CodeCogs

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

You're viewing an older version of this page (#3373). View the current version.

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 the WS is returned to it's original state.

Expansion from $V_1$ to $V_2$ and a Work Ouput of $W_1$

Compression from $V_2$ to $V_1$ and a Work input of $W_2$

$\text{Net Work Output} = W_1 - W_2 = \text{Area of Cycle}$

(1)

The First Law Applied to a Cycle

By Definition:

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

(2)

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

(3)

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

(4)

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.

Drawing the cycle on a T $\Phi$ diagram

Heat supplied, $\Phi$ , increasing

$\text{Heat supplied} = \text{area under 1,B,2} = Q_S$

(5)

Heat rejected, $\Phi$ decreasing

$\text{Heat rejected} = \text{area under 1,B,2} = Q_R$

(6)

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

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

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

(7)

therefore

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

(8)

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.

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

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

(9)

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

(10)

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

(11)

$Q_R=\text{Heat Rejected}$

(12)

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

(13)

therefore

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

(14)

But from the T Q Diagram

$\Phi _4 - \Phi _1) = }(\Phi_ 3 - \Phi _2)}$

(15)

Therefore

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

(16)

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

$\left(\frac{T_2\;-\;T_1}{T_2} \right)Q_S$

(17)

Example 1

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