Thermo, heat transfer rate, power plant

[Confusedby]

Homework Statement

A power plant generates 1300MW and operates at 315 °C a 20°C heat sink is available.
what is the minimum rate at which heat must be discarded?

Homework Equations

not really sure how to begin this one, have a feeling it is related to dq=Tds at constant pressure giving (dq/dT)p=T(dS/dT)p=(dH/dT)p=Cp but that seems to go no where. tried relating it to Enthalpy Changes along the path from 315->100 then condense to liquid water and go 100->20 but none of these seem to use the power out put. honestly completely lost on this one.

The Attempt at a Solution

 

[TSny]

Hi, Confusedby. Welcome to PF!

This looks like a basic heat engine problem. Have you studied the concept of efficiency of a heat engine and how the second law implies a maximum possible efficiency for given hot and cold reservoir temperatures?

 

[Confusedby]

Thank You

Well I feel a little silly, for some reason "maximum Thermal efficiency" became "maximum residual heat". If I were to use this arrangement:
(W/e)-W=q residual. were e is the efficiency and W is the work. it should give the minimal heat flow if I assume Carnot Efficiency correct?

 

[TSny]

Yes. q is the heat discarded rather than the rate at which heat is discarded. But if you let W be the work done each second, then q will be the heat discarded each second.

 

[Nickie]

I would approach this problem by first understanding the basic principles of thermodynamics and heat transfer. The power plant is generating 1300MW of power, which is a measure of the rate at which energy is being produced. The heat transfer rate, or the rate at which heat is being transferred, is a crucial factor in the operation of a power plant. In order to maintain a stable and efficient operation, the power plant must be able to effectively transfer heat from the hot source (315 °C) to the cool sink (20°C).

To determine the minimum rate at which heat must be discarded, we can use the equation Q = mCΔT, where Q is the heat transferred, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature. In this case, we are interested in the heat transfer from the power plant to the cool sink, so we can use the specific heat capacity of water (4.186 J/g°C).

First, we need to calculate the amount of heat that needs to be transferred. We know that the power plant is generating 1300MW, which is equivalent to 1,300,000,000 J/s. We also know that the temperature difference between the hot source and the cool sink is 315°C - 20°C = 295°C. Plugging these values into our equation, we get:

Q = (1,300,000,000 J/s) x (1 s) x (295°C) x (4.186 J/g°C) = 5.13 x 10^12 J

This means that the power plant must transfer at least 5.13 x 10^12 J of heat to the cool sink in order to maintain its operation. This corresponds to a heat transfer rate of 5.13 x 10^12 J/s or 5.13 TW (terawatts).

In conclusion, the minimum rate at which heat must be discarded from the power plant is 5.13 TW. This is an important factor to consider in the design and operation of power plants, as it affects their efficiency and overall performance. By understanding the principles of thermodynamics and heat transfer, we can accurately calculate and optimize the heat transfer processes in power plants.