Torricelli's Theorem

A reservoir, shown in section view, extends a great distance to the left. Its water is constrained by banks (not shown), close to us and of the far side, at a distance into the page. At its right boundary the water is constrained by the left face of a dam. Near the base of the dam, at a depth of "H" feet below the surface, a pipe closed by a valve passes through. While the valve is closed, no water flows. When the valve is opened water flows from the reservoir through the pipe to the space to the right. An obvious question: With the valve full-open, what flow of exiting water is expected?

Basic Thermodynamics ~ J. Pohl ©

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Torricelli's Theorem

The sketch is a section view of a very large body of water. This reservoir extends a great distance to the left. The water is constrained by the banks close to us and a distance into the page. To the right, the water is constrained by the dam (by the left face of that dam). Near the base of the dam, at a depth, H feet below the surface, a pipe closed by a valve passes through. While the valve is closed, no water flows. When the valve is opened water flows from the reservoir through the pipe to the space to the right. An obvious question is: with the valve full-open, what flow of exiting water is expected?

Since friction is involved, the actual flow cannot determined by thermodynamics alone. However, the greatest flow can be calculated. Here our interest is less-so the answer than the steps of analysis. The answer was known to Torricelli in 1644.

(1) The Torricelli Formula (1660) predicts the greatest, that is, frictionless exit velocity of the water.

Confirm Torricelli's Theorem
The Theorem will be confirmed twice, with two distinct system models. For each model it is assumed that the depth of water in the reservoir H is constant. This condition is reasonable provided the reservoir is vastly large to the left or if as much water enters the reservoir from adjacent streams as leaves it through the relatively small, opened pipe. Either way we assume: dH(t)/dt = 0.

SYSTEM I:  This system (sketch to the right) is selected as that water in the reservoir (at any instant in time) bounded by two imaginary, vertical and stationary planes notated as "in - in" and "out - out." These planes identify boundaries of entrance of water and of its exit, respectively. The other boundaries of the water, its surface, the banks and the "water-side" surface of the dam, are natural; they don't move. Below, in succession, we apply the mass equation and energy equation to this system.

MASS EQUATION:  The mass equation is general; it applies to any system by reduction from "general" to the essential, application-specific, terms needed.

(2) This mass equation applies to every possible system description.

Working left-to-right, the system mass, m(t), is constant by the assumption that the reservoir is infinite in size. There is only one entrance "(in)" and a single exit plane "out". The mass equation is reduced and the flows are written in terms of local properties as:

(3) 3

For the moment let's consider the exiting flow ("out") as a constant flow rate. A simplification that results is shown.

(4) Our model tells us that the entering velocity is zero. Of course it is not zero; but it is VERY small.

The exiting volume rate is constant and A_in is very large. Consequently the velocity V_in is very small. Also note that this velocity is an average over the " in - in" plane. These are the conclusions of the mass equation for our system.

ENERGY EQUATION:  Just begin with the general form and reduce it. We expect no "fancy terms" to arise; the general form is the place to begin:

(5) The general energy equation is a bit long.

The system is the water within the boundaries. We expect no temperature changes and its center of mass will not move ~ hence no dKE/dt nor dPE/dt. Thus the expression below equals zero not as a sum of terms but by each term itself.

(6) For "steady behavior, dm(t)/dt must also equal zero.

Often, the heat of a system event is the easily specified. The reservoir water was initially in thermal equilibrium (0 = ΣQ) and no aspect of the event creates a temperature change (of consequence); thus sum of heats remains zero. Next, the system experiences no change of volume and there are no work devices or mechanisms, so the sum of works equals zero. We choose to retain the frictional work term to be considered last.

(7) 7

To proceed, there is one entering stream and one exiting stream. The energy equation becomes:

(8) 8

The entering and leaving mass rates are equal. It is convenient to represent the enthalpies of these flowingly static streams as: h = u + pv.

(9) We continue to carry the friction work.

Next, looking at internal energy, u = u(T,v), we realize its change depends upon change of either of its properties, T or v, (or both). Temperature and specific volume for the water are the same for water entering as leaving (and for the body of water as a whole). Hence the entering and leaving specific internal energy flow rates of the water add to zero.

We represent the specific kinetic and potential energies explicitly; using speeds and elevations.

(10) 10

The "out" terms pose no difficulty. The pressure within a sizable flowing jet equals the surrounding atmospheric pressure (p_out = 1 atm). The "out" potential energy datum is negative, (-H), because Z is positive upward and equal to zero at the reservoir surface.

For water entering, the average velocity is easiest to prescribe. It is negligibly small: V_in,avg ~ 0. We see the other properties of the water, its pressure and elevation, will have different values upon entering depending upon the depth of entry, that is, the elevation of entry, z. As a convenience below, we apply the identity, v = 1/ρ, to have the term, p(z)/ρ appear in the energy equation.

(11) Usually friction is assumed zero at the start. But we keep it to see.

Next we look at the following two terms of entrance ("in") over the range of their values:

(12) This group appears with the energy rate into the system term.

Below we build a table that lists the values of each of these terms and their sum. Elevations are notated, z*, in the left column. The second column lists the pressure-related term, the third column lists their sum - at that level of depth.

z*(elevation)	pin(z*)/ρ	pin(z*)/ρ + goz*
z* = 0 (surface)	patm/ρ	patm/ρ + go· 0 = patm/ρ
z* = - H/3 (1/3 depth)	patm/ρ + goH/3	patm/ρ + goH/3 + go(-H/3) = patm/ρ
z* = - H/2 (1/2 depth)	patm/ρ + goH/2	patm/ρ + goH/2 + go(- H/2) = patm/ρ
z* = - H (full depth)	patm/ρ + goH	patm/ρ + goH + go(- H) = patm/ρ

From the right column of the table we conclude that the energy of the flow of water at all depths equals p_atm/ρ. The sum of the enthalpy (pressure-specific volume term only) and the local potential energy, are the same at every elevation of the entrance, over the "in-in" plane. Therefore we make that substitution into the energy equation (just above the table) and solve to confirm Torricelli's Theorem:

(13) Equations represent physics. It is best to have system-type equations then to leave them in their form as a solution progresses.

Our energy equation has become three terms. Left of equality represents the rate of change of energy of our system. Being infinite, the energy loss of the system is not zero but the rate of energy change is zero. Right of equality the terms within the parentheses subscripted "out" are positive as is the multiplied, mass-rate. Thus the aggregate term is negative meaning there and the sign on "out" is negative... there is energy decrease of the system. And there is friction.

Friction is difficult to prescribe. By the supposition of no friction, the above solves to yield Torricell's Theorem.

(14) The famous TORRICELLI FORMULA predicts the greatest theoretical, that is frictionless exit velocity of the water.

♦  SYSTEM II:  A second system to confirm Torricelli's Theorem is shown. This system is the reservoir again but we will not define our system by vertical cutting planes. There will be an entrance geometry and an exit plane which is the same as above.

The "in" or entrance of this system is a narrow annular area perpendicular to the water surface. All of the water that enters the system passes through this thin, ribbon-like area which has a small dimension, the depth, d and is circular with a great diameter,D. Consequences of the mass equation are the same as concluded for SYSTEM I.

The energy equation for this system is the same as before. The elevation of entering flow is zero (the free surface was set as zero and the annulus depth is very short so z_in = 0. The velocity of flow into the annulus is negligible because the annulus area is very great. The pressure divided by ρ is p_atm/ρ. Hence as above and as Torricelli knew:

(15) The famous TORRICELLI FORMULA predicts the greatest or frictionless exit velocity of the water.

Our first conclusion is the correctness of Torricelli's Theorem. Secondly we have developed the idea "free surface," of a reservoir of water (or other fluid). Such surfaces can be uses as planes of " in" or " out" for a system.