### Power Plant Efficiency Calculation**Problem Statement:**The electric output of a power plant is 800 MW. Cooling water flows through the power plant at the rate of \(1.00 \times 10^8\) L/hr. The cooling water enters the plant at 11.0 °C and exits at 29.0 °C.#### Part A**Question:**What is the power plant's thermal efficiency?**Instructions:**Express your answer as a percentage.---In this scenario, we are tasked with calculating the thermal efficiency of a power plant. Thermal efficiency is a measure of how well the plant converts heat from fuel into electrical energy. Before calculating the efficiency, the heat energy absorbed by the cooling water must be determined.1. **Electrical Power Output (P):** - Given: \( P = 800 \text{ MW} \)2. **Cooling Water Flow Rate:** - Given: \( \dot{V} = 1.00 \times 10^8 \, \text{L/hr} = 1.00 \times 10^5 \, \text{m}^3\text{/hr} \) - Conversion: \(1 \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)3. **Temperature of Cooling Water:** - Entrance: \( T_{\text{in}} = 11.0 \, ^\circ \text{C} \) - Exit: \( T_{\text{out}} = 29.0 \, ^\circ \text{C} \)4. **Specific Heat Capacity of Water (\( c \)):** - Given: Approx. \( c = 4.186 \, \text{kJ/kg} \cdot \, \text{K} \)5. **Heat Absorbed by Cooling Water (\( Q \)):** \[ Q = \dot{m} \cdot c \cdot \Delta T \] where \( \dot{m} \) is the mass flow rate of cooling water. Given the density of water \( \rho = 1000 \, \text{kg/m}^3 \): \[ \dot{m} = \dot{V} \cdot \rho = 1.00 \times 10^

The electric output of a power plant is 800 MW. Cooling water flows through the power plant at the rate 1.00 x 108 L/hr. The cooling water enters the plant at 11.0 °C and exits at 29.0 °C. What is the power plant's thermal efficiency? Express your answer as a percentage.

The electric output of a power plant is 800 MW. Cooling water flows through the power plant at the rate 1.00 x 108 L/hr. The cooling water enters the plant at 11.0 °C and exits at 29.0 °C. What is the power plant's thermal efficiency? Express your answer as a percentage.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Heat Engines, Entropy, And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 22.28P

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Question

### Power Plant Efficiency Calculation

**Problem Statement:**
The electric output of a power plant is 800 MW. Cooling water flows through the power plant at the rate of \(1.00 \times 10^8\) L/hr. The cooling water enters the plant at 11.0 °C and exits at 29.0 °C.

#### Part A

**Question:**
What is the power plant's thermal efficiency?

**Instructions:**
Express your answer as a percentage.

---

In this scenario, we are tasked with calculating the thermal efficiency of a power plant. Thermal efficiency is a measure of how well the plant converts heat from fuel into electrical energy. Before calculating the efficiency, the heat energy absorbed by the cooling water must be determined.

1. **Electrical Power Output (P):**
- Given: \( P = 800 \text{ MW} \)

2. **Cooling Water Flow Rate:**
- Given: \( \dot{V} = 1.00 \times 10^8 \, \text{L/hr} = 1.00 \times 10^5 \, \text{m}^3\text{/hr} \)
- Conversion: \(1 \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)

3. **Temperature of Cooling Water:**
- Entrance: \( T_{\text{in}} = 11.0 \, ^\circ \text{C} \)
- Exit: \( T_{\text{out}} = 29.0 \, ^\circ \text{C} \)

4. **Specific Heat Capacity of Water (\( c \)):**
- Given: Approx. \( c = 4.186 \, \text{kJ/kg} \cdot \, \text{K} \)

5. **Heat Absorbed by Cooling Water (\( Q \)):**
\[ Q = \dot{m} \cdot c \cdot \Delta T \]
where \( \dot{m} \) is the mass flow rate of cooling water.

Given the density of water \( \rho = 1000 \, \text{kg/m}^3 \):

\[ \dot{m} = \dot{V} \cdot \rho = 1.00 \times 10^

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