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<br />where K is a constant on a streamline, and the integral has been taken from <br /> <br />Let us consider now the integral on the right hand side of eqn. (3.53). <br /> <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br />I <br /> <br />, ;)';31 - <br /> <br />BE(l - A) <br />-a3Vp x Vp = v . V(c T + Lws>i <br />POllO P <br /> <br />where <br /> <br />dUn eEl <br />BE = dz <br />o <br /> <br />Substituting eqns.{3.3) and (3.48) into eqn. (3.1) gives <br /> <br />. v., <br /> <br />. d: 't' SE <br />Va!! + (1 - A) !. '. V['- (c T + Lws)]i = 0 <br />Pouo P <br /> <br />or, <br /> <br />Po <br />- nv . <br />P - <br /> <br />BE BE <br />V + v . V -(c T + 1M ) - Av . V[ - (0 T + Lw )] = 0 <br />ups ups <br />o 0 <br /> <br />The Bernoulli equation can be ~itten as <br /> <br />. '. 1\ <br /> <br />2 <br />c T +g1; + L + Lw = constant (on a streamline)' <br />P 2 s <br /> <br />Therefore, eqn. (3.5~) becomes <br /> <br />Po BE 2 <br />- n - U ( r + g~) <br />P 0 <br /> <br />BE <br />u <br />o <br /> <br />11; Ad(c T + Lw )] ;I K <br />P s <br />o <br /> <br />zero displacement of the parcel to displacement t. <br /> <br />The 1st law of thermodynamics can be Wl'itten as <br /> <br />(3.48) <br /> <br /> <br />(3.49) <br /> <br />(3,50) <br /> <br />(3.51) <br /> <br />(3.52) <br /> <br />(3.53) <br />