A small scale steam engine might operate between the temperatures $20°\text{C}$and $300°\text{C}$, with a maximum steam pressure of $10$bars. Calculate the efficiency of a Rankine cycle with these parameters.

A small scale steam engine operates between the temperatures $20°\text{C}$and $300°\text{C}$, with a maximum steam pressure of$10$ bars.

The schematic diagram of a steam engine and the associated PV cycle, called the Rankine cycle is given by:

Here the dashed lines show where the fluid is in liquid form, where it is steam, and where it is part water and part steam.

In the Rankine cycle, heat is absorbed at the constant pressure in the broiler and expelled at constant pressure in the condenser.

The efficiency of Rankine cycle is given by

$\text{e}=1-\frac{{\text{H}}_4{\text{-H}}_1}{{\text{H}}_3{\text{-H}}_1}$.

Here, ${\text{H}}_1$is enthalpy at point $1$, ${\text{H}}_3$is enthalpy at point $3$, and ${\text{H}}_4$is enthalpy at point $4$in the below Rankine cycle.

At point $1$, there is saturated liquid water at $20°\text{C}$. Here the pressure is $0.023$bars and enthalpy is $84\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{Kg}$}\right.$.

At point $3$, there is superheated steam at $300°\text{C}$and $10$bars pressure. the enthalpy is $3051\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{Kg}$}\right.$and the entropy is $7.123\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{Kg}$}\right.$.

As, the expansion in the turbine is approximately adiabatic, so the entropy at point $4$must be equal as point $3$.

At point $4$, the mixture of saturated water role="math" localid="1646940722459" $\left({\text{S}}_{\text{Sw}}\text{=}0.297\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{Kg}$}\right.\right)$and saturated steam role="math" localid="1646940739322" $\left({\text{S}}_{\text{St}}\text{=}8.667\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{Kg}$}\right.\right)$at $20°\text{C}$.

From the properties of superheated steam table, all the above values are taken.

The entropy of the superheated steam is expressed as follows:

$\text{S=x}\left({\text{S}}_{\text{Sw}}\right)\text{+}\left(1-\text{x}\right){\text{S}}_{\text{St}}$

where,

localid="1646940682150" $$\begin{gathered} {\text{S}}_{\text{Sw}}\text{=Entropyofsaturatedwater} \\ {\text{S}}_{\text{St}}\text{=Entropyofsaturatedsteam} \\ \text{x=fractionofwater} \end{gathered}$$

Substituting the above values,

$$\begin{gathered} 7.123=\text{x}\left(0.297\right)\text{+}\left(1-\text{x}\right)\left(8.667\right) \\ \text{⇒x=}\frac{8.667-7.123}{8.667-0.297} \\ \text{⇒x=0.184} \end{gathered}$$

Again enthalpy $\left({\text{H}}_4\right)$of superheated steam is:

$$\begin{gathered} {\text{H}}_4\text{=x}\left({\text{H}}_1\right)\text{+}\left(1-\text{x}\right){\text{H}}_2 \\ \text{where} \\ {\text{H}}_1\text{=Enthalpyofsaturatedwater} \\ {\text{H}}_2\text{=Enthalpyofsaturatedsteam} \\ \text{x=Fractionofwater} \end{gathered}$$

Putting respective values,

localid="1646941622267" $$\begin{gathered} {\text{H}}_4\text{=}\left(0.184\right)\left(84\text{KJ}\right)\text{+}\left(1-0.184\right)\left(2538\text{KJ}\right) \\ {\text{H}}_4\text{=2086KJ} \end{gathered}$$

Here,

$$\begin{gathered} {\text{H}}_1\text{=84} \\ {\text{H}}_3\text{=3051} \\ {\text{H}}_4\text{=2086} \end{gathered}$$

So,$$\begin{gathered} \text{e=1-}\frac{{\text{H}}_4{\text{-H}}_1}{{\text{H}}_3{\text{-H}}_1} \\ \text{⇒e=1-}\frac{2086-84}{3051-84} \\ \text{⇒e=0.33} \end{gathered}$$

Thus, efficiency of engine is$0.33$.