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368

WORK REQUIRED TO

piston is worked just past the communication with the receiver.

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263. Denoting by A the volume of the receiver and by B the volume of the barrel of the pump swept out by the passage of the piston in a single stroke; then, in the absence of any clearance, the air which occupied the volume A at the beginning will occupy the volume A+B at the end of the stroke; or denoting by pn-1 and Pn the densities of the air in the receiver at the beginning and end of this the nth stroke

(A+B)pn= Apn-1,

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264. Supposing the temperature constant, the piston valve opens in the return stroke when the air of density Pn occupying the volume B of the barrel is compressed to atmospheric density p; and therefore at a fraction p/p of the stroke, distances diminishing in G.P.

If ẞ denotes the cross section of the pump barrel in ft2, and kp the atmospheric pressure in lb/ft2, the force in lb required to move the pistons in Hauksbee's air pump, measured by the difference of the tensions of the rods, is

kpß-kpn-1ẞ

A

A+Bx¬kpẞ+kpn-1ẞ

A

B-Bx

at a fraction x of the nth stroke, until x=pn-1/p; after which the valve in the descending piston opens, and the effective tension of this rod is zero.

These tensions can be represented graphically by hyperbolas, as in the hydrometer of fig. 36, p. 113.

265. Work required to exhaust the receiver.

In the nth stroke, the ascending piston in Hauksbee's pump pushes back the atmosphere through a volume B, while the air underneath, originally at pressure kpn-1 and volume A, expands to volume A+B.

Therefore the work done by the ascending piston is, in ft-lb (§ 233),

kpB—kpn-14 log(1+2).

The descending piston yields to the atmospheric pressure kp through a volume B-Вpn-1/p, and at the same time compresses a volume B of air at pressure kpn-1 to a volume

G.H.

Bpn-1.

2A

370

EFFECT OF CLEARANCE

Therefore the work done on the descending piston is

k(p − în - 1)B — kpn - 1B log p/pn - 1.

The work done in the nth stroke is the difference,

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the work done in the n strokes is

kp(A+B)Σ{p2−1(1−p+log p)− (1 − p)np2 - 1log p} = kp(A+B)(1 − p2+polog p")

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if the air is exhausted to one-qth of the atmospheric density; this reduces, when the exhaustion is complete, to kp(A+B) ft-lb, the work required to force back the atmospheric pressure kp lb/ft2 through a volume A+B ft3, as is evident à priori.

A similar result holds for Smeaton's and Tate's air pump, where there is no clearance; the investigation when the effect of clearance is taken into account is left as an exercise.

266. Clearance.

Suppose the piston does not completely sweep out the cylinder, but leaves an untraversed space C, at the bottom of the barrel in Hauksbee's air pump; this space C is called the clearance (espace nuisible, schädlicher Raum).

A volume C of atmospheric air is now left in the barrel at the end of each stroke; and therefore the volume A+B of air at density pn is equivalent to a volume A at density pn-1 and a volume C at density p; so that

(A+B)pn= A pn-1+ Cp.

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shows that, according to the laws of Geometrical Progression,

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Thus, after an infinite number of strokes, when n=∞,

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which gives the ultimate exhaustion when there is a clearance C; and it is only when C=0 that p=0, or the theoretical exhaustion is complete.

267. With a clearance C at the bottom and C' at the top of the stroke in Smeaton's Air Pump, and denoting the density of the air in the barrel in the (n-1)th down stroke by σn - 1,

Bon-1=(B-C')pn-1+C'p,

while in the nth up stroke

(A+B–C)pn=Apn-1+Con -1;

so that, eliminating σn – 1,

B(A+B−C')pn=ABpn-1+C(B—C'') pn-1+CC′p

Pn-μp=λ(pn-1—μp)=\"(1—μ)p,

or

where

B(A+C)-CC'

CC'

λ=

B(A+B−C')' (B-C) (B-C')'

μπ

ρ,

and up is the ultimate density in the receiver, after a large number of strokes.

Similarly, the elimination of leads to

Pn

σn—vp=X(σn-1 —vp)=\"(1 − v)p,

where

y =

C'
B-C

372

EFFECT ON THE EXHAUSTION

In the (n-1)th down stroke the piston valve opens when the volume C' of atmospheric air has expanded to a density σn-1 and therefore to a volume C'p/on-1.

C

In the nth up stroke the lower fixed valve opens when the volume of air of density σn-1 has expanded to density P-1 and therefore to volume Con-1/pn-1; and the upper fixed valve opens when the air of volume B-C and density σn-1 has been compressed to atmospheric density p, and therefore to volume

(B-C)σn-1/p.

268. The weight of the valves is another cause tending to limit the rarefaction; suppose then that and denote the pressures required to lift the fixed and piston valve in Hauksbee's pump.

So long as the valves operate, the pressures in the barrels at the end of the nth in and out stroke are respectively

During the receiver A at pressure kp+

kp+ and kpn—D.

nth stroke the air which occupied the pressure kpn-1 and the clearance Cat has expanded to air of volume A and pressure kon and of volume B and pressure kpn—☎; and therefore

Akpn+B(kpn—☎) = Akpn-1+C'(kp+☎');

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The piston valve is lifted in the down stroke when the volume B of air of pressure kpn-1- is compressed to pressure kp+' and volume to xB, given by

kpn-1-☎=x(hp+w');

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