1. **Stating the problem:** We are given the formula for heat transfer rate $$Q = \dot{m} \cdot c_p \cdot (T_{in} - T_{out})$$ where: - $$Q$$ is the heat transfer rate in kJ/s, - $$\dot{m}$$ is the mass flow rate of exhaust gas in kg/s, - $$c_p$$ is the specific heat capacity in kJ/kg·K, - $$T_{in}$$ and $$T_{out}$$ are the inlet and outlet temperatures respectively. We want to calculate the electrical power output $$P$$ of the ORC system using the formula: $$P = \eta_{ORC} \cdot Q$$ where $$\eta_{ORC}$$ is the efficiency of the ORC system. 2. **Formula explanation:** - The heat transfer rate $$Q$$ represents the amount of heat energy transferred per second. - The power output $$P$$ is the useful electrical power generated, which is a fraction of $$Q$$ determined by the efficiency $$\eta_{ORC}$$. 3. **Intermediate work:** - First, calculate $$Q$$ using the given values of $$\dot{m}$$, $$c_p$$, $$T_{in}$$, and $$T_{out}$$. - Then multiply $$Q$$ by $$\eta_{ORC}$$ to find $$P$$. 4. **Example:** If $$\dot{m} = 2$$ kg/s, $$c_p = 1.005$$ kJ/kg·K, $$T_{in} = 500$$ K, $$T_{out} = 300$$ K, and $$\eta_{ORC} = 0.25$$, Calculate $$Q$$: $$Q = 2 \cdot 1.005 \cdot (500 - 300) = 2 \cdot 1.005 \cdot 200 = 402\ \text{kJ/s}$$ Calculate $$P$$: $$P = 0.25 \cdot 402 = 100.5\ \text{kJ/s}$$ 5. **Summary:** - Use $$Q = \dot{m} \cdot c_p \cdot (T_{in} - T_{out})$$ to find heat transfer rate. - Use $$P = \eta_{ORC} \cdot Q$$ to find electrical power output. This method allows you to calculate the power output of the ORC system from the exhaust gas parameters and system efficiency.