TERMODINAMIKA

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

3.3 The Carnot Cycle

A Carnot cycle is shown in Figure 3.4. It has four processes. There are two adiabatic reversible legs and two isothermal reversible legs. We can construct a Carnot cycle with many different systems, but the concepts can be shown using a familiar working fluid, the ideal gas. The system can be regarded as a chamber enclosed by a piston and filled with this ideal gas.

**Figure 3.4:** Carnot cycle -- thermodynamic diagram on left and schematic of the different stages in the cycle for a system composed of an ideal gas on the right

The four processes in the Carnot cycle are:

The system is at temperature at state . It is brought in contact with a heat reservoir, which is just a liquid or solid mass of large enough extent such that its temperature does not change appreciably when some amount of heat is transferred to the system. In other words, the heat reservoir is a constant temperature source (or receiver) of heat. The system then undergoes an isothermal expansion from to , with heat absorbed .
At state , the system is thermally insulated (removed from contact with the heat reservoir) and then let expand to . During this expansion the temperature decreases to . The heat exchanged during this part of the cycle, $Q_{bc}=0.$ )
At state the system is brought in contact with a heat reservoir at temperature . It is then compressed to state , rejecting heat in the process.
Finally, the system is compressed adiabatically back to the initial state . The heat exchange $Q_{da}=0$ .

The thermal efficiency of the cycle is given by the definition

$\displaystyle \eta = 1 - \frac{Q_R}{Q_A}=1+\frac{Q_1}{Q_2}.$

(3..4)

In this equation, there is a sign convention implied. The quantities , as defined are the magnitudes of the heat absorbed and rejected. The quantities , on the other hand are defined with reference to heat received by the system. In this example, the former is negative and the latter is positive. The heat absorbed and rejected by the system takes place during isothermal processes and we already know what their values are from Eq. (3.1):

$\displaystyle Q_2 = W_{ab} =N\mathbf{R}T_2 [\ln({V_b}/{V_a})],$

$\displaystyle Q_1 = W_{cd} =N\mathbf{R}T_1 [\ln({V_d}/{V_c})] = -N\mathbf{R}T_1 [\ln({V_c}/{V_d})].\quad \textrm{ ($Q_1$ is negative.)}$

The efficiency can now be written in terms of the volumes at the different states as

$\displaystyle \eta = 1+ \frac{T_1[\ln(V_d /V_c)]}{T_2[\ln(V_b /V_a)]}.$

(3..5)

The path from states to and from to are both adiabatic and reversible. For a reversible adiabatic process we know that $PV^\gamma= \textrm{constant}$ . Using the ideal gas equation of state, we have $T V^{\gamma-1} = \textrm{constant}$ . Along curve - , therefore, $T_2 V_b^{\gamma-1}=T_1 V_c^{\gamma-1}$ . Along the curve - , $T_2 V_a^{\gamma-1}=T_1 V_d^{\gamma-1}$ . Thus,

$\displaystyle \left(\frac{V_d}{V_c}\right)^{\gamma-1} = \frac{(T_2/T_1)}{(T_2/T... ...right)^{\gamma-1},\textrm{ which means that } \frac{V_d}{V_c}=\frac{V_a}{V_b}.$

Comparing the expression for thermal efficiency Eq. (3.4) with Eq. (3.5) shows two consequences. First, the heats received and rejected are related to the temperatures of the isothermal parts of the cycle by

$\displaystyle \frac{Q_1}{T_1}+\frac{Q_2}{T_2}= 0.$

(3..6)

Second, the efficiency of a Carnot cycle is given compactly by

$\displaystyle \eta = 1 - \frac{T_1}{T_2}.\qquad\textbf{Carnot cycle efficiency.}$

(3..7)

The efficiency can be 100% only if the temperature at which the heat is rejected is zero. The heat and work transfers to and from the system are shown schematically in Figure 3.5.

**Figure 3.5:** Work and heat transfers in a Carnot cycle between two heat reservoirs

Muddy Points

Since $\eta = 1- {T_1}/{T_2}$ , looking at the - graph, does that mean the farther apart the , isotherms are, the greater efficiency? And that if they were very close, it would be very inefficient? (MP 3.2)