Question 5.5: For a combined-cycle power plant with dual-pressure HRSG, th......
For a combined-cycle power plant with dual-pressure HRSG, the following condi-tions are given:
• High and low pressure: p_\mathrm{HP} = 125 bars and p_\mathrm{LP} = 9 bars
• Gas turbine exhaust gas flow rate m_{g} = 480 kg/s and temperature t_{1 } = 675°C
• Feedwater temperature (state 6): t_{6} = 130°C
• Superheated steam parameters (state 10): p_{10} = 125 bars and temperature t_{10} = 550°C
• Pinch point PP = 10 K
• Specific heat of exhaust gas c_{pg} = 1.05 kJ/(kg K)
Calculate (a) the general steam production rate of the HRSG, (b) gas turbine exhaust heat rate, and (c) the stack gas temperature.
1. The steam and water parameters at various locations in the HRSG, taken from the h–s diagram, Appendix 11 and steam table, are given below.
At pressure p_\mathrm{LP} = p_\mathrm{2} = 9 bars, the saturation temperature is t_\mathrm{7} = t_\mathrm{8} = 175.36°C, vaporization enthalpy is h_\mathrm{fg,LP} = 2029.5 kJ/kg, saturated water enthalpy is h_\mathrm{7} = 742.64 kJ/kg, and saturated steam enthalpy is h_\mathrm{8} = 2772.1 kJ/kg. The feedwater enthalpy is h_\mathrm{6} = c_\mathrm{pw}t_\mathrm{6} = 4.187 × 130 = 544.3 kJ/kg.
At p_\mathrm{10} = 125 bars and t_\mathrm{10} = 550°C, the live steam enthalpy is h_\mathrm{10} = 3475.2 kJ/kg, feedwater temperature is t_\mathrm{fw,HP} = 175.36°C, and its enthalpy is h_\mathrm{fw,HP} = c_\mathrm{pw}t_\mathrm{fw,HP} = 4.187 × 175.36 = 734.2 kJ/kg. At p_\mathrm{ HP} = p_\mathrm{1} = 125 bar, saturated water temperature is t_\mathrm{9} = 327.7°C, and its enthalpy is h_\mathrm{9} = 1511.9 kJ/kg.
2. Flue gas temperature at states 2 and 4:
t_{2}=t_{9}+\mathrm{PP}=327.7^{\circ}{\mathrm{C}}+10\ K=337.7^{\circ}{\mathrm{C}}
t_{4}=t_{7}+\mathrm{PP}=175.36^{\circ}C+10\mathrm{~K=185.36^{\circ}C}
3. Energy balance for the HP superheater and HP evaporator yields the HP steam production rate of the HRSG
m_{s,\mathrm{HP}}\,=\,m_{\mathfrak{g}}c_{\mathrm{pg}}(t_{1}-t_{2})/(h_{10}-h_{9})m_{s,\mathrm{HP}}\,=\,480\times1.05\times(675-337.7)/(3475-151\,1.9)=86.6\,\,\mathrm{kg/s}
4. Energy balance for the HP economizer yields the gas temperature at HRSG HP section exit
t_{3}\,=\,t_{2}\,-\,m_{\mathrm{HP}}(h_{9}\,-\,h_{\mathrm{fw,HP}})/(m_{\mathrm{g}}c_{\mathrm{pg}})t_{3}\,=\,3\,37.7\,-\,86.6\times(1511 {.9\,-\,734.2})/(480\times1.05)=204\,{}^{\circ}C
5. LP steam production rate of the HRSG
m_{\mathrm{s.LP}}\,=\,m_{\mathrm{g}}c_{\mathrm{pg}}(t_{3}-t_{4})/h_{\mathrm{fg,LP}}
m_{\mathrm{s,LP}}\;=\;480 × 1.05 × (204 – 185.36)/2029.5 = 4.87 kg/s
6. Total steam production rate of the HRSG
m_{s}\,=\,m_{s,\mathrm{HP}}\,+\,m_{s,\mathrm{LP}}\,=\,86.6\,+\,4.87\,=\,9\,1.47\,\mathrm{kg/s}7. Heat output rate of the HRSG
Q_{\mathrm{HRSG}}\,=\,m_{\mathrm{s,HP}}(h_{10}\,-\,h_{\mathrm{fw,HP}})\,+\,m_{\mathrm{s,LP}}(h_{8}\,-\,h_{6})Q_{\mathrm{HRSG}}=86.6\times(3475-734.2)+4.87\times(2772.1-544.3)=248,190\ {\mathrm{kJ}} /{\mathrm{s}}
8. Stack gas temperature is
t_{5}\,=\,t_{1}-Q_\mathrm{HRSG}/(m_{\mathrm{g}}c_{\mathrm{pg}})=675-248,190/(480\times1.05)=\,182.5^{\circ}CAlternatively
t_{5}\,=\,t_{3}\,-\,m_{\mathrm{s,LP}}(h_{8}\,-\,h_{6})/(m_{\mathrm{g}}c_{\mathrm{pg}})t_{5}\,=\,204-4.87\times(2772.1-544.3)/(480\times1.05)=\,182.5^{\circ}C
Example 5.6 presents the calculated steam requirements of a dearator (refer to Figure 5.10).
