Askiitians Tutor Team
Last Activity: 5 Months ago
To determine the least rate of heat rejection per kilowatt of net output from a cyclic heat engine operating between two temperatures, we can use the principles of thermodynamics, specifically the Carnot efficiency. The Carnot efficiency provides the maximum possible efficiency for a heat engine operating between two thermal reservoirs.
Understanding the Carnot Efficiency
The Carnot efficiency (\( \eta \)) is given by the formula:
η = 1 - (Tcold / Thot)
Where:
- Tcold is the absolute temperature of the cold reservoir (sink).
- Thot is the absolute temperature of the hot reservoir (source).
Converting Temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin:
- Hot reservoir (source) temperature: 800 °C = 800 + 273.15 = 1073.15 K
- Cold reservoir (sink) temperature: 30 °C = 30 + 273.15 = 303.15 K
Calculating the Carnot Efficiency
Now, we can plug these values into the Carnot efficiency formula:
η = 1 - (303.15 / 1073.15)
Calculating this gives:
η ≈ 1 - 0.282 = 0.718
This means the maximum efficiency of the engine is approximately 71.8%.
Determining the Heat Rejection Rate
Next, we need to find the rate of heat rejection per kilowatt of net output. The relationship between the heat input (\( Q_{in} \)), work output (\( W \)), and heat rejection (\( Q_{out} \)) can be expressed as:
W = Q_{in} - Q_{out}
Given that the net output is 1 kW, we can express the heat input in terms of the work output and efficiency:
Q_{in} = W / η
Substituting the values:
Q_{in} = 1 kW / 0.718 ≈ 1.39 kW
Calculating Heat Rejection
Now, we can find the heat rejection:
Q_{out} = Q_{in} - W
Substituting the known values:
Q_{out} = 1.39 kW - 1 kW ≈ 0.39 kW
Final Result
Thus, the least rate of heat rejection per kilowatt of net output for the engine operating between the given temperatures is approximately 0.39 kW. This means that for every kilowatt of useful work produced, the engine rejects about 0.39 kW of heat to the cold reservoir.