Problem 77
Question
At a coal-burning power plant a steam turbine operates with a power output of \(550 \mathrm{MW}\). The thermal efficiency of the power plant is \(32 \%\). (a) At what rate is heat discarded to the environment by this power plant? (b) At what rate must heat be supplied to the power plant by burning coal?
Step-by-Step Solution
Verified Answer
(a) 1168.75 MW; (b) 1718.75 MW.
1Step 1: Understanding Thermal Efficiency
The thermal efficiency (\(e\)) of a power plant is given by the formula \( e = \frac{W}{Q_H} \), where \(W\) is the work output (power output in this case) and \(Q_H\) is the heat input. Given that the efficiency is 32%, we can write \( 0.32 = \frac{550 \text{ MW}}{Q_H} \).
2Step 2: Calculate Heat Input
Rearrange the efficiency equation to solve for \(Q_H\). Thus, \(Q_H = \frac{550}{0.32} \text{ MW}\). Calculate \(Q_H\) to find the amount of heat that needs to be supplied to the plant.
3Step 3: Calculation: Heat Input
Calculate using the values: \[ Q_H = \frac{550}{0.32} = 1718.75 \text{ MW}. \] Thus, the rate at which heat must be supplied is \(1718.75 \text{ MW}\).
4Step 3: Calculate Heat Discarded
The heat discarded to the environment, \(Q_C\), can be found using the relation \( Q_H = W + Q_C \). By rearranging, \( Q_C = Q_H - W \). Use the calculated \(Q_H\) and given \(W\) (550 MW) to find \(Q_C\).
5Step 5: Calculation: Heat Discarded
Substitute the known values into the equation: \[ Q_C = 1718.75 \text{ MW} - 550 \text{ MW} = 1168.75 \text{ MW}. \] Thus, the rate at which heat is discarded to the environment is \(1168.75 \text{ MW}\).
Key Concepts
Heat InputHeat DiscardedPower Output
Heat Input
When we talk about heat input in the context of a power plant, we mean the amount of thermal energy that must be provided to produce electricity. The burning of coal provides this necessary energy in a coal-burning power plant. Let's break it down into simple steps:
Understanding this lays the foundation for grasping other concepts related to energy efficiency and power management.
- Thermal efficiency is a measurement of how well a power plant converts heat into work, or electricity. In our example, the thermal efficiency is 32%.
- This means that only 32% of the energy provided through heat input is converted into usable power.
- The rest of the energy is not lost; instead, it is expelled in other forms, such as waste heat.
Understanding this lays the foundation for grasping other concepts related to energy efficiency and power management.
Heat Discarded
Once we have supplied the heat energy to the power plant, not all of it will be converted into electrical power. The leftover energy is expelled as heat discarded. This is a crucial aspect of power plants that affects their environmental impact. Let's analyze how this works:
- The thermal efficiency indicates how effectively the input heat energy is utilized. Here, with 32% efficiency, the remainder of energy must be discarded as heat.
- Using the formula \(Q_C = Q_H - W\), we determine the amount of heat expelled. It's derived from subtracting the power output from the total heat input.
- The calculation reveals that about 1168.75 MW of heat is discarded into the environment because only 550 MW is converted into electrical power.
Power Output
Power output refers to the usable electrical energy generated by the power plant. It is a direct measure of how well the plant converts heat input into electricity.
- In our example, the power output is given as 550 MW. This is the electricity generated and supplied to the grid.
- The efficiency plays a vital role because it determines the relation between the power output and the heat input. Higher efficiency means more electricity for the same amount of heat.
- Understanding the power output helps in planning and managing energy resource utilization effectively.
Other exercises in this chapter
Problem 75
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Think \& Calculate A heat engine does 2700 J of work with an efficiency of \(0.18\). What is (a) the heat taken in from the hot reservoir and (b) the heat given
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Think \& Calculate A heat engine does \(2700 \mathrm{~J}\) of work with an efficiency of \(0.18\). What is (a) the heat taken in from the hot reservoir and (b)
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