Question

Carnots refrigerator cycle. Plz.give

Answer 1

IsothermalExpansion. Heat is transferred reversibly from high temperature reservoir at constant temperature TH (isothermal heat addition or absorption). During this step (1 to 2 on Figure 1, A to B in Figure 2) the gas is allowed to expand, doing work on the surroundings by pushing up the piston (stage 1 figure, right). Although the pressure drops from points 1 to 2 (figure 1) the temperature of the gas does not change during the process because it is in thermal contact with the hot reservoir at Th, and thus the expansion is isothermal. Heat energy Q1 is absorbed from the high temperature reservoir resulting in an increase in the entropy of the gas by the amount {\displaystyle \Delta S_{1}=Q_{1}/T_{h}}.



Isentropic(reversible adiabatic) expansion of the gas (isentropic work output). For this step (2 to 3 on Figure 1, B to C in Figure 2) the gas in the engine is thermally insulated from both the hot and cold reservoirs. Thus they neither gain nor lose heat, an 'adiabatic' process. The gas continues to expand by reduction of pressure, doing work on the surroundings (raising the piston; stage 2 figure, right), and losing an amount of internal energy equal to the work done. The gas expansion without heat input causes it to cool to the "cold" temperature, Tc. The entropy remains unchanged.



Isothermal Compression. Heat transferred reversibly to low temperature reservoir at constant temperature TC. (isothermal heat rejection)(3 to 4 on Figure 1, C to D on Figure 2) Now the gas in the engine is in thermal contact with the cold reservoir at temperature Tc. The surroundings do work on the gas, pushing the piston down (stage 3 figure, right), causing an amount of heat energy Q2 to leave the system to the low temperature reservoir and the entropy of the system to decrease by the amount {\displaystyle \Delta S_{2}=Q_{2}/T_{c}}. (This is the same amount of entropy absorbed in step 1, as can be seen from the Clausius inequality.)



Adiabatic reversible compression.(4 to 1 on Figure 1, D to A on Figure 2) Once again the gas in the engine is thermally insulated from the hot and cold reservoirs, and the engine is assumed to be frictionless, hence reversible. During this step, the surroundings do work on the gas, pushing the piston down further (stage 4 figure, right), increasing its internal energy, compressing it, and causing its temperature to rise back to Th due solely to the work added to the system, but the entropy remains unchanged. At this point the gas is in the same state as at the start of step 1.



Figure 1: A Carnot cycle illustrated on a PV diagram to illustrate the work done.

In this case,

{\displaystyle \Delta S_{1}=\Delta S_{2}},

or,

{\displaystyle Q_{1}/T_{h}=Q_{2}/T_{c}}.

This is true as {\displaystyle Q_{2}} and {\displaystyle T_{c}} are both lower and in fact are in the same ratio as {\displaystyle Q_{1}/T_{h}}.

The pressure-volume graphEdit

When the Carnot cycle is plotted on a pressure volume diagram (Figure 1), the isothermal stages follow the isotherm lines for the working fluid, the adiabatic stages move between isotherms, and the area bounded by the complete cycle path represents the total work that can be done during one cycle. From point 1 to 2 and point 3 to 4 the temperature is constant. Heat transfer from point 4 to 1 and point 2 to 3 are equal to zero.