# Coal Power Plant Problem Solution ## Given Data -Plant Capacity:** \( 120) MW (awatts) **Steam Temperature( T_{} \)):** \( 540circ\text{C} \) - **Cooling Water Temperature (\( T_{cold} \)):** \( 35^\circ\text{C} \) - **Plant Efficiency:** \( 35\% \) (\( \eta = .35 \)) - **Coal Energy Density:** \( 24 \) MJ/kg - **Capacity Factor:** \( 90\% \) (\( .90 \)) - **Period:** 1 day --- ## 1. **How much coal (in kg) is used on average per day?** ### **Step 1: Find Daily Average Output Power** \[ \text{Average Power} = (\text{Plant Capacity}) \times (\text{Capacity Factor}) = 120\, \text{MW} \times .90 = 108\, \text{MW} \] ### **Step 2: Find Total Electrical Energy Produced per Day** \[ \text{Energy (per day)} = \text{Power} \times \text{time} = 108\, \text{MW} \times 24\, \text{h} \] Convert MW to MJ/h: \[ 1\, \text{MW} = 1\, \text{MJ/s} \Rightarrow 1\, \text{MW} \times 360\, \text{s/h} = 360\, \text{MJ/h} \] So, \[ \text{Energy} = 108\, \text{MW} \times 24\, \text{h} \times 360\, \text{MJ/h/MW} \] \[ = 108 \times 24 \times 360\, \text{MJ} \] \[ = 25,920 \times 360\, \text{MJ} \] \[ = 93,312,000\, \text{MJ} \] ### **Step 3: Find Chemical Energy Needed** \[ \text{Plant Efficiency} = \frac{\text{Electrical Output}}{\text{Fuel Input}} \] \[ \Rightarrow \text{Fuel Input} = \frac{\text{Electrical Output}}{\text{Efficiency}} \] \[ \text{Fuel Input} = \frac{93,312,000\, \text{MJ}}{.35} = 266,605,714\, \text{MJ} \] ### **Step 4: Calculate Mass of Coal Used** \[ \text{Coal used (kg)} = \frac{\text{Chemical Energy Needed}}{\text{Energy Density}} = \frac{266,605,714\, \text{MJ}}{24\, \text{MJ/kg}} = 11,108,571\, \text{kg} \] ### **Final Answer (Coal Used):** \[ \boxed{11,110,000\, \text{kg (rounded to 3 significant figures)}} \] --- ## 2. **What is the Carnot efficiency of the plant?** ### **Step 1: Convert Temperatures to Kelvin** \[ T_{hot} = 540^\circ\text{C} = 540 + 273.15 = 813.15\, \text{K} \] \[ T_{cold} = 35^\circ\text{C} = 35 + 273.15 = 308.15\, \text{K} \] ### **Step 2: Carnot Efficiency Formula** \[ \eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}} \] \[ \eta_{Carnot} = 1 - \frac{308.15}{813.15} = 1 - .379 = .621 \] ### **Final Answer (Carnot Efficiency):** \[ \boxed{62.1\%} \] --- ## 3. **What is the 2nd law efficiency of the plant?** **2nd Law Efficiency (\( \eta_{II} \))** is the ratio of actual efficiency to Carnot efficiency: \[ \eta_{II} = \frac{\text{Actual Efficiency}}{\text{Carnot Efficiency}} \] \[ \eta_{II} = \frac{.35}{.621} = .564 \] \[ = 56.4\% \] ### **Final Answer (Second Law Efficiency):** \[ \boxed{56.4\%} \] --- ## 4. **Winter vs. Summer: Does the plant run more or less efficiently (1st law efficiency) in winter?** - In **winter**, the cooling water temperature (\( T_{cold} \)) is **lower**. - Carnot efficiency increases as \( T_{cold} \) decreases (because \( 1 - T_{cold}/T_{hot} \) increases). - **Real-world plants** can reject more heat at a lower temperature, potentially increasing actual (1st law) efficiency. ### **Conclusion:** > **The plant runs more efficiently (higher 1st law efficiency) in winter when the cooling water is colder.** --- ## **Summary Table** | Quantity | Value | |---------------------------------|-----------------| | Coal used per day (kg) | 11,110,000 | | Carnot efficiency | 62.1% | | 2nd law efficiency | 56.4% | | 1st law efficiency (winter) | Increases | --- ### **Image: Coal Power Plant Schematic**  *Alt text: Schematic diagram of a coal-fired power plant showing the steam generation, turbine, condenser, and cooling water loop.* ---