## DV Sentence Examples

- At heights from 1500 to 6000 metres his observations agreed well with the formula
**dV**/dh= 34 - o o06 h, V denoting the potential, h the height in metres. - Linke's mean value for
**dV**/dh at the ground was 125. - If V be the potential, p the density of free electricity at a point in the atmosphere, at a distance r from the earth's centre, then assuming statical conditions and neglecting variation of V in horizontal directions, we have r2 (d/dr) (r 2
**dV**/dr) - - 4.rp = o. - In section, and suppose it cut by equipotential surfaces at heights h i and h 2 above the ground, we have for the total charge M included in the specified portion of the tube 47rM = (
**dV**/dh)h i - (**dV**/dh)h2. **Dv**, Velar area or cephalic dome.- If we denote the critical volume, pressure and temperature by Vk, Pk and Tk, then it may be shown, either by considering the characteristic equation as a perfect cube in v or by using the relations that dp/
**dv**=o, d 2 p/**dv**2 =o at the critical point, that Vk = 3b, Pk= a/27b2, T ic = 8a/27b. **Dv**, vessel.**Dv**, dorsal vessel passing into central sinus (bs).- Tsm l, Tergo-sternal muscle (labelled
**dv**in fig. **Dv**' to**dv**s, Dorso-ventral muscles (same as the series labelled tsm in fig.- Tsm, Tergo-sternal muscles, six pairs as in Scorpio (labelled
**dv**in fig. - (1), the effect at B is l abX 2 a+b - cos 2 T t f cos 27rv 2 .
**dv**+sin 27t f sin27rv .**dv**(3), the limits of integration depending upon the disposition of the diffracting edges. - The intensity I 2, the quantity with which we are principally concerned, may thus (be expressed I 2 = 3 fcos27rv 2 .
**dv**} 2 2 t 2 These integrals, taken from v =o, are (known as Fresnel's integrals; we will denote them by C and S, so that C = fo cos 27rv 2 .**dv**, S = fjsinv 2 .**dv**. - From the series for G and H just obtained it is easy to verify that dH = - 7rvG,
**dv**av - dG _ 7rvH -1. - (19), 1 abA) ' ' we may write 12= (cos 27rv 2 .
**dv**) 2 + (f sin zirv 2 .**dv**) 2 (20), or, according to our previous notation, 12 = (2 - C 2 +(z - Sv)2= G2 +H2 Now in the integrals represented by G and H every element diminishes as V increases from zero. - 19, where, according to the definition (5) of C, S, x =i v cos 27rv 2 .
**dv**, y = f v sin ?7rv 2 .**dv**.. - For dx=cos airv 2 .
**dv**, dy= sin 271-v2.**dv**; so that s = f (dx 2 +dy 2) =v, (30), 0= tan1 (dyldx) =171-v 2 (31). - For by (30) do =
**dv**, and by (2)**dv**is proportional to ds. - Ignoring temperature effect, and taking the density as a function of the pressure, surfaces of equal pressure are also of equal density, and the fluid is stratified by surfaces orthogonal to the lines of force; n ap, dy, P d z, or X, Y, Z (4) are the partial differential coefficients of some function P, =fdplp, of x, y, z; so that X, Y, Z must be the partial differential coefficients of a potential -V, such that the force in any direction is the downward gradient of V; and then dP
**dV**(5) ax + Tr=0, or P+V =constant, in which P may be called the hydrostatic head and V the head of potential. - Dp dpu dpv dpw -z)' reducing to the first line, the second line vanishing in consequence of the equation of continuity; and so the equation of motion may be written in the more usual form du du du du d dt +udx+vdy +wdz =X -n dx' with the two others
**dv****dv****dv****dv**i dp dt +u dx +v dy +w dz - Y -P d y' dw dw dw Z w dw i d p dt +u dx +v dy +wd - -P dz. - (5) (8) (I) The components of acceleration of a particle of fluid are consequently Du dudu du du dt = dt +u dx +v dy + wdz' Dr
**dv****dv****dv****dv**dt -dt+udx+vdy+wdz' dt v = dtJ+udx+vdy +w dx' leading to the equations of motion above. - Eliminating H between (5) and (6) DS du
**dv**dw (du**dv**d1zv dt u dx n dx udx' 5 -, dzi =°' and combining this with the equation of continuity Dp du**dv**dw p iit dx+dy+ dz = °' (10) D i du n**dv**dw_ dt (p p dx p dx p dx - o, with two similar equations. - Taking the axis of x for an instant in the normal through a point on the surface H = constant, this makes u = o, = o; and in steady motion the equations reduce to dH/
**dv**=2q-2wn = 2gco sin e, (4) where B is the angle between the stream line and vortex line; and this holds for their projection on any plane to which**dv**is drawn perpendicular. - If r denotes the radius of curvature of the stream line, so that I dp +
**dV**- dH _ dq 2 q2 (6) p**dv****dv****dv****dv**- r ' the normal acceleration. - - In the uniplanar motion of a homogeneous liquid the equation of continuity reduces to du
**dv**dx' dy-O' u= -d,y/dy, v = d i t/dx, (2) surface containing so that we can put _ (6) (9) we have (I) (2) (5) (I) where 4 is a function of x, y, called the streamor current-function; interpreted physically, 4-4c, the difference of the value of 4, at a fixed point A and a variable point P is the flow, in ft. - 1 = yv24, (2) y 2 y y y suppose; and in steady motion, + y 2 dx v-t ' = o, dH +y 2dy0 2P = o, so that 2 "/ y = - y2, 7 2 1,G = dH/d is a function of 1,G, say f'(> '), and constant along a stream line; dH/
**dv**= 2qi', H -f (1/.) = constant, throughout the liquid. - The kinetic energy of the liquid inside a surface S due to the velocity function 4' f i (s given by T=2p + (d) 2+ (t) dxdydz, pff f 75 4 dS (I) by Green's transformation,
**dv**denoting an elementary step along the normal to the exterior of the surface; so that d4ldv = o over the surface makes T = o, and then (d4 2 d4) 2 'x) + (dy) + (= O, dd? - Again, since d4)/
**dv**=d /ds, d4)/ds= - d4y/**dv**, (13) T = 1 p f(1 9 d = - 2 p f4' d (14) With the Stokes' function, y for motion symmetrical about an axis. - ,In a fluid, the circulation round an elementary area dxdy is equal to
**dv**du udx + (v+dx) dy- (u+dy) dx-vdy= () dxdy, so that the component spin is**dv**du (5) 2 dx - dy) in the previous notation of § 24; so also for the other two components and n. - ZI /t = - (a - s) M'Q 2 sine cos ° - EQ sin() =[ - (a - (3)M'U+E]V (8) Now suppose the cylinder is free; the additional forces acting on the body are the components of kinetic reaction of the liquid - aM' (Ç_vR), - (3M' (-- E -FUR), - EC' dR, (9) so that its equations of motion are M (Ç - vR) _ - aM' (_vR) - (a - $) M'VR, (io) M (Ç+uR) = - OM' (
**dV**+U R) - (a - ()M'UR - R, '(II) C dR = dR + (a - Q)M'UV+0V; (12) and putting as before M+aM'=ci, M+13M' = c2, C+EC'=C3, ci dU - c2VR=o,**dV**+(c1U+E)R=o, c 3 dR - (c 1 U+ - c 2 U)V =o; showing the modification of the equations of plane motion, due to the component E of the circulation. - The resultant electric force E at that point is then obtained by differentiating V, since E = -
**dV**/dx, and E is in the direction in which V diminishes fastest. - Let V be the potential at the centre of the prism, then the normal forces on the two faces of area dy.dx are respectively RI dx2 d xl and (dx 2 d x),
**dV**d2 and similar expressions for the normal forces to the other pairs of faces dx.dy, dz.dx. - Hence the total flux through the surface considered is - {(
**dV**i /dn l)-}-(**dV**2 /dn 2)}dS, and this by a previous theorem must be equal to 47radS, or the total included electric quantity. - The electric force outward from that point is -
**dV**/dn, where do is a distance measured along the outwardly drawn normal, and the force within the surface is zero. - Hence we have -
**dV**/dn = 4lra or a = - (1 /47r)**dV**/dn = E/42r. - (22) where
**dV**/dn means differentiation along the normal, and v stands d 2 d 2 d2 for the operator a x2 - P dy2 -{- D. - Then bearing in mind that a= (I/4x1-)
**dV**/dn, and p =-(1/4xr)VV, we have finally E 2 c/v=2 f f v.-dS+ 2J J J Vp**dv**. - In the notation of the calculus the relations become - dH/dp (0 const) = odv /do (p const) (4) dH/
**dv**(0 const) =odp/do (v const) The negative sign is prefixed to dH/dp because absorption of heat +dH corresponds to diminution of pressure - dp. The utility of these relations results from the circumstance that the pressure and expansion co efficients are familiar and easily measured, whereas the latent heat of expansion is difficult to determine. - It is often impossible to observe the pressure-coefficient dp/de directly, but it may be deduced from the isothermal compressibility by means of the geometrically obvious relation, BE = (BEÃ†C) XEC. The ratio BEÃ†C of the diminution of pressure to the increase of volume at constant temperature, or - dp/
**dv**, is readily observed. - We thus obtain the expressions dH = sdo +0 (dp I dO)
**dv**= Sd0 - o (**dv**/do) dp.. - If we put dH=o in equations (8), we obtain the relations between
**dv**and do, or dp and do, under the condition of no heat-supply, i.e. - The change of energy at constant volume is simply sdo, the change at constant temperature is (odp/de - p)
**dv**, which may be written dE/de (v const) =s, dE/**dv**(0 const) =odp/do - p . - We thus obtain the relation ds/
**dv**(o const) =od 2 p/d0 2 (v const),. - The equation to these lines in terms of v and 0 is obtained by integrating dE=sd0+(Odp/de - p)
**dv**= o . - The isothermal elasticity - v(dp/
**dv**) is equal to the pressure p. The adiabatic elasticity is equal to y p, where -y is the ratio S/s of the specific heats. - The heat absorbed in isothermal expansion from vo to v at a temperature 0 is equal to the work done by equation (8) (since d0 =o, and 0(dp/d0)
**dv**=pdv), and both are given by the expression RO log e (v/vo). - In thiscase the ratio of the specific heats is constant as well as the difference, and the adiabatic equation takes the simple form, pv v = constant, which is at once obtained by integrating the equation for the adiabatic elasticity, - v(dp/
**dv**) =yp. - The value of the angular coefficient d(pv)/dp is evidently (b - c), which expresses the defect of the actual volume v from the ideal volume Re/p. Differentiating equation (17) at constant pressure to find
**dv**/do, and observing that dcldO= - nc/O, we find by substitution in (is) the following simple expression for the cooling effect do/dp in terms of c and b, Sdo/dp= (n+I)c - b.. - The values of the partial differential coefficients in terms of n and c are as follows: - Substituting these values in equations already given, we find, from (6) S - s =R(I +nc/V)2 (24) „ (9) dE/
**dv**(o const) =ncp/V . - Since by definition E_ - v(dp/
**dv**) =p(dp/dp) equation (6) becomes U = (dp / dp) (7) The value U = / (E/p) was first virtually obtained by Newton (Principia, bk.