The changes in plane’s exergy when plane lands and parks at the terminal.

Question

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Answer 1

Question

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

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Answer 2

Question

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

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Answer 3

Question

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

Answer 6

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).

Answer 7

Chapter 7.7, Problem 1E

To determine

The changes in plane’s exergy when plane lands and parks at the terminal.

Answer 8

Expert Solution & Answer

Explanation of Solution

Exergy is defined as the maximum theoretical work obtainable from an overall system comprising of a system and environment at equilibrium.

The following expression shows the equation of exergy.

$E=(U−U0)+p0(V−V0)−T0(S−S0)+KE+PE$ (I)

Here, exergy is $E$ , the internal energies is $U$ , the volume is $V$ , initial pressure is $p0$ , entropy is $S$ , the kinetic energy is $KE$ and potential energy is $PE$ .

If the plane is landed and the fuel of plane is burnt, then the kinetic and potential energy of fuel is decreased.

The Equation (I) depicts that when the kinetic and potential energy is decreased, the exergy of plane is also decreased.

Although, the kinetic and potential energy of the plane is zero when the plane is parked at the terminal, but the internal energy of the fuel exist at the rest.

Thus, when the plane is parked at the terminal, the exergy of plane exists as it's internal energy of the fuel (chemical energy).