Open cycle gas turbine on heavy oil

Assumptions: calorific value of the heavy oil is not given; a typical value CV = 42 000 kJ/kg is assumed. Heat released by the fuel equals the heat received by the gases in the combustion chamber.

Data: $p_2 = p_3 = 500$ kPa, $T_3 = 650 + 273 = 923$ K, $p_1 = p_4 = 100$ kPa, $T_1 = 27 + 273 = 300$ K, $\eta_c = 0.80$, $\eta_t = 0.85$, $c_p = 1$ kJ/kg K, $\gamma = 1.4$; $r_p = 5$, $5^{0.2857} = 1.5838$.

Compressor

$$\begin{aligned} T_{2s} &= 300 \times 1.5838 = 475.15\ \text{K} \\ T_2 &= 300 + \frac{475.15 - 300}{0.80} = 518.93\ \text{K} \\ W_c &= 1 \times (518.93 - 300) = 218.93\ \text{kJ/kg} \end{aligned}$$

Turbine

$$\begin{aligned} T_{4s} &= \frac{923}{1.5838} = 582.77\ \text{K} \\ T_4 &= 923 - 0.85(923 - 582.77) = 633.80\ \text{K} \\ W_t &= 1 \times (923 - 633.80) = 289.20\ \text{kJ/kg} \end{aligned}$$

Thermal efficiency

$$\begin{aligned} W_{net} &= 289.20 - 218.93 = 70.26\ \text{kJ/kg} \\ Q_s &= c_p (T_3 - T_2) = 923 - 518.93 = 404.07\ \text{kJ/kg} \\ \eta_{th} &= \frac{70.26}{404.07} = 0.1739 = 17.39\% \end{aligned}$$

Power generating capacity

Heat supplied by fuel:

$$\dot Q = \dot m_f \times CV = 5 \times 42\,000 = 210\,000\ \text{kW}$$ $$P = \eta_{th} \times \dot Q = 0.1739 \times 210\,000 = 36\,518\ \text{kW} \approx 36.5\ \text{MW}$$

Corresponding air–fuel ratio (from $\dot m_f CV = (\dot m_a + \dot m_f)c_p(T_3 - T_2)$): $\dfrac{\dot m_a}{\dot m_f} = \dfrac{42\,000 - 404.07}{404.07} = 102.9$, i.e. about 515 kg/s of air.

Answer: Thermal efficiency = 17.39 %; power capacity ≈ 36.5 MW (for CV = 42 MJ/kg; in general $P = 0.1739 \times 5 \times CV$).