of Arts, 1876). Fig. 75 shows a valve of this kind suitable for a 6-in. water main. The water arriving by the main C passes through an equilibrium valve D into the chamber A, and thence through a sluice O, which can be set for any required area of opening, into the discharging main B. The object of the arrangement is to secure a constant difference of pressure between the chambers A and B, so that a constant discharge flows through the stop valve O. The equilibrium valve D is rigidly connected with a plunger P loosely fitted in a diaphragm, separating A from a chamber B2 connected by a pipe B1 with the discharging main B. Any increase of the difference of pressure in A and B will drive the plunger up and close the equilibrium valve, and conversely a decrease of the difference of pressure will cause the descent of the plunger and open the equilibrium valve wider. Thus a constant difference of pressure is obtained in the chambers A and B. Let ω be the area of the plunger in square feet, p the difference of pressure in the chambers A and B in pounds per square foot, w the weight of the plunger and valve. Then if at any moment pω exceeds w the plunger will rise, and if it is less than w the plunger will descend. Apart from friction, and assuming the valve D to be strictly an equilibrium valve, since ω and w are constant, p must be constant also, and equal to w/ω. By making w small and ω large, the difference of pressure required to ensure the working of the apparatus may be made very small. Valves working with a difference of pressure of 12 in. of water have been constructed.
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| Fig. 75.—Scale 124. |
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| Fig. 76. |
§ 61. External Work during the Expansion of Air.—If air expands without doing any external work, its temperature remains constant. This result was first experimentally demonstrated by J. P. Joule. It leads to the conclusion that, however air changes its state, the internal work done is proportional to the change of temperature. When, in expanding, air does work against an external resistance, either heat must be supplied or the temperature falls.
To fix the conditions, suppose 1 ℔ of air confined behind a piston of 1 sq. ft. area (fig. 76). Let the initial pressure be p1 and the volume of the air v1, and suppose this to expand to the pressure p2 and volume v2. If p and v are the corresponding pressure and volume at any intermediate point in the expansion, the work done on the piston during the expansion from v to v + dv is pdv, and the whole work during the expansion from v1 to v2, represented by the area abcd, is
Amongst possible cases two may be selected.
Case 1.—So much heat is supplied to the air during expansion that the temperature remains constant. Hyperbolic expansion.
Then
Work done during expansion per pound of air
| = p1v1 logε v2/v1 = p1v1 logε p1/p2. | (1) |
Since the weight per cubic foot is the reciprocal of the volume per pound, this may be written
| (p1/G1) logε G1/G2. | (1a) |
Then the expansion curve ab is a common hyperbola.
Case 2.—No heat is supplied to the air during expansion. Then the air loses an amount of heat equivalent to the external work done and the temperature falls. Adiabatic expansion.
In this case it can be shown that
where γ is the ratio of the specific heats of air at constant pressure and volume. Its value for air is 1.408, and for dry steam 1.135.
Work done during expansion per pound of air.
| = −{p1v1γ / (γ − 1)} {1/v2γ−1 − 1/v1γ−1} |
| ={p1v1 / (γ − 1)} {1 − (v1/v2) γ−1}. | (2) |
The value of p1v1 for any given temperature can be found from the data already given.
As before, substituting the weights G1, G2 per cubic foot for the volumes per pound, we get for the work of expansion
| (p1/G1) {1/(γ − 1)} {1 − (G2/G1) γ−1}, | (2a) |
| = p1v1 {1/(γ − 1)} {1 − (p2/p1) γ−1/γ}. | (2b) |
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| Fig. 77. |
§ 62. Modification of the Theorem of Bernoulli for the Case of a Compressible Fluid.—In the application of the principle of work to a filament of compressible fluid, the internal work done by the expansion of the fluid, or absorbed in its compression, must be taken into account. Suppose, as before, that AB (fig. 77) comes to A′B′ in a short time t. Let p1, ω1, v1, G1 be the pressure, sectional area of stream, velocity and weight of a cubic foot at A, and p2, ω2, v2, G2 the same quantities at B. Then, from the steadiness of motion, the weight of fluid passing A in any given time must be equal to the weight passing B:
Let z1, z2 be the heights of the sections A and B above any given datum. Then the work of gravity on the mass AB in t seconds is
where W is the weight of gas passing A or B per second. As in the case of an incompressible fluid, the work of the pressures on the ends of the mass AB is
The work done by expansion of Wt ℔ of fluid between A and B is Wtpdv The change of kinetic energy as before is (W/2g) (v22 − v12) t. Hence, equating work to change of kinetic energy,
| ∴ z1 + p1/G1 + v12/2g = z2 + p2/G2 + v22/2g − pdv. | (1) |
Now the work of expansion per pound of fluid has already been given. If the temperature is constant, we get (eq. 1a, § 61)
But at constant temperature p1/G1 = p2/G2;
| ∴ z1 + v12/2g = z2 + v22/2g − (p1/G1) logε (p1/p2), | (2) |
or, neglecting the difference of level,
| (v22 − v12) / 2g = (p1/G1) logε (p1/p2). | (2a) |
Similarly, if the expansion is adiabatic (eq. 2a, § 61),
| z1 + p1/G1 + v12/2g = z2 + p2/G2 + v22/2g − (p1/G1) {1/(γ − 1) } {1 − (p2/p1)(γ−1)/γ}; | (3) |
or, neglecting the difference of level,
| (v22 − v12)/2g = (p1/G1) [1 + 1/(γ − 1) {1 − (p2/p1)(γ−1)/γ)} ] − p2/G2. | (3a) |
It will be seen hereafter that there is a limit in the ratio p1/p2 beyond which these expressions cease to be true.
§ 63. Discharge of Air from an Orifice.—The form of the equation of work for a steady stream of compressible fluid is
