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We might get some simple derives by the folowings.
In a box(being analysed) is forced by fluids,P represents
pressure,set P=P(x,y,z,t)
(Note:Pressure can be easily to image.)
First we make total differential
d
P
=
∂
P
∂
t
d
t
+
∂
P
∂
x
d
x
+
∂
P
∂
y
d
y
+
∂
P
∂
z
d
z
{\displaystyle dP={\frac {\partial {P}}{\partial {t}}}dt+{\frac {\partial {P}}{\partial {x}}}dx+{\frac {\partial {P}}{\partial {y}}}dy+{\frac {\partial {P}}{\partial {z}}}dz}
The rate of pressure change is
d
P
d
t
=
∂
P
∂
t
+
∂
P
∂
x
d
x
d
t
+
∂
P
∂
y
d
y
d
t
+
∂
P
∂
z
d
z
d
t
{\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+{\frac {\partial {P}}{\partial {x}}}{\frac {dx}{dt}}+{\frac {\partial {P}}{\partial {y}}}{\frac {dy}{dt}}+{\frac {\partial {P}}{\partial {z}}}{\frac {dz}{dt}}}
Hence,
d
P
d
t
=
∂
P
∂
t
+
V
x
∂
P
∂
x
+
V
y
∂
P
∂
y
+
V
z
∂
P
∂
z
{\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+V_{x}{\frac {\partial {P}}{\partial {x}}}+V_{y}{\frac {\partial {P}}{\partial {y}}}+V_{z}{\frac {\partial {P}}{\partial {z}}}}
by
D
D
t
=
∂
∂
t
+
V
⋅
∇
{\displaystyle {\frac {D}{Dt}}={\frac {\partial }{\partial t}}+{\mathbf {V} }\cdot \nabla }
d
P
d
t
=
∂
P
∂
t
+
V
x
∂
P
∂
x
+
V
y
∂
P
∂
y
+
V
z
∂
P
∂
z
=
D
P
D
t
{\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+V_{x}{\frac {\partial {P}}{\partial {x}}}+V_{y}{\frac {\partial {P}}{\partial {y}}}+V_{z}{\frac {\partial {P}}{\partial {z}}}={\frac {DP}{Dt}}}
where
V
{\displaystyle {\mathbf {V} }}
velocity ,
V
(
x
,
y
,
z
)
{\displaystyle V_{(x,y,z)}}
speed , and
∇
{\displaystyle \nabla }
del . ^^
(I'm happy on what achived:Mathematics technology from Taiwanese educations)^^
--HydrogenSu 15:42, 23 December 2005 (UTC)[ reply ]
Reference:James R. Welty,Charles E. Wicks,Robert E. Wilson,Gregory Rorrer Foundamentals of Momentum,Heat,and Mass Transfer ISBN 0-471-38149-7
convective derivative . Linuxlad 14:23, 22 June 2006 (UTC)[ reply ]
