Textbook 3



Textbook Example #3: Potter and Foss


2.32 Air flows from a source through a pipe, impacts a flat disc which weighs 2 pounds, and exits radially to the atmosphere. Determine the spacing h at which the disc will just be raised. Assume that no losses occur as the air moves radially outward so that Bernoulli's equation is applicable. The flow, 200 fps, is essentially uniform from a radius of 2 inches out to the edge at a radius of one foot. At low speeds the air is incompressible with density of .0024 slug per cubic foot. Answer: .575 inches.

Typical Textbook Case

Problem 2.32 from the classic textbook, Fluid Mechanics, by Potter and Foss, John Wiley, 1975.

Note: The diagram indicates the pipe diameter is 2 inches while the associated text indicates the radius is 2 inches. The book gives the answer for h as 0.575 inches.




Bernoulli's Principle. Kinetic energy plus pressure energy is constant, T. Assume density is constant.

T = v2 / 2 + p / d


The valve is the smallest area normal to flow. In this case it is the area around the central orifice, the height of the gap. The valve exit velocity in terms of delivery tube velocity, V0 (Note: the diagram shows this as V1):

VE = V0r0 / 2h


Velocity in the gap as a function of radius, again assuming non-compressible flow:

V(r) = VE r0 / r


Combine (2) and (3)

V(r) = V0 r02/ 2hr


From (3), velocity at the circumference:

VR = VEr0 / R


From (1) and (5) and since T is constant, T at the circumference as a function of orifice, disk radius and ambient pressure:

TR = (VEr0 / R)2 + PR/d


From (1), pressure as a function of radius:

P(r) = d (TR -  (Vr2 / 2))


Combine (3) and (7) and reduce.

P(r) = PR + d(VE2 /2)(r02 / R2)(1 - R2/r2)


Continuing from false premises (8), force is the integration of pressure:

F = Integral[r0 to R](2 pi P(r) rdr


Continuing, integrating, from the center [r0] to the circumference [R]:

F = K[(R2 - r02)/2 - r02 ln(R/r0) + PRA]


K = pi d V02 r04 / 4h2R2




This textbook problem applies Bernoulli Principle to explain a decrease in pressure. It assumes constant fluid density, makes no distinction between radial flow and non-radial flow and does not include information about the motivation for the fluid flow, other than that it is uniform.

The Potter and Foss strategy applies Bernoulli's Principle to explain low pressure as a result of high velocity. Accordingly, since the cross section normal to the airflow increases with radius, velocity decreases with increasing radius and the velocity at the center must be at a maximum. To enable application of the Bernoulli Principle, Potter and Foss assume constant density.

The solution implies that by increasing the velocity one could induce an arbitrarily large, even negative, vacuum. It further implies the motive force driving the flow from the center must, therefore, be a vacuum pump that somehow discharges an air stream against a higher downstream pressure. In the limit, such a pump, as a virtual point air source, would have to deliver an infinitely fast flow of molecules from an infinitely negative pressure; in effect, a pneumatically radiant black hole, discharging high-energy air particles.

Equation (1) claims that T remains constant from the center all the way out to the ambient air under the plate. These events may not be in the same flow stream so might not belong in the same equation. If they were, the air would have to emerge at zero velocity and then curve around under the plate and recycle above it again. Equation (3) seems to contradict this. Also Equations (1) and (3) seem to to qualify for treatment by Bernoulli's Principle by assuming constant density.

For r = R, Equation (8) seems, correctly, to imply that the pressure P(r) is ambient. At r = R, however, the orifice equals the circumference so there is no plate, no air mass and density is moot. For r << R, however, equation (8) leads seems to indicate that pressure drops, and continues dropping, toward the center. Thus, as r ==> 0, to avoid an intractable discontinuity, the air source would have to be a vacuum pump that delivers a higher and higher mass flux at lower and lower pressures. For large VE and/or small r, equation (8) gives that the pressure can be arbitrarily small, even negative. There appears to be no term to account for the pump pressure directly under the orifice. If such a region were to be included, and normalized to the center, it would have to acknowledge infinitely negative pressure and infinitely positive molecular velocity.

The textbook values for this problem are R = 12 inches, V0= 200 ft/sec, p = .0024 slugs/cu-ft and F = 2 pounds. The solution, appearing in the book, [h] = .575 inches.