Question 10.3: Effect of Boiler Pressure and Temperature on Efficiency Cons...

A steam power plant operating on the ideal Rankine cycle is considered. The effects of superheating the steam to a higher temperature and raising the boiler pressure on thermal efficiency are to be investigated.
Analysis The T-s diagrams of the cycle for all three cases are given in Fig. 10–10.

(a) This is the steam power plant discussed in Example 10–1, except that the condenser pressure is lowered to 10 kPa. The thermal efficiency is determined in a similar manner:

State 1:    \left.\begin{array}{l} P_{1}=10 kPa \\ \text { Sat. liquid } \end{array}\right\} \begin{aligned} &h_{1}=h_{f @ 10 kPa }=191.81 kJ / kg \\ &v_{1}=v_{f @ 10 kPa }=0.00101 m ^{3} / kg \end{aligned}

State 2:        \begin{aligned} 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(0.00101 m ^{3} / kg \right)[(3000-10) kPa ]\left(\frac{1 kJ }{1 kPa \cdot m ^{3}}\right) \\ &=3.02 kJ / kg \\ h_{2} &=h_{1}+w_{\text {pump,in }}=(191.81+3.02) kJ / kg =194.83 kJ / kg \end{aligned}

State 3:    \left.\begin{array}{l} P_{3}=3 MPa \\ T_{3}=350^{\circ} C \end{array}\right\} \quad \begin{aligned} &h_{3}=3116.1 kJ / kg \\ &s_{3}=6.7450 kJ / kg \cdot K \end{aligned}

State 4:      \begin{aligned} P_{4} &=10 kPa \quad \text { (sat. mixture) } \\ s_{4} &=s_{3} \end{aligned}

x_{4}=\frac{s_{4}-s_{f}}{s_{f g}}=\frac{6.7450-0.6492}{7.4996}=0.8128

Thus,

\begin{aligned} h_{4} &=h_{f}+x_{4} h_{f g}=191.81+0.8128(2392.1)=2136.1 kJ / kg \\ q_{\text {in }} &=h_{3}-h_{2}=(3116.1-194.83) kJ / kg =2921.3 kJ / kg \\ q_{\text {out }} &=h_{4}-h_{1}=(2136.1-191.81) kJ / kg =1944.3 kJ / kg\end{aligned}

and

\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{1944.3 kJ / kg }{2921.3 kJ / kg }=0.334 \text { or } 33.4 \%

Therefore, the thermal efficiency increases from 26.0 to 33.4 percent as a result of lowering the condenser pressure from 75 to 10 kPa. At the same time, however, the quality of the steam decreases from 88.6 to 81.3 percent (in other words, the moisture content increases from 11.4 to 18.7 percent).
(b) States 1 and 2 remain the same in this case, and the enthalpies at state 3 (3 MPa and 600°C) and state 4 (10 kPa and s_{4}=s_{3} ) are determined to be

\begin{aligned}&h_{3}=3682.8 kJ / kg \\&h_{4}=2380.3 kJ / kg \quad\left(x_{4}=0.915\right)\end{aligned}

Thus,

\begin{aligned}q_{\text {in }} &=h_{3}-h_{2}=3682.8-194.83=3488.0 kJ / kg \\q_{\text {out }} &=h_{4}-h_{1}=2380.3-191.81=2188.5 kJ / kg\end{aligned}

and

\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{2188.5 kJ / kg }{3488.0 kJ / kg }=0.373 \text { or } 37.3 \%

Therefore, the thermal efficiency increases from 33.4 to 37.3 percent as a result of superheating the steam from 350 to 600°C. At the same time, the quality of the steam increases from 81.3 to 91.5 percent (in other words, the moisture content decreases from 18.7 to 8.5 percent).
(c) State 1 remains the same in this case, but the other states change. The enthalpies at state 2 (15 MPa and S_{2}=S_{1} ), state 3 (15 MPa and 600°C), and state 4 (10 kPa and s_{4}=s_{3} ) are determined in a similar manner to be

\begin{aligned}&h_{2}=206.95 kJ / kg \\&h_{3}=3583.1 kJ / kg \\&h_{4}=2115.3 kJ / kg \quad\left(x_{4}=0.804\right)\end{aligned}

Thus,

\begin{aligned}q_{\text {in }} &=h_{3}-h_{2}=3583.1-206.95=3376.2 kJ / kg \\q_{\text {out }}=h_{4}-h_{1} &=2115.3-191.81=1923.5 kJ / kg\end{aligned}

and

\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{1923.5 kJ / kg }{3376.2 kJ / kg }=0.430 \text { or } 43.0 \%

Discussion The thermal efficiency increases from 37.3 to 43.0 percent as a result of raising the boiler pressure from 3 to 15 MPa while maintaining the turbine inlet temperature at 600°C. At the same time, however, the quality of the steam decreases from 91.5 to 80.4 percent (in other words, the moisture content increases from 8.5 to 19.6 percent).