# Expiration through a large hole, Expiration through...

## Expiration through a large hole

A large hole is known as having a vertical size ( a ) more than , i.e. , where is the depth of immersion of the center of gravity of the hole under the free surface.

When flowing out of a large hole in the side wall of the vessel, the head Н is not the same across the entire section of the hole. For the points of the lower part of the section, it will be larger, and in the upper part it will be smaller. However, the pressure at all points of the outflowing jet will be the same (when it flows into the atmosphere it will be equal to atmospheric pressure), which does not correspond to the pressure distribution by the hydrostatic law. In this connection, the Bernoulli equation can be applied here not to the entire jet as a whole, but only to individual elementary streams. To determine the average flow velocity and fluid flow, the cross-sectional area of ​​the hole is divided into elementary areas and an elementary flow is found for each of them. The total flow is determined by summing up elementary expenditures throughout the section.

Let a and b - be the height and width of the lateral hole in the thin wall of some capacity (Figure 8.3 ). We divide the cross-sectional area of ​​the hole into strips with a height dz. The elementary flow through such a strip of section bdz according to formula (8.1) will be where μ - is the flow rate for the small hole. Fig. 8.3. The flow pattern of a liquid through a large hole

Assuming and assuming that the speed on the free

surfaces , we get (8.2)

Denoting by the total head over the center of gravity of the hole, we find and . Expanding members of the form using the binomial series formula,

we will have (we restrict ourselves to four terms of the expansion)  Given these relations, the expression enclosed in parentheses in formula (8.2) takes the form (8.3)

The second term in parenthesis is usually small in comparison with unity, and they can be neglected. Then formula (8.2) taking into account equality (8.3) takes the form or (8.4)

Given that , we get where ω is the cross-sectional area of ​​the hole.

This formula has the same form as the formula for determining the flow rate when a liquid flows from a small hole in a thin wall (see paragraph 8.1).

The assumptions made when deriving formula (8.4) are corrected by the refinement of the flow coefficient p. As experiments show, this coefficient depends essentially on the shape, dimensions of the hole and the head. Thus, as the hole dimensions increase, the flow coefficient decreases, and with increasing head the effect of the hole size on the flow coefficient decreases.

## Flow through a flooded hole

Consider an open vessel divided by a partition into two compartments with different fluid levels (Figure 8.4). In the partition there is an opening through which the liquid flows from one part of the vessel to the other. It is required to determine the rate of flow of liquid through the hole and its flow rate. Fig. 8.4. Flow pattern of a liquid through a submerged hole

Let's make the Bernoulli equation for sections 1-1 and c-c (for simplicity ): Given that the hydrostatic pressure formula and taking due to its smallness, we get From here where The fluid flow through the hole is determined by the formula (8.5)

where - the area of ​​the jet in a narrow section.

Given that (see paragraph 8.1), where is the cross-sectional area of ​​the hole, formula (8.5) is written as Since , where is the flow coefficient, then Experience shows that the flow coefficient μ for flooded and unflooded holes is almost the same.

## Liquid loss at variable head

The expiration at variable head is, in fact, unsteady movement of the fluid. We confine ourselves only to the case when such a motion can be approximately assumed to be steady, i.e. neglect the forces of inertia. Let's consider the simplest case of emptying a tank having a living area (Figure 8.5). Fig. 8.5. Fluid flow with variable head

Let the initial head be equal to and the final . Calculation of the emptying of the reservoir consists in determining the time of this process. The amount of fluid flowing during dt, is equal to or where the minus sign " is taken because dz is negative, and dV is positive.

Then From here Integrating, we get The time for complete emptying is obtained by putting : or where - expiration time at constant head ; V - the volume of the tank; Q is the flow rate at the head Thus, the emptying time of a vessel with a variable head is 2 times greater than the time it takes to drain the fluid at the initial pressure in an amount equal to the primary volume .

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