Question

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55 percent. The waste heat from this engine is rejected to a nearby lake at \(60^{\circ} \mathrm{F}\) at a rate of \(800 \mathrm{Btu} / \mathrm{min} .\) Determine \((a)\) the power output of the engine and \((b)\) the temperature of the source.

Short answer

Answer: The power output of the engine is 16.29 Btu/s, and the temperature of the source is 642.69 K.

Step-by-step solution

Convert the Lake's Temperature to Kelvin

We are given the temperature at which the waste heat is rejected at the lake, which is 60°F. We have to convert this temperature to Kelvin. The formula to convert Fahrenheit to Kelvin is: K = (°F + 459.67) x (5/9) So, T₂ is: T₂ = (60 + 459.67) x (5/9) = 288.71 K

Calculate the Carnot Efficiency

We know the engine's actual thermal efficiency (η) is 55%. Since the engine is working on a Carnot cycle, we use Carnot efficiency (η_carnot) in the equation: 1 - (T₂ / T₁) = η_carnot. η_carnot = 1 - (T₂ / T₁) As η is given, η_carnot = η = 0.55 0.55 = 1 - (T₂ / T₁)

Find the Temperature of the Source (T₁)

From the equation in step 2, we can find the temperature of the source (T₁) using: T₁ = T₂ / (1 - η) Inserting the values, we get: T₁ = 288.71 / (1 - 0.55) = 642.69 K The temperature of the source is 642.69 K.

Calculate the Heat Input (Q₁) and Heat Rejected (Q₂)

The thermal efficiency (η) is given by: η = W / Q₁ The amount of heat rejected (Q₂) is given as 800 Btu/min. We'll convert it to Btu/s by dividing by 60, as there are 60 seconds in a minute. Q₂ = 800 / 60 = 13.33 Btu/s Since the efficiency η = 1 - (Q₂ / Q₁), we can find the heat input using the formula: Q₁ = Q₂ / (1 - η) Inserting the values, we get: Q₁ = 13.33 / (1 - 0.55) = 29.62 Btu/s

Calculate the Power Output (W)

To calculate the power output of the engine, use the thermal efficiency formula: η = W / Q₁ Rearranging the formula for power output (W), we get: W = η × Q₁ W = 0.55 × 29.62 Btu/s W = 16.29 Btu/s The power output of the engine is 16.29 Btu/s. To summarize, we calculated: (a) The power output of the engine is 16.29 Btu/s. (b) The temperature of the source is 642.69 K.

Key concepts

Thermal Efficiency

Thermal efficiency is a measure of a heat engine's ability to convert the energy from a heat source into useful work. It is expressed as a percentage, calculated by the formula \( \eta = \frac{W}{Q_1} \) where \( \eta \) is the thermal efficiency, \( W \) is the work output, and \( Q_1 \) is the heat input.

With a higher thermal efficiency, an engine will be more effective at converting heat into work, leading to less wasted energy. However, due to the second law of thermodynamics, no heat engine can be 100% efficient, as some waste heat will always be produced.

For the given exercise, the thermal efficiency is 55%. This tells us that for every unit of heat energy put into the Carnot engine, 55% is converted into work, while the remaining 45% is rejected, in this case, into a lake.

Carnot Engine

The Carnot engine is an idealized heat engine with the maximum possible efficiency allowed by physical laws. It is based on a theoretical thermodynamic cycle, known as the Carnot cycle. This cycle consists of four reversible processes: two isothermal and two adiabatic. No actual engines achieve the Carnot efficiency due to real-world frictions and non-reversible processes, but it serves as a benchmark to gauge the performance of actual heat engines.

In practice, while no engine can operate with Carnot efficiency, it remains a vital concept as it sets the absolute limit for the efficiency of all heat engines. For the exercise, understanding the Carnot engine is crucial because it provides the foundation for determining the maximum efficiency given the temperatures of the heat source and sink.

Heat Engine Performance

Evaluating heat engine performance involves looking at various parameters, such as thermal efficiency, power output, and heat transfer rates. The engine's performance is directly linked to how efficiently it can convert heat into work and how much waste heat is produced.

To improve an engine's performance, one might focus on operational parameters like minimizing energy losses, optimizing temperatures, and reducing friction. In our example, the Carnot cycle gives us an ideal to compare with, revealing that our engine is relatively performant with a 55% thermal efficiency. By knowing the efficiency and heat rejection rates, we can deduce important performance indicators like power output and the source temperature, which paint a complete picture of the engine's capability.

Temperature Conversion

Temperature conversion is crucial for calculations involving heat engines, as different units are used in various scientific applications. For instance, engineers typically use Kelvin for thermodynamic equations, while Fahrenheit or Celsius may be more common in everyday contexts.

To convert Fahrenheit to Kelvin, the formula \( K = (\text{°F} + 459.67) \times \frac{5}{9} \) is used. This step is essential in solving the exercise to ensure that the temperatures of the sink (lake) and source are in the same units, allowing for accurate computation of efficiency and power output. In the provided solution, the lake's temperature is converted to Kelvin to apply the Carnot efficiency formula correctly and find the heat source temperature.