Combined Heat and Power - University of Dayton
Based on this calculation, at 500 kW of electrical power output, the useful thermal output would be 450 kW. However on a continuous basis, the plant can only use 30 kW of thermal power. Thus, useful power of 300 kW becomes the constraint. Solving Equation 1 for the electrical power output, E, at the useful thermal power constraint of 300 kW gives: ................
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