Chapter 8, Problem 8.1P

Interpretation Introduction

Interpretation:

The thermal efficiency of the cycle and the turbine efficiency for the simple power plant shown should be determined.

Concept introduction:

The steam power plant has the turbine operating adiabatically with the given inlet steam pressure and temperature. We used the values shown in the figure to determine the thermal efficiency and turbine efficiency. Turbine efficiency is determined to be the ratio of enthalpy change in the process to the enthalpy change in the reversible process while thermal efficiency is obtained by taking the ratio of energy released by the system to the heat applied to the system.

Answer to Problem 8.1P

The thermal efficiency of the cycle is determined to be 31.1 % and the turbine efficiency is 79 %.

Explanation of Solution

Given information:

The basic cycle for a steam power plant is shown by Fig. 8.1.

Suppose the turbine operates adiabatically with inlet steam at 6800 kPa and 550°C, and the exhaust steam enters the condenser at 50°C with a quality of 0.96. Saturated liquid water leaves the condenser, and is pumped to the boiler. Neglecting pump work and kinetic- and potential-energy changes, we need to determine the thermal efficiency of the cycle and the turbine efficiency.

Turbine efficiency is determined to be the ratio of enthalpy change in the process to the enthalpy change in the reversible process while thermal efficiency is obtained by taking the ratio of energy released by the system to the heat applied to the system.

Calculation:

Turbine efficiency is given by the formula

  ${\eta }_{turb}=\frac{\Delta H_{process}}{\Delta H_{reversible}}=\frac{H_3-H_2}{H_3{}^{\text{'}}-H_2}$

Where $H_3{}^{\text{'}}=(1-x\text{'})H_{liq}+x\text{'}{H}_{vap}$

Here $x\text{'}$is obtained by the formula

$\begin{array}{l}{S}_2=(1-x\text{'})S_{liq}+x\text{'}{S}_{vap}\\ =>6.9636=(1-x\text{'})0.7035+(x\text{'})(8.0776)\\ =>x\text{'}=0.845\end{array}$

We know that

$x=0.96$represents the vapour fraction of steam and the enthalpy of steam is $H_2=3531.5kJ/kg$

$\begin{array}{l}{H}_3=(1-x)H_{liq}+x{H}_{vap}=(1-0.96)209.3+(0.96)(2592.2)\\ =2496.9kJ/kg\end{array}$

We know that

$H_3{}^{\text{'}}=(1-x\text{'})H_{liq}+x\text{'}{H}_{vap}$ = $(1-0.845)209.3+(0.845)(2592.2)=2222.9kJ/kg$

Enthalpy of steam after neglecting work done is $H_2=H_{liq}=209.3kJ/kg$ .

Therefore, the turbine efficiency is

${\eta }_{turb}=\frac{\Delta H_{process}}{\Delta H_{reversible}}=\frac{H_3-H_2}{H_3{}^{\text{'}}-H_2}$ = $\frac{2496.9-3531.5}{2222.9-3531.5}=79\%$

And the thermal efficiency of the cycle is

${\eta }_{therm}=\frac{H_2-H_3}{H_2-H_1}=\frac{3531.5-2496.9}{3531.5-209.3}=31.1\%$

Conclusion

The thermal efficiency of the cycle is determined to be 31.1 % and the turbine efficiency is 79 %.