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A description of thermodynamic cycles and in particular the Carnot Cycle and the proof of it's thermal efficiency

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THERMODYNAMIC CYCLES

A series of operations carried out on the Working Substance (WS) during which heat is supplied(Q) .There is a work output(W) after which thew WS is returned to it's original state.

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$Expansion\;from\;V_1\;to\;V_2\;\;\;\;\;and\;\;\;\;a\;Work\;out put\;of\;W_1$

(1)

$Compression\;from\;V_2\;to\;V_1\;\;\;\;and\;\;\;\;a\;Work\;in put\;of\;W_2$

(2)

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$Nett\;Work\;out put\;=\;W_1\;-\;W_2\;=\;Area\;of\;Cycle$

(3)

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The First Law Applied to a Cycle.

By Definition:-

$P_1\;\geq \;P_2\;\;\;\;\;\;\;\;E_1\;=\;E_2$

(4)

$V_1\;= \;V_2\;\;\;\;\;\;\;\;H_1\;=\;H_2$

(5)

$T_1\;= \;T_2\;\;\;\;\;\;\;\;Q_1\;=\;Q_2$

(6)

The Energy of the WS at the start of the cycle + Heat supplied = Energy of the WS at the end of the cycle + WD;+Heat losses.

or Work output = Heat supplied - Heat lost and rejected.

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Now let us draw the cycle on a $T\;\Phi$ diagram

$Heat\; supplied\;;\;\Phi \;increasing$

(7)

$heat\;supplied\;=\;area\;under\;1,B,2\;=\;Q_S$

(8)

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$Heat\; rejected\;;\;\Phi \;decreasing$

(9)

$heat\;rejected\;=\;area\;under\;1,B,2\;=\;Q_R$

(10)

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$Area \;of\; diagram\;=\;Q_S\;-\;Q_R\;=\;Work\;Done$

(11)

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NOTE the area of the P V diagram gives the Work Done in FT lbs The area of the T $\Phi$ diagram gives the work Done in BTUs

The Thermal efficiency of a Cycle.

$\eta \;=\;\frac{Work\; Done}{Heat\;Suppplied}\;=\;\frac{Q_S\;-\;Q_R}{Q_S}$

(12)

$\therefore\;\;\;\;\;\eta \;=\;1\;-\;\frac{Q_R}{Q_S}$

(13)

Carnot Cycle.

We can obtain the efficiency of the Carnot Cycle and hence the efficiency of any Reversible cycle operating between the temperatures of $T_1\;and\;T_2$ . This then represents the Ultimate thermal efficiency. This is then used to compare the efficiencies of other cycles operating between the same two temperatures. The importance of the Carnot Cycle in this role can not be under estimated.

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Stage 1 to 2

Reversible adiabatic i.e.isentropic, compression of WS from $T_1\;to\;T_2$

Stage 2 to 3

Isothermal heating with expansion.

Stage 3 to 4

Adiabatic (reversible) isentropic expansion of WS from $T_2\;to\;T_1$

Stage 4 to 1

Isothermal cooling with Compression.

The cycle can also be expressed on a T $\Phi$ diagram.

The Efficiency of the Carnot Cycle

$\eta \;=\;1\;-\;\frac{Q_R}{Q_S}$

(14)

$Q_S\;=\;Heat\; Supplied\;\;\;\;(2\;to\;3)$

(15)

$=\;area\;a\;2\;3\;b\;=\;T(\Phi _3\;-\;\Phi _2)$

(16)

$Q_R\;=\;Heat\;Rejected$

(17)

$=\;area\;a\;1\;4\;b \;=\;T_1(\Phi _4\;-\;\Phi _1)$

(18)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{Q_R}{Q_S}\;=\;1\;-\;\frac{T_1(\Phi _4\;-\;\Phi _1)}{T_2(\Phi_ 3\;-\;\Phi _2)}$

(19)

$But\; from\; the\; T\; Q\; Diagram\;(\Phi _4\;-\;\Phi _1)\;=\;}(\Phi_ 3\;-\;\Phi _2)}$

(20)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{T_1}{T_2}$

(21)

To Improve Efficiency

Increase the value of $(T_2\;-\;T_1)$

Lower the general level of Temperature.

It is more advantageous to lower the temperature at which heat is rejected than to raise the temperature at which it is supplied.

The available Energy is $\left(\frac{T_2\;-\;T_1}{T_2} \right)Q_S$

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EXAMPLE

The efficiency of a steam Engine is 52% of the Carnot Efficiency if the steam is supplied at 350^0,F and condensed at 150^0.F. Find the heat required to produce 1 HP for 1 minute.