/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 A reversible heat engine operate... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A reversible heat engine operates between two reservoirs at temperatures of \(650^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). A reversible refrigerator operates between reservoirs at temperatures of \(35^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\). The refrigerator is driven by the heat engine. The heat transfer to the heat engine is \(1500 \mathrm{~kJ}\). A net work output of \(300 \mathrm{~kJ}\) is obtained from the combined enginerefrigerator plant. Determine the net heat transfer to the reservoir at \(35^{\circ} \mathrm{C}\) and the heat transfer to the refrigerant from cold reservoir.

Short Answer

Expert verified
The net heat transfer to the reservoir at 35°C is 5989.55 kJ and the heat transfer to the refrigerant from the cold reservoir is 4789.5 kJ.

Step by step solution

01

- Convert Temperatures to Kelvin

Convert all given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^{\circ}\text{C}) + 273.15 \] For the heat engine: \[ T_1 = 650 + 273.15 = 923.15 \text{ K} \]\[ T_2 = 35 + 273.15 = 308.15 \text{ K} \] For the refrigerator: \[ T_3 = 35 + 273.15 = 308.15 \text{ K} \]\[ T_4 = -10 + 273.15 = 263.15 \text{ K} \]
02

- Calculate Efficiency of the Heat Engine

Use the Carnot efficiency formula for the heat engine: \[ \text{Efficiency}_{\text{HE}} = 1 - \frac{T_2}{T_1} \] Plug in the values: \[ \text{Efficiency}_{\text{HE}} = 1 - \frac{308.15}{923.15} \] \[ \text{Efficiency}_{\text{HE}} = 1 - 0.3337 = 0.6663 \] So, the efficiency of the heat engine is \(0.6663\).
03

- Calculate Work Produced by Heat Engine

The work produced by the heat engine is given by: \[ W_{\text{HE}} = \text{Efficiency}_{\text{HE}} \times Q_{\text{in}} \] Given that the heat transfer to the heat engine \(Q_{\text{in}}\) is \(1500 \, \text{kJ}\), \[ W_{\text{HE}} = 0.6663 \times 1500 = 999.45 \, \text{kJ} \]
04

- Determine Work Provided to Refrigerator

The net work output of the combined engine-refrigerator plant is \(300 \, \text{kJ}\), implying the work provided to the refrigerator is: \[ W_{\text{ref}} = W_{\text{HE}} - 300 \, \text{kJ} \] \[ W_{\text{ref}} = 999.45 - 300 = 699.45 \, \text{kJ} \]
05

- Calculate the COP of the Refrigerator

Use the Carnot Coefficient of Performance (COP) formula for the refrigerator: \[ \text{COP}_{\text{ref}} = \frac{T_3}{T_3 - T_4} \] Plug in the values: \[ \text{COP}_{\text{ref}} = \frac{308.15}{308.15 - 263.15} \] \[ \text{COP}_{\text{ref}} = \frac{308.15}{45} \] \[ \text{COP}_{\text{ref}} \approx 6.8478 \]
06

- Calculate Heat Transfer to Refrigerant from Cold Reservoir

Use the COP of the refrigerator to find the heat transfer: \[ Q_{\text{c}} = \text{COP}_{\text{ref}} \times W_{\text{ref}} \] Plug in the values: \[ Q_{\text{c}} = 6.8478 \times 699.45 \] \[ Q_{\text{c}} \approx 4789.5 \, \text{kJ} \]
07

- Calculate Heat Transfer to the Reservoir at 35°C

The heat rejection from the heat engine to the 35°C reservoir is: \[ Q_{\text{out,HE}} = Q_{\text{in}} - W_{\text{HE}} \] \[ Q_{\text{out,HE}} = 1500 - 999.45 = 500.55 \, \text{kJ} \] The total heat transfer to the 35°C reservoir includes this and the heat rejected by the refrigerator: \[ Q_{\text{35}} = Q_{\text{out,HE}} + Q_{\text{in,ref}} \] From first law for refrigerator: \[ Q_{\text{in,ref}} = Q_{\text{c}} + W_{\text{ref}} \] \[ Q_{\text{in,ref}} = 4789.5 + 699.45 \] \[ Q_{\text{in,ref}} = 5489 \, \text{kJ} \] So, \[ Q_{\text{35}} = 500.55 + 5489 = 5989.55 \, \text{kJ} \]

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 91Ó°ÊÓ!

Key Concepts

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

Carnot efficiency
The Carnot efficiency is a measure of the maximum possible efficiency of a reversible heat engine operating between two temperature reservoirs. It tells us how well the engine converts input heat into useful work. The Carnot efficiency is defined as: \[ \text{Efficiency}_{\text{HE}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \]. This formula incorporates absolute temperatures (in Kelvin), where \( T_{\text{hot}} \) is the temperature of the hot reservoir and \( T_{\text{cold}} \) is the temperature of the cold reservoir. For example, in this exercise, the temperatures of 650°C (hot) and 35°C (cold) were converted to Kelvin (923.15 K and 308.15 K, respectively). Plugging these values into the formula yielded a Carnot efficiency of approximately 0.6663 or 66.63%. This means that 66.63% of the heat energy put into the engine can be converted into work under ideal conditions.
Coefficient of Performance (COP)
The Coefficient of Performance (COP) of a refrigerator indicates its efficiency and is different from the efficiency of a heat engine. The COP is defined as the ratio of the heat removed from the cold reservoir to the work input: \[ \text{COP} = \frac{Q_{\text{c}}}{W} \]where \( Q_{\text{c}} \) is the heat extracted from the cold reservoir, and \( W \) is the work input. For a Carnot refrigerator, the COP can be computed using: \[ \text{COP}_{\text{ref}} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} \]. This formula uses temperatures in Kelvin, just like the efficiency calculation. For our example (35°C and -10°C converted to 308.15 K and 263.15 K), the COP is approximately 6.8478. This means that for every unit of work, the refrigerator extracts about 6.8478 units of heat from the cold reservoir, demonstrating high efficiency.
Thermodynamic cycles
Thermodynamic cycles are sequences of processes that involve energy transfer through heat and work. These cycles are fundamental in understanding heat engines, refrigerators, and heat pumps. In a thermodynamic cycle, a fluid (often a gas) undergoes various state changes that make the cycle repeatable. The Carnot cycle is a special case of a thermodynamic cycle and serves as a standard of comparison for real-world cycles. It includes:
  • Two isothermal processes (constant temperature) - where heat is added and removed.
  • Two adiabatic processes (no heat exchange) - where the fluid expands and contracts.
The Carnot cycle achieves maximum theoretical efficiency, demonstrating the upper limits of a heat engine or refrigerator's performance. Reversible processes within the cycle imply no net entropy change, ideal but not practically achievable due to real-world irreversibilities.
First law of thermodynamics
The first law of thermodynamics is essentially the law of energy conservation, declaring that energy cannot be created or destroyed, only transferred or converted from one form to another. In mathematical terms, for a closed system, it is expressed as: \[ \text{ΔU} = Q - W \]Here, \( \text{ΔU} \) represents the change in internal energy of the system, \( Q \) the net heat added to the system, and \( W \) the work done by the system. In the exercise, the first law helps explain how the net work output and the heat transfers relate to the internal energy changes in the heat engine and the refrigerator. For the heat engine, the heat input minus the work done equals the heat rejected. For the refrigerator, the work input plus the heat extracted results in the heat rejected to the higher temperature reservoir. This relationship underscores how energy dynamics determine the functional efficiency and performance of thermodynamic systems.

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

Two reversible heat pump cycles operate in series. The first cycle receives energy by heat transfer from a cold reservoir at \(260 \mathrm{~K}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\) greater than 260 K. The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher- temperature reservoir at \(1200 \mathrm{~K}\). If the heat pump cycles have the same coefficient of performance, determine (a) \(T\), in \(\mathrm{K}\), and (b) the value of each coefficient of performance.

Insulin and several other pharmaceuticals required daily by those suffering from diabetes and other medical conditions have relatively low thermal stability. Those living and traveling in hot climates are especially at risk by heatinduced loss of potency of their pharmaceuticals. Design a wearable, lightweight, and reliable cooler for transporting temperature-sensitive pharmaceuticals. The cooler also must be solely powered by human motion. While the long-term goal is a moderately priced consumer product, the final project report need only provide the costing of a single prototype.

For each \(\mathrm{kW}\) of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in \(\mathrm{kg} / \mathrm{h}\), from liquid water at \(0^{\circ} \mathrm{C}\) Assume that \(333 \mathrm{~kJ} / \mathrm{kg}\) of energy must be removed by heat transfer to freeze water at \(0^{\circ} \mathrm{C}\), and that the surroundings are at \(20^{\circ} \mathrm{C}\).

Answer the following true or false. (a) A process that violates the second law of thermodynamics violates the first law of thermodynamics. (b) When a net amount of work is done on a closed system undergoing an internally reversible process, a net heat transfer of energy from the system also occurs. (c) A closed system can experience an increase in entropy only when a net amount of entropy is transferred into the system. 5.3 Answer the following true or false. (a) A process that violates the second law of thermodynamics violates the first law of thermodynamics. (b) When a net amount of work is done on a closed system undergoing an internally reversible process, a net heat transfer of energy from the system also occurs. (c) A closed system can experience an increase in entropy only when a net amount of entropy is transferred into the system.

According to an inventor of a refrigerator, the refrigerator can remove heat from the freezer compartment at the rate of \(13,000 \mathrm{~kJ} / \mathrm{h}\) by net input power consumption of \(0.65 \mathrm{~kW}\). Heat is discharged into the room at \(23^{\circ} \mathrm{C}\). The temperature of freezer compartment is \(-15^{\circ} \mathrm{C}\). Evaluate this claim.

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.