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共 207 题- THE
- ENCYCLOPÆDIA BRITANNICA
- ELEVENTH EDITION
- COPYRIGHT
- THE
- ENCYCLOPÆDIA BRITANNICA
- ELEVENTH EDITION
- New York
- VOLUME XIV, SLICE I
- Husband to Hydrolysis
- ARTICLES IN THIS SLICE:
- Insanity: _Hospital Treatment._
- Incense: _Ritual Use._
- India: _History (in part); Geography and Statistics (in part)._
- PRINCIPAL UNSIGNED ARTICLES
- FOOTNOTE:
- ENCYCLOPÆDIA BRITANNICA
- ELEVENTH EDITION
- VOLUME XIV
- FOOTNOTE:
- FOOTNOTE:
- FOOTNOTE:
- FOOTNOTE:
- FOOTNOTE:
- I. THE DATA OF HYDRAULICS[1]
- R = [kappa][omega]V, (1)
- R = [kappa]´[omega](nV). (2)
- R = [mu][omega]dV/dt, (3)
- RELATION OF PRESSURE, DENSITY, AND TEMPERATURE OF LIQUIDS
- RELATION OF PRESSURE, TEMPERATURE, AND DENSITY OF GASES
- G0 = 1/V0 = 2116.8/26214 = .08075. (2)
- G = p[tau]0G0/p0[tau]. (5)
- If p0 = 2116.8, G0 = .08075, [tau]0 = 460.6 + 32 = 492.6, then
- II. KINEMATICS OF FLUIDS
- Q = [omega]V. (1)
- V1A1 = V2A2;
- V1/V2 = A2/A1 (2)
- V1/V2 = A2/A1,
- Q = [omega]v = [omega][root](2gh) nearly. (2)
- The relation between c_v and c_r for any orifice is easily found:--
- Then
- IV. THEORY OF THE STEADY MOTION OF FLUIDS.
- DISTRIBUTION OF ENERGY IN INCOMPRESSIBLE FLUIDS.
- Let h = H1 - H be termed the Venturi head, then
- 28 × 12.57 = 351 cub. ft. per sec.
- Differentiating, we get
- CURRENTS
- Comparing this with (1) we see that
- Q = 2[pi]r1[delta]v1 = 2[pi]r2[delta]v2
- DISSIPATION OF HEAD IN SHOCK
- V. THEORY OF THE DISCHARGE FROM ORIFICES AND MOUTHPIECES
- But
- 0 + p_a/G + h1 - h1 = v²/2g + p_a/G - h;
- 0 + h1 + p_a/G - h1 = v²/2g + h3 - h2 + p_a/G;
- Q = c[omega]v = c[omega] [root](2gh), (5)
- SIMPLE ORIFICES--HEAD CONSTANT
- Q = (2/3)cB [root](2g) [H2^(3/2) - H1^(3/2)], (8)
- From (8) the discharge is
- Let H1/H2 = [sigma], then
- For
- NOTCHES AND WEIRS
- Q = (2/3)cB [root](2g) H^(3/2), (11)
- But
- Hence discharge of lamina
- SPECIAL CASES OF DISCHARGE FROM ORIFICES
- Q = (2/3)cb [root](2g) {(h2 + [h])^(3/2) - (h1 + [h])^(3/2)}. (1)
- And for a rectangular notch for which h1 = 0, the discharge is
- Q = (2/3)cb [root](2g) {(h2 + [h])^(3/2) - [h]^(3/2)}. (2)
- Q = cb[root](2gh) (h2 - h/3), (4)
- Q = mlh [root](2gh), (1)
- Bazin arrives at the following values of _m_:--
- The original simple equation can be used if
- Values of [mu]--
- MOUTHPIECES--HEAD CONSTANT
- Q = c_v c_c[Omega] [root](2gh), (4)
- If the jet discharges into the air, p1 = p_a; and
- Let [Omega]/[omega] = m. Then
- DISCHARGE WITH VARYING HEAD
- PRACTICAL USE OF ORIFICES IN GAUGING WATER
- Q = (2/3)cbh [root](2gh)
- VI. STEADY FLOW OF COMPRESSIBLE FLUIDS.
- Then
- In this case it can be shown that
- But
- It has already (§ 9, eq. 4a) been seen that
- Let
- The weight of air at pressure p1 and temperature [tau]1 is
- Hence the weight of air discharged is
- VII. FRICTION OF LIQUIDS.
- The laws of fluid friction are generally stated thus:--
- R = f[omega]v² (1)
- R = [xi]G[omega]v²/2g, (2)
- M = {2^(n+2) [pi]^(n+1) N^n/(n + 3)} fR^(n+3),
- VIII. STEADY FLOW OF WATER IN PIPES OF UNIFORM SECTION.
- Then
- The equation above may be put in the form
- Weisbach proposed the formula
- 4[zeta] = [alpha] + [beta]/[root]v = 0.003598 + 0.004289/[root]v. (8)
- V - v = 11.3(r^(3/2)/R) [root]i,
- Q = ¼[pi]d²v. (3)
- By combining equations (1) and (2), v is obtained directly:--
- Using these values we get the very simple expressions--
- For practical purposes, the approximate equations
- Then a very approximate value of [zeta] is
- If
- IX. FLOW OF COMPRESSIBLE FLUIDS IN PIPES
- G[Omega]u = W = constant, (3)
- But from (3a)
- WDH dt = -(c[tau]W/p) dp dt - [zeta](H/m)W dl dt
- But
- Hence the initial velocity in the pipe is
- When l is great, log p0/p1 is comparatively small, and then
- Approximately
- W = (.6916 p0 - .4438 p1)(d^5/[zeta]l[tau])^½. (13a)
- From (1) and (3)
- But
- The following table gives some results:--
- X. FLOW IN RIVERS AND CANALS
- This may be put in the form
- The flow down the stream per second, or discharge of the stream, is
- Q = [Omega]v = [Omega]c [root](mi). (4)
- This gives the following values at different velocities:--
- To obtain a new expression for the coefficient in the formula
- Let
- Then
- M = n(41.6 + 0.00281/i);
- But
- Q = [Omega]v. (3)
- But
- Since
- Let
- The above result may be obtained thus (fig. 116):--
- This will be a minimum for
- Let
- The condition to be satisfied is that
- GzQ[theta]. (2)
- Since the motion is steady--
- Consider the possible changes of value of the fraction
- 1 - [zeta]u²/2giH = 0,
- STANDING WAVES
- But, since [Omega]0u0 = [Omega]1u1, we have
- H0² - H1² = (2u0²/g)(H0²/H1 - H0). (2)
- XI. ON STREAMS AND RIVERS
- But
- Taking a mean value for [zeta], we get
- INSTRUMENTS FOR MEASURING THE VELOCITY OF WATER
- M = E_t I[alpha]/l;
- R = b =
- PROCESSES FOR GAUGING STREAMS
- Consequently the discharge is
- HYDRAULIC MACHINES
- XII. IMPACT AND REACTION OF WATER
- SPECIAL CASES
- Three cases may be distinguished:--
- L = R cos ½[phi] = (G/g) bt (v - u)² sin [phi].
- That due to reaction is
- The energy carried away by the water
- The useful work done on the ship
- PV = (G/g)[Omega]v (v - V)V. (2)
- Hence the efficiency of the jet propeller is
- PV/W = 2V/(v + V). (3)
- Then
- V = 2[omega] [root](hh1).
- L = N cos [phi].
- A simpler and more convenient expression given by Colonel Duchemin is
- Consequently, the total pressure between the fluid and plane is
- L = 2P sin [phi] cos [phi]/(1 + sin² [phi]).
- WATER MOTORS
- F = k p d
- D = H - 1(1/3) to H - 2¼ ft.
- Grashof takes e = (1/6)H, and then
- Let
- If the flow is horizontal, z2 = z1; and
- M = (W/g)(v2p2 - v1p1).
- T = Ma = (GQ/g)(w1r1 - w2r2) [alpha] foot-pounds per second. (5)
- H = H_t - [h]_p
- And the hydraulic efficiency of the turbine alone is
- From equations (5) and (7) we get
- The values most commonly assumed in practice are these:--
- V_i = 0.56 [root]{2g(H - [h])}
- Let
- For
- Let
- The variation of pressure due to rotation in a forced vortex is
- It will be shown immediately that
- Let
- The areas of the outlet and inlet surface of the wheel are then
- Then the velocity is
- The guide-blade angle is then given by the equation
- The tangential velocity of the entering water is
- The circumferential velocity of the wheel may be (at mean radius)
- Hence the vane angle at inlet surface is given by the equation
- If the final velocity of the water is axial, then
- PUMPS
- Also in practice
- Usually
- Hence
- M = (GQ/g)(w_o r_o - w_i r_i).
- H = V_o²/2g - u_o² cosec² [phi]/2g
- V_o = [root](2gH + u_o² cosec² [phi]). (8)
- But
- The discharge per square foot of outlet of fan is--
- The total discharge is
- Q = [pi] dt q = 47 to 56 dt [root]p
- For
- FOOTNOTES:
- C6H5N2·SO3K + 2H = C6H5·NH·NH·SO3K,