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A heat engine receives heat from a source at \(1500 \mathrm{K}\) at a rate of \(600 \mathrm{kJ} / \mathrm{s}\) and rejects the waste heat to a sink at \(300 \mathrm{K} .\) If the power output of the engine is \(400 \mathrm{kW}\), the second-law efficiency of this heat engine is \((a) 42 \%\) (b) \(53 \%\) \((c) 83 \%\) \((d) 67 \%\) \((e) 80 \%\)

Short Answer

Expert verified
Answer: The second-law efficiency of the heat engine is approximately 83%.

Step by step solution

01

Calculate the actual efficiency

The actual efficiency of the heat engine can be calculated using the formula: Efficiency = (Power Output) / (Rate of Heat Received) Efficiency = (400 kW) / (600 kJ/s) Note that 1 kJ/s = 1 kW, so we can simplify the units for the calculation: Efficiency = (400 kW) / (600 kW) = 2/3 ≈ 66.67%
02

Calculate the Carnot efficiency

The maximum possible efficiency for a heat engine operating between the given source and sink temperatures is given by the Carnot efficiency formula: Carnot Efficiency = 1 - (T_sink / T_source) T_source = 1500 K, and T_sink = 300 K, so the Carnot Efficiency will be: Carnot Efficiency = 1 - (300 K / 1500 K) = 1 - 1/5 = 4/5 = 80%
03

Calculate the second-law efficiency

The second-law efficiency is the ratio of the actual efficiency to the Carnot efficiency: Second-law Efficiency = (Actual Efficiency) / (Carnot Efficiency) Second-law Efficiency = (66.67%) / (80%) Second-law Efficiency ≈ 83.33%
04

Select the correct option

Based on our calculations, the closest answer to the second-law efficiency we calculated is: (c) 83% So, the correct option is (c).

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Key Concepts

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

Carnot efficiency
The concept of Carnot efficiency is pivotal when studying thermodynamics and heat engines. It represents the maximum theoretical efficiency that a heat engine operating between two temperatures can achieve. This value is based on the principles set by the Carnot cycle, a model for the most efficient engine possible.

The formula for Carnot efficiency is given as \(\text{Carnot Efficiency} = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}\) where \(T_{\text{sink}}\) and \(T_{\text{source}}\) are the absolute temperatures of the cold and hot reservoirs, respectively. Since these temperatures must be in Kelvin, it ensures that we're speaking in terms of absolute thermal energy available.

The Importance of Kelvin Scale

Using the Kelvin scale is crucial because it starts at absolute zero, a point where there are no heat energy and molecular motion ceases. This scale gives us a clear understanding of the thermodynamic efficiency because the calculations are based on the absolute energy available and removed. It's also important in practical terms because the Carnot efficiency sets an upper bound to the performance of real-world engines—no real engine can be more efficient than a Carnot engine operating between the same two temperatures.

Heat Engine
A heat engine is a device that converts thermal energy into mechanical work. It typically operates by bringing a working substance, like a gas or liquid, into contact with an energy source to absorb heat and then expanding it to do work, after which it releases some heat to a sink and is compressed, ready to start another cycle.

Different types of heat engines exist, such as internal combustion engines in cars and steam turbines in power plants. The performance of a heat engine is often expressed in terms of its efficiency, which is the ratio of the work it outputs to the heat energy it takes in.

Understanding Efficiency in Heat Engines

When we talk about efficiency for the heat engines, we refer to what percentage of the energy added as heat is converted to useful work. The leftover energy is usually lost as waste heat to the environment. Therefore, the actual efficiency is always less than 100%, constrained by the second law of thermodynamics. Examining the efficiency helps engineers and scientists to make improvements in engine designs to save energy and reduce costs.

Thermodynamic Cycles
Thermodynamic cycles are the series of processes that certain systems, like heat engines, undergo to return to their original state. During these cycles, properties such as temperature, pressure, and volume might change, but at the end of the cycle, the system comes back to its starting condition.

Common examples of thermodynamic cycles include the Carnot cycle, the Rankine cycle (used in steam power plants), and the Otto cycle (used in internal combustion engines). These cycles are theoretical models that help us understand how heat engines can convert heat into work.

Significance of Thermodynamic Cycles

In studying these cycles, we can analyze the efficiency of different engine designs and identify the possibility for work production and efficiency enhancements. Understanding the principles of each cycle aids engineers and scientists in optimizing heat engines' performance, leading to more sustainable and cost-effective energy production.

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Most popular questions from this chapter

Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains \(1 \mathrm{m}^{3}\) of \(\mathrm{N}_{2}\) gas at \(500 \mathrm{kPa}\) and \(80^{\circ} \mathrm{C}\) while the other side contains \(1 \mathrm{m}^{3}\) of \(\mathrm{He}\) gas at \(500 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine \((a)\) the final equilibrium temperature in the cylinder and ( \(b\) ) the wasted work potential during this process. What would your answer be if the piston were not free to move? Take \(T_{0}=25^{\circ} \mathrm{C}\)

How much exergy is lost in a rigid vessel filled with 1 kg of liquid \(R-134\) a, whose temperature remains constant at \(24^{\circ} \mathrm{C},\) as \(\mathrm{R}-134 \mathrm{a}\) vapor is released from the vessel? This vessel may exchange heat with the surrounding atmosphere, which is at \(100 \mathrm{kPa}\) and \(24^{\circ} \mathrm{C}\). The vapor is released until the last of the liquid inside the vessel disappears.

Refrigerant-134a is condensed in a refrigeration system by rejecting heat to ambient air at \(25^{\circ} \mathrm{C} .\) R-134a enters the condenser at \(700 \mathrm{kPa}\) and \(50^{\circ} \mathrm{C}\) at a rate of \(0.05 \mathrm{kg} / \mathrm{s}\) and leaves at the same pressure as a saturated liquid. Determine (a) the rate of heat rejected in the condenser, ( \(b\) ) the COP of this refrigeration cycle if the cooling load at these conditions is \(6 \mathrm{kW}\), and \((c)\) the rate of exergy destruction in the condenser.

Nitrogen gas enters a diffuser at \(100 \mathrm{kPa}\) and \(110^{\circ} \mathrm{C}\) with a velocity of \(205 \mathrm{m} / \mathrm{s}\), and leaves at \(110 \mathrm{kPa}\) and \(45 \mathrm{m} / \mathrm{s}\) It is estimated that \(2.5 \mathrm{kJ} / \mathrm{kg}\) of heat is lost from the diffuser to the surroundings at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\). The exit area of the diffuser is \(0.04 \mathrm{m}^{2} .\) Accounting for the variation of the specific heats with temperature, determine ( \(a\) ) the exit temperature, \((b)\) the rate of exergy destruction, and \((c)\) the second-law efficiency of the diffuser.

Steam is to be condensed on the shell side of a heat exchanger at \(120^{\circ} \mathrm{F}\). Cooling water enters the tubes at \(60^{\circ} \mathrm{F}\) at a rate of \(115.3 \mathrm{lbm} / \mathrm{s}\) and leaves at \(73^{\circ} \mathrm{F}\). Assuming the heat exchanger to be well insulated, determine ( \(a\) ) the rate of heat transfer in the heat exchanger and \((b)\) the rate of exergy destruction in the heat exchanger. Take \(T_{0}=77^{\circ} \mathrm{F}\)

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