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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 1/2 in. of water have been constructed.

Fig. 75.—Scale 1/24.

VI. STEADY FLOW OF COMPRESSIBLE FLUIDS.

 

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

$\int _{v_{1}}^{v_{2}}pdv.$

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

$pv=p_{1}v_{1}$.

Work done during expansion per pound of air

= $\int _{v_{1}}^{v_{2}}pdv=p_{1}v_{1}\int _{v_{1}}^{v_{2}}dv/v$

⁠= 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

pv^γ = p₁v₁^γ,

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.

= $\int _{v_{1}}^{v_{2}}$pdv = p₁v₁^γ $\int _{v_{1}}^{v_{2}}$ dv/v^γ

= −{p₁v₁^γ / (γ − 1)} {1/v₂^γ−1 − 1/v₁^γ−1}

= {p₁v₁^γ / (γ − 1)} {1/v₁^γ−1 − 1/v₂^γ−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)

 

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:

G₁ω₁v₁t = G₂ω₂v₂t.

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

G₁ω₁v₁t (z₁ − z₂) = W (z₁ − z₂) t,

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

p₁ω₁v₁t − p₂ω₂v₂t,

= (p₁/G₁ − p₂/G₂) Wt.

The work done by expansion of Wt ℔ of fluid between A and B is Wt$\int _{v_{1}}^{v_{2}}$pdv The change of kinetic energy as before is (W/2g) (v₂² − v₁²) t. Hence, equating work to change of kinetic energy,

W(z₁ − z₂) t + (p₁/G₁ − p₂/G₂)Wt + Wt $\int _{v_{1}}^{v_{2}}$ pdv = (W/2g) (v₂² − v₁²) t;

∴ z1 + p1/G1 + v12/2g = z2 + p2/G2 + v22/2g − $\int _{v_{1}}^{v_{2}}$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)

Z₁ + P₁/G₁ + v₁²/2g = z₂ + p²/G₂ + v₂²/2g − (p₁/G₁) log_ε (G₁/G₂).

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

z₁ + p₁/G₁ + v₁²/2g = z₂ + p₂/G₂ + v₂²/2g−(p₁/G₁) {1/(γ − 1)} {1 − (p₂/p₁^(γ−1)/γ},