Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A patent application describes a device that at steady state receives a heat transfer at the rate \(1 \mathrm{~kW}\) at a temperature of \(167^{\circ} \mathrm{C}\) and generates electricity. There are no other energy transfers. Does the claimed performance violate any principles of thermodynamics? Explain.

Short Answer

Expert verified
Yes, the claimed performance violates the second law of thermodynamics as it assumes 100% conversion of heat to work, which is impossible.

Step by step solution

01

- Identify the Working Temperatures

Convert given temperature to Kelvin. Temperature in Celsius is given as 167°C. Use the formula: \[ T(K) = T(°C) + 273.15 \] So, the temperature in Kelvin is \[ 167 + 273.15 = 440.15 \text{ K} \]
02

- Determine the Heat Transfer Rate

Given the heat transfer rate is 1 kW (which is equivalent to 1000 W).
03

- Understand the Claimed Performance

The device claims to generate electricity by receiving heat at the given rate.
04

- Apply the Second Law of Thermodynamics

According to the second law of thermodynamics, no device can convert heat completely into work without any other effect. An ideal heat engine's efficiency is given by: \[ \text{Efficiency} = 1 - \frac{T_C}{T_H} \] where \( T_H \) is the high-temperature reservoir (in this case, 440.15 K) and \( T_C \) is the low-temperature reservoir (usually, the ambient temperature). Assume ambient temperature \( T_C \) is 300 K.
05

- Calculate the Maximum Efficiency

Using the temperatures: \[ \text{Efficiency} = 1 - \frac{300}{440.15} = 1 - 0.6815 = 0.3185 \text{ or } 31.85\text{\text%} \]
06

- Evaluate the Performance

The maximum theoretical efficiency is 31.85%, meaning the generated electricity can be at most 31.85% of the heat received. For a 1 kW heat transfer: \[ \text{Max Electricity Generation} = 1 \text{ kW} \times 0.3185 = 0.3185 \text{ kW} \text{ or } 318.5 \text{ W} \]
07

- Conclusion

If the patent claims converting 1 kW of heat directly into 1 kW of electricity, it violates the second law of thermodynamics because it exceeds the maximum possible efficiency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

heat engine efficiency
Heat engine efficiency is a key concept in thermodynamics. It measures how well a heat engine converts heat into useful work. The efficiency of a heat engine is defined as the ratio of work output to heat input. The formula for the efficiency of an ideal heat engine, also known as the Carnot efficiency, is given by:
\[ \text{Efficiency} = 1 - \frac{T_C}{T_H} \]
Here, \(T_H\) represents the temperature of the high-temperature reservoir, and \(T_C\) denotes the temperature of the low-temperature reservoir.

According to the second law of thermodynamics, no heat engine can have an efficiency of 100%, meaning that some heat must always be rejected to the surroundings. In the example given, the temperatures are 440.15 K (from 167°C) and 300 K (ambient temperature).

Using the formula, we find that the maximum efficiency is approximately 31.85%. Thus, the device in the problem can theoretically convert only up to 31.85% of the input heat into electricity. This translates to a maximum electricity generation of around 318.5 W from an input of 1 kW.
thermodynamic principles
Thermodynamic principles are the laws that govern the transfer of energy and the direction of these processes. The Second Law of Thermodynamics states that heat cannot spontaneously flow from a colder body to a hotter body, and also that it is impossible to convert all the heat from a heat source into work.

For any heat engine, this law implies that its efficiency will always be less than 100%. This principle is crucial when assessing the performance of heat engines and other devices that convert energy.

In the provided exercise, the second law directly addresses the claimed performance of the device. If a patent claims that a device can convert 1 kW of heat into 1 kW of electricity without any heat rejection, it violates this law. The actual efficiency calculated (31.85%) showcases that the claimed performance is not feasible and emphasizes the importance of the second law in evaluating such systems.
heat transfer
Heat transfer is the process of thermal energy moving from one body or system to another. It occurs via conduction, convection, or radiation. In the context of thermodynamics, heat transfer plays a significant role in the performance of heat engines.

In heat engines, heat is usually transferred from a high-temperature source to a working substance, which then performs work by moving a piston or turning a turbine. The remaining heat is transferred to a low-temperature sink. The efficiency of a heat engine is determined by how well it can utilize the heat transfer processes.

In the described exercise, the device receives heat at a rate of 1 kW. This heat input is essential for generating electricity. However, laws of thermodynamics stipulate that not all heat can be used for generating work; some of it must be rejected to the surroundings. This highlights the inherent limitations due to heat transfer inefficiencies, further underscoring the impossibility of the claims made by the patent.
  • High-temperature reservoir: Where heat is absorbed.
  • Working substance: Converts heat into work.
  • Low-temperature sink: Where excess heat is rejected.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Answer the following true or false. If false, explain why. (a) The change of entropy of a closed system is the same for every process between two specified states. (b) The entropy of a fixed amount of an ideal gas increases in every isothermal compression. (c) The specific internal energy and enthalpy of an ideal gas are each functions of temperature alone but its specific entropy depends on two independent intensive properties. (d) One of the \(T d s\) equations has the form \(T d s=d u-p d v\). (e) The entropy of a fixed amount of an incompressible substance increases in every process in which temperature decreases.

Steam at \(0.7 \mathrm{MPa}, 355^{\circ} \mathrm{C}\) enters an open feedwater heater operating at steady state. A separate stream of liquid water enters at \(0.7 \mathrm{MPa}, 35^{\circ} \mathrm{C}\). A single mixed stream exits as saturated liquid at pressure \(p\). Heat transfer with the surroundings and kinetic and potential energy effects can be ignored. (a) If \(p=0.7 \mathrm{MPa}\), determine the ratio of the mass flow rates of the incoming streams and the rate at which entropy is produced within the feedwater heater, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of liquid exiting. (b) Plot the quantities of part (a), each versus pressure \(p\) ranging from \(0.6\) to \(0.7 \mathrm{MPa}\).

A compressor operating at steady state takes in atmospheric air at \(20^{\circ} \mathrm{C}, 1\) bar at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) and discharges air at 5 bar. Plot the power required, in \(\mathrm{kW}\), and the exit temperature, in \({ }^{\circ} \mathrm{C}\), versus the isentropic compressor efficiency ranging from 70 to \(100 \%\). Assume the ideal gas model for the air and neglect heat transfer with the surroundings and changes in kinetic and potential energy.

Air enters an insulated compressor operating at steady state at 1 bar, \(350 \mathrm{~K}\) with a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and exits at 4 bar. The isentropic compressor efficiency is \(82 \%\). Determine the power input, in \(\mathrm{kW}\), and the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), using the ideal gas model with (a) data from Table A-22. (b) \(I T\). (c) a constant specific heat ratio, \(k=1.39\).

Ammonia enters a valve as a saturated liquid at 7 bar with a mass flow rate of \(0.06 \mathrm{~kg} / \mathrm{min}\) and is steadily throttled to a pressure of 1 bar. Determine the rate of entropy production in \(\mathrm{kW} / \mathrm{K}\). If the valve were replaced by a power-recovery turbine operating at steady state, determine the maximum theoretical power that could be developed, in \(\mathrm{kW}\). In each case, ignore heat transfer with the surroundings and changes in kinetic and potential energy. Would you recommend using such a turbine?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free