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For each of the cases below, determine if the heat engine satisfies the first law (energy equation) and if it violates the second law. a. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=4 \mathrm{~kW}, \quad \dot{W}=2 \mathrm{~kW}\) b. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=0 \mathrm{~kW}, \quad \dot{W}=6 \mathrm{~kW}\) c. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=2 \mathrm{~kW}, \quad \dot{W}=5 \mathrm{~kW}\) d. \(\dot{Q}_{H}=6 \mathrm{~kW}, \quad \dot{Q}_{L}=6 \mathrm{~kW}, \quad \dot{W}=0 \mathrm{~kW}\)

Short Answer

Expert verified
(a) Satisfies both laws; (b) Violates second law; (c) Violates first law; (d) Satisfies both laws.

Step by step solution

01

Understanding the First Law of Thermodynamics

The first law of thermodynamics for a heat engine relates heat added to work output and heat rejected: \( \dot{Q}_H = \dot{Q}_L + \dot{W} \). We will use this to check if the energy equation is satisfied.
02

Checking First Law for Case (a)

For case (a), \( \dot{Q}_H = 6 \text{kW}, \dot{Q}_L = 4 \text{kW}, \dot{W} = 2 \text{kW} \). Checking the energy balance: \( \dot{Q}_H = 6 = 4 + 2 = \dot{Q}_L + \dot{W} \). The first law is satisfied.
03

Checking Second Law for Case (a)

The second law requires that \( \dot{Q}_H \geq \dot{W} + \dot{Q}_L \). For (a), \( 6 \geq 4 + 2 \); both the first and second laws are satisfied, as no laws are violated.
04

Checking First Law for Case (b)

For case (b), \( \dot{Q}_H = 6 \text{kW}, \dot{Q}_L = 0 \text{kW}, \dot{W} = 6 \text{kW} \). Checking the energy balance: \( \dot{Q}_H = 6 = 0 + 6 = \dot{Q}_L + \dot{W} \). The first law is satisfied.
05

Checking Second Law for Case (b)

For (b), \( \dot{Q}_H = \dot{W} \), implying \( \dot{Q}_L = 0 \). This suggests a perfect conversion of heat to work without any waste heat, violating the second law's requirement. The second law is violated.
06

Checking First Law for Case (c)

For case (c), \( \dot{Q}_H = 6 \text{kW}, \dot{Q}_L = 2 \text{kW}, \dot{W} = 5 \text{kW} \). Checking energy balance: \( \dot{Q}_H = 6 eq 2 + 5 = \dot{Q}_L + \dot{W} \). The first law is violated.
07

Evaluating the Second Law for Case (c)

Because the first law is not satisfied, the second law check is redundant. Violating the first law automatically means the engine setup is not feasible.
08

Checking First Law for Case (d)

For case (d), \( \dot{Q}_H = 6 \text{kW}, \dot{Q}_L = 6 \text{kW}, \dot{W} = 0 \text{kW} \). Checking the energy balance: \( \dot{Q}_H = 6 = 6 + 0 = \dot{Q}_L + \dot{W} \). The first law is satisfied.
09

Checking Second Law for Case (d)

For (d), since all input heat is rejected and no work is done, it satisfies the second law. In a reversible process, efficiency can be zero without violating any law.

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Key Concepts

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

First Law of Thermodynamics
The first law of thermodynamics is all about the conservation of energy. In simple terms, it means that energy cannot be created or destroyed, only transformed from one form to another. For a heat engine, which is a device that transforms thermal energy into mechanical work, this law is expressed as: \[ \dot{Q}_H = \dot{Q}_L + \dot{W} \]where:
  • \( \dot{Q}_H \) is the heat energy supplied to the engine.
  • \( \dot{Q}_L \) is the heat energy rejected by the engine.
  • \( \dot{W} \) is the work done by the engine.
Whenever evaluating a heat engine, checking if this equation holds is the first step. If the equation balances, it means the system adheres to the first law, affirming that energy is conserved. For instance, in case (a), with \( \dot{Q}_H = 6 \text{kW} \), \( \dot{Q}_L = 4 \text{kW} \), and \( \dot{W} = 2 \text{kW} \), the energy relationship holds because \( 6 = 4 + 2 \). This indicates that the first law is satisfied.
Second Law of Thermodynamics
The second law of thermodynamics is centered on the direction of energy transfers and transformations. It states that processes occur in a direction that increases entropy, or disorder, and that energy has quality as well as quantity. In the context of a heat engine, it implies that not all input energy can be converted into work, as some heat will always be "wasted" to a cold reservoir or environment.According to this law, for a heat engine:\[ \dot{Q}_H \geq \dot{Q}_L + \dot{W} \]This means the heat added to the system must be greater than or equal to the sum of the heat rejected and the work produced, ensuring some energy is unavailable for work. If this condition is not met, the engine is said to violate the second law. For example, in case (b), where \( \dot{Q}_H = 6 \text{kW} \), \( \dot{Q}_L = 0 \text{kW} \), and \( \dot{W} = 6 \text{kW} \), it appears as though all input heat is converted to work with no waste. This scenario constitutes a "perfect" engine, which is impossible in reality, thus violating the second law.
Heat Engine Analysis
When analyzing a heat engine, it's crucial to check both the first and second laws of thermodynamics to ensure its feasibility. The first law confirms energy conservation, while the second law ensures proper energy transformation direction and accounts for efficiency realism.Considerations during analysis include:
  • Energy Balance: Verifying the sum of heat rejected and work matches the heat supplied confirms adherence to the first law.
  • Efficiency Check: Calculating the efficiency and checking against theoretical limits ensures the heat engine respects the second law. For example, efficiency, \( \eta \), of an ideal heat engine is calculated as:\[ \eta = \frac{\dot{W}}{\dot{Q}_H} \]
In case (d), \( \dot{Q}_H = 6 \text{kW} \), \( \dot{Q}_L = 6 \text{kW} \), and \( \dot{W} = 0 \text{kW} \), the engine does no work, merely transferring heat completely to the sink, thus aligning with second law's expectations as efficiency becomes effectively zero.

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Most popular questions from this chapter

A window air-conditioner unit is place on a laboratory bench and tested in cooling mode using \(0.75 \mathrm{Btu} / \mathrm{s}\) of electric power with a \(\mathrm{COP}\) of \(1.75 .\) What is the cooling power capacity, and what is the net effect on the laboratory?

An air conditioner on a hot summer day removes 8 Btu/s of energy from a house at \(70 \mathrm{~F}\) and pushes energy to the outside, which is at \(88 \mathrm{~F}\). The house has 30000 lbm mass with an average specific heat of \(0.23 \mathrm{Btu} / \mathrm{lbm}^{\circ} \mathrm{R}\). In order to do this, the cold side of the air conditioner is at \(40 \mathrm{~F}\) and the hot side is at \(100 \mathrm{~F}\). The air conditioner (refrigerator) has a COP that is \(60 \%\) that of a corresponding Carnot refrigerator. Find the actual COP of the air conditioner and the power required to run it.

In a remote location, you run a heat engine to provide the power to run a refrigerator. The input to the heat engine is \(800 \mathrm{~K}\) and the low \(T\) is \(400 \mathrm{~K}\); it has an actual efficiency equal to half of that of the corresponding Carnot unit. The refrigerator has \(T_{L}=-10^{\circ} \mathrm{C}\) and \(T_{H}=35^{\circ} \mathrm{C},\) with a COP that is one-third that of the corresponding Carnot unit. Assume a cooling capacity of \(2 \mathrm{~kW}\) is needed and find the rate of heat input to the heat engine.

After you have driven a car on a trip and it is back home, the car's engine has cooled down and thus is back to the state in which it started. What happened to all the energy released in the burning of gasoline? What happened to all the work the engine gave out?

A \(10-\mathrm{m}^{3}\) tank of air at \(500 \mathrm{kPa}, 600 \mathrm{~K}\) acts as the high-temperature reservoir for a Carnot heat engine that rejects heat at \(300 \mathrm{~K}\). A temperature difference of \(25^{\circ} \mathrm{C}\) between the air tank and the Carnot cycle high temperature is needed to transfer the heat. The heat engine runs until the air temperature has dropped to \(400 \mathrm{~K}\) and then stops. Assume constant specific heat for air and find how much work is given out by the heat engine.

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