Internal Energy

Energy has many forms, and they are interconvertible, subject to some restriction (the restrictions apply only to the direction of conversion). Let us consider a 1-kg ball of steel.

A. Internal Energy

If the steel is at a temperature of 20^oC, you can hold it in your hand. If it is at a temperature of 200^oC, you cannot hold it in your hand very long. Clearly, the ball at 200^oC produces effects which the ball at 20^oC cannot. Yet, if we measure the mass of the ball, it is the same at 20^oC as it is at 200^oC (within the precision of current measuring techniques). If we could label the atoms when the ball was at 20^oC and take a census of them when the ball was at 200^oC, we would find exactly the same atoms present. Therefore, the difference between what the ball will do at 20^oC and what it will do at 200^oC is not dependent on changing the mass or identify of the matter present. Something else obviously is involved. For now we say that a body which is hot possesses more internal energy that the same body does when cold.

Now suppose that instead of an iron ball we have a balloon which contain a mixture of gasoline and oxygen with a total mass of 1 kg at 20^oC. Now we can introduce a small spark, and the contents of the balloon become very hot (explosively so). After a moment the contents will be much hotter than at the start, and they will have a different chemical composition; instead of being oxygen and gasoline, they will be carbon dioxide and water vapor. Clearly the oxygen-gasoline mixture at 20^oC can produce effects that the mixture of carbon dioxide and water (when cooled to 20^oC) could not. Therefore, there must be a different in energy. This we classify also as a change of internal energy.

Thus, an approximate rule (with exceptions to be seen later) is the internal energy is a measure of hotness plus the ability to cause heat releasing chemical reactions.

Bernoulli's Law

Bernoulli's equation is steady flow equation; however, it can be applied successfully for some unsteady flows, if the changes in flow rate are slow enough to be ignored. To decide how slow the change must be to be ignored, we reason as follows.

For a steady flow, (dV/dt) is zero. This means that although an observer riding with the fluid would observe a changing velocity, an obeserver watching a specific point in the system would observe no change in velocity with respect to time. Then we are safe in ignoring unsteady flow effects if (dV/dt), for all points in the system is small compared with the acceleration we are considering, i.e. the acceleration of gravity or the acceleration due to pressure forces (dP/dL)/p. If (dV/dt) at any point in the system is comparable to the largest of the other acceleration terms in the system, then we can't safely apply Bernoulli's equation to the system.

Most flow problems in which the unsteady flow can not be neglected and hence Bernoulli's equation cannot be applied involve starting the flow from rest or a sudden stopping of the flow. Consider the pipe and valve, initially the pipe is practically full of fluid, and the valve is closed. Then the valve is suddenly opened. If the friction effect is negligible, the fluid will fall freely, maintaining its cylindrical shape, just as a solid rod would. In the whole fluid is still accelerating when the last practicle of fluid leaves the pipe. Friction and surface tension complicate the picture, but for low viscosity fluids in short, large diameter pipe, the result describe above is observed experimentally.

In the case, all fluid have the same velocity. and thus (dV/dt), is the same at all points where there is fluid. Here it is equal to g, so it is same size as the largest acceleration in Bernoulli's equation, and the test indicates that we cannot safely apply Bernoulli's equation to this problem.

Polygon of Forces

It is seldom necessary in a practical problem to consider the resultant of more than two concurrent forces. However, the characteristics of such a resultant can be determined graphically by extending the principle of the triangle of forces. The procedure is illustrated on the picture below, (a) and (b).

In (a) are shown the characteristics of four concurrent forces, F1, F2, F3 and F4, which act on a body. In (b), vectors representing the given forces are drawn in sequence in such a manner that each vector is parallel to the line of action of the corresponding force. The length of each vector indicates the magnitude of the force to a selected scale, and the tail of each vector after the first one coincides with the tip of the preceding vector.

Thus the vector OA represents the force F1 the vector AB represent the force F2, the vector BC represents the force F3, and the vector CD represents the force F4. The vector representing the resultant R of the given forces is drawn from the initial point O to the tip D of the vector for the last given force F4. The characteristic of can be determined in the manner described in step 4. For the forces in the figure above, the magnitude of the resultant is about 170 lb, the angle between the horizontal reference line and the line of action of the resultant is about 18^o, and the direction of the resultant along its line of action is toward the right and downward.

A diagram like that shown in (b) is called as polygon of forces. In some other case, the vector representing the resultant may across a vector representing one of the given forces, as ilustrated in the example above.