Question

8. A refrigerator may be regarded as the inverse of a steam engine, in that work is performed in order to remove heat from a system. It may be regarded as a steam engine run in reverse (i.e. the first step is a reversible adiabatic expansion from Th to Tc.) Heat is removed from the gas at the lower temperature in the succeeding isothermal expansion. The efficiency of a refrigerator is: Er = Aqoutput/AWinput Assuming a refrigerator can be built using a reversible Carnot cycle of a van der Waals gas, find an expression for the efficiency er. If Es is the efficiency of a Carnot steam engine, what is its relationship to €r? {i.e. find a relationship between Er and es} Let T = 2Ty. Find the value of that maximizes the efficiency. What is the maximum value of Er? Can a temperature of absolute zero be achieved by repeating the cycle with temperature lowered by the maximum efficiency each cycle? Explain using part c.

Answer 1

Carnot cycle is a theoretical ideal thermodynamic cycle It provides an upper limit on the efficiency that any classical thermodynamic engine can achieve during the conversion of heat into work, or conversely, the efficiency of a refrigeration system in creating a temperature difference by the application of work to the system. It is not an actual thermodynamic cycle but is a theoretical construct.

The efficiency of a heat engine relates how much useful work is output for a given amount of heat energy input.

From the laws of thermodynamics, after a completed cycle:

{} ${\displaystyle W\ =\ Q_{c}\ -\ (-Q_{h})}$

where

${\displaystyle W=-\oint PdV}$ is the work extracted from the engine. (It is negative since work is done by the engine.)

${\displaystyle Q_{h}=T_{h}\Delta S_{h}}$ is the heat energy taken from the high temperature system. (It is negative since heat is extracted from the source, hence

${\displaystyle Q_{c}=T_{c}\Delta S_{c}}$ is the heat energy delivered to the cold temperature system. (It is positive since heat is added to the sink.)

In other words, a heat engine absorbs heat energy from the high temperature heat source, converting part of it to useful work and delivering the rest to the cold temperature heat sink.

In general, the efficiency of a given heat transfer process (whether it be a refrigerator, a heat pump or an engine) is defined informally by the ratio of "what is taken out" to "what is put in".

In the case of an engine, one desires to extract work and puts in a heat transfer.

${\displaystyle \eta ={\frac {-W}{-Q_{h}}}={\frac {-Q_{h}-Q_{c}}{-Q_{h}}}=1-{\frac {Q_{c}}{-Q_{h}}}}$

The theoretical maximum efficiency of any heat engine depends only on the temperatures it operates between. This efficiency is usually derived using an ideal imaginary heat engine such as the Carnot heat engine, although other engines using different cycles can also attain maximum efficiency. Mathematically, this is because in reversible processes, the change in entropy of the cold reservoir is the negative of that of the hot reservoir

${\displaystyle \Delta S_{c}=-\Delta S_{h}}$ ), keeping the overall change of entropy zero. Thus:

${\displaystyle \eta _{\text{max}}=1-{\frac {T_{c}\Delta S_{c}}{-T_{h}\Delta S_{h}}}=1-{\frac {T_{c}}{T_{h}}}}$

where   $T_{h}$ is the absolute temperature of the hot source and

$T_{c}$ that of the cold sink, usually measured in kelvins. Note that $dS_{c}$ is positive while $dS_{h}$ is negative; in any reversible work-extracting process, entropy is overall not increased, but rather is moved from a hot (high-entropy) system to a cold (low-entropy one), decreasing the entropy of the heat source and increasing that of the heat sink.

The reasoning behind this being the maximal efficiency goes as follows. It is first assumed that if a more efficient heat engine than a Carnot engine is possible, then it could be driven in reverse as a heat pump. Mathematical analysis can be used to show that this assumed combination would result in a net decrease in entropy. Since, by the second law of thermodynamics, this is statistically improbable to the point of exclusion, the Carnot efficiency is a theoretical upper bound on the reliable efficiency of any thermodynamic cycle.

A) Coefficient of performance of Refrigerator

Ratio of work done on the engine to the heat absorbed at lower temperature is the coefficient of performance (∈ )

Thus ε = work done the engine/ heat absorbed from sink … (1)

ε = w/ q1

ε = q2 – q1 / q1 …(2)

In terms of temperature ε = T2 – T1 / T1 …(3)

B)Carnot engine efficiency is:

η w o r k = 1 − T c T h ηwork=1−TcTh

Carnot refrigeration efficiency is

η c o o l = T c T h − T c ηcool=TcTh−Tc η c o o l = 1 T h T c − 1 ηcool=1ThTc−1

Simple multiplication should give me the efficiency where both the engine and the refrigeration share the same hot and cold reservoirs:

η c o m b i n e d = 1 − T c T h T h T c − 1 ηcombined=1−TcThThTc−1

C) LAMBDA =1

D)PRACTICALLY ABSOLUTE ZERO TEMPERATURE CANNOT BE ACHIEVED