Thermal Efficiency of a power plant

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SUMMARY

The thermal efficiency of a power plant can be calculated using the formula n = Wout / Qh, where Wout is the electric output and Qh is the heat input. In this case, the electric output is 750 MW. To determine Qh, one must first calculate Qc, the heat rejected to the cooling water, which flows at a rate of 1.0 x 10^8 L/hr and experiences a temperature rise from 16 degrees C to 27 degrees C. The calculation of Qc involves determining the heat required to raise the temperature of the cooling water, which is essential for finding Qh and subsequently the thermal efficiency.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first law of thermodynamics.
  • Familiarity with power plant operations and energy conversion processes.
  • Knowledge of heat transfer calculations, particularly for fluids.
  • Ability to convert flow rates from liters per hour to megawatts.
NEXT STEPS
  • Calculate the heat transfer (Qc) required to raise the temperature of 1.0 x 10^8 L of water from 16 to 27 degrees C.
  • Determine the heat input (Qh) using the relationship Qh = Wout + Qc.
  • Research the concept of thermal efficiency in power plants and its significance in energy production.
  • Explore the impact of cooling water temperature on overall power plant efficiency.
USEFUL FOR

Engineers, thermodynamics students, and power plant operators seeking to understand and optimize the thermal efficiency of power generation systems.

jalamb90
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Homework Statement


The electric output of a power plant is 750 MW. Cooling water flows through the power plant at the rate 1.0*10^8 L/hr. The cooling water enters the plant at 16 degrees C and exits at 27 degrees C. What is the poer plant's thermal efficiency?


Homework Equations


W = Qh - Qc
n=Wout/Qh


The Attempt at a Solution


w=Qh-Qc
790MW=Qh-Qc
tried to solve for Qc in order to find Qh but couldn't figur out how to get Qc from 1.0*10^8 L/hr to MW.
 
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jalamb90 said:
...
The electric output of a power plant is 750 MW. Cooling water flows through the power plant at the rate 1.0*10^8 L/hr. The cooling water enters the plant at 16 degrees C and exits at 27 degrees C. What is the poer plant's thermal efficiency?
...
w=Qh-Qc
790MW=Qh-Qc
tried to solve for Qc in order to find Qh but couldn't figur out how to get Qc from 1.0*10^8 L/hr to MW.
What is the rate of heat flow to the cold reservoir (how much heat does it take to raise 10^8 L of water from 16 to 27 degrees C. ?)

That will give you Qc which will then enable you to find Qh and the efficiency.

AM
 

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