By introducing the enthalpy

. . . . .[tex]h\, =\, e\, +\, \dfrac{p}{\rho}[/tex]

into the energy equation

. . . . .[tex]\dfrac{De}{Dt}\, +\, \dfrac{pD(\rho^{-1})}{Dt}\, =\, 0[/tex]

show that, for steady flow,

. . . . .[tex]h_x\, -\, \dfrac{1}{\rho}p_x\, =\, 0[/tex]

Prove that

. . . . .[tex]h\, +\, \dfrac{1}{2}v^2\, =\, \mbox{constant}[/tex]

Any helpis appreciated

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