maximum possible efficiency for the power

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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**Question:**

Geothermal power plants use underground sources of hot water or steam to produce electricity. If a plant receives hot water at 350°F and rejects energy by heat transfer to the atmosphere at 60°F, what is the maximum possible efficiency for the power cycle?

**Options:**

- ○ 64%
- ○ 100%
- ○ 83%
- ○ 17%
- ○ 36%

**Explanation:**

This question is about calculating the maximum possible efficiency of a power cycle using the temperatures of heat source and heat sink. The efficiency can be calculated using the Carnot efficiency formula, which is based on the temperatures of the heat source (\(T_{hot}\)) and the heat sink (\(T_{cold}\)):

\[
\text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}}
\]

It's important to convert the temperatures from Fahrenheit to Kelvin before using them in the equation. 

\[ 
T_{hot} = (350 - 32) \times \frac{5}{9} + 273.15 
\]
\[ 
T_{cold} = (60 - 32) \times \frac{5}{9} + 273.15 
\] 

Using these values in the equation will give the maximum possible efficiency.
Transcribed Image Text:**Question:** Geothermal power plants use underground sources of hot water or steam to produce electricity. If a plant receives hot water at 350°F and rejects energy by heat transfer to the atmosphere at 60°F, what is the maximum possible efficiency for the power cycle? **Options:** - ○ 64% - ○ 100% - ○ 83% - ○ 17% - ○ 36% **Explanation:** This question is about calculating the maximum possible efficiency of a power cycle using the temperatures of heat source and heat sink. The efficiency can be calculated using the Carnot efficiency formula, which is based on the temperatures of the heat source (\(T_{hot}\)) and the heat sink (\(T_{cold}\)): \[ \text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}} \] It's important to convert the temperatures from Fahrenheit to Kelvin before using them in the equation. \[ T_{hot} = (350 - 32) \times \frac{5}{9} + 273.15 \] \[ T_{cold} = (60 - 32) \times \frac{5}{9} + 273.15 \] Using these values in the equation will give the maximum possible efficiency.
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