The Simple Ideal Rankine Cycle
Consider a steam power plant operating on the simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this cycle.
A steam power plant operating on the simple ideal Rankine cycle is considered. The thermal efficiency of the cycle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig. 10–3. We note that the power plant operates on the ideal Rankine cycle. Therefore, the pump and the turbine are isentropic, there are no pressure drops in the boiler and condenser, and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure.
First we determine the enthalpies at various points in the cycle, using data from steam tables (Tables A–4, A–5, and A–6):
\text{State 1:}\quad\quad\left.\begin{array}{l}P_{1}=75 \mathrm{kPa} \\\text { Sat. liquid }\end{array}\right\} \begin{array}{l}h_{1}=h_{ f @ 75 \mathrm{kPa}}=384.44 \mathrm{~kJ} / \mathrm{kg} \\v_{1}=v_{ f @ 75 \mathrm{kPa}}=0.001037 \mathrm{~m}^{3} / \mathrm{kg}\end{array}\begin{aligned}\text{State 2:}\quad\quad P_{2} &= 3 MPa\\s_{2} &= s_{1}\end{aligned}
\begin{aligned}w_{\text {pump,in }} &=v_{1}\left(P_{2}-P_{1}\right)=\left(0.001037 \mathrm{~m}^{3} / \mathrm{kg}\right)[(3000-75) \mathrm{kPa}]\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^{3}}\right) \\&=3.03 \mathrm{~kJ} / \mathrm{kg} \\h_{2} &=h_{1}+w_{\text {pump,in }}=(384.44+3.03) \mathrm{kJ} / \mathrm{kg}=387.47 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
\text{State 3:}\quad\quad\left.\begin{array}{c}P_{3}=3 \mathrm{MPa} \\T_{3}=350{ }^{\circ} \mathrm{C}\end{array}\right\} \begin{array}{l}h_{3}=3116.1 \mathrm{~kJ} / \mathrm{kg} \\s_{3}=6.7450 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{array}
\begin{aligned}\text{State 4:}\quad\quad P_{4} &= 75 kPa \quad \text{( sat. mixture )}\\s_{4} &= s_{3}\\x_{4}&=\frac{s_{4}-s_{f}}{s_{f g}}=\frac{6.7450-1.2132}{6.2426}=0.8861 \\h_{4}&=h_{f}+x_{4} h_{f g}=384.44+0.8861(2278.0)=2403.0 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Thus,
\begin{array}{l}q_{\text {in }}=h_{3}-h_{2}=(3116.1-387.47) \mathrm{kJ} / \mathrm{kg}=2728.6 \mathrm{~kJ} / \mathrm{kg} \\q_{\text {out }}=h_{4}-h_{1}=(2403.0-384.44) \mathrm{kJ} / \mathrm{kg}=2018.6 \mathrm{~kJ} / \mathrm{kg}\end{array}
and
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{2018.6 \mathrm{~kJ} / \mathrm{kg}}{2728.6 \mathrm{~kJ} / \mathrm{kg}}=0.260 \text { or } 26.0 \%
The thermal efficiency could also be determined from
\begin{array}{c}w_{\text {turb }, \text { out }}=h_{3}-h_{4}=(3116.1-2403.0) \mathrm{kJ} / \mathrm{kg}=713.1 \mathrm{~kJ} / \mathrm{kg} \\w_{\text {net }}=w_{\text {turb }, \text { out }}-w_{\text {pump,in }}=(713.1-3.03) \mathrm{kJ} / \mathrm{kg}=710.1 \mathrm{~kJ} / \mathrm{kg}\end{array}
or
w_{\text {net }}=q_{\text {in }}-q_{\text {out }}=(2728.6-2018.6) \mathrm{kJ} / \mathrm{kg}=710.0 \mathrm{~kJ} / \mathrm{kg}
and
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{710.0 \mathrm{~kJ} / \mathrm{kg}}{2728.6 \mathrm{~kJ} / \mathrm{kg}}=0.260 \text { or } 26.0 \%
That is, this power plant converts 26 percent of the heat it receives in the boiler to net work. An actual power plant operating between the same temperature and pressure limits will have a lower efficiency because of the irreversibilities such as friction.
Discussion Notice that the back work ratio \left(r_{b w}=w_{\text {in }} / w_{\text {out }}\right) of this power plant is 0.004, and thus only 0.4 percent of the turbine work output is required to operate the pump. Having such low back work ratios is characteristic of vapor power cycles. This is in contrast to the gas power cycles, which typically involve very high back work ratios (about 40 to 80 percent).
It is also interesting to note the thermal efficiency of a Carnot cycle operating between the same temperature limits
\eta_{\text {th,Carnot }}=1-\frac{T_{\min }}{T_{\max }}=1-\frac{(91.76+273) \mathrm{K}}{(350+273) \mathrm{K}}=0.415
Here T_{min} is taken as the saturation temperature of water at 75 kPa. The difference between the two efficiencies is due to the large external irreversibility in the Rankine cycle caused by the large temperature difference between steam and the heat source.
