2:** Arrangement of heat engines to demonstrate the thermodynamic temperature scale

We choose the cycles so is the same for A and C. Also is the same for B and C. For a Carnot cycle

$\displaystyle \eta =1+\frac{Q_L}{Q_H} = F(T_L,\;T_H);\;\eta\textrm{ is only a function of temperature.}$

Also

$\displaystyle \frac{Q_1}{Q_2}= F(T_1,\; T_2),$

$\displaystyle \frac{Q_2}{Q_3}= F(T_2,\; T_3),$

$\displaystyle \frac{Q_1}{Q_3}= F(T_1,\; T_3).$

But

$\displaystyle \frac{Q_1}{Q_3}=\frac{Q_1}{Q_2}\frac{Q_2}{Q_3}.$

Hence

$\displaystyle \underbrace{F(T_1,\;T_3)}_\textrm{Not a function of $T_2$} = \underbrace{F(T_1,\;T_2)\times F(T_2,\;T_3)}_\textrm{Cannot be a function of $T_2$}.$

We thus conclude that $F(T_1,\;T_2)$ has the form , and similarly $F(T_2,\;T_3)=f(T_2)/f(T_3)$ . The ratio of the heat exchanged is therefore

$\displaystyle \frac{Q_1}{Q_3}=F(T_1,\;T_3)=\frac{f(T_1)}{f(T_3)}.$

In general,

$\displaystyle \frac{Q_H}{Q_L} =\frac{f(T_H)}{f(T_L)},$

so that the ratio of the heat exchanged is a function of the temperature. We could choose any function that is monotonic, and one choice is the simplest:

. This is the thermodynamic scale of temperature,

. The temperature defined in this manner is the same as that for the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent.

Disusun Ulang Oleh:

Arip Nurahman

Pendidikan Fisika, FPMIPA. Universitas Pendidikan Indonesia

Follower Open Course Ware at MIT-Harvard University. Cambridge. USA.

Materi kuliah termodinamika ini disusun dari hasil perkuliahan di departemen fisika FPMIPA Universitas Pendidikan Indonesia dengan Dosen:

1. Bpk. Drs. Saeful Karim, M.Si.

2. Bpk. Insan Arif Hidayat, S.Pd., M.Si.

Dan dengan sumber bahan bacaan lebih lanjut dari :

Massachusetts Institute of Technology, Thermodynamics

Professor Z. S. Spakovszk, Ph.D.

Office: 31-265

Phone: 617-253-2196

Email: zolti@mit.edu

Aero-Astro Web: http://mit.edu/aeroastro/people/spakovszky

Gas Turbine Laboratory: home

Ucapan Terima Kasih:

Kepada Para Dosen di MIT dan Dosen Fisika FPMIPA Universitas Pendidikan Indonesia

Semoga Bermanfaat

Senin, 02 November 2009

Termodinamika

6.1 Limitations on the Work that Can be Supplied by a Heat Engine

The second law enables us to make powerful and general statements concerning the maximum work that can be derived from any heat engine which operates in a cycle. To illustrate these ideas, we use a Carnot cycle which is shown schematically in Figure 6.1.

**Figure 6.1:** A Carnot cycle heat engine

The engine operates between two heat reservoirs, exchanging heat with the high temperature reservoir at and with the reservoir at . The entropy changes of the two reservoirs are

0.$">

The same heat exchanges apply to the system, but with opposite signs: the heat received from the high temperature source is positive, and conversely. Denoting the heat transferred to the engines by subscript ``

'',

$\displaystyle Q_{He} =-Q_H;\quad Q_{Le} =-Q_L.$

The total entropy change during any operation of the engine is

$\displaystyle \Delta S_\textrm{total} = \underbrace{\Delta S_H}_\textrm{Reservo... ...elta S_L}_\textrm{Reservoir at $T_L$} +\underbrace{\Delta S_e}_\textrm{Engine}.$

For a cyclic process, the third of these ( $\Delta S_e$ )is zero, and thus (remembering that > ),

$\displaystyle \Delta S_\textrm{total} =\Delta S_H +\Delta S_L = \frac{Q_H}{T_H} +\frac{Q_L}{T_L}.$

(6..1)

For the engine we can write the first law as

$\displaystyle \Delta U_e =0\; \textrm{(cyclic process)} =Q_{He} +Q_{Le} -W_e.$

Or,

$\displaystyle W_e$	$\displaystyle =Q_{He} +Q_{Le}$
	$\displaystyle = -Q_H -Q_L.$

Hence, using (6.1)

$\displaystyle W_e$	$\displaystyle =-Q_H -T_L\Delta S_\textrm{total} +Q_H \left(\frac{T_L}{T_H}\right)$
	$\displaystyle =-Q_H\left[1-\left(\frac{T_L}{T_H}\right)\right]-T_L\Delta S_\textrm{total}.$

The work of the engine can be expressed in terms of the heat received by the engine as

$\displaystyle W_e=Q_{He}\left[1-\left(\frac{T_L}{T_H}\right)\right]-T_L\Delta S_\textrm{total}.$

The upper limit of work that can be done occurs during a reversible cycle, for which the total entropy change ( $\Delta S_\textrm{total}$ ) is zero. In this situation:

$\displaystyle \textrm{Maximum work for an engine working between $T_H$ and $T_L$:}\quad W_e =Q_{He}\left[1-\left(\frac{T_L}{T_H}\right)\right].$

Also, for a reversible cycle of the engine,

$\displaystyle \frac{Q_H}{T_H} +\frac{Q_L}{T_L} =0.$

These constraints apply to all reversible heat engines operating between fixed temperatures. The thermal efficiency of the engine is

$\displaystyle \eta$	$\displaystyle = \frac{\textrm{Work done}}{\textrm{Heat received}}=\frac{W}{Q_{He}}$
	$\displaystyle =1-\frac{T_L}{T_H}=\eta_\textrm{Carnot}.$

The Carnot efficiency is thus the maximum efficiency that can occur in an engine working between two given temperatures.

We can approach this last point in another way. The engine work is given by

$\displaystyle W_e =-Q_H -T_L\Delta S_\textrm{total} +Q_H \frac{T_L}{T_H}$

or,

$\displaystyle T_L \Delta S_\textrm{total} =-Q_H +Q_H \frac{T_L}{T_H}-W_e.$

The total entropy change can be written in terms of the Carnot cycle efficiency and the ratio of the work done to the heat absorbed by the engine. The latter is the efficiency of any cycle we can devise:

$\displaystyle \Delta S_\textrm{total} = \frac{Q_{He}}{T_L}\left[1-\frac{T_L}{T_... ...c{Q_{He}}{T_L}\biggl[\eta_\textrm{Carnot}-\eta_\textrm{Any other cycle}\biggr].$

The second law says that the total entropy change is equal to or greater than zero. This means that the Carnot cycle efficiency is equal to or greater than the efficiency for any other cycle, with the total equality only occurring if $\Delta S_\textrm{total}=0$ .

Muddy Points

So, do we lose the capability to do work when we have an irreversible process and entropy increases? (MP 6.1)

Why do we study cycles starting with the Carnot cycle? Is it because it is easier to work with? (MP 6.2)