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48. To determine the surface of a fluid between two parallel vertical plates.

Let D'APD (fig. 24.) be a section of the surface of the fluid and of the parallel plates, made by a plane perpendicular to their surfaces; AC equidistant from the parallel plates, meeting the surface of the interior fluid in A, and the plane of the surface of the exterior fluid in C. Then, the rest of the construction and the notation being the same as in Art. 47.) we have

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When the fluid is water, and the plates of glass, and very close to each other, B'AB is a semicircle,

... area D, B" ABD = 2ae +2a2 - πa2 = 2ac+3a2 nearly,

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c = -- sin 0 (1-9) very nearly; and

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It appears that the elevation of a fluid between two parallel plates, is nearly half the elevation in a tube whose diameter is equal to the distance between the plates.

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When a single plate is immersed vertically in a fluid, we

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This is the differential equation to the elastic curve.

The investigation of the form of the surface of the fluid when it is depressed, leads to precisely the same equations as when it is elevated, the sign of y being changed.

If V be the space between the surface of the mercury in a vertical glass tube and the plane of the surface of the mercury on the outside, and a the radius of the tube, V = (0,01) πα.

49. To find the tension of a flexible cylindrical vessel containing fluid.

Let MK, PQ, HL (fig. 25.) be equidistant sections of the cylinder made by planes perpendicular to its axis. Draw PE, QE normals at the extremities of the small arc PQ; MPH, KQL perpendicular to PEQ; and let p be the pressure of the fluid at P, t.MH the tension of MH or KL, r the radius of curvature of PQ. Now ML is kept at rest by the pressure of the fluid, and the tensions of its edges; the tensions of MH, KL, and the pressure of the fluid, are the only forces that act in the plane PEQ; the tensions act perpendicular to PE, QE respectively, and the pressure of the fluid acts perpendicular to PQ, therefore ultimately

EP PQ=t.MH: p. (area ML)=t: p. PQ;

:

.. t = pr.

50. To find the tension of a vessel of any form containing fluid.

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Let PCP', QCQ' (fig. 26.) be the normal sections of least and greatest curvature of the vessel at C; PC PC, QC = QC; PE, PE, QF, QF normals at the extremities of the small arcs PCP, QCQ'; MPK, HPL sections of the vessel made by planes perpendicular to PEP'; MQH, KQL sections of the vessel made by planes perpendicular to QFQ. Let p be the pressure of the fluid at C, t. QQ' the tension of HL or MK, v. PP' the tension of MH or KL; r, s the radii of curvature of PCP, QCQ. ML is kept at rest by the pressure of the fluid, and the tensions of its edges; therefore the resultant of the tensions must be equal and opposite to the pressure of the fluid. The resultant of the tensions of MK, HL.

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And the resultants of the tensions act in the direction CE; the pressure of the fluid = p. PP'. QQ', and it acts in the direction EC,

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When the tensions are the same in every direction, or v = t,

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When the vessel is immersed in fluid, p is the difference of the pressures of the interior and exterior fluids.

E

SECTION V.

ON THE MOTION OF FLUIDS.

ART. 51. WHEN an incompressible fluid flows through a tube, the velocities of the fluid at any two points, are inversely proportional to the areas of the perpendicular sections of the tube at those points; supposing the tube to continue always full, and the velocities at all points in the same section, to be equal to one another, and perpendicular to the section.

For equal volumes of fluid must pass through each section in the same time; and if u, v be the velocities at the two sections; H, K the areas of the sections; uHt, v Kt will be the volumes of the fluid that passes through the two sections in the small time t; and these are equal,

.. uH=vK, and .. u : v = K : H.

52. When a fluid is in motion acted on by any forces, to determine the effective accelerating force in the direction of its motion at any point.

Draw the curve APQR (fig. 27.) so that a tangent to it at any point may be in the direction of the motion of the fluid at that point. The motion of the fluid will not be altered if we suppose a portion of it, of the form of a very small cylinder having PQ for its axis, to become solid for an instant.

Let p be the pressure of the fluid at P, S the accelerating force at P resolved in the direction of a tangent to APR at P, APs; then p+d,p. PQ will ultimately be the pressure at Q; and if p be the density of the fluid at P, κ the area of the base of the cylinder PQ; the mass of PQ= pK. PQ, and the moving force

on PQ in the direction PQ = S. (mass PQ) + pressure on the end P – pressure on the end Q = Spк. PQ - кd ̧p. PQ; therefore the effective accelerating force on the fluid at P in the direction PQ

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53. To find the relation between the pressure and the velocity at P, when the velocity at any point is independent of the time.

Let v be the velocity at P, then vd,v = the effective accelerating force at P in the direction PQ,

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When the fluid is non-elastic v2 + - p = fsS.

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SV+C, where V is the velocity acquired by a point in moving from A to P in a tube AP, acted on by the same forces as the fluid.

54. If the fluid be acted on by gravity only, and if ≈ be the depth of P below a given horizontal plane,

S=gd,≈, fS= g≈ + C ;

therefore when the fluid is non-elastic,

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and when the fluid is elastic v2 + μ log€ p = g≈ + C.

55. To find the relation between the pressure and the velocity at P, when the velocity depends upon the time as well as the position of P.

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