THERMO Spoken Here! ~ J. Pohl © | TOC NEXT ~ 185 |

The system of this discussion is a typical military tank that moves only in the horizontal. The perspective of discussion is the energy equation, rate form, (potential energy omitted ~ horizontal motion).

THIS ITEM IS UNDER REVISION

(1)The car is modeled as a "simple compressible substance. |

Consider the following situations:

**Tank Motionless and Inoperative:** The tank sits, its engine has been off for weeks. The energy equation is:

(2)The car is modeled as a "simple compressible substance. |

- Internal energies of the tank components are functions of temperature and specific volume, U = U(T,v), (v = 1/ρ). All changes of volume of the tank parts are negligible, very small. So, tank internal energy is related to temperatures. Having sat a week or so, parts have attained a constant temperature. Hence dU/dt = 0.
- Velocity is zero, meaning dKE/dt = 0.
- Should work occur it would be at the system (tank) boundary and there would be a displacement of the boundary. In our mind's eye, we imaginarily traverse the surface of the tank. Force (atmosheric pressure times area) acts everywhere but there is no displacement; ΣW-dot = 0.
- If a system has been sitting for some time in an environment that is unchanging, it is likely that ΣQ-dot = 0. A next question might be, "Is there any heat." Expanded, ΣQ-dot=0 might be imagined to be three pieces as: Q-dot
_{1}+ Q-dot_{2}+ Q-dot_{3}= 0. Any of these might be non-zero. It is only their sum that is zero.

**Mechanical Transient:** There is a period of time that commences when the engine is cranked and ends once it is runnibg smoothly. In this time the pistons move, and friction is overcome. The energy equation is:

(3)The car is modeled as a "simple compressible substance. |

(2)Work is expanded to represent its power and drag components. |

- (1) Since the vehicle has attained a constant temperature, its "dU/dt" equals zero.
- (2) is zero because the speed is constant.
- (3) is zero because the road is horizontal.
- (4) The "simple compressible" car has a constant volume.
- (5) The sum of heats to the car equals zero.

Our energy equation becomes:

(3)Drag represents "work-like" energy LOSS of the car. |

The "work" term of original energy equation consists of two components. Work arrives to the car delivered by the combustion of the fuel/air mixture that passes through it. Some of the combustion power is expended by moving parts of the car, especially its motor and tires. We assume the combustion work to exceed the losses by 160 kW.

In horizontal travel the power of combustion passed to the car passes to the surround environment. To write the term for "work passage to the environment" is awkward. In general work can be positive or negative. Thus to write "-W" is insufficient to describe loss of energy. Our choice is "-|W|" which means "energy loss of a system by a work mechanism.

(4)Combustion power to the speeding car minus car "inside friction" losses is lost to the friction of moving through air; drag. |

We represent that work rate a being the negative of an p

The system of this discussion is a typical military tank that moves only in the horizontal. The perspective of discussion is the energy equation, rate form, (potential energy omitted ~ horizontal motion).

Premise presently unwritted!