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The Carnot Cycle And Efficiency Of A Perfect Heat Engine


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Strings (SiPjAjk) = S7P3A32     Base Sequence = 12735     String Sequence = 12735 - 3 - 32



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The Carnot Cycle

(a) Derive the equation for the efficiency of a perfect heat engine in terms of the temperatures of the primary heat reservoir and the secondary heat reservoir using the Carnot cycle.

(b) Determine the efficiency of a steam engine (heat engine) operated reversibly between a primary reservoir and a secondary reservoir at 35oC.

The strings:

S7P3A32 (Force - Push)

The math:
Pj Problem of Interest is of type force (force-push). Work problems are generally of type force. It is then left to determine whether the problem is of type pull or push. Usually a heat engine absorbs heat from the primary reservoir and releases (push) the heat to the secondary reservoir. It is in this sense that the primary problem of interest is of type force-push.

The Carnot Cycle

(a) In 1824 A.D. Sardi Carnot (1796-1832) presented a cycle of changes in the state of a gas interacting with its environment. This cycle of changes is now known as the Carnot Cycle (figure 14.28).
The Carnot cycle consists of four stages:
(1) An isothermal expansion of the gas from volume V1 to volume V2 in which work is done on the environment and heat absorbed from the environment at temperature Thot
(2) Expansion of the gas from volume V2 to volume V3 at a lower temperature Tcold without heat transfer (adiabatic expansion).
(3) Isothermal compression of gas at volume V3 to volume V4.
(4) Adiabatic compression of gas from volume V3 to its original volume, V1.

The Carnot Circle assumes that all processes are reversible so that net entropy change in the universe is zero. Processes (2) and (4) (adiabatic processes) have no heat transfer (q = 0), so no change in entropy of the environment. Processes (1) and (3) involve the transfer of heat, so there is change in the entropy of the environment.
Let heat transfer during processes (1) and (3) be qhot and qcold respectively, then:

entropy change in the environment due to process (1):
= -qhot/Thot (heat is absorbed from the environment)

entropy change in the environment due to process (3):
= qcold/Tcold (heat is transfered to the environment)

Since net entropy change in the universe is zero, we have:
-qhot/Thot + qcold/Tcold = 0
So, qhot/Thot = qcold/Tcold ------(5)

by the conservation of energy, we have:
work done by system = w = qhot - qcold
So, w/qhot = (Thot - Tcold)/Thot

So efficiency of a perfect heat engine (reversible heat engine);
= (Thot - Tcold)/Thot
where Thot is temperature of primary reservoir and Tcold is temperature of secondary reservoir.

(b) Temperature of primary reservoir = 100oC = 373oK
Temperature of secondary reservoir = 35oC = 308oK
So, efficiency of steam engine = (373-308)/373 = 0.174 or 17.4%.
Efficiency is less if engine is not perfect.

The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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