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Chapter 13

Q. 13.1

Consider a 20-MW cogeneration power plant in a campus complex. An energy balance analysis indicates the following energy fluxes for the power plant:
• Electricity generation 33 percent
• Condenser losses 30 percent
• Stack losses 30 percent
• Radiation losses 7 percent
It is estimated that all the condenser losses but only 12 percent of the stack losses can be recovered. Determine both the overall thermodynamic efficiency as well as the PURPA efficiency of the power plant.


Verified Solution

First, the recovered thermal energy is determined:

Et = Condenser Losses + Part of the Stack Losses

This thermal energy output can be expressed in terms of the fuel use (FU) of the power plant:

Et = 30% * FU + 12% * [30% * FU] = 0.34 * FU

The energy flow for the campus power plant is summarized in the diagram below:

Thus, the overall thermodynamic efficiency of the power plant can be easily determined using Eq (13.1):

η_{overall}=\frac {E_{e}+E_{t}}{FU}=\frac {0.33*FU+0.34*FU}{FU}=0.67

The PURPA efficiency can be calculated using Eq. (13.2),

η_{PURPA}=\frac {E_{e}+E_{t}/2}{FU}=\frac {0.33*FU+0.34*FU/2}{FU}=0.67

Therefore, the campus power plant meets the PURPA criteria (η_{PURPA} > 45 percent) and is thus a qualified cogeneration facility.