How Much Heat Does a Power Plant Transfer to a River Each Second?

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SUMMARY

The power plant outputs 1246 MW with an efficiency of 34.8%, indicating that 65.2% of the energy is released as heat to the nearby river. The total energy input can be calculated using the equation 1246 MW = 34.8% of total energy, leading to a total energy input of approximately 3585.63 MW. The heat transferred to the river can be determined by subtracting the power output from the total energy input. The river's flow rate of 1.4 * 10^6 kg/s plays a crucial role in how quickly the heat is absorbed, directly impacting the heat transfer rate in J/s.

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  • Understanding of thermodynamic efficiency
  • Knowledge of power output calculations
  • Familiarity with heat transfer principles
  • Basic concepts of fluid dynamics
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  • Calculate total energy input for power plants using efficiency formulas
  • Explore the relationship between flow rate and heat absorption in fluids
  • Learn about specific heat capacity of water and its implications in heat transfer
  • Investigate the impact of river flow rates on thermal pollution
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Engineers, environmental scientists, and students studying thermodynamics and fluid dynamics, particularly those interested in energy efficiency and heat transfer in power generation systems.

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Homework Statement


A power plant has a power output of 1246 MW and operates with an efficiency of 34.8 percent. Excess energy is carried away as heat from the plant to a nearby river that has a flow rate of 1.4 * 10^6 kg/s.
How much energy is transferred as heat to the river each second.
Answer in J/s


Homework Equations


change in entropy = Q/T


The Attempt at a Solution


I'm not sure how to set this one up, is 1246 the total output ie the fuel being consumed expels that much and 34.8 percent of that goes to making power, or is 1246 what goes to making power, and the other 65.2 percent goes to the river, and where does the rate of flow of the river come into play?
 
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1246 must be 34.8% of total fuel power. the flow rate of river is associated with how much heat 1 kg of water absorbs in 1 second. but for that we need other data.
 
I interpreted it as: The power plant has a power output of 1246 MW, which represents 34.8% of the energy put in.

So if the total energy is x, then:

\frac{1246}{x} = \frac{34.8}{100}

And so the amount of energy being released as heat is H = x - 1246.

The flowrate of the river affects how quickly the heat released by the power plant gets absorbed. The quicker the water flow, the quicker we expect the heat to get absorbed. If you use the flowrate of the river (kg/s) along with the specific heat of the water (J/kg*K) and the temperature (K), you will get an answer in J/s.
 

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