Question 6.4: A combined-cycle-based cogeneration plant consisting of a ga......

A combined-cycle-based cogeneration plant consisting of a gas turbine, an HRSG, and a condensing–extraction steam turbine is used for the production of power \mathrm{P_{el}} and useful heat \mathrm{Q_{u}}. The cogeneration plant (refer to Figure 6.8) is operating under the following conditions:

• Gas turbine capacity \mathrm{P_{GT}} is 360 MWe
• Gas turbine efficiency \mathrm{\bf{\eta}_{GT}} is 0.4
• Steam turbine capacity \mathrm{P_{ST}} is 176 MWe
• Fuel in gas turbine combustor is light fuel oil with LHV of 43 MJ/kg
• Plant useful heat production rate \mathrm{Q_{u}} is 254 MJ/s

Calculate

• Overall efficiency of the combined cycle
• Efficiency of the steam turbine
• Overall efficiency (EUF) of the cogeneration plant
• Fuel consumption rate of the cogeneration plant

Screenshot 2023-04-29 005219
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

1. Rate of fuel energy input to the gas turbine combustor

Q_{\mathrm{f}}\,=\,P_{\mathrm{GT}}/\mathbf{h}_{\mathrm{GT}}\,=\,360/0.4\,=\,900\,M\mathrm{J}/s

2. Overall (electrical) efficiency of the combined-cycle power plant

{\bf h}_{\mathrm{CC}}\,=\,P_{\mathrm{el}}/Q_{\mathrm{f}}\,=\,(360\,+\,176)/900=\,0.596

3. Efficiency of the steam turbine

{\bf h}_{ST}\,=\,({\bf h}_{CC}-{\bf h}_{GT})/(1-{\bf h}_{GT})=(0.596-0.4)/({1}-0.4)=0.326

4. Energy utilization factor (overall efficiency) of the cogeneration plant

\mathrm{{E}}\mathrm{{UF}}=(P_{\mathrm{el}}+Q_{\mathrm{u}})/Q_{\mathrm{f}}\,=(360+176+254)/900=0.878

5. Fuel consumption rate of the cogeneration plant

m_{\mathrm{f}}\,=\,Q_{\mathrm{f}}/\,\mathrm{LHV}\,=\,900/43\,=\,20.93\,\mathrm{kg/s}

Performance parameters of a syngas-fired combined-cycle cogeneration plant are calculated in Example 6.5.

Related Answered Questions